prime numbers - gaurish4math€¦ · the rst proof was that of peter gustav lejeune dirichlet...
TRANSCRIPT
Prime Numbers
Gaurish Korpal1
gaurishkorpalniseracin
Summer Internship Project Report
13rd year Int MSc Student National Institute of Science Education and Research Jatni(Bhubaneswar Odisha)
Certificate
Certified that the summer internship project report ldquoPrime Numbersrdquo is the bona fide work ofldquoGaurish Korpalrdquo 3rd Year Int MSc student at National Institute of Science Education andResearch Jatni (Bhubaneswar Odisha) carried out under my supervision during June 05 2017to July 15 2017
Place ChennaiDate July 15 2017
Prof Kotyada SrinivasSupervisorProfessor H
The Institute of Mathematical SciencesTharamani Chennai 600113
Abstract
The fact of being prime (or composite) is just a property of number itself regardless of theway we write it In general the divisibility properties are independent of base system (decimalor binary) or writing system we choose (roman or hindu-arabic) In this report I will scratch
the surface of the vast topic known as distribution of primes
Acknowledgements
Foremost I would like to express my sincere gratitude to my advisor Prof Kotyada Srinivas forhis motivation and immense knowledge I am also thankful to my fellow interns Sayan Kundu1
and Aditya Kumar Shukla2 for the enlightening discussionsLast but not the least I would like to thank
ndash Donald Knuth for TEXndash Michael Spivak for AMS-TEXndash Sebastian Rahtz for TEX Livendash Leslie Lamport for LATEXndash American Mathematical Society for AMS-LATEXndash Han The Thanh for pdfTEX
ndash Christian Feuersanger amp Till Tantau for PGFTikZ interpreter
ndash Heiko Oberdiek for hyperref package
ndash Steven B Segletes for stackengine package
ndash Alan Jeffrey amp Frank Mittelbach for inputenc package
ndash Axel Sommerfeldt for subcaption package
ndash David Carlisle for graphicx package
ndash Javier Bezos for enumitem package
ndash Hideo Umeki for geometry package
ndash Philipp Lehman amp Joseph Wright for csquotes package
ndash Peter R Wilson amp Will Robertson for epigraph package
ndash Sebastian Rahtz for textcomp package
ndash Walter Schmidt for gensymb package
ndash Philipp Kuhl amp Daniel Kirsch for Detexify (a tool for searching LATEX symbols)ndash TeXStackExchange community for helping me out with LATEX related problems
GeoGebra was used to generate all the PGFTikZ codes for this document
12nd year BSc(Hons) Mathematics and Computing IMA-Bhubaneswar23rd year BSc(Hons) Mathematics BHU-Varanasi
Contents
Abstract 1
Introduction 3
1 Prime Number Theorem 411 Entire Functions 4
111 Finite Order without Zeros 4112 Finite Order with Zeros 5113 Order 1 with Zeros 6
12 Gamma Function 8121 Analytic Continuation 8122 Stirlingrsquos Formula 9
13 Zeta Function 10131 Euler Zeta Function 10132 Euler Product 10133 Analytic Properties of Euler Zeta Function 11134 Riemann Zeta Function 12135 Zeros of Riemann Zeta Function 14
14 von Mangoldt Function 1715 Prime Counting Function 1816 Chebyshev Functions 1917 The Proof 21
171 Zero-Free Region for ζ(s) 22172 Counting the Zeros of ζ(s) 24173 The Explicit Formula for ψ(x) 28174 Completing the Proof 34
18 Some Remarks 36181 Lindelof Hypothesis 36182 Elementary Proof 36183 Heuristic Approach 37
2 Primes in Arithmetic Progression 3821 Dirichlet Density 3822 Dirichlet Characters 38
221 Number of Dirichlet Characters 39222 Orthogonality Relations 39
23 L-function 40231 Dirichlet L-function 40232 Product Formula 40233 Logarithm Formula 41234 Analytic Continuation 42
1
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Certificate
Certified that the summer internship project report ldquoPrime Numbersrdquo is the bona fide work ofldquoGaurish Korpalrdquo 3rd Year Int MSc student at National Institute of Science Education andResearch Jatni (Bhubaneswar Odisha) carried out under my supervision during June 05 2017to July 15 2017
Place ChennaiDate July 15 2017
Prof Kotyada SrinivasSupervisorProfessor H
The Institute of Mathematical SciencesTharamani Chennai 600113
Abstract
The fact of being prime (or composite) is just a property of number itself regardless of theway we write it In general the divisibility properties are independent of base system (decimalor binary) or writing system we choose (roman or hindu-arabic) In this report I will scratch
the surface of the vast topic known as distribution of primes
Acknowledgements
Foremost I would like to express my sincere gratitude to my advisor Prof Kotyada Srinivas forhis motivation and immense knowledge I am also thankful to my fellow interns Sayan Kundu1
and Aditya Kumar Shukla2 for the enlightening discussionsLast but not the least I would like to thank
ndash Donald Knuth for TEXndash Michael Spivak for AMS-TEXndash Sebastian Rahtz for TEX Livendash Leslie Lamport for LATEXndash American Mathematical Society for AMS-LATEXndash Han The Thanh for pdfTEX
ndash Christian Feuersanger amp Till Tantau for PGFTikZ interpreter
ndash Heiko Oberdiek for hyperref package
ndash Steven B Segletes for stackengine package
ndash Alan Jeffrey amp Frank Mittelbach for inputenc package
ndash Axel Sommerfeldt for subcaption package
ndash David Carlisle for graphicx package
ndash Javier Bezos for enumitem package
ndash Hideo Umeki for geometry package
ndash Philipp Lehman amp Joseph Wright for csquotes package
ndash Peter R Wilson amp Will Robertson for epigraph package
ndash Sebastian Rahtz for textcomp package
ndash Walter Schmidt for gensymb package
ndash Philipp Kuhl amp Daniel Kirsch for Detexify (a tool for searching LATEX symbols)ndash TeXStackExchange community for helping me out with LATEX related problems
GeoGebra was used to generate all the PGFTikZ codes for this document
12nd year BSc(Hons) Mathematics and Computing IMA-Bhubaneswar23rd year BSc(Hons) Mathematics BHU-Varanasi
Contents
Abstract 1
Introduction 3
1 Prime Number Theorem 411 Entire Functions 4
111 Finite Order without Zeros 4112 Finite Order with Zeros 5113 Order 1 with Zeros 6
12 Gamma Function 8121 Analytic Continuation 8122 Stirlingrsquos Formula 9
13 Zeta Function 10131 Euler Zeta Function 10132 Euler Product 10133 Analytic Properties of Euler Zeta Function 11134 Riemann Zeta Function 12135 Zeros of Riemann Zeta Function 14
14 von Mangoldt Function 1715 Prime Counting Function 1816 Chebyshev Functions 1917 The Proof 21
171 Zero-Free Region for ζ(s) 22172 Counting the Zeros of ζ(s) 24173 The Explicit Formula for ψ(x) 28174 Completing the Proof 34
18 Some Remarks 36181 Lindelof Hypothesis 36182 Elementary Proof 36183 Heuristic Approach 37
2 Primes in Arithmetic Progression 3821 Dirichlet Density 3822 Dirichlet Characters 38
221 Number of Dirichlet Characters 39222 Orthogonality Relations 39
23 L-function 40231 Dirichlet L-function 40232 Product Formula 40233 Logarithm Formula 41234 Analytic Continuation 42
1
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Abstract
The fact of being prime (or composite) is just a property of number itself regardless of theway we write it In general the divisibility properties are independent of base system (decimalor binary) or writing system we choose (roman or hindu-arabic) In this report I will scratch
the surface of the vast topic known as distribution of primes
Acknowledgements
Foremost I would like to express my sincere gratitude to my advisor Prof Kotyada Srinivas forhis motivation and immense knowledge I am also thankful to my fellow interns Sayan Kundu1
and Aditya Kumar Shukla2 for the enlightening discussionsLast but not the least I would like to thank
ndash Donald Knuth for TEXndash Michael Spivak for AMS-TEXndash Sebastian Rahtz for TEX Livendash Leslie Lamport for LATEXndash American Mathematical Society for AMS-LATEXndash Han The Thanh for pdfTEX
ndash Christian Feuersanger amp Till Tantau for PGFTikZ interpreter
ndash Heiko Oberdiek for hyperref package
ndash Steven B Segletes for stackengine package
ndash Alan Jeffrey amp Frank Mittelbach for inputenc package
ndash Axel Sommerfeldt for subcaption package
ndash David Carlisle for graphicx package
ndash Javier Bezos for enumitem package
ndash Hideo Umeki for geometry package
ndash Philipp Lehman amp Joseph Wright for csquotes package
ndash Peter R Wilson amp Will Robertson for epigraph package
ndash Sebastian Rahtz for textcomp package
ndash Walter Schmidt for gensymb package
ndash Philipp Kuhl amp Daniel Kirsch for Detexify (a tool for searching LATEX symbols)ndash TeXStackExchange community for helping me out with LATEX related problems
GeoGebra was used to generate all the PGFTikZ codes for this document
12nd year BSc(Hons) Mathematics and Computing IMA-Bhubaneswar23rd year BSc(Hons) Mathematics BHU-Varanasi
Contents
Abstract 1
Introduction 3
1 Prime Number Theorem 411 Entire Functions 4
111 Finite Order without Zeros 4112 Finite Order with Zeros 5113 Order 1 with Zeros 6
12 Gamma Function 8121 Analytic Continuation 8122 Stirlingrsquos Formula 9
13 Zeta Function 10131 Euler Zeta Function 10132 Euler Product 10133 Analytic Properties of Euler Zeta Function 11134 Riemann Zeta Function 12135 Zeros of Riemann Zeta Function 14
14 von Mangoldt Function 1715 Prime Counting Function 1816 Chebyshev Functions 1917 The Proof 21
171 Zero-Free Region for ζ(s) 22172 Counting the Zeros of ζ(s) 24173 The Explicit Formula for ψ(x) 28174 Completing the Proof 34
18 Some Remarks 36181 Lindelof Hypothesis 36182 Elementary Proof 36183 Heuristic Approach 37
2 Primes in Arithmetic Progression 3821 Dirichlet Density 3822 Dirichlet Characters 38
221 Number of Dirichlet Characters 39222 Orthogonality Relations 39
23 L-function 40231 Dirichlet L-function 40232 Product Formula 40233 Logarithm Formula 41234 Analytic Continuation 42
1
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Acknowledgements
Foremost I would like to express my sincere gratitude to my advisor Prof Kotyada Srinivas forhis motivation and immense knowledge I am also thankful to my fellow interns Sayan Kundu1
and Aditya Kumar Shukla2 for the enlightening discussionsLast but not the least I would like to thank
ndash Donald Knuth for TEXndash Michael Spivak for AMS-TEXndash Sebastian Rahtz for TEX Livendash Leslie Lamport for LATEXndash American Mathematical Society for AMS-LATEXndash Han The Thanh for pdfTEX
ndash Christian Feuersanger amp Till Tantau for PGFTikZ interpreter
ndash Heiko Oberdiek for hyperref package
ndash Steven B Segletes for stackengine package
ndash Alan Jeffrey amp Frank Mittelbach for inputenc package
ndash Axel Sommerfeldt for subcaption package
ndash David Carlisle for graphicx package
ndash Javier Bezos for enumitem package
ndash Hideo Umeki for geometry package
ndash Philipp Lehman amp Joseph Wright for csquotes package
ndash Peter R Wilson amp Will Robertson for epigraph package
ndash Sebastian Rahtz for textcomp package
ndash Walter Schmidt for gensymb package
ndash Philipp Kuhl amp Daniel Kirsch for Detexify (a tool for searching LATEX symbols)ndash TeXStackExchange community for helping me out with LATEX related problems
GeoGebra was used to generate all the PGFTikZ codes for this document
12nd year BSc(Hons) Mathematics and Computing IMA-Bhubaneswar23rd year BSc(Hons) Mathematics BHU-Varanasi
Contents
Abstract 1
Introduction 3
1 Prime Number Theorem 411 Entire Functions 4
111 Finite Order without Zeros 4112 Finite Order with Zeros 5113 Order 1 with Zeros 6
12 Gamma Function 8121 Analytic Continuation 8122 Stirlingrsquos Formula 9
13 Zeta Function 10131 Euler Zeta Function 10132 Euler Product 10133 Analytic Properties of Euler Zeta Function 11134 Riemann Zeta Function 12135 Zeros of Riemann Zeta Function 14
14 von Mangoldt Function 1715 Prime Counting Function 1816 Chebyshev Functions 1917 The Proof 21
171 Zero-Free Region for ζ(s) 22172 Counting the Zeros of ζ(s) 24173 The Explicit Formula for ψ(x) 28174 Completing the Proof 34
18 Some Remarks 36181 Lindelof Hypothesis 36182 Elementary Proof 36183 Heuristic Approach 37
2 Primes in Arithmetic Progression 3821 Dirichlet Density 3822 Dirichlet Characters 38
221 Number of Dirichlet Characters 39222 Orthogonality Relations 39
23 L-function 40231 Dirichlet L-function 40232 Product Formula 40233 Logarithm Formula 41234 Analytic Continuation 42
1
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Contents
Abstract 1
Introduction 3
1 Prime Number Theorem 411 Entire Functions 4
111 Finite Order without Zeros 4112 Finite Order with Zeros 5113 Order 1 with Zeros 6
12 Gamma Function 8121 Analytic Continuation 8122 Stirlingrsquos Formula 9
13 Zeta Function 10131 Euler Zeta Function 10132 Euler Product 10133 Analytic Properties of Euler Zeta Function 11134 Riemann Zeta Function 12135 Zeros of Riemann Zeta Function 14
14 von Mangoldt Function 1715 Prime Counting Function 1816 Chebyshev Functions 1917 The Proof 21
171 Zero-Free Region for ζ(s) 22172 Counting the Zeros of ζ(s) 24173 The Explicit Formula for ψ(x) 28174 Completing the Proof 34
18 Some Remarks 36181 Lindelof Hypothesis 36182 Elementary Proof 36183 Heuristic Approach 37
2 Primes in Arithmetic Progression 3821 Dirichlet Density 3822 Dirichlet Characters 38
221 Number of Dirichlet Characters 39222 Orthogonality Relations 39
23 L-function 40231 Dirichlet L-function 40232 Product Formula 40233 Logarithm Formula 41234 Analytic Continuation 42
1
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
235 Product of L-functions 4224 The Proof 43
241 Trivial Dirichlet Character 43242 Non-trivial Dirichlet Character 43243 Completing the Proof 45
25 Some Remarks 45251 Chebotarev Density Theorem 45252 Prime Number Theorem for Arithmetic Progressions 46253 Tchebychev Bias 46
Conclusion 48
A Abelrsquos Summation Formula 50
Bibliography 51
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Introduction
ldquoPrime numbers are what is left when you have taken all the patterns away I thinkprime numbers are like life They are very logical but you could never work out therules even if you spent all your time thinking about themrdquo
mdash Christopher Boone narrator in Mark Haddonrsquos The Curious Incident of the Dogin the Night Time
Prime numbers are the integers bigger than 1 which are only divisible by 1 and themselvesWe encounter these numbers as soon as we start analysing the integers in form of fundamentaltheorem of arithmetic which says that prime numbers are the ldquobuilding blocksrdquo of integers[21] Moreover in the past reports we saw that these numbers play an important role in solvingDiophantine equations [24 sect15 16] generalising the concept of factorisation [27] and extendingthe concept of number system [28] An exposition on further extension of these three ideas canbe seen in Richard Taylorrsquos lecture [21]
In this report I intend to introduce the basic tools involed in study of distribution of primenumbers These tools will be developed with the help of complex function theory Effectivelywe will be scratching the surface of the multiplicative aspects of analytic number theory Inanalytic number theory one looks for good approximations For the sort of quantity that oneestimates in analytic number theory one does not expect an exact formula to exist exceptperhaps one of a rather artificial and unilluminating kind [9]
The first good guess about the asymptotic estimate for the primes emerged at the beginningof the nineteenth century by Carl Friedrich Gauss at 16 years of age that ldquothe density of primesaround x is about 1 log(x)rdquo This statement was proved by the end of nineteenth century byshowing that Riemann zeta function ζ(s) has no zeros near the line Re(s) = 1 I will discussthis proof in the first chapter of this report Later in 1930s Harald Cramer gave a probabilisticway of interpreting Gaussrsquo prediction The Gauss-Cramer model provides a nice way to thinkabout distribution questions concerning the prime numbers but it does not give proofs For adiscussion on distribution of primes based on this model refer to Andrew Granvillersquos article [9]
We can obtain two types of results about the distribution of primes ndash results of a qualitativeand quantitative kind [2] The second chapter is devoted to a result of qualitative nature Longtime ago it had been asserted that every arithmetic progression a a+m a+ 2m in which aand m has no common factor includes infinitely many primes The first proof was that of PeterGustav Lejeune Dirichlet published in 1837 and to do this he introduced the L-functions whichbear his name But I will discuss that proof with a slight modification introduced by CharlesJean de la Vallee Poussin in 1896 One must note that the use of Dirichlet L-functions extendsbeyond the proof of this theorem [5 sect166]
Unlike the lemma-theorem-proof style followed in my past reports in this report I have triedto follow Harold Davenportrsquos prose style [3] My attempt is to let the reader appreciate howrandom appearing tricks or ideas come together to build this beautiful theory of prime numbersItrsquos similar to the idea that while reading a nicely written story the reader is able to appreciatethe ingenuity of ideas Some steps in proofs may appear to be presented without any properlogical motivation behind them and thatrsquos why it took great imagination to even conjecturethe statements whose proof we try to understand in this report
3
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Chapter 1
Prime Number Theorem
ldquoIt should be pointed out that among the most celebrated and cultivated of theGreeks Eratosthenes was considered an extraordinary man who hurled the javelinwrote poems defeated the great runners and solved astronomical problems Variousworks of his have come down to posterity He presented King Ptolemy III of Egyptwith a table on which the prime numbers were etched on a metal plate with thosenumbers with multiples marked with a small hole And so they give the name ofEratosthenesrsquo sieve to the process the wise astronomer used to draw up his tablerdquo
mdash Malba Tahan The Man who counted
Itrsquos easy to prove that there are infintely many prime numbers [7 sect14] and itrsquos apparentfrom a glance at a list of primes that the sequence is rather irregular but the list seems to thinout at a regular rate even if it never ends [13] Despite not having a good exact formula for thesequence of primes we do have a fairly good inexact formula [20] in the form of prime numbertheorem
11 Entire Functions
An entire function (or integral function) is a complex-valued function that is holomorphic atall finite points over the whole complex plane We call an entire function f(z) to be of finiteorder if there exists a number α such that
f(z) = O(e|z|
α)
as |z| rarr infin
We must have α gt 0 excluding the case when f(z) is just a constant The lower bound of thenumbers α with this property is called the order of f(z) [3 sect11]
111 Finite Order without Zeros
Consider an entire function f(z) of finite order with no zeros Then the function g(z) = log(f(z))can be defined as to be single valued and is itself an entire function It satisfies
Re(g(z)) = log |f(z)| lt 2rα (11)
on any large circle |z| = r Since g(z) is an analytic function we can write
g(z) =
infinsumn=0
(an + ibn)zn
and then for z = reiθ
Re(g(z)) =infinsumn=0
(an cos(nθ)minus bn sin(nθ)) rn
4
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Now we wish to obtain an inequality similar to Cauchyrsquos inequality but involving upper boundof Re(g(z)) [1 sect253] Since the series for Re(g(z)) converges uniformly with respect to θ wemay multiply by cos(nθ) or sin(nθ) and integrate term by term to obtainint 2π
0Re(g(z)) cos(nθ)dθ = πanr
n
int 2π
0Re(g(z)) sin(nθ)dθ = minusπbnrn
for n gt 0 while int 2π
0Re(g(z))dθ = 2πa0 (12)
Hence for n gt 0
(an + ibn)rn =1
π
int 2π
0Re(g(z))eminusinθdθ
But we know that [1 sect231]
|(an + ibn)rn| = |an + ibn| rn le1
π
int 2π
0|Re(g(z))| dθ
because∣∣eik∣∣ = 1 for all k isin R Also |an| le |an + ibn| =
radica2n + b2n therefore
|an|rn le1
π
int 2π
0|Re(g(z))| dθ
Adding (12) to this we get
|an|rn + 2a0 le1
π
int 2π
0
(|Re(g(z))|+ Re(g(z))
)dθ
Without loss of generality we can take g(0) = 0 ie a0 = 0 and then using (11) we get
|an|rn le1
π
int 2π
0(2rα + 2rα) dθ = 8rα
It follows on making r rarrinfin that an = 0 if n gt α and similarly for bn This proves that g(z)is a polynomial where f(z) = eg(z)
An entire function of finite order with no zeros is necessarily of the form eg(z) whereg(z) is a polynomial and its order is simply the order of g(z) and so is an integer
112 Finite Order with Zeros
Consider an entire function f(z) of order ρ with zeros z1 z2 zn in |z| lt R (multiple zerosbeing repeated as appropriate) and no zeros on |z| = R For α gt ρ we have
log∣∣∣f (Reiθ)∣∣∣ lt Rα
for all sufficiently large R Then we can write
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ lt 1
2π
int 2π
0Rαdθ = Rα (13)
Let n(r) be the number of zeros of f(z) in |z| le r and f(0) 6= 0 then using Jensenrsquos formulafor analytic functions [1 sect361]
1
2π
int 2π
0log∣∣∣f (Reiθ)∣∣∣ dθ minus log |f(0)| = log
(Rn
|z1| |zn|
)=
int R
0
n(r)
rdr
5
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
we can rewrite (13) as int R
0
n(r)
rdr + log |f(0)| lt Rα
rArrint R
0
n(r)
rdr lt Rα minus log |f(0)| lt 2Rα (14)
Moreover since n(r) is non-decreasingint 2R
R
n(r)
rdr ge n(R)
int 2R
R
1
rdr = n(R) log(2)
Hence using (14) we conclude that
n(R) = O(Rα)
Now let rk = |zk| for k = 1 2 where zk is a zero1 of f(z) then for R = rk and β gt α gt ρwe get
k lt Arαk =rArr 1
rβklt
Aprime
nβα
Thereforeinfinsumk=1
1
rβkltinfinsumk=1
Aprime
nβα
ltinfin
sincesumnminus` converges for ` gt 1
If r1 r2 are the moduli of the zeros of an entire function f(z) of order ρ with
f(0) 6= 0 then the seriessumrminusβn converges if β gt ρ
The lower bound of the positive numbers β for whichsumrminusβn is convergent is called the exponent
of cenvergence of the zeros and is denoted by ρ1 What we proved above is that ρ1 le ρ Butwe may have ρ1 lt ρ for example if f(z) = ez then ρ = 1 but ρ1 = 0 since there are no zerosMoreover any function with finite number of zeros has ρ1 = 0 Therefore ρ1 gt 0 implies thatthere are infinitely many zeros [1 sect822]
113 Order 1 with Zeros
Consider an entire function f(z) of order 1 with with zeros z1 z2 (multiple zeros beingrepeated as appropriate) and f(0) 6= 0 We can then assert that
sumrminus1minusεn converges for any
ε gt 0 and in particularsumrminus2n converges Hence the product
P (z) =infinprodn=1
(1minus z
zn
)ezzn
converges absolutely for all z and converges uniformly in any bounded domain not containingthe points zn [1 sect144] So P (z) is an entire function with zeros (of the appropriate multiplic-ities) at z1 z2 If we put
f(z) = P (z)F (z) (15)
then F (z) is an integral function without zeros
1An entire function which is not a polynomial may have infinitely many zeros zn
6
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Sincesumrminus2n converges the total length of all the intervals
(rn minus rminus2
n rn + rminus2n
)on the real
line is finite and consequently there exist arbitrarily large values of R with the property that
|Rminus rn| gt1
r2n
for all n (16)
Put P (z) = P1(z)P2(z)P3(z) where these are the sub-products extended over the followingsets of n
P1 |zn| ltR
2
P2 R
2le |zn| le 2R
P3 |zn| gt 2R
For the factors of P1 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (∣∣∣∣ zzn∣∣∣∣minus 1
)eminus |z||zn| gt
(R
R2minus 1
)eminus
Rrn = e
minusRrn
and since sumrnlt
R2
1
rnlt
(R
2
)ε infinsumn=1
1
r1+εn
it follows that
|P1(z)| =
∣∣∣∣∣∣∣prod|zn|ltR
2
(1minus z
zn
)ezzn
∣∣∣∣∣∣∣ gtprod|zn|ltR
2
eminusRrn = exp
minusR sumrnlt
R2
1
rn
gt exp
(minusR
1+ε
2ε
infinsumn=1
1
r1+εn
)
rArr |P1(z)| gt exp(minusR1+2ε
)(17)
For the factors of P2 we have on |z| = R
∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ gt ∣∣∣∣z minus znzn
∣∣∣∣ eminus |z||zn| gt |z minus zn|2Reminus2 gt
∣∣∣|z| minus |zn|∣∣∣2R
eminus2 gteminus2
2r2nR
gtC
R3
where C is a positive constant as per (16) And by the result proved in subsection 112n(R) = O(R1+ε) we know that the number of factors of P2 is less than R1+ε
rArr |P2(z)| gt(C
R3
)R1+ε
gt exp(minusR1+2ε) (18)
For the factors of P3 we have on |z| = R∣∣∣∣(1minus z
zn
)ezzn
∣∣∣∣ ge (1minus∣∣∣∣ zzn∣∣∣∣) eminus |z||zn| gt (1minus 1
2
)eminus
Rrn gt e
minusc(Rrn
)2
for some positive constant c and sincesumrngt2R
1
r2n
lt1
(2R)1minusε
infinsumn=1
1
r1+εn
it follows that
|P3(z)| =
∣∣∣∣∣∣prod
|zn|gt2R
(1minus z
zn
)ezzn
∣∣∣∣∣∣ gtprod
|zn|gt2R
eminus(Rrn
)2= exp
( sumrngt2R
minusR2
r2n
)gt exp
(minusR1+ε
21minusε
infinsumn=1
1
r1+εn
)
7
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
rArr |P3(z)| gt exp(minusR1+2ε
)(19)
From (17) (18) and (19) it follows that
|P (z)| gt exp(minusR1+3ε
)Hence from (15) and f(z) = O
(exp(|z|1+ε)
) it follows that
|F (z)| lt exp(R1+4ε
)As seen in subsection 111 this implies that F (z) = eg(z) where g(z) is a polynomial of
degree at most 1 Finally we have
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
where A and B are constantsWe know that the series
sumrminus1minusεn converges for any ε gt 0 Then series
sumrminus1n may or may
not converge but if it does then f(z) satisfies the inequality
|f(z)| lt eC|z|
for some constant C because for all w isin C we have
|1minus w| le 1 + |w| lt e|w| and |ew| =∣∣∣eRe(w)
∣∣∣ lt eradic
Re(w)2+Im(w)2 = e|w|
rArr |(1minus w)ew| le e2|w|
An entire function f(z) of order 1 necessarily has the form
f(z) = eA+Bzinfinprodn=1
(1minus z
zn
)ezzn
If rn = |zn| where zn are the zeros of f(z) thensumrminus1minusεn converges for any ε gt 0 Ifsum
rminus1n converges then f(z) satisfies |f(z)| lt eC|z|
12 Gamma Function
The gamma-function
Γ(z) =
int infin0
eminusttzminus1dt
is a uniformly convergent integral over any finite region throughout which Re(z) gt 0 [1 sect151]and so a continuous function for Re(z) gt 0 [1 sect152] It is an analytic function regular forRe(z) gt 0 [1 sect285]
121 Analytic Continuation
Consider the function
f(z) =
intCeminusw(minusw)zminus1dw
8
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
where C consists of the real axis from infin to δ the circle |w| = δ described in the positivedirection and the real axis from δ to infin again Hence f(z) is regular2 for all finite values of zand the function
g(z) =1
2if(z)cosec(πz) =
i
2 sin(πz)
intCeminusw(minusw)zminus1dw
is regular for all values of z except at the poles z = 0minus1minus2 [1 sect441] We can thereforetake g(z) as the analytic continuation of Γ(z) over C 0minus1minus2 All the gamma-functionformulae for real z [1 sect186] can now be extended to general complex values of z For examplewe get the functional equations
Integration by Parts Γ(z + 1) = zΓ(z) forallz isin C minusNEulerrsquos Reflection Formula Γ(z)Γ(1minus z) = πcosec(πz) forallz isin C Z
Legendrersquos Duplication Formula Γ(z)Γ(z + 1
2
)= 21minus2zradicπΓ(2z) forallz isin C minusN2
Weierstrassrsquo Formula Γ(z) = eminusγz
z
prodinfinn=1
(1 + z
n
)minus1ezn forallz isin C minusN
where N = 0 1 2 3 and γ is the Euler-Mascheroni constant given by
γ = minus loginfinprodn=1
(1 +
1
n
)eminus1n = lim
Nrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
122 Stirlingrsquos Formula
When we try to observe the asymptotic behaviour of Γ(x) as x rarr infin we get the Stirlinigrsquostheorem [1 sect187]
Γ(x) = xxminus12 eminusx
radic2π (1 + o(1))
where f(x) = o(g(x)) means that limxrarrinfin
f(x)
g(x)= 0 By taking logarithm of the Weierstrassrsquo
formula
log(Γ(z)) =infinsumn=1
( znminus log
(1 +
z
n
))minus γz minus log(z)
with each logarithm having its principal value Then after appropriate manipulations we get[1 sect442]
log(Γ(z)) =
(z minus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)for minusπ + δ le arg(z) le π minus δ which is the extension of Stirlingrsquos formula to complex values ofz Therefore for any constant a we can write
log(Γ(z + a)) =
(z + aminus 1
2
)log(z)minus z +
1
2log(2π) +O
(1
|z|
)as |z| rarr infin uniformly for minusπ + δ le arg(z) le π minus δ And for any fixed value of x as y rarr plusmninfin
|Γ(x+ iy)| sim eminusπ|y|2 |y|xminus
12
radic2π
2The many-valued function (minusw)zminus1 = e(zminus1) log(minusw) is made definite by taking log(minusw) to be real at w = minusδ
9
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
13 Zeta Function
131 Euler Zeta Function
For a given real number k gt 1 consider the series
infinsumn=1
1
nk= 1 +
1
2k+
1
3k+
Note that nminusk decreases steadily as n increases but is always positive and
infinsumn=1
1
nklt 1 + lim
ξrarrinfin
int ξ
1
1
xkdx
So the seriessuminfin
n=1 nminusk converges because
limξrarrinfin
int ξ
1
1
xkdx =
1
k minus 1
is finite Hence we conclude that the seriessuminfin
n=1 nminusk is uniformly convergent for a le k le b if
1 lt a lt b
Now let s = σ + it be a complex number then ζ(s) =infinsumn=1
1
nsconverges for σ gt 1 since
infinsumn=1
1
nsltinfinsumn=1
∣∣nminuss∣∣ =infinsumn=1
∣∣∣eminuss log(n)∣∣∣ =
infinsumn=1
∣∣∣∣ 1
nσ
∣∣∣∣and is called the zeta function Also the series
suminfinn=1 n
minuss is uniformly convergent throughoutany finite region in which σ ge a gt 1 Therefore the function ζ(s) is continuous at all points ofthe region σ gt 1
Since ζ(s) is absolutely convergent for σ gt 1 by Dirichlet multiplication [1 sect166(xi)] weconclude that
(ζ(s))2 =
infinsuma=1
1
as
infinsumb=1
1
bs=
infinsumc=1
1
cs
sumab=c
1 =
infinsumn=1
d(n)
ns σ gt 1
where d(n) denotes the number of divisors of n More generally
(ζ(s))k =
infinsumn=1
dk(n)
ns σ gt 1
where k = 1 2 and dk(n) is the number of ways of expressing n as a product of k factors
132 Euler Product
The product prodp
(1minus 1
ps
)where p runs through the primes 2 3 5 is uniformly convergent in any finite region through-out which σ gt 1 because same thing is true of the series
sump |pminuss| which consists of some of the
terms of the seriessuminfin
n=1 |nminuss| studied in subsection 131 [1 sect144]
10
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
The value of the product is1
ζ(s) Since for
(1minus 1
2s
)ζ(s) = 1 +
1
3s+
1
5s+
all terms containing the factor 2 being omitted on the right3 Next(1minus 1
2s
)(1minus 1
3s
)ζ(s) = 1 +
1
5s+
1
7s+
all terms containing the factors 2 or 3 being omitted So generally if pn is the nth prime(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1 +
1
`s+
where all numbers containing the factors 2 3 pn are omitted Since all numbers upto pnare of this form∣∣∣∣(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s)minus 1
∣∣∣∣ le ∣∣∣∣ 1
pn + 1
∣∣∣∣s +
∣∣∣∣ 1
pn + 2
∣∣∣∣s +
which tends to 0 as pn rarrinfin Hence
limnrarrinfin
(1minus 1
2s
)(1minus 1
3s
)middot middot middot(
1minus 1
psn
)ζ(s) = 1
hence proving the result4 that
ζ(s) =prodp
(1minus 1
ps
)minus1
σ gt 1
This was proved by Leonhard Euler5 in 1737 Since a convergent infinite product of non-zerofactors is non-zero [4 sect11] we conclude that
ζ(s) has no zeros for σ gt 1
133 Analytic Properties of Euler Zeta Function
Since un(s) = nminuss is analytic inside any region D sub C with σ gt 1 and the seriessuminfin
n=1 un(s) =suminfinn=1 n
minuss is uniformly convergent throughout every region Dprime sub D The function ζ(s) =suminfinn=1 n
minuss is an analytic function regular6 for σ gt 1 and all its derivatives can be calculatedby term-by-term differentiation [1 sect28] Therefore in general
ζ(k)(s) = (minus1)kinfinsumn=2
(log(n))k
ns σ gt 1
3Notice the resemblance to the sieve of Eratosthenes4This identity is an analytic equivalent of the fundamental theorem of arithmetic [3 pp 2]5ldquoVariae observationes circa series infinitasrdquo St Petersburg Acad 1737 httpeulerarchivemaaorg
pagesE072html6A one-valued analytic function is regular at any point which is interior to one of the circles used in continuation
from the original element [1 sect42]
11
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
134 Riemann Zeta Function
Consider the following term-by-term integration over an infinite rangeint infin0
xsminus1
ex minus 1dx =
int infin0
( infinsumn=1
xsminus1enx
)dx
=
infinsumn=1
int infin0
xsminus1eminusnxdx
=infinsumn=1
nminussint infin
0ysminus1eminusydy
=infinsumn=1
nminussΓ(s)
Then sincesuminfin
n=1 nminuss is convergent for σ gt 1 we have [1 sect178]
ζ(s) =1
Γ(s)
int infin0
xsminus1
ex minus 1dx σ gt 1
We can use this formula to continue ζ(s) across σ = 1 in the same way that we continue Γ(s)across σ = 0 In fact we can prove it precisely the same way as for gamma-function [1 sect443]Consider the complex integral
I(s) =
intC
wsminus1
ew minus 1dw
where the contour C starts at infinity on the positive real axis encircles the origin once in thepositive direction excluding the points plusmn2iπplusmn4iπ and returns to positive infinity7 Herethe many-valued function (w)sminus1 = e(sminus1) log(w) is made definite by taking log(w) to be real atthe beginning of the contour thus Im(log(w)) varies from 0 to 2π round the contour [4 sect24] Ifs is any integer the integrand in I(s) is one-valued and I(s) can be evaluated by the theoremof residues [1 sect311]
We can take C to consist of the real axis from infin to δ (0 lt δ lt 2π) the circle |w| = δ andthe real axis from δ to infin
δ
7The only difference between this contour and the one used for gamma-function is that C must now excludeall the poles of 1(ew minus 1) other than w = 0 ie the points w = plusmn2iπplusmn4iπ
12
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
We note that|wsminus1| = e(σminus1) log |w|minust arg(w) le |w|σminus1e2π|t|
|ew minus 1| =∣∣∣∣w +
w2
2+w3
6+
∣∣∣∣ gt A|w| for some A gt 0
Hence on the circle |w| = δ for fixed s we have∣∣∣∣∣int|z|=δ
wsminus1
ew minus 1dw
∣∣∣∣∣ le δσminus1e2π|t|
Aδmiddot 2πδ rarr 0
as δ rarr 0 and σ gt 1 Thus on letting δ rarr 0 if σ gt 1 we get
I(s) = minusint infin
0
xsminus1
ex minus 1dx+
int infin0
(xe2πi
)sminus1
ex minus 1dx
= minusζ(s)Γ(s) + e2πi(sminus1)ζ(s)Γ(s) (using the boxed formula on previous page)
=(e2πi(sminus1) minus 1
)ζ(s)Γ(s)
=(e2πi(sminus1) minus 1
)ζ(s)
π
sin(πs)Γ(1minus s)(using Eulerrsquos Reflection Formula)
=e2πis minus e2πi
e2πi
ζ(s)
Γ(1minus s)2πi
eiπs minus eminusiπs
=2πieiπs
Γ(1minus s)ζ(s) (using e2πi = 1)
Hence we get
ζ(s) =eminusiπsΓ(1minus s)
2πi
intC
wsminus1
ew minus 1dw
This formula has been proved for σ gt 1 The integral I(s) however is uniformly convergentin any finite region of the complex plane and so defines an integral function of s Hence theformula provides the analytic continuation of ζ(s) over the whole complex plane The onlypossible singularities are the poles of Γ(1minus s) ie s = 1 2 3 4 From the subsection 133we know that ζ(s) is regular at s = 2 3 and in fact it follows at once from Cauchyrsquos theorem[1 sect233] that I(s) vanishes at these points Hence the only possible singularity is a simple poleat s = 1 For s = 1 the contour integral is equal to
I(1) =
intC
dw
ew minus 1= 2πi
by the residue theorem [1 sect311] and Γ(1minus s) has a pole (as seen in section 12) Hence s = 1is the only pole Hence the above formula is analytic continuation of ζ(s) over C 1
To deduce the functional equation from the analytic continuation formula take the integral
ofwsminus1
ew minus 1along the contour Cn n = 1 2 3 consisting of the positive real axis from infinity
to (2n+1)π then round the square with the corners (2n+1)π(plusmn1plusmni) and then back to infinityalong the positive real axis Between the contours C and Cn the integrand has poles at thepoints plusmn2iπ plusmn2inπ
The residues at 2mπi and minus2mπi are taken together8
limzrarr2mπi
(z minus 2mπi) (2mπi)sminus1
e2mπiminus1+ limzrarrminus2mπi
(z + 2mπi) (minus2mπi)sminus1
eminus2mπiminus1=(
2mπeiπ2
)sminus1+(
2mπe3iπ2
)sminus1
8Can use LrsquoHopitalrsquos rule to evaluate this limit This rule is a local statement ie it is concerned aboutbehaviour of a function (single valuedmulti valued) near a particular point and not the global issues (like branchcuts)
13
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
= (2mπ)sminus1 eiπ(sminus1)2 cos(π(sminus1)
2
)= minus2 (2mπ)sminus1 eiπs sin
(πs2
)
(2n+ 1)π
(2n+ 1)π(1 + i)(2n+ 1)π(minus1 + i)
(2n+ 1)π(minus1minus i) (2n+ 1)π(1minus i)
2πi
2nπi
minus2πi
minus2nπi
Hence by the theorem of residues [1 sect311]intCn
wsminus1
ew minus 1dw =
intC
wsminus1
ew minus 1dw + 2πi
(minus2eiπs sin
(πs2
) nsumm=1
(2mπ)sminus1
)Now let σ lt 0 and make nrarrinfin The function 1(ez minus 1) is bounded on the contours Cn andzsminus1 = O
(|z|σminus1
) Hence the integral round Cn tends to zero and we obtain
I(s) = 4πieiπs sin(πs
2
) infinsumm=1
(2mπ)sminus1
= 4πieiπs sin(πs
2
)(2π)sminus1ζ(1minus s)
Now using the analytic continuation formula we deduce the desired functional equation
ζ(s) = 2sπsminus1 sin(πs
2
)Γ(1minus s)ζ(1minus s)
This was proved by Bernhard Riemann9 in 1859 [4 sect24]
135 Zeros of Riemann Zeta Function
Using the Eulerrsquos reflection formula and Legendrersquos duplication formula we can rewrite thefunctional equation as
ζ(s) = πsminus12
Γ(
1minuss2
)Γ(s2
) ζ(1minus s)
We can rewrite this as
πminuss2 Γ(s
2
)ζ(s) = πminus
1minuss2 Γ
(1minus s
2
)ζ(1minus s)
9ldquoUeber die Anzahl der Primzahlen unter einer gegebenen Grosserdquo Monatsberichte der Berliner Akademie1859 httpwwwmathstcdiepubHistMathPeopleRiemannZeta
14
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
which can be expressed by saying that the function on the left is an even function of s minus 12
[3 sect8] The functional equation allows the properties of ζ(s) for σ lt 0 to be inferred from itsproperties for σ gt 1 In particular the only zeros of ζ(s) for σ lt 0 are at the poles of Γ
(s2
)
except 0 since itrsquos a simple pole of ζ(1 minus s) So the points s = minus2minus4minus6 are called thetrivial zeros
Figure 11 Plot of Riemann zeta function for real values of s ie when t = 0 illustrating the trivialzeros Plotted using plot(zeta(x) (x-13-1) rgbcolor=(000) legend label= prime
$ zeta(x)$prime thickness=2) in SageMath 751
The remainder of the plane where 0 le σ le 1 is called the critical strip To analyse zetafunction in this strip we will consider the function ξ(s) defined by
ξ(s) =1
2s(sminus 1)πminus
s2 Γ(s
2
)ζ(s) (110)
This is an entire function because it has no pole for σ ge 12 and is an even function of s minus 12
Note that ξ(s) = ξ(1 minus s) hence it will suffice to prove the statements about ξ(s) just forσ ge 12 Observe that ∣∣∣∣12s(sminus 1)πminus
s2
∣∣∣∣ lt exp(C1|s|) (111)
for some constant C1 and by Stirlingrsquos formula (subsection 122)∣∣∣Γ(s2
)∣∣∣ lt exp(C2|s| log |s|) (112)
for some constant C2 where minusπ2 lt arg(s) lt π
2 And using Abelrsquos partial summation (Ap-pendix A) we can write for σ gt 1sum
nlex
1
ns=bxcxs
+ s
int x
1
btcts+1
dt
and by taking xrarrinfin we getinfinsumn=1
1
xs= s
int infin1
1
ts+1dt
15
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Therefore we can rewrite ζ(s) for σ gt 1 as
ζ(s) = s
int infin1
bxcxs+1
dx
= s
int infin1
xminus xxs+1
dx
=s
sminus 1minus s
int infin1
xxs+1
dx
(113)
Since the integral on the right is absolutely convergent for σ gt 0 we get the analytic continua-tion10 of ζ(s) for σ gt 0 Hence for σ ge 12 we have
|ζ(s)| lt C3|s| (114)
when |s| is large Combining (111) (112) and (114) we get
|ξ(s)| lt exp(C|s| log |s|)
as |s| rarr infin for some constant C Hence ξ(s) is of order at most 1 Moreover as s rarr +infinthrough real values the above inequality is the best possible (apart from the value of C) sincelog(Γ(s)) sim s log(s) and ζ(s)rarr 1 Hence if fact ξ(s) is of order 1
Consequently ξ(s) does not satisfy the more precise inequality indicated in subsection 113for entire functions of order 1 Hence
sumrminus1n diverges ie ξ(s) has infinitely many zeros
The zeros of ξ(s) are the non-trivial zeros of ζ(s) for in (110) the trivial zeros of ζ(s) arecancelled by the poles of Γ( s2) (subsection 121) and s
2Γ( s2) has no zeros and the zero of sminus 1is cancelled by the pole of ζ(s) (subsection 134) Then recalling that ζ(s) has no zero forσ gt 1 (subsection 132) we get
ζ(s) has infinitely many non-trivial zeros in the critical strip 0 le σ le 1
This was proved by Jacques Hadamard11 in 1893 [3 sect12] We can analyse the function ξ(s) towrite its product formula explicitly as shown in subsection 113
ξ(s) = eA+Bsprodρ
(1minus s
ρ
)esρ (115)
where ρ is a zero of ξ(s) Eliminating ξprime(s)ξ(s) from the logarithmic differentiation of (110)and (115) we get
ζ prime(s)
ζ(s)= B minus 1
sminus 1+
log(π)
2minus 1
2
Γprime(s2 + 1
)Γ(s2 + 1
) +sumρ
(1
sminus ρ+
1
ρ
)
This exhibits the pole of ζ(s) at s = 1 and the non-trivial zeros at s = ρMoreover Riemann conjectured that any non-trivial zero of ζ(s) has σ = 12 (known as
Riemann Hypothesis) This can be illustrated by the following figure [11][6 Chapter 8]
10So unlike previous proof of analytic continuation for all s isin C itrsquos really easy to define ζ(s) as a meromorphicfunction for σ gt 0 with only pole at s = 1
11ldquoEtude sur les proprietes des fonctions entieres et en particulier drsquoune fonction consideree par RiemannrdquoJournal de Mathematiques Pures et Appliquees (1893) 171-216 httpeudmlorgdoc234668
16
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Figure 12 A plot of the real and imaginary parts of the Riemann zeta function ζ(12 + it) for0 lt t lt 30 Plotted using i = CDFgen() v = [zeta(05 + n10 i) for n in range(300)] L = [(zreal() zimag()) for z in v] line(L
rgbcolor=(000) thickness=2) in SageMath 751
14 von Mangoldt Function
We define the von Mangoldt function Λ(n) as
Λ(n) =
log(p) if n = pk for some prime p and integer k ge 1
0 otherwise
Hence we have the identitysum
m|n Λ(m) = log(n)Furthermore from the Euler product we obtain that
log(ζ(s)) = minussump
log
(1minus 1
ps
) σ gt 1 (116)
And for the principal branch of the complex logarithm we have
minus log(1minus z) =
infinsumn=1
zn
n
where |z| le 1 z 6= 1 Hence we can re-write (116) as
log(ζ(s)) =sump
summ
1
mpms=sump
summ
1
meminusms log(p) σ gt 1
where p runs through all primes and m through all positive integers On differentiating bothsides (since analytic) we get
ζ prime(s)
ζ(s)= minus
sump
summ
log(p)
pms σ gt 1
Which can be re-written as
minusζprime(s)
ζ(s)=
infinsumn=1
Λ(n)
ns σ gt 1
and gives us the generating function for Λ(n) [7 sect177]
17
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
15 Prime Counting Function
We define π(x) to be the number of primes which do note exceed x ie
π(x) =sumplex
1
so that π(1) = 0 π(2) = 1 and so on If pn is the nth prime then π(pn) = n Therefore π(x)as a function of x and pn as a function of n are inverse functions [7 sect15] Therefore to askfor an exact formula for π(x) is same as to ask for an exact formula for pn No such formula isknown
The prime number theorem states that
π(x) sim x
log(x)
which is equivalent to
pn sim n log(n)
because log(log(x)) = o(log(x)) [7 sect18]Since there are infinitely many primes π(x) rarr infin as x rarr infin Moreover we can use the
arguments used for proving existence of infinitely many primes to get a lower bound for π(x)and upper bound for pn Following are the two bounds [7 sect22 26]
Bound 1 pn lt 22n for any positive integer n and π(x) ge log(log(x)) for all x ge 2
Proof Using induction we can prove that pn lt 22n for any positive integer n Alsowe observe that for n ge 4 enminus1 gt 2n ie ee
nminus1gt e2n gt 22n (since ex is an increasing
function)
Now we divide the interval x ge 2 into smaller disjoint intervals[2 ee
3]
and(eenminus1
een]
for all n ge 4 Itrsquos enough to prove for x isin(eenminus1
een]
for all n ge 4 since for x isin[2 ee
3]
the inequality follows from the fact that log(log(x)) isin (minus04 3] and log(x) is an increasingfunction
Let x isin(eenminus1
een]
where n ge 4 Since π(x) is a non-decreasing function pn lt 22n lt
eenminus1
lt x implies that π(pn) le π(x) ie n le π(x) for all n ge 4 Since log(x) isan increasing function x lt ee
nimplies that log(log(x)) le n Hence we conclude that
log(log(x)) le n le π(x) for all n ge 4 Thus completing the proof
Bound 2 π(x) ge log(x)
2 log(2)for all x ge 1 and pn le 4n for any positive integer n
Proof Suppose that 2 3 5 pj are the first j primes and let N(x) be the number ofpositive integers n not exceeding x which are not divisible by any prime p gt pj If weexpress such an n in the form n = n2
1m where m is a square free integer ie not divisible
by the square of any prime Hence m = 2b13b2 middot middot middot pbjj with bi is either 0 or 1 There are
just 2j possible choices of the exponents and so not more than 2j different values of mAgain n1 le
radicn leradicx and so there are not more than
radicx different values of n1 Hence
N(x) le 2jradicx
Now we take j = π(x) so that pj+1 gt x and N(x) = x Hence we have
N(x) = x le 2π(x)radicx =rArr log(x)
2 log(2)le π(x)
If we put x = pn so that π(x) = n we get pn le 4n
18
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
The importance of ζ(s) in the theory of prime numbers lies in the fact that it combines twoexpressions one of which contains the primes explicitly while the other does not The theoryof primes is largely concerned with the function π(x) the number of primes not exceeding xWe can transform the Euler product into a relation between ζ(s) and π(x) [4 sect11] For σ gt 1
log (ζ(s)) = minussump
log
(1minus 1
ps
)
= minusinfinsumn=2
(π(n)minus π(nminus 1)) log
(1minus 1
ns
)
= minusinfinsumn=2
π(n)
(log
(1minus 1
ns
)minus log
(1minus 1
(n+ 1)s
))
=
infinsumn=2
π(n)
int n+1
n
(d
dxlog
(1minus 1
xs
))dx
=infinsumn=2
π(n)
int n+1
n
s
x(xs minus 1)dx
= s
int infin2
π(x)
x(xs minus 1)dx
The rearrangement of the series is justified since log(1minus 1
ns
)= O(nminusσ) and π(n) le n Therefore
we have
log (ζ(s)) = s
int infin2
π(x)
x(xs minus 1)dx
16 Chebyshev Functions
The first Chebyshev function ϑ(x) is given by
ϑ(x) =sumplex
log(p) = log
prodplex
p
and the second Chebyshev function ψ(x) is given by
ψ(x) =sumpklex
log(p) =sumnlex
Λ(n)
Therefore if pm is the highest power of p not exceeding x log(p) occurs m times in ψ(x) Alsopm is the highest power of p which divides any number upto x so that
ψ(x) = log (U(x))
where U(x) is the least common multiple of all numbers upto x We can also express φ(x) inthe form
ψ(x) =sumplex
lfloorlog(x)
log(p)
rfloorlog(p)
The definition of ϑ(x) and ψ(x) are more complicated than that of π(x) but they are inreality more lsquonaturalrsquo functions [7 sect221] The function ψ(x) can be thought of as a novel wayof counting primes Instead of adding 1 every time a prime occurs it gives a greater weight tothe primes and their powers namely the weight log(p) With such weights the count of primes
19
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
becomes almost equal to the upper bound of the interval in which they are counted Hence thegraph of ψ(x) is almost a straight line (with jump discontinuities) that makes a 45deg angle withhorizontal axis expecially for large numbers x [13]
Since p2 le x p3 le x are equivalent to p le x12 p le x
13 we have
ψ(x) = ϑ(x) + ϑ(x
12
)+ ϑ
(x
13
)+
The series breaks off when x1m lt 2 ie when
m gtlog(x)
log(2)
And from the definition of ϑ(x) we can conclude that ϑ(x) lt x log(x) for all x ge 2 and in fact
ϑ(x
1m
)lt x
1m log(x) le x
12 log(x)
for m ge 2 Therefore summge2
ϑ(x
1m
)= O
(x
12 (log(x))2
)since there are only O(log(x)) terms in the series Hence proving that
ψ(x) = ϑ(x) +O(x
12 (log(x))2
)(117)
Next we note without proof that ϑ(x) is of the order x [7 sect222]
ϑ(x) x and ψ(x) x for x ge 2 (118)
where f(x) φ(x) means Aφ(x) lt f(x) lt Bφ(x) for some positive constants A and B inde-pendent of x ie f is of the same order of magnitude as φ [7 sect16] Hence using (117) we canconclude that ψ(x) is about the same as ϑ(x) when x is large
Now we can write
ϑ(x) =sumplex
log(p) le log(x)sumplex
1 = π(x) log(x)
and so by (118)
π(x) ge ϑ(x)
log(x)gt
Ax
log(x)(119)
for some positive constant A independent of x On the other hand if 0 lt δ lt 1
ϑ(x) gesum
x1minusδltplex
log(p)
ge (1minus δ) log(x)sum
x1minusδltplex
1
= (1minus δ) log(x)(π(x)minus π
(x1minusδ
))ge (1minus δ) log(x)
(π(x)minus x1minusδ
)and so
π(x) le x1minusδ +ϑ(x)
(1minus δ) log(x)lt
Bx
log(x)(120)
20
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
for some positive constant B independent of x By (119) and (120) we conclude that [7 sect18]
π(x) x
log(x)
Also from (119) and (120) it follows that
1 le π(x) log(x)
ϑ(x)le x1minusδ log(x)
ϑ(x)+
1
1minus δ
For any ε gt 0 we can choose δ = δ(ε) so that
1
1minus δlt 1 +
ε
2
and then choose x0 = x0(δ ε) = x0(ε) so that
x1minusδ log(x)
ϑ(x)ltC log(x)
xδltε
2
for all x gt x0 and some positive constant C independent of x Hence
1 le π(x) log(x)
ϑ(x)lt 1 + ε
for all x gt x0 Since ε is arbitrary it follows that
π(x) sim ϑ(x)
log(x)
Then by (117) and (118) it follows that
π(x) sim ϑ(x)
log(x)sim ψ(x)
log(x)
17 The Proof
We will prove the following version of prime number theorem12
π(x) = Li(x) +O(xeminuscradic
log(x))
where Li(x) =int x
2dt
log(t) and c is some positive constant following the method used by Charles
Jean de la Vallee Poussin (1896 1899) [3 sect18]Following is Terence Taorsquos very informal sketch of the proof [20]
1 Create a ldquosound waverdquo (von Mangoldt function) which is noisy at prime number timesand quiet at other times
2 ldquoListenrdquo (Mellin transform) to this wave and record the notes that you hear (the zeros ofthe Riemann zeta function or the music of the primes) Each such note corresponds to ahidden pattern in the distribution of the primes
3 Show that certain types of notes do not appear in this music
4 From this (and tools such as Fourier analysis) one can prove the prime number theorem
We can illustrate the statement of prime number theorem by the following graph
12A weaker result where the error term in O(xeminusc(log(x))
110)
can be proved without much machinery from
complex analysis (like the theory of entire functions of finite order Hadamard products etc) That proof wasgiven by Edmund Landau (1903 1912) and proceeds directly to the function π(x) but involves a nonabsolutelyconvergent integral For that proof one may refer to the book by Ayoub [2 pp 47ndash72]
21
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Figure 13 The dashed solid and dotted curves represent Li(x) π(x) and xlog(x) respectively for 2 le
x le 10000 Plotted using P = plot(Li(x) (x210000) linestyle=-- thickness=2 rgbcolor=(000)) Q = plot(prime pi(x) (x210000)
thickness=2 rgbcolor=(000)) R =plot(xlog(x) (x210000) linestyle= thickness=3 rgbcolor=(000)) (P+Q+R) in SageMath 751
171 Zero-Free Region for ζ(s)
The vital step in the proof of prime number theorem is to show the existence of a thin zero-freeregion for ζ(s) to the left of σ = 1 We will work with the function ζ prime(s)ζ(s) since its analyticcontinuation is easy and has poles only at the zeros of ζ(s) for σ gt 0 From section 14 we knowthat
minusζprime(s)
ζ(s)=infinsumn=1
Λ(n)
ns σ gt 1
Therefore we have
minus Re
(ζ prime(s)
ζ(s)
)=
infinsumn=1
Λ(n)
nσcos(t log(n)) σ gt 1 (121)
since Re (nminuss) = Re(eminus(σ+it) log(n)
)
Now to proceed we need to use following observation by Franz Mertens13 for all θ isin R
2(1 + cos(θ))2 ge 0
rArr2 + 2 cos2(θ) + 4 cos(θ) ge 0
rArr2 + cos(2θ) + 1 + 4 cos(θ) ge 0
rArr3 + 4 cos(θ) + cos(2θ) ge 0
Applied to (121) with t replaced14 by 0 t 2t in succession it gives
3
(minusζprime(σ)
ζ(σ)
)+ 4
(minusRe
(ζ prime(σ + it)
ζ(σ + it)
))+
(minusRe
(ζ prime(σ + 2it)
ζ(σ + 2it)
))ge 0 (122)
13According to Davenport [3 pp 84] the usage of this trick was discovered in 1898 but I couldnrsquot find thatarticle A less known fact is that the symbol micro for Mobius function was introduce by Mertens in 1874 [6 pp382]
14Note that the real parts are functions of cos(t) etc
22
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
The behaviour of minusζ prime(σ)ζ(σ) as σ rarr 1 in view of the simple pole of ζ(s) at s = 1 we have
minus ζ prime(σ)
ζ(σ)lt
1
σ minus 1+A1 (123)
for 1 lt σ le 2 where A1 denotes a positive absolute constantThe behaviour of the other two functions near σ = 1 is influenced by any zero that ζ(s)
has to the left of σ = 1 at a height near to t or 2t Considered the following partial fractionformula shown at the end of subsection 135
minus ζ prime(s)
ζ(s)=
1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) minussumρ
(1
sminus ρ+
1
ρ
)(124)
Using Stirlingrsquos formula (subsection 122) we know that
log(
Γ(s
2+ 1))
=
(s+ 1
2
)log(s
2+ 1)minus(s
2+ 1)
+1
2log(2π) +O
(1∣∣ s
2 + 1∣∣)
Therefore on differentiating it we get
Γprime(s2 + 1
)Γ(s2 + 1
) =1
2log(s
2+ 1)
+s+ 1
2(s+ 2)minus 1
2+O
(1∣∣ s
2 + 1∣∣)lt C log(s)
for some positive constant C Now if we bound σ such that 1 le σ le 2 but keep t unboundedsuch that 2 le t then we can replace C log(s) by some C prime log(t) Hence for some positiveconstant A2 we have
Re
(1
sminus 1minusB minus log(π)
2+
1
2
Γprime(s2 + 1
)Γ(s2 + 1
) ) lt A2 log(t)
if 1 le σ le 2 and 2 le t Hence in this region using (124) we get
minus Re
(ζ prime(s)
ζ(s)
)lt A2 log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)(125)
The sum over ρ is positive since15
Re
(1
sminus ρ
)=
σ minus β|sminus ρ|2
and Re
(1
ρ
)=
β
|ρ|2(126)
where s = σ + it and ρ = β + i`We obtain a valid inequality when s = σ + 2it by just omitting the sum
minus Re
(ζ prime(σ + 2it)
ζ(σ + 2it)
)lt A2 log(t) (127)
For the case when s = σ+ it we choose t to coincide with the ordinate ` of the zero ρ = β+ i`with ` ge 2 and just the one term 1(sminus ρ) in the sum which corresponds to this zero
minus Re
(ζ prime(σ + it)
ζ(σ + it)
)lt A2 log(t)minus 1
σ minus β(128)
Substituting (123) (127) and (128) in (122) we obtain
3
(1
σ minus 1+A1
)+ 4
(A2 log(t)minus 1
σ minus β
)+A2 log(t) gt 0
15Note that 2 Re(z) = z + z
23
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
rArr 4
σ minus βlt
3
σ minus 1+A3 log(t)
Take σ = 1 + δ log(t) where δ is a positive constant Then
β lt 1 +δ
log(t)minus 4δ
(3 +A3δ) log(t)
and if δ is suitably chosen in relation to A3 this gives
β lt 1minus c
log(t)
where c is a positive constant to which a numerical value could be assigned Thus we have
There exists a positive constant numerical constant c such that ζ(s) has no zero inthe region
σ ge 1minus c
log(t) t ge 2
Hence ζ(s) 6= 0 on σ = 1 This was proved independently by Hadamard16 and de la ValleePoussin17 in 1896 [3 sect13]
172 Counting the Zeros of ζ(s)
Consider a rectangle 0 lt σ lt 1 0 lt t lt T and we want to find an approximate formula forN(T ) the number of zeros of ζ(s) in this rectangle But it is convenient to work initially withξ(s) rather than with ζ(s) because it has a simple functional equation and is an entire function(subsection 135) hence itrsquos an analytic function
2
2 + iTminus1 + iT
minus1
12 + iT
O 1
iT
E F
GH J
K
16ldquoSur la distribution des zeros de la fonction ζ(s) et ses consequences arithmetiquesrdquo Bulletin de la SocieteMathematique de France 24 (1896) 199ndash220 httpeudmlorgdoc85858
17ldquoRecherches analytiques la theorie des nombres premiersrdquo Ann Soc scient Bruxelles 20 (1896) 183ndash256httpsarchiveorgdetailsrecherchesanaly00pousgoog
24
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
If we assume (for simplicity) that T (which we suppose to be large) does not coincide withthe ordinate of a zero of ξ(s) and minus1 lt σ lt 2 then using Cauchyrsquos argument principle [1 sect341]we can say that
N(T ) =1
2π∆R arg (ξ(s)) (129)
where ∆R denotes the variation of arg (ξ(s)) round the contour R which is a rectangle withvertices 2 2 + iT minus1 + iT minus1 described in the positive sense
There is no change in arg(ξ(s)) as s describes the base of the base of the rectangle sinceξ(s) is then real and nowhere 0 Further the change in arg(ξ(s)) as s moves from 1
2 + iT tominus1 + iT and then to minus1 is equal to the change as s moves from 2 to 2 + iT and then to 1
2 + iT since
ξ(σ + it) = ξ(1minus σ minus it) = ξ(1minus σ + it)
Hence we can rewrite (129) as
N(T ) = 2
(1
2π∆L arg (ξ(s))
)=
1
π∆L arg (ξ(s)) (130)
where L consists of the edges of rectangle from 2 to 2 + iT and then to 12 + iT
The definition of ξ(x) from (110) can be written as
ξ(s) = (sminus 1)πminuss2 Γ(s
2+ 1)ζ(s)
Hence we have
∆L arg(ξ(s)) = ∆L arg(sminus 1) + ∆L arg(πminus
s2
)+ ∆L arg
(Γ(s
2+ 1))
+ ∆L arg (ζ(s)) (131)
We note that by using the facts that
arctan(x) + arctan
(1
x
)=π
2 x gt 0
and
arctan(x) = minusinfinsumn=0
(minus1)n
2n+ 1x2n+1 |x| le 1
we get
arctan(x) =π
2minus arctan
(1
x
)=π
2minus 1
x+
1
3x3minus 1
5x5+
for large values of x gt 0 Hence we have
∆L arg(sminus 1) = arg
(1
2+ iT minus 1
)minus arg(2minus 1)
= arg
(minus1
2+ iT
)= π minus arctan(2T )
=π
2+ arctan
(1
2T
)=π
2+
1
2Tminus 1
3(2T )3+
1
5(2T )5minus
=π
2+O
(1
T
)
(132)
25
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
We know that ab = eb log(a) hence using the definition of argument we get
∆L arg(πminus
s2
)= ∆L arg
(exp
(minuss
2log(π)
))= arg
(exp
(minus1 + i2T
4log(π)
))minus arg (exp (minus log(π)))
= minusT2
log(π)
(133)
Using the fact that log(f(z)) = log |f(z)|+ i arg(f(z)) and Stirlingrsquos formula (subsection 122)we get
∆L arg(
Γ(s
2+ 1))
= ∆L Im(
log(
Γ(s
2+ 1)))
= Im
(log
(Γ
(5
4+ i
T
2
)))minus Im (log (Γ (2)))
= Im
((3
4+ i
T
2
)log
(5
4+ i
T
2
)minus 5
4minus iT
2+
1
2log(2π) +O
(1
T
))=
3
4arg
(5
4+ i
T
2
)+T
2log
∣∣∣∣54 + iT
2
∣∣∣∣minus T
2+O
(1
T
)=
3
4arctan
(2T
5
)+T
2log
∣∣∣∣∣radic
25 + 4T 2
4
∣∣∣∣∣minus T
2+O
(1
T
)=
3
4
(π
2minus arctan
(5
2T
))+T
2log
(T
2
)minus T
2+O
(1
T
)=
3π
8+T
2log
(T
2
)minus T
2+O
(1
T
)(134)
Now to find ∆L arg(ζ(s)) we will make use of (125) from the previous section for 1 le σ le 2and 2 le t
minusRe
(ζ prime(s)
ζ(s)
)lt A log(t)minus
sumρ
Re
(1
sminus ρ+
1
ρ
)where A is some positive constant In this formula we take s = 2 + iT Since |ζ prime(s)ζ(s)| isbounded for such s we obtain sum
ρ
Re
(1
sminus ρ+
1
ρ
)lt A log(T )
As seen earlier in (126) all the terms in both series are positive and since
Re
(1
sminus ρ
)=
2minus β(2minus β)2 + (T minus `)2
ge 1
4 + (T minus `)2
we conclude that if ρ = β + i` runs through the non-trivial zeros of ζ(s) then for large Tsumρ
1
1 + (T minus `)2= O(log(T ))
From this it follows that
(a) the number of zeros with T minus 1 lt ` lt T + 1 is O(log(T ))
(b) the sumsum
(T minus `)minus2 extended over the zeros with ` outside the interval T minus1 lt ` lt T + 1is also O(log(T ))
26
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Also by (124) applied at s and 2 + it and subtracted we get
minusζprime(s)
ζ(s)= O(log(t)) +
sumρ
(1
sminus ρminus 1
2 + itminus ρ
)We can split the summation over ρ in two parts namely when |tminus `| lt 1 and when |tminus `| ge 1For the terms with |tminus `| ge 1 we have∣∣∣∣ 1
sminus ρminus 1
2 + itminus ρ
∣∣∣∣ =2minus σ
|(sminus ρ)(2 + itminus rho)|le 3
|`minus t|2
and sum of these is of O(log(t)) by the conclusion (b) above As for the terms with |`minus t| lt 1we have |2 + it minus ρ| ge 1 and the number of terms is O(log(t)) by the conclusion (a) aboveHence we deduce that for large t (not coinciding with the ordinate of a zero) and minus1 le σ le 2
ζ prime(s)
ζ(s)=sumρprime
1
sminus ρprime+O(log(t)) (135)
where ρprime = β + i` are those non-trivial zeros of ζ(s) for which |tminus `| lt 1From this fact we can find ∆L arg(ζ(s)) as
∆L arg(ζ(s)) = arg
(ζ
(1
2+ iT
))minus arg(ζ(2))
= Im
(log
(ζ
(1
2+ iT
)))= O(1)minus
int 2+iT
12
+iTIm
(ζ prime(s)
ζ(s)
)ds
= O(1)minusint 2+iT
12
+iTIm
sumρprime
1
sminus ρprime+O(log(t))
ds
= O(1)minussumρprime
int 2+iT
12
+iTIm
(1
sminus ρprime
)ds+O(log(T ))
= O(1)minussumρprime
∆Lprime arg(sminus ρ) +O(log(T ))
where O(1) term is from the variation along σ = 2 and Lprime is the line joining 12 + iT to 2 + iT
Since |∆Lprime arg(s minus ρ)| is at most π and the number of terms in the sumsum
ρprime ∆Lprime arg(s minus ρ) isO(log(T )) by the conclusion (a) above we get
∆L arg(ζ(s)) = O(1) +O(log(T )) +O(log(T ))
= O(log(T ))(136)
Now using (132) (133) (134) and (136) in (131) we get
∆L arg(ξ(s)) =T
2log
(T
2π
)minus T
2+
7π
8+O(log(T ))
And using this in (130) we finally get
N(T ) =T
2πlog
(T
2π
)minus T
2π+
7
8+O(log(T ))
27
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
This was proved by von Mangoldt first in 1895 with a slightly less good error term and thenfully in 1905 It would seem at first sight that we might as well omit the term 78 but as weshall see later it has a certain significance [3 sect15]
From the following plot for T = 50 we can see the number of zeros of the Riemann zetafunction along the critical line18 is 10 and N(50) asymp 94
Figure 14 The dashed and solid curves represent the plot of arg(ζ(s)) and |ζ(s)| respectively forσ = 12 and 1 le t le 50 Plotted using i = CDF0 p1 = plot(lambda t arg(zeta(05+ti)) 1 50linestyle=-- thickness=2
rgbcolor=(000)) p2 = plot(lambda t abs(zeta(05+ti)) 1 50thickness=2 rgbcolor=(000)) p1+p2 in SageMath 751
173 The Explicit Formula for ψ(x)
Riemannrsquos great accomplishment was transforming the problem of describing prime numbersinto the problem of describing the zeros of the Riemann zeta function - which can be attackeddirectly [13] There is an explicit formula for ψ(x) valid for x gt 1 which consists of a sum overthe complex (non-trivial) zeros ρ of ζ(s) It is astonishing that there can be such a formula anexact expression for the number of primes upto x in terms of the zeros of a complicated function[9 sect3] To avoid some minor complications we shall suppose that x ge 2 though the formulawill be valid for x gt 1
Next we will use the inverse Mellin transform called Perronrsquos Formula [1 sect942] which statesthat if x is not an integer c is any positive number and σ gt σ0 minus c then
sumnltx
annw
=1
2πi
int c+iinfin
cminusiinfinf(s+ w)
xs
sds
where f(s) =suminfin
n=1anns is a Dirichlet series The particular case w = 0 is
sumnltx
an =1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds (137)
18Assuming Riemann hypothesis all zeros in region 0 lt σ lt 1 lie on σ = 12
28
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
and applying this to the result obtained in section 14 we getsumnltx
Λ(n) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds c gt 1
and since x is not an integer we can write
ψ(x) =1
2πi
int c+iinfin
cminusiinfin
(minusζprime(s)
ζ(s)
)xs
sds x 6isin Z c gt 1 (138)
If we can move the vertical line of integration away to negative infinity on the left we shallexpress ψ(x) as the sum of the residues [1 sect311] of the function (minusζ prime(s)ζ(s))xss at its polesThe pole of ζ(s) at s = 1 contributes x the pole of 1s at s = 0 contributes minusζ prime(0)ζ(0) andeach zero ρ of ζ(s) whether trivial or not contributes minusxρρ
The basic idea is to make use of following discontinuous integral [2 Theorem 32]
1
2πi
int c+iinfin
cminusiinfin
ys
sds =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
(139)
where c gt 0 We can obtain (137) from (139) by taking y = xn where n lt x such that x 6isin Zas
1
2πi
int c+iinfin
cminusiinfinf(s)
xs
sds =
1
2πi
int c+iinfin
cminusiinfin
infinsumn=1
annsxs
sds
=1
2πi
infinsumn=1
an
int c+iinfin
cminusiinfin
(xn
)s 1
sds
=sumnltx
an
Moreover the method used to prove19 (139) can be extended to prove [3 sect17] that if
δ(y) =
0 if 0 lt y lt 112 if y = 1
1 if y gt 1
and I(y T ) =1
2πi
int c+iT
cminusiT
ys
sds
for y gt 0 c gt 0 T gt 0 then
|δ(y)minus I(y T )| lt
yc min
(1 1
T | log(y)|
)if y 6= 1
cT if y = 1
(140)
To be able to use this result we will have to rewrite ψ(x) Note that just like π(x) ψ(x) isalso a discontinuous function with jump discontinuities at the points where x is a prime powerSo we modify the definition by taking the mean of the values as
ψ0(x) =1
2
sumnlex
Λ(n) +sumnltx
Λ(n)
=
ψ(x)minus Λ(x)
2 if x = pk for some prime p and k gt 0
ψ(x) otherwise
(141)
19We will have to use ML inequality [1 sect231] after selecting the appropriate contour
29
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
By replacing y by xn δ(y) by ψ0(x) and I(y T ) by
J(x T ) =1
2πi
int c+iT
cminusiT
(minusζprime(s)
ζ(s)
)xs
sds (142)
we get from (140) that
|ψ0(x)minus J(x T )| ltinfinsumn=1n6=x
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
+c
TΛ(x) (143)
where c gt 1 and the term containing Λ(x) is present only if x is a prime powerNow we make a choice of c as per our convenience as
c = 1 +1
log(x)
which is equivalent to xc = ex We want to estimate the series on the right of (143) and we will achieve that by considering
various cases20 (so as to avoid n = x)
Case 1 If n le 34x or n ge 5
4x
Then | log(xn)| has a positive lower bound so we get
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) x
T
infinsumn=1
Λ(n)
nc
=x
T
(minusζprime(c)
ζ(c)
) x
Tlog(x)
Case 2 If 34x lt n lt x
Let x1 be the largest prime power less than x we can suppose that 34x lt x1 lt x since
otherwise the terms under consideration vanish
(a) For the term n = x1 using the series expansion of log function (section 14) we have
log(xn
)= minus log
(1minus xminus x1
x
)ge xminus x1
x
and therefore the contribution of this term issumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x1) min
(1
x
T (xminus x1)
) log(x) min
(1
x
T (xminus x1)
)(b) For other terms we can put n = x1 minus ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have
log(xn
)ge log
(x1
n
)= minus log
(1minus ν
x1
)ge ν
x1
20We will be using Vinogradovrsquos symbolism [3 pp 107] where f(x) g(x) is equivalent to f(x) = O(g(x))
30
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x1 minus ν)
Tmiddot x1
ν
x
T(log(x))2
Case 3 If x lt n lt 54x
Let x2 be the least prime power greater than x we can suppose that x lt x2 lt54x since
otherwise the terms under consideration vanish
(a) For the term n = x2 using the series expansion of log function (section 14) we have∣∣∣log(xn
)∣∣∣ = minus log(xn
)= minus log
(1minus x2 minus x
x2
)ge x2 minus x
x2
and therefore the contribution of this term is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣) Λ(x2) min
(1
x2
T (x2 minus x)
) log(x) min
(1
x
T (x2 minus x)
)(b) For other terms we can put n = x2 + ν where 0 lt ν lt 1
4x and then again usingthe series expansion of log function (section 14) we have∣∣∣log
(xn
)∣∣∣ = minus log(xn
)ge log
(n
x2
)= minus log
(1minus ν
n
)ge ν
n
Hence the contribution of these terms is
sumn
Λ(n)(xn
)cmin
(1
1
T∣∣log
(xn
)∣∣)
sum0ltνlt 1
4x
Λ(x2 + ν)
Tmiddot x2 + ν
ν
x
T(log(x))2
If we define 〈x〉 to be the distance from x to the nearest prime power other than x itself incase x is a prime power and replace21 ψ0(x) by ψ(x) we can rewrite the above estimates for(143) as
|ψ(x)minus J(x T )| x(log(x))2
T+ log(x) min
(1
x
T 〈x〉
)(144)
The next step is to replace the vertical line of integration in (142) by the other three sidesof the rectangle with vertices at cminus iT c+ iTminusU + iTminusU minus iT where U is a large odd integer(in order to achieve what we earlier commented for (138)) Thus the left vertical side passeshalfway between two of the trivial zeros of ζ(s)
Hence we can write
J(x T ) =1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds
21Note that ψ0(x) is the same as ψ(x) except that at its jump discontinuities (the prime powers) it takes thevalue halfway between the values to the left and the right So while analysing the asymptotic behaviour we cansafely make this replacement
31
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
+1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds
The sum of the residues of the integrand at its poles inside the rectangle is (as earliercommented for (138))
1
2πi
intC
(minusζprime(s)
ζ(s)
)xs
sds = xminus ζ prime(0)
ζ(0)minussum|`|ltT
xρ
ρminus
sum0lt2mltU
xminus2m
minus2m(145)
where ρ = β + i` is the non-trivial zero of ζ(s)
cminus iT
c+ iTminusU + iT
minusU minus iT
0 1-2-4
E F
GH
Now we need to carefully choose the T By the conclusion (a) of subsection 172 we knowthat for any large T the number of zeros with |` minus T | lt 1 is O(log(T )) ie N(T ) log(T )Among the ordinates of these zeros there must be a gap of length22 1
log(T ) Hence by varyingT by a bounded amount we can ensure that
1
log(T ) |`minus T |
for all zeros ρ = β + i`Now we will determine the contribution made by horizontal integrals depending upon the
choice of σ
Case 1 If minus1 le σ le 2
Recall (135) from subsection 172 that
ζ prime(s)
ζ(s)=
sum|`minusT |lt1
1
sminus ρ+O (log(T ))
22Recall that f(x) = O(g(x)) lArrrArr existMx0 isin RM gt 0 such that |f(x)| lt M |g(x)|forallx gt x0 and hence1
g(x)= O
(1
f(x)
) In Vinogradov symbolism f(x) g(x) lArrrArr 1
g(x) 1
f(x)
32
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
for s = σ+ iT and minus1 le σ le 2 With the present choice of T each term is log(T ) andthe number of terms is also log(T ) Hence on the new horizontal lines of integrationwe have
ζ prime(s)
ζ(s)= O
((log(T ))2
)The contribution made to the horizontal integrals by this range of σ minus1 le σ le 2 istherefore
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds (log(T ))2
int c
minus1
∣∣∣∣xss∣∣∣∣ dσ
(log(T ))2
T
int c
minusinfinxσdσ
x (log(T ))2
T log(x)(146)
Case 2 If minusU le σ le minus1
We need to estimate for |ζ prime(s)ζ(s)| for σ le minus1 From the functional equation stated atthe end of subsection 134
ζ(1minus s) = 21minussπminuss cos(πs
2
)Γ(s)ζ(s)
we note that if 1minusσ le minus1 the functions on the right have to be considered only for σ ge 2The logarithmic derivative is
minusζprime(1minus s)ζ(1minus s)
= minus log(2)minus log(π)minus π
2tan
(πs2
)+
Γprime(s)
Γ(s)+ζ prime(s)
ζ(s)
Observe that tan(πs2
)is bounded if |s minus (2m + 1))| ge 1
2 that is if |(1 minus s) + 2m| ge 12
Also Γprime(s)Γ(s) log |s| and therefore log (2|1minus s|) for σ ge 2 And ζprime(s)
ζ(s) is bounded Henceit follows that ∣∣∣∣ζ prime(s)ζ(s)
∣∣∣∣ log(2|s|) (147)
in the half-plane σ le minus1 provided that the circles of radius 12 (say) around all thetrivial zeros at s = minus2minus4 are excluded Hence the contribution to the remainder ofthe horizontal integrals for minusU le σ le minus1 is
1
2πi
int c+iT
minusU+iT
(minusζprime(s)
ζ(s)
)xs
sds+
1
2πi
int minusUminusiTcminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2T )
T
int minus1
minusUxσdσ
log(T )
Tx log(x)
which is negligible compared with (146)
Also by using (147) we get the contribution made by the vertical integral at σ = minusU as
1
2πi
int minusU+iT
minusUminusiT
(minusζprime(s)
ζ(s)
)xs
sds log(2U)
U
int T
minusTxminusUdt
T log(U)
UxU
33
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
which vanishes as U rarrinfinMaking U rarr infin and adding the estimate (146) to (145) then using (138) and (144) we
obtain
ψ(x) = xminussum|`|ltT
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)+R(x T ) (148)
where
|R(x T )| x(log(xT ))2
T+ log(x) min
(1
x
T 〈x〉
)(149)
As T rarrinfin for any given x ge 2 we have R(x T )rarr 0 and therefore we get the desired explicitformula
ψ(x) = xminussumρ
xρ
ρminus ζ prime(0)
ζ(0)minus 1
2log
(1minus 1
x2
)where the sum over the non-trivial zeros ρ of ζ(s) is to be understood in the symmetric sense aslimTrarrinfin
sum|`|ltT
xρ
ρ The convergence is uniform in any closed interval of x which doesnrsquot containa prime power but not otherwise since ψ(x) is discontinuous at each prime power value of xThis was proved by von Mangoldt in 1895 [3 sect17]
The result (148) and (149) constitute the more precise form of the explicit formula Weproved them subject to a restriction on T but this can now be removed The effect of varyingT by a bounded amount is to change the sum over ρ by O(log(T )) terms and each term isO(xT ) Hence the variation in thersquo sum is O(x(log(T ))T ) and this is covered by the estimateon the right of (149)
We note for future reference that if x is an integer then 〈x〉 ge 1 and (149) takes thesimpler form
|R(x T )| x
T(log(xT ))2 (150)
The results (148) and (149) continue to hold for 1 lt x lt 2 with a slight modification in theform of the estimate for R(x T )
174 Completing the Proof
We have to estimate the sumsum|`|ltT
xρ
ρ in (148) of subsection 173 Firstly we will estimatexρ using the fact that the real part β of ρ is not too near 1 (subsection 171) It follows fromthe conclusion of subsection 171 that if ` lt T where T is large then
β lt 1minus c1
log(T )
where c1 is a positive absolute constant Hence
|xρ| = xβ lt x exp
(minusc1
log(x)
log(T )
)(151)
Secondly we will estimate 1ρ Since |ρ| ge ` for ` gt 0 so we just need to estimatesum
0lt`ltT1`
If N(t) denotes as in subsection 172 the number of zeros in the critical strip with ordinatesless than t this sum is sum
0lt`ltT
1
`=
int T
0
1
tdN(t) =
1
TN(T ) +
int T
0
1
t2N(t)dt
From the conclusion of subsection 172 we know that N(t) t log(t) for large t hence we havesum0lt`ltT
1
|ρ|ltsum
0lt`ltT
1
` (log(T ))2 (152)
34
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Hence combining (151) and (152) we conclude thatsum|`|ltT
∣∣∣∣xρρ∣∣∣∣ x (log(T ))2 exp
(minusc1
log(x)
log(T )
)(153)
Now without loss of generality we can take x to be an integer Then using (153) and (150)in (148) of subsection 173 it follows that
|ψ(x)minus x| x(log(xT ))2
T+ x(log(T ))2 exp
(minusc1
log(x)
log(T )
)(154)
for large x Next we determine T by a function of x by equating (log(T ))2 to log(x) so that
1
T= exp
(minusradic
log(x))
and using this in (154) we get
|ψ(x)minus x| x(log(x))2 exp(minusradic
log(x))
+ x log(x) exp(minusc1
radiclog(x)
) x exp
(minusc2
radiclog(x)
)provided c2 is a constant that is less than both 1 and c1 This proves
ψ(x) = x+O(x exp
(minusc2
radiclog(x)
))Now from this we will derive the desired relation for π(x) First we consider the function
π1(x) =sumnlex
Λ(n)
log(n)(155)
This can be expresses in terms of the function ψ(x) by
π1(x) =sumnlex
Λ(n)
int x
n
dt
t(log(t))2+
1
log(x)
sumnlex
Λ(n)
=
int x
2
ψ(t)
t(log(t))2dt+
ψ(x)
log(x)
We can replace ψ(t) by t to getint x
2
1
(log(t))2dt+
x
log(x)=
int x
2td
dt
(minus 1
log(t)
)dt+
x
log(x)
=
[minust
log(t)
]x2
+
int x
2
1
log(t)dt+
x
log(x)
= Li(x) +2
log(2)
Thus we have
π1(x) = Li(x) +2
log(2)+ E(x t)
where
|E(x t)| int x
2exp
(minusc2
radiclog(t)
)dt+ x exp
(minusc2
radiclog(x)
)
35
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
The contribution of the range t lt x14 to the integral is trivially less than x
14 and in the rest of
the range we haveradic
log(t) gt 12
radiclog(x)
Henceπ1(x) = Li(x) +O
(x exp
(minusc3
radiclog(x)
))(156)
where c3 = c22
Using the definition of Λ(n) from section 14 we can rewrite (155) as
π1(x) =sumpmlex
log(p)
m log(p)
= π(x) +1
2π(x
12
)+
1
3π(x
13
)+
since π(x) =sum
plex 1 from section 15 Also since π(x
12
)le x
12 π
(x
13
)le x
13 the difference
between π1(x) and π(x) is O(x12 ) Thus using this in (156) we get
π(x) = Li(x) +O(x exp
(minusc3
radiclog(x)
))This was proved by de la Vallee Poussin in 1899 [3 sect18]
18 Some Remarks
181 Lindelof Hypothesis
Besides the Riemann hypothesis another important conjecture involving the Riemann zetafunction is the Lindelof hypothesis given by Ernst Leonard Lindelof in 1908 The conjecturestates that
ζ
(1
2+ it
)= O(tε)
for every positive ε In other words itrsquos concerned with the growth of the Riemann zeta functionon the line σ = 1
2 and conjectures that the modulus of ζ(12+it) grows slower than any positivepower t as t tends to infinity
Riemann hypothesis implies Lindelof hypothesis but Lindelof hypothesis does not implyRiemann hypothesis It was the Lindelof hypothesis that lead to the study of moments ofRiemann zeta function The 2kth moment of the modulus of the Riemann zeta function isdefined as
Ik(T ) =1
T
int T
0
∣∣∣∣ζ (1
2+ it
)∣∣∣∣2k dtThe Lindelof hypothesis is equivalent to the statement that for any k and any positive εIk(T ) = O(T ε)
However moments are now appriciated in their own right becuase they represent meanvalues of the Riemann zeta function on the critical line over a finite interval and estimate ofthese average values can provide information about the zeros of ζ(s) A nice exposition aboutthe moments of Riemann zeta function is given in Jennifer Beineke and Chris Hughesrsquos article[11]
182 Elementary Proof
A simple question like ldquoHow many primes are there up to xrdquo deserves a simple answer onethat uses elementary methods rather than all the methods of complex analysis which may seemfar from the question at hand [9 sect3] In 1948 using a fundamental inequality of Atle Selberg
36
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Selberg and Paul Erdos succeeded in giving an elementary proof of the prime number theoremThe proof is based on Selbergrsquos formulasum
pltx
(log(p))2 +sumpqltx
(log(p))(log(q)) = 2x log(x) +O(x)
which is completely combinatorial in nature For that proof one may refer [7 Ch XXII] Muchabout this elementary proof can be summarised by quoting following lines by Carl Pomerance[17]
Thus far from being an isolated intellectual challenge the elementary proof of theprime number theorem was a signal that good ideas and strong tools are close athand We already had an inkling of this in Riemannrsquos era when Chebyshev usedcombinatorial methods to show that there is a prime in [n 2n] for every naturalnumber n And a century ago the elementary proof of Brun stating that mostprimes are not part of twin-prime pairs opened the door for combinatorial sievemethods and their many glorious consequences
Since the elementary proof some of the most profound and exciting results in thefield have had strong elementary and combinatorial leanings After Roth used the(analytic) circle method to show that dense sets of integers must have 3-term arith-metic progressions Szemeredi used an elementary (and very complicated) proof togeneralize this to k-term arithmetic progressions This result became an intrinsictool in the recent GreenndashTao proof that the set of primes contains arbitrarily longarithmetic progressions
The Green-Tao theorem was proved by Ben Green and Terence Tao23 in 2004 [20][9]
183 Heuristic Approach
A heuristic technique is any approach to problem-solving that employs a practical methodnot guaranteed to be optimal or perfect but sufficient for the immediate goals Commonheuristic techniques are drawing a picture working backward by assuming a solution examininga concrete example if problem is too abstract and examining a more general problem if theproblem is too specific
But for the prime number theorem there is a problem with an attempt at a heuristicexplanation because the sieve of Eratosthenes does not behave as one might guess it would frompure probabilistic considerations [9 sect2] Sieving out the composites under x using primes uptoradicx would lead to (as seen in subsection 132)
xprodpltradicx
(1minus 1
p
)
which turns out to be asymptotic to 2eγ log(x) instead of x
log(x) (proved by Franz Mertens in
1874) Thus the sieve is about 11 more efficient at eliminating composites than one mightexpect In 2006 Hugh L Montgomery and Stan Wagon transformed an old heuristic approachinto a proof of following result[16]
If xπ(x) is asymptotic to an increasing function then π(x) sim x log(x)
The proof involves the use of elementary calculus and Tauberian result instead of complexfunction theory Hence there is a possibility of finding another proof of prime number theoremby proving that xπ(x) is asymptotic to an increasing function
23ldquoThe primes containing arbitrarily long arithmetic progressionsrdquo Annals of Mathematics 167 no 2 (2008)481ndash547 httparxivorgabsmath0404188
37
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Chapter 2
Primes in Arithmetic Progression
ldquoThe great fusion between arithmetic and analysisndashbetween counting and measuringbetween numbers staccato and numbers legatondashcame about as the result of an inquiryinto prime numbers conducted by Lejeune Dirichlet in 1830srdquo
mdash John Derbyshire Prime Obsession
It is an interesting question whether it is possible to define an infinite number of primenumbers using only addition subtraction and multiplication Letrsquos consider the polynomials inone variable whose coefficients are natural numbers Peter Gustav Lejeune Dirichlet proved thatthe linear polynomial ax+ b where a b and x are positive integers with gcd(a b) = 1 definesan infinite number of prime numbers It is easy to prove that no non-constant polynomial cangenerate only prime values but it is still unknown whether there exists a polynomial of degree2 or more that generates infinitely many primes [8][23]
21 Dirichlet Density
The Dirichlet density D(P) of a set P of prime numbers if exists is given by
D(P) = limσrarr1
minus1
log(σ minus 1)
sumpisinP
1
pσ
We can draw the following conclusions from this definition [5 sect161]
(a) If P has finitely many elements then D(P) = 0
(b) If P consists all but finitely many positive primes then D(P) = 1
(c) If P = P1 cup P2 where P1 and P2 are disjoint and D(P1) and D(P2) both exist thenD(P) = D(P1) + D(P2)
22 Dirichlet Characters
Let m be a fixed positive integer Let χprime (ZmZ)times rarr Ctimes be a homomorphism Given χprimedefine χ Zrarr Ctimes as follows
χ(a) =
χprime(a) if gcd(am) = 1
0 if gcd(am) 6= 1
The functions χ defined in this manner are called Dirichlet characters modulo m
38
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
221 Number of Dirichlet Characters
If G is a multiplicatively written finite abelian group then the character on G is a homomor-phism from G to Ctimes [28 Chapter 2] We will denote the set of all such characters by G Ifχ χprime isin G define χχprime to be the function which takes g isin G to χ(g)χprime(g) Then χχprime is also acharacter Letrsquos define χ0 to be the trivial character ie χ0(g) = 1 for all g isin G If χ isin Gdefine χminus1 by χminus1(g) = 1
χ(g) for all g isin G With these definitions G becomes an abelian groupwith χ0 as the identity element
If n is the order of G then for g isin G we have gn = e where e is the identity element
of G So if χ isin G then (χ(g))n = 1 ie the values of χ are nth roots of unity Therefore
χ(g) = 1χ(g) = χminus1(a) where bar denotes the complex conjugation Therefore χminus1 can be
written as χ and called the conjugate character of χIn general G is a direct product of cyclic groups ie there are elements g1 g2 gt isin G
such that order of gk is nk with n = n1n2 middot middot middotnt and every element g isin G can be uniquely writtenin the form g = gm1
1 gm22 middot middot middot gmtt where 0 le mk lt nk for all k Hence we can conclude that G
is generated by χk for all 1 le k le t such that χk(gk) = e2πink and χk(g`) = 1 for k 6= ` And
hence we have G sim= G where G is the direct product of the cyclic subgroups generated by theχk with order of χk being nk [5 sect163] Therefore |G| = |G| = n If we take G = (ZmZ)times
and note that though Dirichlet characters are defined on Z but are induced by the elements inthe character group of (ZmZ)times we can conclude that
There are exactly φ(m) Dirichlet characters modulo m where φ(m) = | (ZmZ)times |
222 Orthogonality Relations
Let χ isin G then we havesum
gisinG χ(g) = n for χ = χ0 But if χ 6= χ0 then there is a gprime isin G suchthat χ(gprime) 6= 1 and hence sum
gisinGχ(g) =
sumgisinG
χ(ggprime) = χ(gprime)sumgisinG
χ(g)
Sosum
gisinG χ(g) = 0 if χ 6= χ0 Therefore for χ χprime isin G we have
sumgisinG
χχprimeminus1(g) =sumgisinG
χ(g)χprime(g) =
n if χ = χprime
0 if χ 6= χprime(21)
Also for g isin G we havesum
χisinG χ(g) = n if g = e But if g 6= e then g = gm11 gm2
2 middot middot middot gmtt with
0 le mk lt nk for all k and at least one mk 6= 0 Since χ(g) = χ1(g)e1χ2(g)e2 middot middot middotχt(g)et with
0 le ek lt nk we have χk(g) = χk(gmkk ) = χk(gk)
mk = e2πimknk 6= 1 and hencesum
χisinG
χ(g) =sumχisinG
χkχ(g) = χk(g)sumχisinG
χ(g)
Sosum
χisinG χ(g) = 0 for a 6= e Therefore for g gprime isin G we have
sumχisinG
χ(ggprimeminus1) =sumχisinG
χ(g)χ(gprime) =
n if g = gprime
0 if g 6= gprime(22)
From (21) and (22) we conclude that
39
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Let χ and χprime be Dirichlet characters modulo m and a b isin Z then
(a)
mminus1suma=0
χ(a)χprime(a) =
φ(m) if χ = χprime
0 if χ 6= χprime
(b)sumχ
χ(a)χ(b) =
φ(m) if a equiv b (mod m)
0 if a 6equiv b (mod m)
23 L-function
231 Dirichlet L-function
Let χ be a Dirichlet character modulo m We define the Dirichlet L-function associated to χby the formula
L(s χ) =
infinsumn=1
χ(n)
ns
where s = σ + it is a complex number Since |χ(n)| = 1 (subsection 221) we have∣∣∣∣χ(n)
ns
∣∣∣∣ le 1
ns
and hence we see that the terms of L(s χ) are dominated in absolute value by the correspondingterms of Euler zeta function ζ(s) (subsection 131) Thus L(s χ) converges and is analytic forσ gt 1
232 Product Formula
Since χ is completely multiplicative we have a product formula for L(s χ) in exactly the sameway as for ζ(s) (subsection 132)
L(s χ) =prodp
(1minus χ(p)
p
)minus1
σ gt 1
Since χ(p) = 0 for p|m the above product is over positive primes not dividing m There isa close connection between L(s χ0) and ζ(s) In fact
L(s χ0) =prodp-m
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)prodp
(1minus 1
ps
)minus1
=prodp|m
(1minus 1
ps
)ζ(s)
(23)
From this we conclude that
L(s χ0) has a pole at s = 1 hence L(1 χ0) 6= 0
40
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
233 Logarithm Formula
The values of L(s χ) are in general complex (even if we restrict s to be real) So we choose theprincipal branch of the logarithm of product formula of L(s χ) and then do its series expansion(section 14) to get
G(s χ) = log(L(s χ)) =sump
sumk
χ(pk)
kpks σ gt 1
where p runs through all primes and k through all positive integers Moreover we have
G(s χ) =sump-m
χ(p)
ps+sump
infinsumk=2
χ(pk)
kpks
and using triangle inequality and (1minus x)minus1 = 1 + x+ x2 + for |x| lt 1 we get∣∣∣∣∣sump
infinsumk=2
χ(pk)
kpks
∣∣∣∣∣ lesump
infinsumk=2
∣∣∣∣∣χ(pk)
kpks
∣∣∣∣∣=sump
infinsumk=2
1
kpks
lesump
infinsumk=2
1
pks
=sump
1
p2s
(1minus 1
ps
)minus1
le(
1minus 1
2s
)minus1sump
1
p2s
le 2ζ(2)
Therefore we have
G(s χ) =sump-m
χ(p)
ps+Rχ(s) σ gt 1 (24)
where Rχ(s) remains bounded around s = 1 Now letrsquos multiply both sides of (24) by χ(a)where a isin Z with gcd(am) = 1 and then sum over all Dirichlet characters modulo m to getsum
χ
χ(a)G(s χ) =sump-m
1
ps
sumχ
χ(a)χ(p) +sumχ
χ(a)Rχ(s)
Using the result (b) from subsection 222 we seesumχ
χ(a)G(s χ) = φ(m)sum
pequiva (mod m)
1
ps+Rχa(s) σ gt 1 (25)
where Rχa(s) remains bounded around s = 1Now restricting s to be real ie t = 0 and s = σ in (24) and (25) we get
41
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Let σ gt 1 be a real number then
(a) G(σ χ) =sump-m
χ(p)
pσ+Rχ(σ)
where Rχ(σ) remains bounded as σ rarr 1
(b)sumχ
χ(a)G(σ χ) = φ(m)sum
pequiva (mod m)
1
pσ+Rχa(σ)
where Rχa(σ) remains bounded as σ rarr 1
234 Analytic Continuation
We want to analytically continue L(s χ) to σ gt 0 from σ gt 1 when χ is a non-trivial Dirichletcharacter modulo m We have already seen such analytic continuation of ζ(s) in subsection 135as (113) We can use same technique to extend L(s χ) Let A(x) =
sumalex χ(a) then by taking
xrarrinfin in Abelrsquos partial summation (Appendix A) we get
L(s χ) = s
int infin1
A(x)
xs+1dx
Since χ(a + m) = χ(a) for all a isin Z (χ 6= χ0) we note that if N = bxc = qm + r for someq r isin Z with 0 le r lt m then
|A(x)| =
∣∣∣∣∣Nsuma=0
χ(a)
∣∣∣∣∣ =
∣∣∣∣∣q(mminus1suma=0
χ(a)
)+
rsuma=0
χ(a)
∣∣∣∣∣=
∣∣∣∣∣rsum
a=0
χ(a)
∣∣∣∣∣le
mminus1suma=0
|χ(a)|
= φ(m)
using the result (a) of subsection 222 which sayssummminus1
a=0 χ(a) = 0 and that |χ(a)| = 1 From thiswe conclude that |A(x)| le φ(m) for all x and hence we get that the above integral convergesTherefore if χ is non-trivial Dirichlet character modulo m then L(s χ) can be continued to aanalytic function in the region s isin C|σ gt 0
235 Product of L-functions
Let F (s) =prodχ
L(s χ) where the product is over all Dirichlet character modulo m Assume s
is real and s gt 1 ie s = σ gt 1 From subsection 233 we know that
G(σ χ) =sump
infinsumk=1
χ(pk)
kpkσ
Summing over χ and using the conclusion (b) of subsection 222 we getsumχ
G(σ χ) = φ(m)sump
sumk
1
kpkσ(26)
where the sum is over all primes p and integers k such that pk equiv 1 (mod m) The right-handside of (26) is non-negative hence taking the exponential of both sides we get
For s real and s gt 1 ie s = σ gt 1 we have F (s) = F (σ) ge 1
42
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
24 The Proof
Let P(am) be the set of prime numbers p such that p equiv a (mod m) We wish to provethat P(am) has infinite number of elements We will divide the proof into two parts whilerestricting s to real ie t = 0 we will write s = σ [5 sect164]
241 Trivial Dirichlet Character
Let χ0 denote the trivial character modulo m then L(σ χ0) is a real function of positive realnumbers From (23) it follows that
G(σ χ0) =sump|m
log
(1minus 1
pσ
)+ log(ζ(σ)) (27)
As seen in subsection 131 for s = σ
limσrarr1
(σ minus 1)ζ(σ) = 1
which implies that [5 sect161]
limσrarr1
minus log(ζ(σ))
log(σ minus 1)= 1 (28)
Using (28) in (27) we get
limσrarr1
minusG(σ χ0)
log(σ minus 1)= 1
242 Non-trivial Dirichlet Character
Let χ be a non-trivial Dirichlet character modulo m We need to ensure that L(1 χ) 6= 0 andto prove this we will consider following two cases
Complex Character
Let χ be a complex character modulo m ie a character which takes only non-real valuesFrom the series defining L(s χ) we see that for s = σ real σ gt 1
L(σ χ) = L(σ χ)
Assume L(1 χ) = 0 then L(1 χ) = 0 Hence the functions L(σ χ) and L(σ χ) are distinct andboth have a zero at σ = 1 In the product F (σ) =
prodχ L(σ χ) we know L(σ χ0) has a simple
pole at σ = 1 (subsection 232) and all other factors are analytic about σ = 1 (subsection 234)It follows that F (1) = 0 But from subsection 235 we know that F (σ) ge 1 for all σ gt 1 Thiscontradicts our assumption Therefore L(1 χ) 6= 0 when χ is a non-trivial complex charactermodulo m
Real Character
Let χ be a non-trivial real character ie χ(a) = 0 1 or minus1 for all a isin Z We will make use of fol-lowing result whose proof is analogous to the proof of Eulerrsquos product formula (subsection 132)[5 sect165]
Suppose f is a non-negative multiplicative function of Z+ ie for all a b gt 0 withgcd(a b) = 1 f(ab) = f(a)f(b) Assume there is a constant c such that f
(pk)lt c
for all prime powers pk Thensuminfin
n=1 f(n)nminuss converges for all σ gt 1 Moreover
infinsumn=1
f(n)
ns=prodp
(1 +
infinsumk=1
f(pk)
pks
)(29)
43
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Assume L(1 χ) = 0 and consider the function
ω(s) =L(s χ)L(s χ0)
L(2s χ0)(210)
The zero of L(s χ) at s = 1 cancels the simple pole of L(s χ0) so the numerator is analyticon σ gt 0 (subsection 234) The denominator is non-zero and analytic for σ gt 1
2 Thus ω(s)is analytic on σ gt 1
2 (compare with subsection 135) Moreover since L(2s χ0) has a pole ats = 1
2 (subsection 232) we have for real s ie s = σ ω(σ)rarr 0 as σ rarr 12
We can re-write (210) to get the infinite product expansion (subsection 232)
ω(s) =prodp
(1minus χ(p)
ps
)minus1(1minus χ0(p)
ps
)minus1(1minus χ0(p)
p2s
)
=prodp-m
(1minus χ(p)
ps
)minus1(1minus 1
ps
)minus1(1minus 1
p2s
)
=prodp-m
(1 + pminuss)
(1minus χ(p)pminuss)
If χ(p) = minus1 the pminusfactor is equal to 1 Thus
ω(s) =prod
χ(p)=1
1 + pminuss
1minus pminuss(211)
where the product over all p such that χ(p) = 1 Moreover
1 + pminuss
1minus pminuss=
(1 +
1
ps
)( infinsumk=0
1
pks
)
= 1 +2
ps+
2
p2s+
Using this and applying (29) in (211) we find that
ω(s) =infinsumk=1
akks
(212)
where ak ge 0 and the series converges for σ gt 1 Note that a1 = 1Since ω(s) is analytic for σ gt 1
2 expanding ω(s) in a power series about s = 2 using Taylorrsquostheorem [1 sect243] we get
ω(s) =
infinsumj=0
ω(j)(2)
j(sminus 2)j (213)
where ω(j)(s) is the jth derivative of ω(s) with the radius of convergence at least 32 [1 sect71]
But from (212) we get
ω(j)(2) =infinsumk=1
ak(minus log(k))j
k2= (minus1)jcj (214)
where cj ge 0 Now using (214) in (213) we get
ω(s) =
infinsumj=0
cj(2minus s)j
44
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
where cj is non-negative and
c0 = ω(2) =
infinsumk=1
akk2ge a1 = 1
It follows that for real s ie s = σ in the interval 12 lt σ lt 2 we have ω(σ) ge 1 This
contradicts our assumption that ω(σ)rarr 0 as σ rarr 12 and so L(1 χ) 6= 0
Combining the above two cases we can say that
If χ is a non-trivial Dirichlet character modulo m then G(σ χ) remains bounded as σ rarr 1
This was proved by de la Vallee Poussin1 in 1896 [3 sect4]
243 Completing the Proof
Now we simply divide all terms on both sides of the conclusion (b) of subsection 233 byminus log(σ minus 1) and take the limit σ rarr 1
limσrarr1
sumχ
χ(a)minusG(σ χ)
log(σ minus 1)= φ(m) lim
σrarr1
minus1
log(σ minus 1)
sumpisinP(am)
1
pσ+ limσrarr1
minusRχa(σ)
log(σ minus 1)
From the results of subsection 241 and subsection 242 we get the limit on left-hand side is1 (because all except trivial-character will vanish) whereas the limit on the right-hand side isφ(m)D(P(am)) (section 21) Thus D(P(am)) = 1φ(m) and we are done
If am isin Z with gcd(am) = 1 then
D(P(am)) =1
φ(m)
where φ(m) is the number of positive integers n less than m with gcd(nm) = 1
Since D(P(am)) 6= 0 this implies that P(am) has infinite number of elements
25 Some Remarks
251 Chebotarev Density Theorem
Chebotarevrsquos density theorem may be regarded as the least common generalization of Dirich-letrsquos theorem on primes in arithmetic progressions (1837) and a theorem of Frobenius (1880published 1896) [28 sect11] Generally Frobeniusrsquos theorem for f(X) = Xm minus 1 is implied byDirichletrsquos theorem for the same m but not conversely One can formulate a sharper version ofFrobeniusrsquos theorem that for f(X) = Xmminus1 does come down to Dirichletrsquos theorem To do thiswe need do something called Frobenius substitution of p σp Once the Frobenius substitutionhas been defined one can wonder about the density of the set of primes p for which σp is equalto a given element of the Galois group G of f(X) corresponding to the field K generated bythe zeros of f(X) Note that the Frobenius map Frobp is an automorphism of the field Fp ofcharacteristic p and the Frobenius substitution σp is going to be an automorphism of the fieldK of characteristic zero
Theorem (Reformulation of Dirichletrsquos Theorem) If f(X) = Xmminus1 for some positive integerm then the set of prime numbers p for which σp is equal to a given element of the Galois groupG of f(X) has a density and this density equals 1|G|
1ldquoRecherches analytiques sur la theorie des nombres premiersrdquo Deuxieme partie Ann Soc Sci Bruxelles20 281-362 (1896)
45
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Thus the Frobenius substitution is equidistributed over the Galois group if p varies over allprimes not dividing m This leads to the desired generalization of the theorems of Dirichlet andFrobenius It was formulated as a conjecture by Ferdinand Georg Frobenius and was ultimatelyproved by Nikolaı Chebotarev in 1922 Chebotarevrsquos theorem extends this to all f(X)
Theorem (Chebotarevrsquos Density Theorem) Let f(X) be a polynomial with integer coefficientsand with leading coefficient 1 Assume that the discriminant ∆(f) of f(X) does not vanish LetC be a conjugacy class of the Galois group G of f(X) Then the set of primes p not dividing∆(f) for which σp belongs to C has a density and this density equals |C||G|
For details refer to the article by P Stevenhagen and H W Lenstra Jr[14]
252 Prime Number Theorem for Arithmetic Progressions
We can combine the the results in the two chapters to talk about quantitative aspects of distri-bution of primes in arithmetic progression
π(xm a) sim π(x)
φ(m)
The analytic proof of this statement is analogous to that of prime number theorem with ζ(s)replaced by L(s χ) For that proof one can refer to Davenportrsquos book [3 sect20] For discus-sion regarding the elementary proof on may refer to this MathOverflow discussion initiated byQiaochu Yuan Is a ldquonon-analyticrdquo proof of Dirichletrsquos theorem on primes known or possibleURL (version 2011-09-29) httpsmathoverflownetq16735
One can make enormous improvements in the error term of this estimate provided thereare no Siegel zeros And the Siegel zeros are rare as a consequence of Deuring-Heilbronn phe-nomenon which states that the zeros of L-functions repel each other just like in Rothrsquos theorem[26] different algebraic numbers repel each other There is a close relation between Siegel zerosand class numbers [27 sect41] For details refer to Andrew Granvillersquos article [9]
253 Tchebychev Bias
This phenomenon was first observed in a letter written by Pafnuty Tchebychev to M Fuss on23 March 1853
There is a notable difference in the splitting of the prime numbers between the twoforms 4k + 3 4k + 1 the first form contains a lot more than the second
This bias is perhaps unexpected because as per the the prime number theorem for arithmeticprogressions discussed above the primes tend to be equally split amongst the various forms p equiv a(mod m) with gcd(am) = 1 for any given modulus m Hence we have
π(x 4 1) sim π(x 4 3) sim x
2 log(x)
ie half the primes are of the form 4k + 1 and half of the form 4k + 3 This asymptotic resultdoes not inform us about any of the fine details of these prime number counts so neither verifiesnor contradicts the observation that π(x 4 3) gt π(x 4 1)
But as we count the number of primes for various values of x we observe that from timeto time more primes of the form 4k + 1 than of the form 4k + 3 but this lead is held onlyvery briefly and then relinquished for a long stretch Now one might guess that 4n + 1 willoccasionally take the lead as we continue to watch for bigger x Indeed this is the case as JohnEdensor Littlewood discovered in 1914
46
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Theorem (Littlewood 1914) There are arbitrarily large values of x for which there are moreprimes of the form 4k + 1 up to x than primes of the form 4k + 3 In fact there are arbitrarilylarge values of x for which
π(x 4 1)minus π(x 4 3) geradicx log(log(log(x)))
2 log(x)
At first sight this seems to be the end of the story But in 1962 Stanislaw Knapowskiand Pal Turan made a conjecture that is consistent with Littlewoodrsquos result but also bears outTchebychevrsquos observation
As X rarr infin the percentage of integers x le X for which there are more primes ofform 4k + 3 up to x than of the form 4k + 1 goes to 100
This conjecture may be paraphrased as ldquoTchebychev was correct almost all of the timerdquo Fordetails and recent develpoments refer to the article by Andrew Granville and Greg Martin [15]
47
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Conclusion
ldquoRiemann discovered that prime numbers too could be studied by harmonic analysisalbeit of a slightly different kind He realized that the psi function can be thought ofas a sum of elementary waveforms To me that the distribution of primenumbers can be so accurately represented in a harmonic analysis is absolutelyamazing and incredibly beautiful It tells of an arcane music and a secret harmonycomposed by the prime numbersrdquo
mdash Enrico Bombieri Prime Territory
One very important tool for our understanding of primes that I couldnrsquot discuss in thisreport is the computational tool which enables us to establish faith in the conjectures thatwe are trying to prove For example Alan Turing2 (1912ndash1954) was very much interested indisproving Riemann hypothesis by finding a counter-example using an automatic computer Fordetails about Turingrsquos attempts in this direction refer to the lecture by Yuri Matiyasevich [22]
As pointed in the beginning of second chapter there canrsquot be any polynomial in one variablethat gives only prime number values But we have following remarkable result by Julia Robinsonand Hilary Putnam
There exists an exponential polynomial R(x0 xk) such that all of its positivevalues for positive integer variables are prime numbers and every prime number isso presented
One can find a good discussion on this theorem by Yuri Matiyasevich [8] who in 1971extended this result[23] This result played an important role in solving Hilbertrsquos Tenth Problemwhose statement I have discussed earlier [24 Conclusion] But this result is of no practical usesince it will be computationally difficult to calculate prime numbers by solving this system ofequations
Luckily we have better ways of determining whether a given number is a prime or not In1970s Gary Miller3 showed how a generalization of the Riemann Hypothesis allows a determin-istic polynomial-time procedure for recognizing primes a few decades later in 2002 ManindraAgrawal Neeraj Kayal and Nitin Saxena4 showed the same with completely elementary (andrigorous) methods [17] But so far this new polynomial-time primality test is not good inpractice But there exist ways for primanlity testing like using the arithmetic of elliptic curveswhich are conjectured to run in polynomial time but we have not even proved that they alwaysterminates [10]
Practical importance of prime numbers lies in the modern public key encryption [25] which isbased on the belief that there canrsquot exist a polynomial-time algorithm for factorising compositenumbers (unlike the efficient Euclidrsquos algorithm for finding the greatest common divisor) Mostpractical factoring algorithms are based on unproved but reasonable-seeming hypotheses about
2He is widely considered to be the father of theoretical computer science and artificial intelligence3ldquoRiemannrsquos Hypothesis and tests for primalityrdquo Proceedings of the seventh annual ACM symposium of
Theory of Computing 1975 pp 234ndash239 doi1011458001168037734ldquoPRIMES is in Prdquo Annals of Mathematics 160 no 2 (2004) 781ndash793 doi 104007annals2004160781
48
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
the natural numbers Although we may not know how to prove rigorously that these methodswill always produce a factorisation or do so quickly in practice they do Broadly speaking thereare two types of factorisation methods sieve-based methods [12] and elliptic curve method Anice exposition about computational aspects of analytic number theory (the distribution ofprimes and the Riemann hypothesis) Diophantine equations (Fermatrsquos last theorem and theabc conjecture) and elementary number theory (primality and factorisation) can be found inCarl Pomerancersquos article [10]
I will end my report by stating the two recent developments towards our understanding ofprime numbers
In 2013 Yitang Zhang5 proved that
lim infnrarrinfin
(pn+1 minus pn) lt 7times 106
where pn denotes the nth prime This was a result of Zhangrsquos attempt to prove theTwin Primes conjecture The proof was inspired by idea of Bombieri-Friedlander-Iwaniec6
and Goldston-Pintz-Yıldırım7 sieve (which makes the connection between prime gaps andprimes in arithmetic progression) Few months later James Maynard and Terence Tao (in-dependently) gave a different simpler approach than Zhangrsquos which reduced the boundedgaps between primes to 246 [23] One interesting aspect of the better bounds is that thebest bounds were obtained by using important theorems in Algebraic Geometry whatare known as the lsquoWeil Conjecturesrsquo which at first glance appear to be far removed fromTwin Primes [18]
In 2016 Kannan Soundararajan and Robert Lemke Oliver8 discovered that that primesseem to avoid being followed by another prime with the same final digit They presentedboth numerical and theoretical evidence that prime numbers repel other would-be primesthat end in the same digit and have varied predilections for being followed by primesending in the other possible final digits [19]
5ldquoBounded gaps between primesrdquo Annals of Mathematics 179 no 3 (2014) 1121ndash1174 doi 104007an-nals201417937
6Bombieri E Friedlander J B and Iwaniec H ldquoPrimes in arithmetic progressions to large modulirdquoActa Mathematica 156 (1986) 203ndash251 doi101007BF02399204 httpprojecteuclidorgeuclidacta
14858904167Goldston D A Pintz J and Yıldırım C Y ldquoPrimes in Tuples Irdquo Annals of Mathematics 170 no 2
(2009) 819ndash862 doi104007annals2009170819 httpsarxivorgabsmath05081858ldquoUnexpected Biases in the Distribution of Consecutive Primesrdquo Proceedings of the National Academy of
Sciences 113 no 31 (2016) E4446ndashE4454 doi101073pnas1605366113 httpsarxivorgabs160303720
49
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Appendix A
Abelrsquos Summation Formula
This formula is based on our understanding of RiemannndashStieltjes integral We will follow [7sect225]
Theorem Suppose that a1 a2 is a sequence of numbers such that A(t) =sumnlet
an and f(t)
is any function of t Thensumnlex
anf(n) =sum
nlexminus1
A(n)(f(n)minus f(n+ 1)
)+A(x)f(bxc) ()
If in addition aj = 0 for all j less than some positive integer m and f(t) has a continuousderivative for t ge m then sum
nlexanf(n) = A(x)f(x)minus
int x
mA(t)f prime(t)dt ()
Proof We rewrite the left-hand-side of () assumnlex
anf(n) = A(1)f(1) +(A(2)minusA(1)
)f(2) + +
(A(bxc)minusA(bxc minus 1)
)f(bxc)
= A(1)(f(1)minus f(2)
)+ +A(bxc minus 1)
(f(bxc minus 1)minus f(bxc)
)+A(bxc)f(bxc)
Since A(bxc) = A(x) this proves () To deduce () we observe that A(t) = A(n) a constantwhen n le t lt n+ 1 and so
A(n)(f(n)minus f(n+ 1)
)= minus
int n+1
nA(t)f prime(t)dt
Also A(t) = 0 when t lt m Hence proving ()
If we put an = 1 and f(t) = 1t we have A(x) = bxc and () becomessumnlex
1
n=bxcx
+
int x
1
btct2dt
=xminus x
x+
int x
1
(tminus t
) 1
t2dt
= 1minus xx
+ log(x)minusint x
1
tt2dt
50
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Observe that ∣∣∣∣int x
1
tt2dt
∣∣∣∣ le int x
1|t| 1
t2dt
leint x
1
1
t2dt
= 1minus 1
x
Hence the integral
int x
1
tt2dt converges to a limit as xrarrinfin Thus we can write
int x
1
tt2dt =
int infin1
tt2dtminus
int infinx
tt2dt
Therefore we havesumnlex
1
n= log(x) +
(1minus
int infin1
tt2dt
)+
(int infinx
tt2dtminus x
x
)= log(x) + γ +
(int infinx
O(1)
t2dt+O
(1
x
))Therefore sum
nlex
1
n= log(x) + γ +O
(1
x
)where γ is the Euler-Mascheroni constant given by
γ = limNrarrinfin
(1 +
1
2+ +
1
Nminus logN
)asymp 057721
51
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
Bibliography
[1] Titchmarsh E C The Theory of Functions London Oxford University Press 1939httpsarchiveorgdetailsTheTheoryOfFunctions
[2] Ayoub R An Introduction to the Analytic Theory of Numbers (Mathematical Surveysand Monographs 10) Providence American Mathematical Society 1963 httpdoiorg101090surv010
[3] Davenport H Multiplicative Number Theory New York Springer-Verlag 1980 httpdoiorg101007978-1-4757-5927-3
[4] Titchmarsh E C The Theory of the Riemann Zeta-Function Oxford Oxford UniversityPress 1986
[5] Ireland K and Rosen M A Classical Introduction to Modern Number Theory NewYork Springer-Verlag 1990 httpdoiorg101007978-1-4757-2103-4
[6] Derbyshire J Prime Obsession Bernhard Riemann and the Greatest Unsolved Prob-lem in Mathematics Washington DC Joseph Henry Press 2003 httpwww
johnderbyshirecomBooksPrimepagehtml
[7] Hardy G H and Wright E M An Introduction to the Theory of Numbers OxfordOxford University Press 2008
[8] Matiyasevich Y ldquoFormulas for Prime Numbersrdquo in Kvant Selecta Algebra and Analysis- II edited by S Tabachnikov pp 13ndash24 Providence The American MathematicalSociety 1999 (The Russian original was published in Kvant 1975 no 5 pp 5-13 httpkvantmccmeru197505formuly_dlya_prostyh_chiselhtm Translated by N KKulman)
[9] Granville A ldquoAnalytic Number Theoryrdquo in The Princeton Companion for Mathematicsedited by T Gowers J Barrow-Green and I Leader pp 332ndash348 Princeton and OxfordPrinceton University Press 2008 Preprint available at httpwwwdmsumontrealca
~andrewPDFPrinceComppdf
[10] Pomerance C ldquoComputational Number Theoryrdquo in The Princeton Companion forMathematics edited by T Gowers J Barrow-Green and I Leader pp 348ndash362Princeton and Oxford Princeton University Press 2008 Preprint available at https
mathdartmouthedu~carlpPDFpcm0049pdf
[11] Beineke J and Hughes C ldquoGreat Moments of the Riemann zeta Functionrdquo in Biscuitsof Number Theory edited by A T Benjamin and E Brown pp 199ndash215 ProvidenceThe Mathematical Association of America 2009
[12] Pomerance C ldquoThe Search for Prime Numbersrdquo Scientific American 247 no 6(1982) 136ndash147 doi 101038scientificamerican1282-136 httpsmathdartmouth
edu~carlpPDFsearchpdf
52
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-
[13] Bombieri E ldquoPrime Territoryrdquo The Sciences 32 no 5 (1992) 30ndash36 httpdoiorg101002j2326-19511992tb02416x
[14] Stevenhagen P and Lenstra H W ldquoChebotarev and His Density Theoremrdquo The Mathe-matical Intelligencer 18 no 2 (1996) 26ndash37 httpdxdoiorg101007bf03027290
[15] Granville A and Martin G ldquoPrime Number Racesrdquo The American MathematicalMonthly 113 no 1 (2006) 1ndash33 httpdoiorg10230727641834
[16] Montgomery H L and Wagon S ldquoA Heuristic for the Prime Number Theoremrdquo TheMathematical Intelligencer 28 no 3 (2006) 6ndash9 httpdoiorg101007BF02986877
[17] Spencer J and Graham R ldquoThe Elementary proof of Prime Number TheoremrdquoThe Mathematical Intelligencer 31 no 3 (2009) 18ndash23 httpdoiorg101007
s00283-009-9063-9
[18] Sreekantan R ldquoYitang Zhang and The Twin Primes Conjecturerdquo At RightAngles 2 no 3 (2013) 14ndash17 httpteachersofindiaorgenarticle
yitang-zhang-and-twin-primes-conjecture
[19] Klarreich E ldquoMathematicians Discover Prime Conspiracyrdquo QuantaMagazine March 13 2016 httpswwwquantamagazineorg
mathematicians-discover-prime-conspiracy-20160313
[20] Tao T (22 Jan 2009) Structure and Randomness in the Prime Numbers [VideoFile] University of California Los Angeles Retrieved from httpswwwyoutubecom
watchv=PtsrAw1LR3E Lecture slides available at httpswwwmathuclaedu~taopreprintsSlidesprimespdf
[21] Taylor R (27 April 2012) Primes and Equations [Video file] Institute for AdvancedStudy Princeton Retrieved from httpswwwyoutubecomwatchv=5pTaZu3C--s
[22] Matiyasevich Y (10 Dec 2012) Alan Turing and Number Theory [Video File] Alan Tur-ing Centenary Conference Manchester 2012 Retrieved from httpswwwyoutube
comwatchv=5ADOT72v_9U Lecture slides available at httpvideolecturesnet
turing100_matiyasevich_number_theory
[23] Granville A (20 August 2015) The Patterns in the Primes [Video file] Yale UniversityNew Haven Retrieved from httpswwwyoutubecomwatchv=pO7Egc5Dtqs Lec-ture slides available at httpwwwdmsumontrealca~andrewPrimeLecthtml
[24] Korpal G ldquoDiophantine Equationsrdquo Summer Internship Project Report guided by ProfS A Katre (18 May 2015 ndash 16 June 2015)
[25] Korpal G ldquoEnigma Cryptanalysisrdquo Summer Internship Project Report guided by ProfGeetha Venkataraman (6 July 2015 ndash 26 July 2015)
[26] Korpal G ldquoDiophantine Approximationsrdquo Winter Internship Project Report guided byProf R Thangadurai (13 December 2015 ndash 8 January 2016)
[27] Korpal G ldquoNumber Fieldsrdquo Summer Internship Project Report guided by Prof RameshSreekantan (1 June 2016 ndash 31 July 2016)
[28] Korpal G ldquoReciprocity Lawsrdquo Winter Internship Project Report guided by Prof Chan-dan Dalawat (9 December 2016 ndash 7 January 2017)
Prepared in LATEX 2ε by Gaurish Korpal
53
- Abstract
- Introduction
- Prime Number Theorem
-
- Entire Functions
-
- Finite Order without Zeros
- Finite Order with Zeros
- Order 1 with Zeros
-
- Gamma Function
-
- Analytic Continuation
- Stirlings Formula
-
- Zeta Function
-
- Euler Zeta Function
- Euler Product
- Analytic Properties of Euler Zeta Function
- Riemann Zeta Function
- Zeros of Riemann Zeta Function
-
- von Mangoldt Function
- Prime Counting Function
- Chebyshev Functions
- The Proof
-
- Zero-Free Region for Riemann Zeta Function
- Counting the Zeros of Riemann Zeta Function
- The Explicit Formula for Chebyshev Function
- Completing the Proof
-
- Some Remarks
-
- Lindeloumlf Hypothesis
- Elementary Proof
- Heuristic Approach
-
- Primes in Arithmetic Progression
-
- Dirichlet Density
- Dirichlet Characters
-
- Number of Dirichlet Characters
- Orthogonality Relations
-
- L-function
-
- Dirichlet L-function
- Product Formula
- Logarithm Formula
- Analytic Continuation
- Product of L-functions
-
- The Proof
-
- Trivial Dirichlet Character
- Non-trivial Dirichlet Character
- Completing the Proof
-
- Some Remarks
-
- Chebotareumlv Density Theorem
- Prime Number Theorem for Arithmetic Progressions
- Tcheacutebychev Bias
-
- Conclusion
- Abels Summation Formula
- Bibliography
-