the riemann zetafunction chandrasekharan tata

8/22/2019 The Riemann Zetafunction Chandrasekharan Tata

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Lectures on

The Riemann ZetaFunction

By

K. Chandrasekharan

Tata Institute of Fundamental Research, Bombay

1953


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Lectures on the Riemann Zeta-Function

By

K. Chandrasekharan

Tata Institute of Fundamental Research

1953


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The aim of these lectures is to provide an intorduc-

tion to the theory of the Riemann Zeta-function for stu-dents who might later want to do research on the subject.

The Prime Number Theorem, Hardys theorem on the

Zeros of (s), and Hamburgers theorem are the princi-

pal results proved here. The exposition is self-contained,

and required a preliminary knowledge of only the ele-

ments of function theory.


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Contents

1 The Maximum Principle 1

2 The Phragmen-Lindelof principle 9

3 Schwarzs Lemma 17

4 Entire Functions 25

5 Entire Functions (Contd.) 35

6 The Gamma Function 45

1 Elementary Properties . . . . . . . . . . . . . . . . . . . 45

2 Analytic continuation of(z) . . . . . . . . . . . . . . . 49

3 The Product Formula . . . . . . . . . . . . . . . . . . . 51

7 The Gamma Function: Contd 55

4 The Bohr-Mollerup-Artin Theorem . . . . . . . . . . . . 55

5 Gausss Multiplication Formula . . . . . . . . . . . . . 58

6 Stirlings Formula . . . . . . . . . . . . . . . . . . . . . 59

8 The Zeta Function of Riemann 63

1 Elementary Properties of(s) . . . . . . . . . . . . . . . 63

9 The Zeta Function of Riemann (Contd.) 692 Elementary theory of Dirichlet Series . . . . . . . . . . 69

v


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vi Contents

10 The Zeta Function of Riemann (Contd) 75

2 (Contd). Elementary theory of Dirichlet series . . . . . . 75


3 Analytic continuation of(s). First method . . . . . . . 87

4 Functional Equation (First method) . . . . . . . . . . . . 90

5 Functional Equation (Second Method) . . . . . . . . . . 92


6 Some estimates for (s) . . . . . . . . . . . . . . . . . . 97

7 Functional Equation (Third Method) . . . . . . . . . . . 99


8 The zeros of (s) . . . . . . . . . . . . . . . . . . . . . 105


9 Riemann-Von Magoldt Formula . . . . . . . . . . . . . 113


10 Hardys Theorem . . . . . . . . . . . . . . . . . . . . . 121



12 The Prime Number Theorem . . . . . . . . . . . . . . . 13513 Prime Number Theorem and the zeros of(s) . . . . . . 145

14 Prime Number Theorem and the magnitude ofpn . . . . 146


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Lecture 1

The Maximum Principle

Theorem 1. If D is a domain bounded by a contour C for which Cauchys 1

theorem is valid, and f is continuous on C regular in D, then |f| Mon C implies |f| M in D, and if|f| = M in D, then f is a constant.Proof. (a) Let zo D, n a positive integer. Then

|f(zo)|n = 12i

C

{f(z)}ndzz zo

lc M

n

2,

where lc = length ofC, = distance ofzo from C. As n

|f(z)| M.

(b) If|f(zo)| = M, then f is a constant. For, applying Cauchys inte-gral formula to

d

dz

{f(z)}n, we get|n{f(zo)}n1 f(zo)| =

1

2i

C

fndz

(z zo)2 lCMn

22,

1


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2 1. The Maximum Principle

so that

|f(zo)| l

cM

22

1

n 0, as n

Hence

|f(zo)| = 0.

(c) If|f(zo)| = M, and |f(zo)| = 0, then f(zo) = 0, for2

d2

dz2{f(z)}n = n(n 1){f(z)}n2{f(z)}2 + n{f(z)}n1f(z).

At zo we have

d2

dz2{f(z)}nz=zo = n fn1(Z0)f(z0),

so that

|nMn1f(z0)| = 2!2i

C

{f(z)}ndz(z zo)3

2!lc

23Mn,

and letting n

, we see that f(zo) = 0. By a similar reasoning

we prove that all derivatives of f vanish at z0 (an arbitrary pointofD). Thus f is a constant.

Remark. The above proof is due to Landau [12, p.105]. We shall now

show that the restrictions on the nature of the boundary C postulated in

the above theorem cab be dispensed with.

Theorem 2. If f is regular in a domain D and is not a constant, and

M = maxzD

|f|, then

|f(zo)

|< M, zo

D.

For the proof of this theorem we need a


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1. The Maximum Principle 3

Lemma. If f is regular in |z z0| r, r > 0, then

|f(z0)| Mr,

where Mr = max|zzo|=r|f(z)|, and |f(zo)| = Mr only if f(z) is constant for3

|z zo| = r.Proof. On using Cauchys integral formula, we get

f(zo) =1

2i

|zzo|=n

f(z)

z z0dz

=

1

2

20

f(zo + rei

)d. (1)

Hence

|f(zo)| Mr.Further, if there is a point (such that) | zo| = r, and |f()| 0 and we should have|f(zo)| < Mr; so that |f(zo)| = Mr implies |f(z)| = Mr everywhere on|z zo| = r. That is |f(zo + rei)| = |f(zo)| for 0 2, or

f(zo + rei

) = f(zo)ei

, 0 2On substituting this in (1), we get

1 =1

2

20

cos d

Since is a continuous function ofand cos 1, we get cos 1 forall , i.e. = 0, hence f(s) is a constant.

Remarks. The Lemma proves the maximum principle in the case of a 4

circular domain. An alternative proof of the lemma is given below [7,Bd 1, p.117].


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Aliter. If f(z0 + rei) = () (complex valued) then

f(z0) =1

2

20

()d

Now, if a and b are two complex numbers, |a| M, |b| M anda b, then |a + b| < 2M. Hence if() is not constant for 0 < 2,then there exist two points 1, 2 such that

|(1) + (2)| = 2Mr 2, say, where > 0

On the other hand, by regularity,

|(1 + t) (1)| < /2, for 0 < t < |(2 + t) (2)| < /2, for 0 < t <

Hence

|(1 + t) + (2 + t)| < 2Mr , for 0 < t < .

Therefore

2

0

()d =

1

0

+

1+

1

+

2

1+

+

2+

2

+

2

2+

=

1

0

+

21+

+

22+

()d+

0

[(1 + t) + (2 + t)] dt

Hence

|2

0

() d| Mr(2 2) + (2Mr ) = 2Mr

|1

2

20

()d| < Mr


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Proof of Theorem 2. Let zo D. Consider the set G1 of points z such5that f(z) f(zo). This set is not empty, since f is non-constant. Itis a proper subset of D, since zo D, zo G1. It is open, because f iscontinuous in D. Now D contains at least one boundary point ofG1. For

if it did not, G1 D would be open too, and D = (G1 G1)D would be

disconnected. Let z1 be the boundary point ofG1 such that z1 D. Thenz1 G1, since G1 is open. Therefore f(z1) = f(zo). Since z1 BdG1and z1 D, we can choose a point z2 G1 such that the neighbourhoodofz1 defined by

|z z1| |z2 z1|lies entirely in D. However, f(z2) f(z1), since f(z2) f(zo), and

f(zo) = f(z1). Therefore f(z) is not constant on

|z

z1

|= r, (see

(1) on page 3) for, if it were, then f(z2) = f(z1). Hence if M =max

|zz1|=|z2z1||f(z)|, then

|f(z1)| < M M = maxzD

|f(z)|i.e. |f(zo)| < M

Remarks. The above proof of Theorem 2 [7, Bd I, p.134] does not make

use of the principle of analytic continuation which will of course provide

an immediate alternative proof once the Lemma is established.

Theorem 3. If f is regular in a bounded domain D, and continuous

in D, then f attains its maximum at some point in Bd D, unless f is a

constant.

Since D is bounded, D is also bounded. And a continuous function

on a compact set attains its maximum, and by Theorem 2 this maximum 6

cannot be attained at an interior point of D. Note that the continuity of

|f| is used.Theorem 4. If f is regular in a bounded domain D, is non-constant, and

if for every sequence {zn}, zn D, which converges to a point Bd D,we have

lim supn

|f(zn)| M,

then |f(z)| < M for all z D. [17, p.111]


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Proof. D is an F, for define sets Cn by the property: Cn consists of

all z such that |z| n and such that there exists an open circular neigh-bourhood of radius 1/n with centre z, which is properly contained in D.

Then Cn Cn+1, n = 1, 2, . . .; and Cn is compact.

Define

maxZCn

|f(z)| mn.

By Theorem 2, there exists a zn Bd Cn such that |f(zn)| = mn. Thesequence {mn} is monotone increasing, by the previous results; and thesequence {zn} is bounded, so that a subsequence {znp} converges to alimit

Bd D. Hence

lim|f(znp )| Mi.e. lim mnp M

or mn < M for all n

i.e. |f(z)| < M. z D.

N.B. That Bd D can be seen as follows. If D, then thereexists an No such that CNo D, and |f()| < mNo lim mn, whereas7we have f(znp ) f(), so that |f(znp )| |f()|, or lim mn = |f()|.

Corollaries. (1) If f is regular in a bounded domain D and contin-uous in D, then by considering ef(z) and ei f(z) instead of f(z),it can be seen that the real and imaginary parts of f attain their

maxima on Bd D.

(2) If f is an entire function different from a constant, then

M(r) l.u.b|z|=r

|f(z)|, and

A(r) l.u.b|z|=r

Re{f(z)}

are strictly increasing functions of r. Since f(z) is bounded if M(r)is, for a non-constant function f , M(r) with r.


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(3) If f is regular in D, and f 0 in D, then f cannot attain a

minimum in D.Further if f is regular in D and continuous in D, and f 0 in D,

and m = minzBd D

|f(z)|, then |f(z)| > m for all z D unless f is aconstant. [We see that m > 0 since f 0, and apply the maximum

principle to 1/f ].

(4) If f is regular and 0 in D and continuous in D, and |f| is aconstant on Bd D, then f = in D. For, obviously 0, and,

by Corollary 3, |f(z)| for z D, se that f attains its maximumin D, and hence f = .

(5) If f is regular in D and continuous in

D, and |f| is a constant on 8Bd D, then f is either a constant or has a zero in D.

(6) Definition: If f is regular in D, then is said to be a boundary

value of f at Bd D, if for every sequence zn Bd D,zn D, we have

limn f(zn) =

(a) It follows from Theorem 4 that if f is regular and non - con-

stant in D, and has boundary-values {}, and || H foreach , then |f(z)| < H, z D. Further, if M = max

b.v|| then

M=

M maxzD |f(z)|For, || is a boundary value of |f|, so that || M, henceM M. On the other hand, there exists a sequence {zn},zn D, such that |f(zn)| M. We may suppose thatzn zo (for, in any case, there exists a subsequence {znp}with the limit zo, say, and one can then operate with the

subsequence). Now zo D, for otherwise by continuity

f(zo) = limn f(zn) = M, which contradicts the maximum

principle. Hence zo Bd D, and M is the boundary-valueof |f| at zo. Therefore M M, hence M = M.

(b) Let f be regular and non-constant in D, and have boundary-

values {} on Bd D. Let m = minzD

|f|; m = minb.v.

|| > 0.


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Then, if f 0 in D, by Corollary (6)a, we get

maxzD

1

|f(z)| =1

m=

1

m;

so that|f| > m = m in D. Thus if there exists a point zo D9such that |f(zo)| m, then f must have a zero in D, so thatm > 0 = m. We therefore conclude that, if m > 0, thenm = m if and only if f 0 in D [6, Bd. I, p.135].


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Lecture 2

The Phragmen-Lindelof

principle

We shall first prove a crude version of the Phragmen-Lindelof theorem 10

and then obtain a refined variant of it. The results may be viewed as

extensions of the maximum-modulus principle to infinite strips.

Theorem 1. [6, p.168] We suppose that

(i) f is regular in the strip < x < ; f is continuous in x

(ii) |f| M on x = and x = (iii) f is bounded in x

Then, |f| M in < x < ; and |f| = M in < x < only if f isa constant.

Proof. (i) If f(x + iy) O as y uniformly in x, x ,then the proof is simple. Choose a rectangle x , |y| with sufficiently large to imply |f|(x i) M for x

. Then, by the maximum-modulus principle, for any zo in the

interior of the rectangle,

|f(zo)| M.

9


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10 2. The Phragmen-Lindelof principle

(ii) If |f(x + iy)| 0, as y , consider the modified functionfn(z) = f(z)e

z2/n

. Now

|fn(z)| 0 as y uniformly in x, and|fn(z)| Mec

2/n

on the boundary of the strip, for a suitable constant c.

Hence11

|fn(zo)| Mec2/n

for an interior point zo of the rectangle, and letting n , we get

|f(zo)

| M.

If |f(zo)| = M, then f is a constant. For, if f were not a constant, in theneighbourhood ofzo we would have points z, by the maximum-modulus

principle, such that |f(z)| > M, which is impossible. Theorem 1 can be restated as:

Theorem 1: Suppose that

(i) f is continuous and bounded in x (ii) |f( + iy)| M1, |f( + iy)| M2 for all y

Then|f(xo + iyo)| ML(xo)1 M

1L(xo)2

for < xo < , |yo| < , where L(t) is a linear function oftwhich takesthe value 1 at and 0 at . If equality occurs, then

f(z) = cML(z)

1M

1L(z)2

; |c| = 1.

Proof. Consider

f1(z) =f(z)

ML(z)

1M

1L(z)2

and apply Theorem 1 to f1

(z).12

More generally we have [12, p.107]


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2. The Phragmen-Lindelof principle 11

Theorem 2. Suppose that

(i) f is regular in an open half-strip D defined by:

z = x + iy, < x < , y >

(ii) limzin D

|f| M, for Bd D

(iii) f = Oexp

e

|z|

, < D

uniformly in D.

Then, |f| M in D; and |f| < M in D unless f is a constant.Proof. Without loss of generality we can choose

= 2

, = +/2, = 0

Set

g(z) = f(z). exp(eikz) f(z).{(z)}, say,

where > 0, < k < 1. Then, as y +,

g = O expe|z| eky cos k2 uniformly in x, and 13

e|z| eky cos k2

e(y+/2) eky cos k2

, as y ,

uniformly in x, since < k. Hence

|g| 0 uniformly as y ,

so that |g| M, for y > y, < x < .


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Let zo D. We can then choose H so that

|g(z)| M

for y = H, < x < , and we may also suppose that H > y0 = im(zo).

Consider now the rectangle defined by < x < , y = 0, y = H.

Since || 1 we havelim

zin D

|g| M

for every point on the boundary of this rectangle. Hence, by the

maximum-modulus principle,

|g(zo)| < M, unless g is a constant;or |f(zo)| < M|ee

ikzo |

Letting 0, we get14

|f(zo)| < M.

Remarks. The choice of in the above proof is suggested by the critical

case = 1 of the theorem when the result is no longer true. For take

= 1, = /2, = /2, = 0, f = eeiz

Then

|eeiz | = eRe(eiz) = eRe{eyix} = eey cos x

So

|f| = 1 on x = /2, and f = eey on x = 0.

Corollary 1. If f is regular in D, continuous on Bd D, |f| M on BdD, and

f = O ee

|z| , < 1,

then |f| M in D.


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Corollary 2. Suppose that hypotheses (i) and (iii) of Theorem 2 hold;

suppose further that f is continuous on Bd D,

f = O(ya) on x = , f = O(yb) on x = . Then

f = O(yc) on x = ,

uniformly in , where c = p+ q, and px + q is the linear function which

equals a at x = and b at x = .

Proof. Let > 0. Define (z) = f(z)(z), where is the single-valued 15

branch of (zi)(pz+q) defined in D, and apply Theorem 2 to . We firstobserve that is regular in D and continuous in D. Next

|(z)| = |(y ix)(px+q)ipy |

= |y ix|(px+q) exp(py im logy ixy

)

= y(px+q)|1 + O

1

y

|O(1) exp

py

x

y+ O

1

y2

= y(px+q)epx{1 + O

1

y

},

the Os being uniform in the xs.Thus = O(1) on x = and x = . Since = O(yk) = O(|z|k), we

see that condition (iii) of Theorem 2 holds for if is suitably chosen.Hence = O(1) uniformly in the strip, which proves the corollary.

Theorem 3. [12, p.108] Suppose that conditions (i) and (iii) of Theorem

2 hold. Suppose further that f is continuous in Bd D, and

limy|f| M on x = , x =

Then

limy

|f| M uniformly in x .

Proof. f is bounded in

D, by the previous theorem. Let 0, > 0. 16Let H = H() be the ordinate beyond which |f| < M+ on x = , x = .


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Let h > 0 be a constant. Then zz + hi

< 1in the strip. Choose h = h(H) so large thatf(z) z

z + hi

< M+ on y = H(possible because

f zz + hi

|f| < M). Then the functiong(z) = f z

z + hi

satisfies the conditions of Theorem 2 (with M + in place of M) in the

strip above y = H. Thus

|g| M+ in this stripand so lim

y|g| M+ uniformly.That is,

lim|f| M+ ,since

z

z + hi 1 uniformly in x. Hence

lim|f| M.

Corollary 1. If conditions (i) and (iii) of Theorem 2 hold, if f is contin-17

uous on Bd D, and if

lim|f| a on x = ,b on x = ,

where a 0, b 0, then

limy|f| e

px+q,

uniformly, p and q being so chosen that epx+q

=a for x

= and

=b for

x = .


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Proof. Apply Theorem 3 to g = f e(pz+q)

Corollary 2. If f = O(1) on x = , f = o(1) on x = , then f = o(1) on

x = , < .[if conditions (i) and (iii) of Theorem 2 are satisfied].

Proof. Take b = in Corollary 1, and note that ep+q 0 as 0 forfixed , provided p and q are chosen as specified in Corollary 1.


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Lecture 3

Schwarzs Lemma

Theorem 1. Let f be regular in |z| < 1, f(o) = 0. 18If, for |z| < 1, we have |f(z)| 1, then

|f(z)| |z|, for |z| < 1.

Here, equality holds only if f(z) c z and |c| = 1. We further have|f(o)| 1.

Proof.f(z)

zis regular for |z| < 1. Given o < r < 1 choose such that

r < < 1; then since

|f(z)

| 1 for

|z

|= , it follows by the maximum

modulus principle that f(z)z

1

,

also for |z| = r. Since the L.H.S is independent of 1, we let andobtain |f(z)| |z|, for |z| < 1.

If for z0, (|z0| < 1), we have |f(z0)| = |z0|, then f(z0)

z0

= 1, (by themaximum principle applied to

f(z)

z) hence f = cz, |c| = 1.

Since f(z) = f(o)z + f(o)z2 + in a neighbourhood of the origin,and since f(z)

z 1 in z 1, we get |f(o)| 1. More generally, we

have

17


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18 3. Schwarzs Lemma

Theorem 1: Let f be regular and |f| M in |z| < R, and f(o) = 0.Then

|f(z)| M|z|R

, |z| < R.In particular, this holds if f is regular in |z| < R, and continuous in19|z| R, and |f| M on |z| = R.Theorem 2 (Caratheodorys Inequality). [12, p.112] Suppose that

(i) f is regular in |z| < R f is non-constant(ii) f(o) = 0

(iii) Re f

in

|z

|< R. (Thus

> 0, since f is not a constant).

Then,

|f| 2 R , for|z| = < R.

Proof. Consider the function

(z) =f(z)

f(z) 2 .

We have Re{f 2} < 0, so that is regular in |z| < R.If f = u + iv, we get

|| =

u2 + v2

(2 u)2 + v2

so that

|| 1, since 2 u |u|.But (o) = o, since f(o) = o. Hence by Theorem 1,

|(z)| |z|R

, |z| < R.

But f =

2

1 . Now take

|z|

= < R.

Then, we have,20


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3. Schwarzs Lemma 19

|f

|

2 ||

1 ||

2

R

N.B. If |f(zo)| =2

R , for |zo| < R, then

f(z) =2U cz1 cz , |c| = 1.

More generally, we have

Theorem 2: Let f be regular in |z| < R; f non-constant

f(o) = ao = + i

Re f (|z| < R), so that Then

|f(z) ao| 2( )

R for |z| = < RRemark . If f is a constant, then Theorem 2 is trivial. We shall now

prove Borels inequality which is sharper than Theorem 2.

Theorem 3 (Borels Inequality). [12, p.114] Let f(z) =

n=0anz

n, be

regular in |z| < 1. LetRe f . f(o) = ao = + i.Then

|an| 2( ) 2 1, say, n > 0.Proof. We shall first prove that |a1| < 1, and then for general an.

(i) Let f1

n=1anz

n. Then Re f1 1. 21

For |z| = < 1, we have, by Theorem 2,

|f1| 21

1 i.e.

f1(z)z 211


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Now letting z 0, we get

|a1| 21

(ii) Define = e2i/k, ka +ve integer.

Then

kv=1

m = o

ifm 0& ifm a multiple ofk= k

ifm = 0

or if m = a multiple ofk.

We have

1

k

k1r=0

f1(rz) =

n=1

ankznk

= g1(zk) g1(), say.

The series for g1 is convergent for |z| < 1, and so for || < 1. Henceg1 is regular for || < 1. Since Re g1 1, we have

| coefficient of | 21 by the first part of the proof i.e. |ak| 2221. Corollary 1. Let f be regular in |z| < R, andRe{f(z)f(o)} 1. Then

|f(z)| 2 1 R(R )2 , |z| = < R

Proof. Suppose R = 1. Then

|f(z)|

n|an|n1 2 1

nn1

This argument can be extended to the nth derivative.

Corollary 2. If f is regular for every finite z, and

ef(z) =

O(e|z|

k

), as

|z

| ,

then f(z) is a polynomial of degree k.


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Proof. Let f(z) = anzn. For large R, and a fixed , || < 1, we have

Re{f(R)} < cRk.By Theorem 3, we have

|anRn| 2(2Rk Re ao), n > 0.Letting R , we get

an = 0 ifn > k.

Apropos Schwarzs Lemma we give here a formula and an inequal-

ity which are useful.

Theorem 4. Let f be regular in |z zo| r 23f = P(r, ) + Q(r, ).

Then

f(zo) =1

r

2o

P(r, )eid.

Proof. By Cauchys formula,

(i) f(zo) =1

2i |zzo|=rf(z)dz

(z

zo

)2=

1

2r

2

0

(P + iQ)eid

By Cauchys theorem,

(ii) 0 =1

r2 1

2i

|zzo|=r

f(z)dz =1

2r

20

(P + iQ)eid

We may change i to i in this relation, and add (i) and (ii).Then

f(zo) =1

r

2

0

P(r, )eid


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Corollary. If f is regular, and | Re f M in |z z0| r, then

|f(zo)| 2Mr

.

Aliter: We have obtained a series of results each of which depended

on the preceding. We can reverse this procedure, and state one general

result from which the rest follow as consequences.

Theorem 5. [11, p.50] Let f(z)

n=0

cn(zz0)n be regular in |zz0| < R,24andRe f < . Then

|en| 2( Re co)

Rn, n = 1, 2, 3, . . . (5.1)

and in |z zo| < R, we have

|f(z) f(z0)| 2

R { Re f(zo)} (5.2) f(n)(z)

n!

2R(R )n+1 { Re f(zo)} n = 1, 2, 3, . . . (5.3)Proof. We may suppose zo = 0.

Set (z) = f(z) = c0 1

cnzn

0

bnzn, |z| < R.

Let denote the circle |z| = < R. Then

bn =1

2i

(z)dz

zn+1=

1

2n

(P + iQ)eind, n 0, (5.4)

where (r, ) = P(r, ) + iQ(r, ). Now, ifn 1, then (z)zn1 is regularis , so that

0 =

n

2r

(P + iQ)ein

d, n 1 (5.5)


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Changing i to i in (5.5) and adding this to (5.4) we get

bnn

=1

Penid, n 1.

But P = Re f 0 in |z| < R and so in . Hence, ifn 1, 25

|bn|n 1

|Peni|d = 1

Pd = 2 Re bo, (5.6)

on using (5.4) with n = 0. Now letting R, we get

|bn|Rn 2o 2 Re bo.

Since bo = co, and bn = cn, n 1, we at once obtain (5.1). Wethen deduce, for |z| < R

|(z) (o)| = |1

bnzn|

1

20

R

n=

2o

R (5.7)

and ifn 1,

|(n)(z)

|

r=n

r(r

1) . . . (r

n + 1)

20rn

Rr

=

d

d

n n=0

20

R

r

=

d

d

n2eR

R = 2

oR n!(R )n+1 (5.8)

(5.7) and (5.8) yield the required results on substituting = f and0 = Re f(0).


30/154


31/154

Lecture 4

Entire Functions

An entire function is an analytic function (in the complex plane) which 26

has no singularities except at . A polynomial is a simple example. Apolynomial f(z) which has zeros at z1, . . . ,zn can be factorized as

f(z) = f(0)

1 z

z1

. . .

1 z

zn

The analogy holds for entire functions in general. Before we prove this,

we wish to observe that ifG(z) is an entire function with no zeros it can

be written in the form eg(z) where g is entire. For considerG(z)G(z)

; every

(finite) point is an ordinary point for this function, and so it is entireand equals g1(z), say. Then we get

log

G(z)

G(z0)

=

z1z0

g1()d = g(z) g(z0), say,

so that

G(z) = G(z0)eg(z)g(z0) = eg(z)g(z0)+log G(zo)

As a corollary we see that if G(z) is an entire function with n zeros,

distinct or not, then

G(z) = (z z1) . . . (z zn)eg(z)

25


32/154

26 4. Entire Functions

where g is entire. We wish to uphold this in the case when G has an

infinity of zeros.

Theorem 1 (Weierstrass). [15, p.246] Given a sequence of complex27

numbers an such that

0 < |a1| |a2| . . . |an| . . .

whose sole limit point is , there exists an entire function with zeros atthese points and these points only.

Proof. Consider the function

1

z

an

eQ(z)

,

where Q(z) is a polynomial of degree q. This is an entire function whichdoes not vanish except for z = an. Rewrite it as

1 zan

eQ(z) = eQ(z)+log

1 zan

= e z

an z2

2a2...+Q(z),

and choose

Q(z) =

z

an+ . . . +

z

an

so that 1 z

an

eQ(z) = e

zan

+11

+1

...

1 + n(z), say.

We wish to determine in such a way that

1

1 zan

eQ(z)

is absolutely and uniformly convergent for |z| < R, however large R may28


33/154

4. Entire Functions 27

be.

Choose an R > 1, and an such that 0 < < 1; then there exists aq (+ ve integer) such that

|aq| R

, |aq+1| >

R

.

Then the partial product

q1

1 z

an

eQ(Z)

is trivially an entire function of z. Consider the remainder

q+1

1 z

an

eQ(z)

for |z| R.We have, for n > q, |an| >

R

or

zan < < 1. Using this fact we

shall estimate each of the factors {1 + n(z)} in the product, for n > q.

| n (z)| =e 1

+1

z

an

+1... 1 e

1+1 zan

+1... 1

Since |em 1| e|m| 1, for em 1 = m + m2

2!+ . . . or |em 1| 29

|m| + |m2|

2. . . = e|m| 1.

|Un(z)| e zan

+11+ zan+... 1,

e

zan

+1

(1++2+ ) 1

= e1

1 zan +1 1


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1

1 z

an+1

e

11

zan

+1

, since if x is

real, ex 1 xex, for ex 1 = x

1 +x

2!+

x2

3!+

x(1 + x + x

2

2!+

) = xexHence

|Un(z)| e1

1

1 zan

+1Now arise two cases: (i) either there exists an integer p > 0 such30

that

n=11

|an|p < or (ii) there does not. In case (i) take = p 1, so

that

|Un(z)| Rp

|an|p e

11a

1 , since |z| R;

and hence |Un(z)| < for |z| R, i.e.

q+1

(1 + Un(z)) is absolutely and

uniformly convergent for |z| R.In case (ii) take = n 1, so that

|Un(z)| q

zan < 1|z| R|an|

Then by the root-test |Un(z)| < , and the same result follows

as before. Hence the product

1

1 z

an

eQ(z)

is analytic in |z| R; since R is arbitrary and does not depend on R we31see that the above product represents an entire function.


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Remark. If in addition z = 0 is also a zero ofG(z) thenG(z)

zm

G(z)

has no

zeros and equals eg(z), say, so that

G(z) = eg(z) zm

n=1

1 z

an

eQ(z)

In the above expression for G(z), the function g(z) is an arbitrary

entire function. IfG is subject to further restrictions it should be possible

to say more about g(z); this we shall now proceed to do. The class of

entire functions which we shall consider will be called entire functions

of finite order.

Definition. An entire function f(z) is of finite order if there exists a con-stant such that |f(z)| < er for |z| = r > r0. For a non-constant f , of

finite order, we have > 0. If the above inequality is true for a certain

, it is also true for > . Thus there are an infinity ofs o satisfyingthis. The lower bound of such s is called the order of f . Let usdenote it by . Then, given > 0, there exists an ro such that

|f(z)| < er+

for |z| = r > r0.This would imply that

M(r) = max|z|=r |f(z)| < er+

, r > r0,

while 32

M(r) > er

for an infinity of values ofrtending

to +Taking logs. twice, we get

log log M(r)

log r< +

and

log log M(r)log r

> for an infinity of values ofr


36/154


Hence

= limr log log M(r)logR

Theorem 2. If f(z) is an entire function of order < , and has aninfinity of zeros, f(0) 0, then given > 0, there exists an R such that

for R Ro, we have

n

R

3

1

log2 log e

R+

|f(o)|

Here n(R) denotes the number of zeros of f(z) in |z| R.

Proof. We first observe that if f(z) is analytic in |z| R, and a1 . . . anare the zeros of f inside |z| < R/3, then for

g(z) =f(z)

1 za1

. . .

1 z

an

we get the inequality33

|g(z)| M2n

where M is defined by: |f(z)| M for |z| = R. For, if |z| = R, then since

|ap| R

3 , p = 1 . . . n, we get zap 3 or 1

z

ap

2By the maximum modules theorem,

|g(o)| M2n

, i.e. |f(o)| M2n

or

n nR

3

1

log2 log

M

f(o)

If, further, f is of order , then for r > r, we have M < er+

which

gives the required results.N.B. The result is trivial if the number of zeros is finite.


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Corollary.

n(R) = O(R+

).

For

n

R

3

1

log2 R+ log |f(o)| ,

and log |f(0)| < R+, say.Then for R R

n

R

3

, then

1|an|

<

Proof. Arrange the zeros in a sequence:

|a1| |a2| . . .Let

p = |ap|Then in the circle |z| r = n, there are exactly n zeros. Hence

n < c

+

n , for n Nor

1

n>

1

c 1

+n

Let > + (i.e. choose 0 < < ). Then1

n/+>

1

c/+ 1

n,

or1

n< c/+ 1

n/+for n N

Hence

1n

< .


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Remark. (i) There cannot be too dense a distribution of zeros, since

n < c < +

n . Nor can their moduli increase too slowly, since, forinstance,

1(log n)p

does not converge.

(ii) The result is of course trivial if there are only a finite number of35

zeros.

Definition. The lower bound of the numbers for which 1

|an|< ,

is called the exponent of convergence of{an}. We shall denote it by 1.Then

1|an|1+ < , 1

|an|1

=

, > 0

By Theorem 1, we have 1

N.B. If the an s are finite in number or nil, then 1 = 0. Thus 1 > 0implies that f has an infinity of zeros. Let f(z) be entire, of order < ;f(o) 0, and f(zn) = 0, n = 1, 2, 3, . . .. Then there exists an integer

(p + 1) such that 1

|zn|+1<

(By Theorem 1, any integer > will serve for + 1). Thus, by

Weierstrasss theorem, we have

f(z) = eQ(z)

1

1 z

zn

e

z

zn +

z2

2zn2 ++z

zn ,

where = p (cf. proof of Weierstrasss theorem).

Definition. The smallestinteger p for which 1

|zn|p+1 < is the genusof the canonical product

1 z

zn

e

zzn

++ zppz

pn , with zeros zn. If z

n s36

are finite in number, we (including nil) define p = 0, and we define the

product as

1 z

zn

Examples. (i) zn=

n, p=

1 (ii) zn=

en

, p=

0(iii) z1 =

12

log 2, zn = log n, n 2, no finite p.


39/154


Remark. If > 1, then 1|zn| < Also, 1|zn|p+1

<

But 1|zn|p

=

Thus, if1 is not an integer, p = [1], and if1 is an integer, then either 1|zn|1 0, let N be defined by the property:

|zN| R, |zN+1| > R.We need the following

Lemma. Let0 < < 1. Thenf(z)f(z)

N

n+1

1

z zn

< cR1+

for |z| = R2

> R0, and |z| = R2

is free from any zn.

35


42/154

36 5. Entire Functions (Contd.)

For we can choose an R0 such that for |z| = R > R0,(i) |f(z)| < eR+ ,38

(ii) no zn lies on |z| = R2

, and

(iii) log1

|f(0)| < (2R)+

For such an R > R0, define

gR(z) =f(z)

f(0)

N

n=1

1 zzn

1

Then for |z| = 2R, we have

|gR(z)| 1.43

Hence, for all u, we have

|E(u)| e|2u|+log(1+|u|),

if p p + 1.It is possible to choose in such a way that

(i) p p + 1,

(ii) > 0,

(iii) 1

|zn|< , (if there are an infinity of zn s) and

(iv) 1 1 +

For, if 1

|zn|1< (if the series is infinite this would imply that

1 > 0), we have only to take = > 0. And in all other cases we must

have p 1 < p + 1, [for, if 1 = p + 1, then 1

|zn|1< ] and so we

choose such that1 < < p + 1; this implies again that1

|zn|

0. For such a we can write the above inequality as

|E(u)| ec1 |u| , c1 is a constant.

Hence44

|f(z)| e|Q(z)|+c1

n=1

zzn

i.e. M(r) < ec2rq

+c3r

, |z| = r > 1, and c1

|1/zn| = c3.

Hence

log M(r) = O(rq) + O(r), as r ,


47/154

5. Entire Functions (Contd.) 41

= O(rmax(q,)), as r (2.1)

so that

max(q, )Since is arbitrarily near1, we get

max(q, 1),

and this proves the theorem.

Theorem 3. If the order of the canonical product G(z) is, then = 1

Proof. As in the proof of Theorem 2, we have

|E(u)| ec1|u|

p p + 1; > 0 1

|zn| 0

Hence

c1

n=1

z

zn

|G(z)| e ec3r , |z| = r.

If M(r) = max|z|=r

|G(z)| , then 45

M(r) < ec3r

Hence

which implies

1; but 1

. Hence = 1

Theorem 4. If, either


48/154

42 5. Entire Functions (Contd.)

(i) 1 < q or

(ii) 1

|zn|1 < (the series being infinite),

then

log M(r) = O(r), for = max(1, q).

Proof. (i) If1 < q, then we take < q and from (2.1) we get

log M(r) = O(rq) = O(rmax(1,q))

(ii) If

n=11

|zn

|1

< , then we may take = 1 > 0 and from (2.1) we

get

log M(r) = O(rq) + O(r1 )

= O(r)

Theorem 5. (i) We have

= max(1, q)

(The existence of either side implies the other)

(ii) If > 0, and if log M(r) = O(r) does not hold, then 1 = , f(z)46

has an infinity of zeros zn, and |zn| is divergent.

Proof. The first part is merely an affirmation of Hadamards theorem,

1 and Theorem 2.For the second part, we must have 1 = ; for if1 < , then = q

by the first part, so that 1 < q, which implies, by Theorem 4, that

logM(r) = O(rq) = O(r) in contradiction with our hypothesis. Since

> 0, and 1 = we have therefore 1 > 0, which implies that f(z) has

an infinity of zeros. Finally

1|zn|1 is divergent, for if

|zn|1 < ,

then by Theorem 4, we would have log M(r)=

O(r

) which contradictsthe hypothesis.


49/154

5. Entire Functions (Contd.) 43

Remarks. (1) 1 is called the real order of f(z), and the apparent

order Max (q, p) is called the genus of f(z).

(2) If is not an integer, then = r, for q is an integer and =

max(1, q). In particular, if is non-integral, then f must have an

infinity of zeros.


50/154


51/154

Lecture 6

The Gamma Function

1 Elementary Properties [15, p.148]47

Define the function by the relation.

(x) =

0

tx1etdt, x real.

(i) This integral converges at the upper limit , for all x, sincetx1et = t2tx+1et = O(t2) as t ; it converges at the lowerlimit o for x > 0.

(ii) The integral converges uniformly for 0 < a x b. For

0

tx1etdt =

10

+

1

= O

1

0

ta1dt

+ O

1

tb1etdt

= O(1), independently of x.

Hence the integral represents a continuous function for x > 0.

(iii) Ifz is complex,

0

tz1etdtis again uniformly convergent over any

finite region in which Rez

a > 0. For ifz = x + iy, then

|tz1| = tx1,

45


52/154

46 6. The Gamma Function

and we use (ii). Hence (z) is an analytic function forRez > 0

(iv) Ifx > 1, we integrate by parts and get

(x) =tx1et

0

+ (x 1)

0

tx2etdt

= (x 1)(x 1).

However,48

(1) =

0

etdt = 1; hence

(n) = (n 1)!,

ifn is a +ve integer. Thus (x) is a generalization ofn!

(v) We have

log (n) = (n 12

) log n n + c + O(1),

where c is a constant.

log (n) = log(n 1)! =n1r=1

log r.

We estimate log rby an integral: we have

r+ 12

r 12

log t dt =

12

12

log(s + r)ds =

12

0

+

0 1

2

=

1/2

0

log(t+ r) + log(r t) dt


53/154

1. Elementary Properties 47

=

1/2

0

log r

2

+ log

1 t2

r2

dt

= log r+ cr, where cr = O

1

r2

.

Hence

log (n) =

n1r=1

r+ 1

2r 1

2

log t dt cr

=

n 12

12

log t dtn

1

r=1

cr

=

n 1

2

log

n 1

2

1

2log

1

2

(n 1

2) +

1

2

r=1

cr +

n

Cr

= (n 12

) log n n + c + o(1), where c is a constant.49

Hence, also(n!) = a.ennn+

12 eo(1),

where a is a constant such that c = log a

(vi) We have

(x)(y)

(x +y)=

O

ty1

(1 + t)x+ydt, x > 0, y > 0.

For

(x)(y) =

0

tx

1

et

dt

0

sy

1

es

ds, x > o, y > 0


54/154


Put s = tv; then

(x)(y) =

0

tx1et dt

0

tyvy1etv dv

=

0

vy1dv

0

tx+y1et(1+v)dt

=

0

vy1dv

0

ux+y1eu

(1 + v)x+ydu

= (x +y)

0

vy1

(1 + v)x+y dv.

The inversion of the repeated integral is justified by the use of the50

fact that the individual integrals converge uniformly for x > 0,y > 0.

(vii) Putting x = y =1

2in (vii), we get [using the substitution t =

tan2 ]

{(

1

2

)

}2

= 2(1)

/2

0

d = .

Since ( 12

) > 0, we get ( 12

) =

Putting y = 1 x, on the other hand, we get the important relation

(x)(1 x) =

0

ux

1 + udu =

sin(1 x) , 0 < x < 1,

since

0

xa1

1 + x dx =

sin a for 0 < a < 1, by contour


55/154

2. Analytic continuation of (z) 49

integration. Hence

(x)(1 x) = sin x

, 0 < x < 1.

2 Analytic continuation of(z) [15, p. 148]

We have seen that the function

(z) =

o

ettz1 dt

is a regular function for Rez > 0. We now seek to extend it analytically

into the rest of the complex z-plane. Consider

I(z) C

e()z1d,

where C consists of the real axis from to > 0, the circle || = in 51the positive direction and again the real axis from to .

We define

()z1 = e(z1) log()by choosing log() to be real when =

The integral I(z) is uniformly convergent in any finite region of the

Z-plane, for the only possible difficulty is at =

, but this was covered

by case (iii) in 1. Thus I(z) is regular for all finite values of z.We shall evaluate I(z). For this, set

= ei

so that

log() = log + i( ) on the contour[so as to conform with the requirement that log() is real when = ]

Now the integrals on the portion of C corresponding to (, ) and(,) give [since = 0 in the first case, and = 2 in the second]:

e+(z

1)

(log

i)

d +

e+(z

1)(log+i)

d


56/154


=

e+(z1)(logi) + e+(z1)(log+i) d=

e+(z1)loge(z1)i e(z1)i

d

= 2i sinz

e z1d

On the circle || = , we have

|()z

1

|=

|e(z

1)

log()|=

|e(z

1)[log +i(

)]

|= e(x1)log y()

= O(x1).

52

The integral round the circle || = therefore gives

O(x) = O(1), as 0, if x > 0.

Hence, letting 0, we get

I(z) = 2i sin z

0

ez1d, Rez > 0

= 2i sin z(z), Rez > 0.

We have already noted that I(z) is regular for all finite z; so

1

2iI(z) cosec z

is regular for all finite z, except (possibly) for the poles of cosec z

namely, z = 0, 1, 2, . . .; and it equals (z) for Rez > 0.Hence -

1

2 iI(z) cosecz is the analytic continuation of

(z) all overthe z-plane.


57/154

3. The Product Formula 51

We know, however, by (iii) of 1, that (z) is regular for z =1, 2, 3, . . .. Hence the only possible poles of 1

2iI(z) cosec x are z =

0, 1, 2, . . .. These are actually poles of (z), for if z is one of thesenumbers, say n, then ()z1 is single-valued in 0 and I(z) can be eval-uated directly by Cauchys theorem. Actually

(1)n+1C

e

n+1d =

2i

n!(1)n+n+1 = 2i

n!

i.e. I(n) = 2in!

.

So the poles of cosec z, at z = 0,

n, are actually poles of(z). The 53

residue at z = n is therefore

limzn

2in!

z + n

2i sinz

=(1)n

n!

The formula in (viii) of 1 can therefore be upheld for complex values.Thus

(z) (1 z) = cosec zfor all non-integral values of z. Hence, also,

1

(z)is an entire function.

(Since the poles of(1 z) are cancelled by the zeros of sin z)

3 The Product Formula

We shall prove that1

(z)is an entire function of order 1.

Since I(z) = 2i sinz (z), and

(z) (1 z) = sin z

,

we get

I(1 z) = 2i sin( z)(1 z)


58/154


=

2i sinz

sin z

1

(z)

=

2i

(z)

or1

(z)=

I(1 z)2i .

Now54

I(z) = O

e|z|

10

ettx1dt +

1

ettx1dt

= O

e|z|

1 +

1

ett|z|dt

= O

e|z|1 +

1+|z|1

+

1+|z|

= Oe|z|(|z| + 1)|z|+1

= O

e|z|

1+

, > 0.

Hence1

(z)is of order 1. However, the exponent of convergence

of its zeros equals 1. Hence the order is = 1, and the genus of the

canonical product is also equal to 1.

Thus1

(z)= z, eaz+b

n=1

(1 +z

n)e

zn

by Hadamards theorem. However, we know that limz0+

1

z(z)= 1 and

(1) = 1. These substitutions give b = 0, and

1 = ea

1 +

1

e1/n

or,

0 = a +

1

log

1 + 1

n 1n


59/154

3. The Product Formula 53

= a

limm

m

1log 1 + 1

n

1

n

= a + limm

m1

log(n + 1) log n m

1

1

n

= a + lim

m

log(m + 1) m1

1

n

= a , where is Eulers constant.

55

Hence

1(z)

= ezz n=1

1 + z

n

ez/n

As a consequence of this we can derive Gausss expression for (z).

For

(z) = limn

ez(log n1

12... 1

n )1

z e

z/1

1 + z1

. . .ez/n

1 + zn

= limn

nz 1

z 1

z + 1 2

z + 2. . .

n

z + n

= lim

n nz n!

z(z + 1) . . . (z + n)

We shall denote

nz n!z(z + 1) . . . (z + n)

n(z),

so that 56

(z) = limn n(z).

Further, we get by logarithmic-differentiation,

(z)

(z)=

1

z

n=1

1z + n

1

n

.


60/154


Thus

d2

dz2[log (z)] =

n=0

1(n +z)2

;

Henced2

dz2[log (z)] > 0 for real, positive z.


61/154

Lecture 7

The Gamma Function:

Contd

4 The Bohr-Mollerup-Artin Theorem

[7, Bd.I, p.276]

We shall prove that the functional equation of (x), namely (x + 1) = 57

x(x), together with the fact thatd2

dx2[log (x)] > 0 for x > 0, deter-

mines (x) essentially uniquely-essentially, that is, except for a con-

stant of proportionality. If we add the normalization condition (1) = 1,

then these three properties determine uniquely. N.B. The functional

equation alone does not define uniquely since g(x) = p(x)(x), where

p is any analytic function of period 1 also satisfies the same functional

equation.

We shall briefly recall the definition of Convex functions. A real-

valued function f(x), defined for x > 0, is convex, if the corresponding

function

(y) =f(x +y) f(x)

y,

defined for all y > x, y 0, is monotone increasing throughout therange of its definition.

If 0 < x1 < x < x2 are given, then by choosing y1 = x1 x and

55


62/154

56 7. The Gamma Function: Contd

y2 = x2 x, we can express the condition of convexity as:

f(x) x2 xx2 x1

f(x1) +x x1

x2 x1f(x2)

Letting x x1, we get f(x1 + 0) f(x1); and letting x2 x, we58get f(x) f(x + 0) and hence f(x1 + 0) = f(x1) for all x1. Similarlyf(x1 0) = f(x1) for all x1.

Thus a convex function is continuous.

Next, if f(x) is twice (continuously) differentiable, we have

(y) =y f(x +y) f(x +y) + f(x)

y2

and writing x +y = u, we get

f(x) = f(u y) = f(u) y f(u) + y2

2f[u (1 )y], 0 1

which we substitute in the formula for (y) so as to obtain

(y) =1

2f[x + y].

Thus if f(x)

0 for all x > 0, then (y) is monotone increasing and fis convex. [The converse is also true].

Thus log (x) is a convex function; by the last formula of the previ-

ous section we may say that (x) is logarithmically convex for x > 0.

Further (x) > 0 for x > 0.

Theorem. (x) is the positive function uniquely defined for x > 0 by the

conditions:

(1.) (x + 1) = x(x)

(2.) (x) is logarithmically convex

(3.) (1) = 1.


63/154

4. The Bohr-Mollerup-Artin Theorem 57

Proof. Let f(x) > 0 be any function satisfying the above three condi-

tions satisfied by (x).Choose an integer n > 2, and 0 < x 1.59Let

n 1 < n < n + x n + 1.By logarithmic convexity, we get

log f(n 1) log f(n)(n 1) n 0, we then have

dx

dx2[log f(x)] > 0.

Further

f(x + 1) = x f(x).

Hence by the Bohr-Artin theorem, we get

f(x) = ap(x).

Put x = 1. Then we get

ap = p

1

p

2

p

. . .

p

p

(5.1)

[a1 = 1, by definition]. Put k = 1, 2, . . . p, in the relation

n k

p =nk/pn!pn+1

k(k+ p) . . . (k+ np),

and multiply out and take the limit as n . We then get from (5.1),

ap = p. limn

np+1

2 (n!)ppnp+p

(np + p)!

However

(np + p)! = (np)!(np)p

1 +1

np

1 +

2

np

. . .

1 +

p

np

Thus, for fixed p, as n , we get

ap = p limn

(n!)p

pnp

(np)!n(p1)/2


65/154

6. Stirlings Formula 59

Now by the asymptotic formula obtained in (v) of 1, [see p.49]61

(n!)p = apnnp+p/2enpeo(1)

(np)! = annp+12pnp+

12 enp eo(1)

Hence

ap =

p ap1.Putting p = 2 and using (5.1) we get

a =12

, a2 =12

2

1

2

(1) =

2 (5.2)

Hence

ap = p(2)(p1)/2.Thus

px

x

p

. . .

x + p 1

p

=

p(2)(p1)/2(x)

For p = 2 and p = 3 we get:

x

2

x + 1

2

=

2x1(x);

x

3

x + 1

3

x + 2

3

=

2

3x12

(x).

6 Stirlings Formula [15, p.150]

From (v) of 1, p.49 and (5.2) we get

(n!) =

2 en nn+ 12 eo(1),

which is the same thing as

log(n 1)! =

n 12

log n n + log

2 + o(1) (6.1)

Further 62

1 + 12

+ + 1N 1 logN = + 0(1), as N (6.2)


66/154


and

log(N +z) = logN + log

1 +

z

N

= logN +

z

N+ O

1

N2

, as N . (6.3)

We need to use (6.1)-(6.3) for obtaining the asymptotic formula for

(z) where z is complex.

From the product-formula we get

log (z) =

n=1 z

n log 1 +

z

n z logz, (6.4)each logarithm having its principal value.

Now

N0

[u] u + 12

u +zdu =

N1n=0

n+1n

n + 12 +zu +z 1 du

=

N1n=0

(n +1

2+z)[log(n + 1 +z) log(n z)] N

=

N1n=0

n[log(n+

1+

z) log(n +z)] + z + 12 N1n=0

log(n + 1 +z) log(n +z) N= (N 1) log(N +z)

N1n=1

log(n +z) +

z +

1

2

log(N +z)

z +

1

2

logz N

=

N 1

2+z

log(N +z)

z +

1

2

logz

NN1n=1

log n + log

1 +

z

n


67/154

6. Stirlings Formula 61

= N1

2

+z log(N +z) z +1

2 logz N log(N 1)!

+

N1n=1

z

n log

1 +

z

n

z

N11

1

n

63

Now substituting for log(N +z) etc., from (6.1)-(6.3), we get

N0

[u] u + 12

u +zdu = (N 1

2+z)

logN +

z

N+ O

1

N2

z +

1

2

logz N N1

2 logN + N c + O(1)+N1n=1

z

n log

1 +

z

n

z

N11

1

n

= z

logNN1

1

1

n

+z + O(1) log 2z 1

2

logz logz +

N1n=1

z

n log

1 +

z

n

Now letting N , and using (6.4), and (6.2), we get 64

0

[u] u + 12u +z

du + 12

log 2 z +z 1

2

logz = log (z) (6.5)

If(u) =u

0

([v] v + 12

)dv, then

(u) = O(1), since (n + 1) = (n),

ifn is an integer.

Hence

0

[u]

u + 1

2

u +z du =

0

d(u)

u +z =

0

(u)

(u +z)2


68/154


=O

0

dv

|u +z|2 If we write z = rei, and u = ru1, then

0

[u] u + 12

u +zdu = O

1r

0

du1

|u1 + i|2

The last integral is finite if . Hence

0

[u] u + 12

u+

z

du = O 1

r , uniformly for |

arg z

|

<

Thus we get from (6.5):

log (z) =

z 1

2

logz z + 1

2log2 + O

1

|z|

for |arg z| < Corollaries. (1) log (z + ) = (z + 1

2)logz z + 1

2log2 + O

1|z|

as |z| uniformly for|arg z| < . bounded.65(2) For any fixed x, as y

|(x + iy)| e 12 |y||y|x 12 2

(3)

(z)(z)

= logz 12z

+ O

1

|z|2, |arg z| < [11, p.57]

This may be deduced from (1) by using

f(z) =1

2i

C

f()

(z)2 d,

where f(z) = log (z)

z

12 logz +z

12

log2 and C is a circle

with centre at = z and radius |z| sin /2.


69/154

Lecture 8

The Zeta Function of

Riemann

1 Elementary Properties of(s) [16, pp.1-9]

We define (s), for s complex, by the relation 66

(s) =

n=1

1

ns, s = + it, > 1. (1.1)

We define xs

, for x > 0, as es log x

, where log x has its real determination.Then |ns| = n, and the series converges for > 1, and uniformlyin any finite region in which 1 + > 1. Hence (s) is regularfor 1 + > 1. Its derivatives may be calculated by termwisedifferentiation of the series.

We could express (s) also as an infinite product called the Euler

Product:

(s) =

p

1 1

ps

1, > 1, (1.2)

where p runs through all the primes (known to be infinite in number!).

It is the Euler product which connects the properties of with the prop-erties of primes. The infinite product is absolutely convergent for > 1,

63


70/154

64 8. The Zeta Function of Riemann

since the corresponding series

p

1ps =

p

1

p< , > 1

Expanding each factor of the product, we can write it as

p

1 +

1

ps+

1

p2s+

Multiplying out formally we get the expression (1.1) by the unique67

factorization theorem. This expansion can be justified as follows:

pp

1 +

1

ps+

1

p2s+

= 1 +

1

ns1

+1

ns2

+ ,

where n1, n2, . . . are those integers none of whose prime factors exceed

P. Since all integers P are of this form, we get

n=1

1

nspP

1 1

p2

1 =

n=1

1

ns 1 1

ns1

1ns

2

1

(P + 1)+

1

(P + 2)+

0, as P , if > 1.Hence (1.2) has been established for > 1, on the basis of (1.1).

As an immediate consequence of (1.2) we observe that (s) has no zeros

for > 1, since a convergent infinite product of non-zero factors is

non-zero.

We shall now obtain a few formulae involving (s) which will be of

use in the sequel. We first have

log (s) = p

log

1 1

p2

, > 1. (1.3)


71/154

1. Elementary Properties of(s) 65

If we write (x) = px1, then

log (s) =

n=2

{(n) (n 1)} log

1 1ns

=

n=2

(n)

log

1 1

ns

log

1 1

(n + 1)s

=

n=2

(n)

n+1n

s

x(xs 1) dx

68

Hence

log (s) = s

2

(x)

x(xs 1) dx, > 1. (1.4)

It should be noted that the rearrangement of the series preceding the

above formula is permitted because

(n) n and log

1 1ns

= O(n).

A slightly different version of (1.3) would be

log (s) =

p

m

1

mpms, > 1 (1.3)

where p runs through all primes, and m through all positive integers.

Differentiating (1.3), we get

(s)

(s)=

p

ps log p1 ps =

p,m

logpms

,

or

(s)(s) =

n=1

(n)

ns , > 1 , (1.5)


72/154


where

(n) = log p, ifn is a + ve power of a prime p0, otherwise.

69

Again

1

(s)=

p

1 1

ps

=

n=1

(n)

ns, > 1 (1.6)

(1.6) where (1) = 1, (n) = (1)k if n is the product of k differentprimes, (n) = 0 ifn contains a factor to a power higher than the first.

We also have

2(s) =

n=1

d(n)

ns, > 1

where d(n) denotes the number of divisors ofn, including 1 and n. For

2(s) =

m=1

1

ms

n=1

1

ns=

=1

1

s

mn=

1

More generally

k(s) =

n=1dk(n)

ns, > 1 (1.7)

where k = 2, 3, 4, . . ., and dk(n) is the number of ways of expressing n70

as a product ofkfactors.

Further

(s) (s a) =

m=1

1

ms

n=1

na

ns,

=

=1

1

s

mn=

na,

so that

(s) (s a) =

=1

2()

(1.8)


73/154

1. Elementary Properties of(s) 67

where a() denotes the sum of the ath powers of the divisors of .

If a 0, we get, from expanding the respective terms of the Eulerproducts,

(a)(s a) =

p

1 +

1

ps+

1

p2s+

1 +

pa

ps+

p2a

p2a+

=

p

1 +

1 + pa

ps+

1 + pa + p2a

p2s+

=

p

1 +

1 p2a1 pa

1

ps+

Using (1.8) we thus get

2(n) =1 p(m1+1)a

1

1 pa1

. . .1 p(mr+1)ar

1 par(1.9)

ifn = pm11

, pm22

. . . pmrr by comparison of the coefficients of

1

ns. 71

More generally, we have

(s) (s a) (s b) (s a b)(2s a b) =

p

1 p2s+a+b(1 ps)(1 ps+a)(1 ps+b)

(1 ps+a+b)

for > max {1, Re a + 1, Re b + 1, Re(a + b) + 1}. Putting ps = z, weget the general term in the right hand side equal to

1 pa+bz2(1 z) (1 paz) (1 pbz)(1 pa+bz)

=1

(1 pa)(1 pb)

1

1 z pa

1 paz pb

1 pbz +pa+b

1 pa+bz

=1

(1 pa)(1 pb)

m=0

1 p(m+1)a p(m+1)b + p(m+1)(a+b)

zm

= 1(1 pa)(1 pb)

m=o

1 p(m+1)a 1 p(m+1)bzm


74/154


Hence 72

(s) (s a)(s b) (s a h)(2s a b)

=

p

m=0

1 p(m+1)a1 pa

1 p(m+1)b1 pb

1

pms

Now using (1.9) we get

(s) (s a) (s b)(s a b)(2s a b) =

n=1

a(n)b(n)

ns(1.10)

> max{1, Re a+

1, Re b+

1, Re(a+

b)+

1}Ifa = b = 0, then

{(s)}4(2s)

=

n=1

{d(n)2}ns

, > 1. (1.11)

If is real, and 0, we write i for a and i for b in (1.10), andget

2(s)(s i)(s + i)(2s)

=

n=1

|i(n)|2ns

, > 1, (1.12)

where i(n)=

d/n di

.


75/154

Lecture 9


Riemann (Contd.)

2 Elementary theory of Dirichlet Series [10]73

The Zeta function of Riemann is the sum-function associated with the

Dirichlet series 1

ns. We shall now study, very briefly, some of the

elementary properties of general Dirichlet series of the form

n=1

anen s, o 1 < 2 < . . .

Special cases are: n = log n; n = n, es

= x. We shall first prove a

lemma applicable to the summation of Dirichlet series.

Lemma. Let A(x) =

nxan, where an may be real or complex. Then, if

x 1, and (x) has a continuous derivative in (0, ), we have

n a

n(n) =

1

A(x)(x)dx + A()().

69


76/154

70 9. The Zeta Function of Riemann (Contd.)

If A()() 0 as , then

n=1

an(n) =

1

A(x)(x)dx,

provided that either side is convergent.

Proof.

A()()

nan(n)

= n

an{()

(n)

}

=

n

n

an(x)dx

=

1

A(x)(x)dx

74

Theorem 1. If

n=1

anen s converges for s = s0

0 + ito, then it

converges uniformly in the angular region defined by

|am(s s0)| < /2, for 0 < < /2

Proof. We may suppose that so = 0. Forane

n s =

anen s0 en(ss0)

bne

n s , say,

and the new series converges at s = 0. By the above lemma, we get

+1

anen s =

1

1

anens


77/154

2. Elementary theory of Dirichlet Series 71

=

A(x) exs

sdx + A()e s

A()e s

= s

{A(x) A()}exs dx + [A() A()]e s

We have assumed that

an converges, therefore, given > 0, there 75

exists an n0 such that for x > n0, we have

|A(x) A()| <

Hence, for such , we have

+1

anen s

|s|

exdx + e

2 |s|

+

< 2 (sec + 1),

if 0, since|s|

< sec , and this proves the theorem. As a conse-

quence of Theorem 1, we deduce

Theorem 2. If

anens converges for s = s0, then it converges for

Re s > o, and uniformly in any bounded closed domain contained in

the half-plane > o. We also have

Theorem 3. If

anen s converges for s = s0 to the sum f(s0), then

f(s) f(s0) as s s0 along any path in the region |am(s s0)| 1, and diverges for Re s 1.

If a Dirichlet series converges for some, but not all, values of s, and76

if s1 is a point of convergence while s2 is a point of divergence, then,

on account of the above theorems, we have 1 > 2. All points on the

real axis can then be divided into two classes L and U; U if theseries converges at , otherwise L. Then any point ofU lies to theright of any point ofL, and this classification defines a number o such

that the series converges for every > o and diverges for < o,

the cases = 0 begin undecided. Thus the region of convergence is a

half-plane. The line = 0 is called the line of convergence, and the

number o is called the abscissa of convergence. We have seen that 0may be . On the basis of Theorem 2 we can establish

Theorem 4. A Dirichlet series represents in its half-plane of conver-

gence a regular function of s whose successive derivatives are obtained

by termwise differentiation.

We shall now prove a theorem which implies the uniqueness of

Dirichlet series.

Theorem 5. If

ane

sn converges for s = 0, and its sum f(s) = 0 foran infinity of values of s lying in the region:

> 0, |am s| < /2,

then an = 0 for all n.

Proof. f(s) cannot have an infinite number of zeros in the neighbour-

hood of any finite point of the giver region, since it is regular there.

Hence there exists an infinity of valves sn = n + itn, say, with n+1 >

, lim n = such that f(sn) = 0.However,77

g(s) e1 sf(s) = a1 +2

ane(n1)s,


79/154

2. Elementary theory of Dirichlet Series 73

(here we are assuming 1 > 0) converges for s = 0, and is therefore

uniformly convergent in the region given; each term of the series on theright 0, as s and hence the right hand side, as a whole, tendsto a1 as s . Thus g(s) a1 as s along any path in the givenregion. This would contradict the fact that g(sn) = 0 for an infinity of

sn , unless a1 = 0. Similarly a2 = 0, and so on. Remark. In the hypothesis it is essential that > 0, for if = 0, the

origin itself may be a limit point of the zeros of f, and a contradiction

does not result in the manner in which it is derived in the above proof.

The above arguments can be applied to

ane

n s so as to yield theexistence of an abscissa of absolute convergence , etc. In general,

0. The strip that separates from o may comprise the wholeplane, or may be vacuous or may be a half-plane.

Ex.1. 0 = 0, = 1

(1 21s)(s) =

1

1s+

1

2s+

1

3s+

2

1

2s+

1

4s+

=

1

1s 1

2s+

1

3s 1

4s+

Ex.2. (1)n

n(log n)s converges for all s but never absolutely. In the 78

simple case n = log n, we have

Theorem 6. 0 1.For if

ann

s < , then |an| n = O(1), so that |an|

ns+1+<

for > 0

While we have observed that the sum-function of a Dirichlet series

is regular in the half-plane of convergence, there is no reason to assume

that the line of convergence contains at least one singularity of the func-

tion (see Ex 1. above!). In the special case where the coefficients are

positive, however we can assert the following

Theorem 7. If an 0, then the point s = 0 is a singularity of thefunction f(s).


80/154

74 9. The Zeta Function of Riemann (Contd.)

Proof. Since an 0, we have 0 = , and we may assume, withoutloss of generality, that 0 = 0. Then, if s = 0 is a regular point, theTaylor series for f(s) at s = 1 will have a radius of convergence > 1.

(since the circle of convergence of a power series must have at least one

singularity). Hence we can find a negative value of s for which

f(s) =

=0

(s 1)!

f()(1) =

=0

(1 s)1

n=1

anne

n

Here every term is positive, so the summations can be interchanged and

we get79

f(s) =

n=1

anen

=0

(1 s)n!

=

n=1

anesn

Hence the series converges for a negative value of s which is a contra-

diction.


81/154

Lecture 10


Riemann (Contd)

2 (Contd). Elementary theory of Dirichlet series80

We wish to prove a formula for the partial sum of a Dirichlet series. We

shall give two proofs, one based on an estimate of the order of the sum-

function [10], and another [14, Lec. 4] which is independent of such an

estimate. We first need the following

Lemma 1. If > 0, then

1

2i

+ii

eus

sds =

1, if u > 0

0, if u < 012

, if u = 0

where+i

1= lim

T

+iTiT

.

Proof. The case u = 0 is obvious. We shall therefore study only the

cases u 0. Now

+iTiT

eusds

s =

eus

us

+iTiT

+

+iTiT

eus

us2 ds

75


82/154

76 10. The Zeta Function of Riemann (Contd)

However

|eu(+iT)

| = eu

and |s| > T; hence, letting T , we get+i

i

eusds

s=

+ii

eus

us2ds I, say.

We shall evaluate I on the contour suggested by the diagram. If81

u < 0 , we use the contour L + CR; and ifu > 0, we use L + CL.

Ifu < 0, we have, on CR,

eus

s2 eu

r2, since Re s > .

Ifu > 0, we have, on CL,euss2 eur2 , since Re s < .

Hence

CR,CL

eusds

us2

2reu

|u|2 0, as r .

By Cauchys theorem, however, we have

L

= CR

;


83/154

2. (Contd). Elementary theory of Dirichlet series 77

hence

(1)1

2i

+ii

eus

us2ds = 0, ifu < 0.

If, however, u > 0, the contour L + CL encloses a pole of the second

order at the origin, and we getL

= CL

+1,

so that

(2)1

2i

+ii

eus

us2ds = 1, ifu > 0.

Remarks. We can rewrite the Lemma as: if a > 0, 82

1

2i

a+i

ai

e(n)s

sds =

0, if < n

1, if > n1

2 , if = n

Formally, therefore, we should have

1

2i

a+iai

f(s)es

sds =

1

2i

1

an

a+iai

e(n)s

sds

=

n

an

the dash denoting that if = n the last term should be halved. We wish

now to establish this on a rigorous basis.

We proceed to prove another lemma first.


84/154


Lemma 2. Let anen s converge for s = real. Then

n1

ae s = o(|t|),

the estimate holding uniformly both in + > and in n. Inparticular, for n = , f(s) = o(|t|) uniformly for + > Proof. Assume that = 0, without loss of generality. Then

a = O(1) and A() A() = O(1) for all and If 1 < N < n, then

n1

ae s =N1

+

nN+1

ae s

=

N1

ae s + s

nN

{A(x) A(N)}exsdx

{A(n) A(N)}esn

= O(N) + O(1) + O

|s|

nN

exdx

= O(N) + O |s| eN

= O(N) + O|t|eN

since |s|2 = t2 + 2 and .83

Hence

n1

ae s = O(N) + O(|t|eN), if 1 < N < n

IfN n, then, trivially,n1

ae s = O(N).


85/154


If we choose N as a function of|t|, tending to more slowly than |t|, weget

n1

ae s = O(|t|)

in either case. 84

Theorem (Perrons Formula). Let0 be the abscissa of convergence ofane

n s whose sum-function is f(s). Then for > 0 and > 0, wehave

an =1

2i

+i

i

f(s)

sesds,

where the dash denotes the fact that if = n the last term is to be

halved.

Proof. Let m < m+1, and

g(s) = es

f(s) m1

anen s

(This has a meaning since Re s > 0). Now

1

2i

+ii

g(s)

sds

=1

2i

+ii

f(s)es

sds

n

an

by the foregoing lemma. We have to show that the last expression is

zero.

Applying Cauchys theorem to the rectangle

iT1, + iT2, + iT2, iT1,


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where T1, T2 > 0, and > , we get

1

2i

+ii

g(s)

sds = 0,

for85

1

2i

+iT2iT1

=1

2i

iT1

iT1

+

+iT2iT1

+

+iT2+iT2

= I1 + I2 + I3, say.

Now, if we fix T1 and T2 and let

, then we see that I2

0,

for g(s) is the sum-function of 1

bn sn , where bn = am+n, n = m+n ,

with 1 > 0, and hence g(s) = o(1) as s in the angle |ams| 2

.Thus

1

2i

+iT2iT1

g(s)

sds =

1

2i

iT1

iT1

+iT2

+iT2

if the two integrals on the right converge.

Now if

h(s)

g(s)e(m+1)s = am+1 + am+2e(m+2m+1)s +

then by lemma 2, we have|h(s)| < T2

for s = + iT2, 0 + , T2 T0. Therefore for the second integralon the right we have86

+iT2+iT2

g(s)

sds

0. This proves not only that the integral converges for a finite T2,

but also that as T2 , the second integral tends to zero. Similarly forthe first integral on the right.


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N.B. An estimate for g(s) alone (instead ofh(s)) will not work!

Remarks. More generally we have for > 0 and >

n

ane

n s =1

2i

+ii

f(s)

s s e(ss)ds,

for

anenS en(ss) =

ane

n s = f(s),

and writing s = s s and bn = anen s , we have f(s) f(s) =bne

n s .We shall now give an alternative proof of Persons formula without

using the order of f(s) as s [14, Lec. 4]

Aliter. If f(s) =1

anen s has 0 as its abscissa of convergence,

and ifa > 0, a > 0, then for n, we have

n 0.By Lemma 1, as applied to the first integral on the right hand side, we


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get

(B) limT

1

2i

a+iTaiT

f(s)es

sds

n . For all such m the last relation holds.

Now write bn = anen , (x) = ex(s

), so that

anen s =

bn(). Write B(x) =

nxbn.

Then applying the Lemma of the 9th lecture, we get

Nm+1

anen s =

N1

m1

= (s )N

m

B(x)ex(s) dx

+ B(N)eN(s) B(m)em(s)

= {B(N) B(m)}eN(s)

+(s )

N

m

{B(x) B(m)}ex(s

)

dx.

88

Now choose > 0 and a > > 0. [If o 0 the secondcondition implies the first; if0 < 0, then since a > 0, choose 0 <

. Hence, lettingN , we get

rm(s) = (s )

m

{B(x) B(m)}ex(s

)

dx,


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or

|rm(s)| c |s

| em()

Now consider1

2i

L+CR

rm(s)es

sds

where L + CR is the contour indicated in the diagram.

rm(s) is regular in this contour, and the integral is therefore zero. 89

Hence L

=

CR

On CR, however, s =

+ rei and

a, so that

|rm(s)| c ra e

mrcos

|es| = e(+rcos ), |s| r.

Hence1

2i

CR

rm(s)es

sds

cr

2(a ) e

/2/2

e(m)rcos d

=

cr

(a ) e

/2

0

e(

m)rsin

d


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However, sin

2

for 0

/2; hence

1

2i

a+iTaiT

rm(s)es

sds

cre

(a )

/20

e2

(m)rd

cre

2(a )(m )r,

for all T. Hence

(C) limT

a+iT

aiT

rm(s)es

sds

c1

(m

)

for all m mo.90Now reverting to relation (A), we get

(D) limT

1

2i

a+iTaiT

f(s)es

sds 1

2i

a+iTaiT

fm(s)es

sds

= lim

T

1

2i

a+iTaiT

rm(s)es

sds

for every m. However,

limT

1

2i

a+iTaiT

fm(s)es

sds =

n


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limT

1

2i

a+iTaiT

f(s)es

sds

n mo.

Letting m we get the result. 91Remarks. (i) If = n the last term in the sum

n 0) then in the contour-integration the residue

at the origin has to be taken, and this will contribute a term - f(0).

In the proof, the relation between |s| and rhas to be modified.


92/154


93/154

Lecture 11


Riemann (Contd)

3 Analytic continuation of (s). First method [16,

p.18]92

We have, for > 0,

(s) =

0

xs1exdx.

Writing nx for x, and summing over n, we get

n=1

(s)

ns=

n=1

n=0

xs1enxdx

=

0

n=1

enxxs1dx, if > 1.

since

n=1

xs

1

enx

dx

n=1

0

x

1

enx

dx =

n=1

()

n < ,

87


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if > 1. Hence

n=1

(s)

ns=

0

ex

1 ex xs1dx =

0

xs1dxex 1

or

(s)(s) =

0

xs1

ex 1 dx, > 1

In order to continue (s) analytically all over the s-plane, consider

the complex integral

I(s) =C

zs1

e 1 dz

where C is a contour consisting of the real axis from + to , 0 < A|z|;Hence, for fixed s,

|

|z|=| 2

1

A e2|t| 0 as 0, if > 1.

Thus, on letting 0, we get, if > 1,

I(s) =

0

xs1

ex 1 dx +

0

(xe2i)s1

ex 1 dx


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3. Analytic continuation of (s). First method 89

= (s)(s) + e2is (s)(s)= (s)(s)(e2s 1).

Using the result

(s)(1 s) = sin s

we get 94

I(s) =(s)

(1 s) 2i.e2is 1

eis eis

=(s)

(1 s) 2i eis,

or

(s) = eis

(1 s)2i

C

zs

1

dzez 1 , > 1.

The integral on the right is uniformly convergent in any finite region of

the s-plane (by obvious majorization of the integrand), and so defines

an entire function. Hence the above formula, proved first for > 1,

defines (s), as a meromorphic function, all over the s-plane. This only

possible poles are the poles of(1 s), namely s = 1, 2, 3, . . .. We knowthat (s) is regular for s = 2, 3, . . . (As a matter of fact, I(s) vanishes at

these points). Hence the only possible pole is at s = 1.

Hence

I(1) =C

dz

ez 1 = 2i,

while

(1 s) = 1s 1 +

Hence the residue of(s) at s = 1 is 1.

We see in passing, since

1

ez 1 =1

z 1

2+ B1

z

2! B2

z3

4!+

that 95

(0) = 12

, (2m) = 0, (1 2m) = (1)m

Bm2m

, m = 1, 2, 3, . . .


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4 Functional Equation (First method) [16, p.18]

Consider the integral zs1dzez 1

taken along Cn as in the diagram.

Between C and Cn, the integrand has poles at

2i , . . . , 2ni.

The residue at

2mi is (2me2

i)s1

while the residue at 2mi is (2me3/2i)s1; taken together they amountto

(2m)ei(s1)e

i2 (s 1) + e i2 (s1)

= (2m)s1ei(s1)2cos

2(s 1)

= 2(2m)s1eis sin 2

s


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4. Functional Equation (First method) 91

Hence96

I(s) =Cn

zs

1

dzez 1 + 4i sin s2 e

is

nm=1

(2m)s1

by the theorem of residues.

Now let < 0, and n . Then, on Cn,

|zs1| = O(|z|1) = O(n1),

and1

ez 1 = O(1),

for

|ez 1|2 = |ex+iy 1|2 = |ex(cosy + i siny) 1|2= e2x 2ex cosy + 1,

which, on the vertical lines, is (ex 1)2 and, on the horizontal lines,= (ex + 1)2. (since cosy = 1 there).

Also the length of the square-path is O(n). Hence the integral round

the square 0 as n .Hence

I(s) = 4ie

is

sin s

2

m=1

(2m)s1

= 4ieis sins

2 (2)s1(1 s), if < 0.

or 97

(s)(s)(e2is 1) = (4i)(2)s1eis sin s2

(1 s)

= 2ieis(s)

(1 s)Thus

(s) = (1 s)(1 s)2ss1 sin s2

,


98/154


for < 0, and hence, by analytic continuation, for all values of s (each

side is regular except for poles!).This is the functional equation.

Since

x

2

x + 1

2

=

2x1(x),

we get, on writing x = 1 s,

2s

1 s2

1 s

2

=

(1 s);

also

s

2 1 s

2 =

sin s

2Hence

(1 s) = 2s

1 s2

s

2

1 {sin s

2}1

Thus

s/2(s/2)(s) = (1s)

2

1 s

2

(1 s)

If98

(s) 12

s(s 1)s/2(s/2)(s)

1

2 s(s 1)(s),then (s) = (1 s) and (s) = (1 s). If (z) = ( 1

2+ iz), then

(z) (z).

5 Functional Equation (Second Method) [16, p.13]

Consider the lemma given in Lecture 9, and write n = n, an = 1,

(x) = xs in it. We then get

nx n

s

= s

X1

[x]

xs+1 dx +

[X]

Xs , if X 1.


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5. Functional Equation (Second Method) 93

=

s

s 1 s

(s 1)Xs1 s

X

1

x

[x]

xs+1 dx +

1

Xs1 X

[X]

Xs

Since 1Xs1 = 1/X1, and

X [X]Xs 1/X,

we deduce, on making X ,

(s) =s

s 1 s

1

x [x]xs+1

dx, if > 1

or

(s) = s

1

[x] x + 12

xs+1dx +

1

s 1 +1

2, if > 1

5.1 Since [x] x + 12

is bounded, the integral on the right hand side 99

converges for > 0, and uniformly in any finite region to the right of

= 0. Hence it represents an analytic function of s regular for > 0,

and so provides the continuation of(s) up to = 0, and s = 1 is clearly

a simple pole with residue 1.

For 0 < < 1, we have, however,

10

[x] xxs+1

dx = 1

0

xsdx =1

s 1 ,

and

s

2=

1

dx

xs+1=

1

2

Hence


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5.2

(s) = s

0

[x] xxs+1

dx, 0 < < 1

We have seen that (5.1) gives the analytic continuation up to = 0.

By refining the argument dealing with the integral on the right-hand side

of (5.1) we can get the continuation all over the s-plane. For, if

f(x) [x] x + 12

, f1(x) x

1

f(y)dy,

then f1(x) is also bounded, since

n+1n

f(y)dy = 0 for any integer n.

Hence100

x2x1

f(x)

xs+1dx =

f1(x)

xs+1

x2

x1

+ (s + 1)

x2x1

f1(x)

xs+2dx

0, as x1 , x2 ,if > 1.

Hence the integral in (5.1) converges for > 1.

Further

s

10

[x] x + 12

xs+1dx =

1

s 1 +1

the riemann zetafunction chandrasekharan tata

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