extended higher herglotz functions i. functional equations

EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL

EQUATIONS

ATUL DIXIT, RAJAT GUPTA AND RAHUL KUMAR

Dedicated to Professor Don Zagier on the occasion of his 70th birthday

Abstract. In 1975, Don Zagier obtained a new version of the Kronecker limit formula for a realquadratic field which involved an interesting function F (x) which is now known as the Herglotzfunction. As demonstrated by Zagier, and very recently by Radchenko and Zagier, F (x) satisfiesbeautiful properties which are of interest in both algebraic number theory as well as in analyticnumber theory. In this paper, we study Fk,N (x), an extension of the Herglotz function whichalso subsumes higher Herglotz function of Vlasenko and Zagier. We call it the extended higherHerglotz function. It is intimately connected with a certain generalized Lambert series. We derivetwo different kinds of functional equations satisfied by Fk,N (x). Radchenko and Zagier gave a

beautiful relation between the integral

∫ 1

0

log(1 + tx)

1 + tdt and F (x) and used it to evaluate this

integral at various rational as well as irrational arguments. We obtain a relation between Fk,N (x)and a generalization of the above integral involving polylogarithm. The asymptotic expansions ofFk,N (x) and some generalized Lambert series are also obtained along with other supplementaryresults.

Contents

1. Introduction 12. Main results 52.1. Functional equations for extended higher Herglotz functions Fk,N (x) 62.3. Asymptotics of the extended higher Herglotz functions 82.4. An interesting relation between a generalized polylogarithm function and the extended

higher Herglotz functions 92.5. Transformations involving higher Herglotz functions and generalized Lambert series 103. Proofs of functional equations satisfied by Fk,N (x) 123.1. Proof of the equivalence of Theorems 2.1 and 2.2 for odd natural numbers k and N

such that 1 < k ≤ N 173.2. Proof of a relation between Fk,N (x) and a generalized polylogarithm 204. Proofs of transformations and asymptotic expansions of the extended higher Herglotz

functions and generalized Lambert series 215. Concluding Remarks 26References 28

1. Introduction

The Kronecker limit formula is one of the celebrated theorems in number theory having applica-tions not only in other areas of Mathematics such as in geometry but also in Physics, for example,in string theory for the one-loop computation in the Polyakov’s perturbative approach [22]. Let

2010 Mathematics Subject Classification: Primary 30D05 Secondary 30E15, 33E20Keywords: Herglotz function, functional equations, polylogarithm, Lambert series, asymptotic expansions.

1

EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 2

K denote a number field and ζK(s) the Dedekind zeta-function of K. If A denotes an ideal classgroup of K, then we can write ζK(s) =

∑A ζ(s,A), where, for Re(s) > 1, ζ(s,A) =

∑a∈A

1N(a)s ,

with N(a) being the norm of the fractional ideal a. Then the Kronecker limit formula is concernedwith the evaluation of the constant term %(A) in the Laurent series expansion of ζ(s,A). This wasdone by Kronecker [18] in the case of an imaginary quadratic field K. While Hecke [15] obtaineda corresponding formula for a real quadratic field K, it was not explicit as one of the expressionsin it was left as an integral containing the Dedekind eta-function.

In a seminal paper [35], Zagier obtained an explicit form of the Kronecker limit formula for realquadratic fields. Zagier’s formula gives a new representation for the value

%(B) := lims→1

(Ds/2ζ(s,B)− log ε

s− 1

),

where B is an element of the group of narrow ideal classes of invertible fractional OD-ideals1 of the

quadratic order OD = Z + ZD+√D

2 of discriminant D > 0, ζ(s,B), for Re(s) > 1, is the sum ofN(a)−s over all invertible OD-ideals a in the class B, and ε = εD is the smallest unit > 1 in OD ofnorm 1. The reason such a representation is worth studying is because the special value at s = 1of the Dirichlet series LK(s, χ) =

∑a χ(a)N(a)−s equals D−1/2

∑B χ(B)%(B) for any character χ

on the narrow ideal class group. Zagier’s formula [35, p. 164] reads

%(B) =∑

w∈Red(B)

P (w,w′), (1.1)

where the function P (x, y) for x > y > 0 is defined as

P (x, y) := F (x)− F (y) + Li2

(yx

)− π2

6+ log

(x

y

)(γ − 1

2log(x− y) +

1

4log

(x

y

)), (1.2)

with Red(B) being the set of larger roots w = (−b +√D)/(2a) of all reduced primitive quadratic

forms Q(X,Y ) = aX2 + bXY + cY 2(a, c > 0, a+ b+ c < 0) of discriminant D belonging to the classB. Here γ is Euler’s constant, Li2(t) is the dilogarithm function

∞∑n=1

tn

n2(0 < t < 1), (1.3)

and

F (x) :=∞∑n=1

ψ(nx)− log(nx)

n(x ∈ C\(−∞, 0]) , (1.4)

where ψ(a) = Γ′(a)/Γ(a) is the logarithmic derivative of the gamma function. The function F (x)is similar to a function that Herglotz studied in [16] and hence Radchenko and Zagier [26] call itthe Herglotz function.

The Herglotz function F (x) satisfies several interesting and important properties as has beendemonstrated by Zagier in [35], and very recently by Radchenko and Zagier in [26] where theystudy, among other things, the relation of this function with the Dedekind eta-function, func-tional equations satisfied by F (x) in connection with Hecke operators, the cohomological aspectsof F (x), and its special values at positive rationals and quadratic units. Prior to this, Zagier [35,Equations (7.4), (7.8)] had obtained beautiful functional equations that F (x) satisfies, namely, forx ∈ C\(−∞, 0],

F (x)− F (x+ 1)− F(

x

x+ 1

)= −F (1) + Li2

(1

1 + x

), (1.5)

1Here two ideals belong to the same narrow class if their quotient is a principal ideal λOD with N(λ) > 0.


F (x) + F

(1

x

)= 2F (1) +

1

2log2(x)− π2

6x(x− 1)2, (1.6)

where [35, Equation (7.12)]

F (1) = −1

2γ2 − π2

12− γ1,

and γ1 denotes the first Stieltjes constant.Apart from their intrinsic beauty, these functional equations are very useful, for example, Zagier

[35] used them to prove Meyer’s theorem and Radchenko and Zagier [26] have used them to obtainasymptotic expansions of F (x) as x→ 0 and x→ 1. Functional equations of the type in (1.5) occurin a variety of areas of mathematics such as period functions for Maass forms [19, Equation (0.1)](see also [5, Equation (1)]), cotangent functions, double zeta functions etc. We refer the reader toa beautiful article of Zagier [37] on this topic.

Several generalizations of the Herglotz function (also known sometimes as the Herglotz-Zagierfunction) have been studied in the literature. Ishibashi [17] studied the jth order Herglotz functiondefined by

Φj(x) :=∞∑n=1

R′j(nx)− logj(nx)

n(j ∈ N),

where Rj(x) is a generalization of the function R(x) studied by Deninger [8], and is defined as thesolution to the difference equation

f(x+ 1)− f(x) = logj(x), (f(1) = (−1)j+1ζ(j)(0)),

with ζ(s) denoting the Riemann zeta function. It is clear that Φ1(x) = F (x). Ishibashi’s functionΦj(x) appears in his explicit representation [17, Theorem 3] of the Laurent series coefficients ofZQ`(s), the zeta function associated to indefinite quadratic forms Q`, where

Ds/2ζ(s,B) =∑`

ZQ`(s) (Re(s) > 1).

Zagier [35] had earlier obtained the constant coefficient and the coefficient of (s−1)−1 in the Laurent

series expansion of Ds/2ζ(s,B). The Herglotz function F (x) makes its conspicuous presence inZagier’s expression of the constant coefficient as can be seen from (1.1) and (1.2).

Masri [21] generalized F (x) in a different direction by considering

F (α, s, x) :=∞∑n=1

α(n) (ψ(nx)− log(nx))

ns(Re(s) > 0, x > 0),

where α : Z → C is a periodic function with period M . Vlasenko and Zagier [30] studied thefollowing special case of Masri’s F (α, s, x) when α(n) ≡ 1 and s > 1 is a natural number butwithout considering log(nx) term (which is not needed for convergence of the series since s > 1):

Fk(x) :=∞∑n=1

ψ(nx)

nk(k ∈ N, k > 1, x ∈ C\(−∞, 0]) . (1.7)

They termed it the higher Herglotz function2 and obtained following beautiful functional relationsfor it [30, Equations (11), (12)]:

Fk(x) + (−x)k−1Fk

(1

x

)= −γζ(k)

(1 + (−x)k−1

)−k−1∑r=2

ζ(r)ζ(k + 1− r)(−x)r−1

2It is important to note that Vlasenko and Zagier [30] define their higher Herglotz function by∑∞n=1 ψ(nx)/nk−1

for k > 2.


+ ζ(k + 1)

((−x)k − 1

x

), (1.8)

and3

Fk(x)− Fk(x+ 1) + (−x)k−1Fk

(x+ 1

x

)= (−x)k−1 (ζ(k, 1) + ζ(k + 1)− γζ(k))

−k−1∑r=1

ζ(k + 1− r, r)(−x)r−1 + ζ(k + 1)

((−x)k

x+ 1− 1

x

),

(1.9)

where

ζ(m,n) :=∑p>q>0

1

pmqn(m ≥ 2, n ≥ 1).

The function Fk(x) plays an instrumental role in the higher Kronecker “limit” formulas developedin [30].

The goal of our paper is to study a novel generalization of F (x), which we call the extendedhigher Herglotz function. We define it by

Fk,N (x) :=

∞∑n=1

ψ(nNx)− log(nNx)

nk(x ∈ C\(−∞, 0]) , (1.10)

where k and N are positive real numbers such that k+N > 1. It is clear that when k > 1, one canwrite

Fk,N (x) =∞∑n=1

ψ(nNx)

nk−∞∑n=1

log(nNx)

nk

=∞∑n=1

ψ(nNx)

nk+Nζ ′(k)− ζ(k) log x. (1.11)

However, we keep (1.10) as the definition of the extended higher Herglotz function so as to keepmany special cases under one roof. Indeed, from (1.7) and (1.11), we have for k ∈ N,

Fk,1(x) = Fk(x) + ζ ′(k)− ζ(k) log x. (1.12)

Also F1,1(x) is the Herglotz function defined in (1.4).We were naturally led to the idea of considering the extended higher Herglotz function in (1.10)

from our previous work [9] on a generalized Lambert series. Indeed, the extended higher Herglotzfunction turns up after differentiating a transformation for the generalized Lambert series

∞∑n=1

n−2Nm−1exp(−a(2n)Nα

)1− exp (−(2n)Nα)

(1.13)

with respect to a and then letting a = 1 as can be seen from (4.8) below.An application of our extended higher Herglotz functions is in obtaining the asymptotic expansion

of the Lambert series∞∑n=1

n−2Nm−1+N

exp ((2n)Nα)− 1(N odd)

as α → 0+ which can be seen in Theorem 2.13. Obtaining asymptotic expansions for Lambertseries and its generalizations has received a lot of attention since last century because of theirapplications in the theory of the Riemann zeta function. For example, Wigert [32] (see also [29,

3The interpretation of ζ(1, k − 1) given in [30, p. 28] has a minus sign missing in front of the whole term, that is,the correct form is − (ζ(k − 1, 1) + ζ(k)− γζ(k − 1)).


p. 163, Theorem 7.15]) obtained the asymptotic expansion of∑∞

n=1 1/(enz − 1) as z → 0 whichwas used by Lukkarinen [20] to not only study the moments of the Riemann zeta function but to

also obtain meromorphic continuation of the modified Mellin transform of |ζ(1/2 + ix)|2 defined

for Re(s) > 1 by∫∞1

∣∣ζ (12 + ix)∣∣2 x−s dx. For odd N ∈ N, Wigert himself [33, p. 9-10] obtained the

asymptotic expansion of the generalized Lambert series∞∑n=1

1

enNx − 1=∞∑j=1

∞∑n=1

e−nN jx.

as x→ 0. Note that the inner sum in the above double series is a generalized partial theta function.Wigert not only wrote a paper on this partial theta function [34] but also on the above generalizedLambert series [33].

Zagier [36] had conjectured that the asymptotic expansion of the Lambert series associated withthe square of the Ramanujan tau function τ(n), that is of

∑∞n=1 τ

2(n)e−nx, as x → 0+, can beexpressed in terms of the non-trivial zeros of the Riemann zeta function. This conjecture wasproved by Hafner and Stopple [14]. Thus obtaining asymptotic expansions of Lambert series is auseful endeavor.

In this paper, we obtain two different kinds of functional equations satisfied by Fk,N (x). SeeTheorems 2.1 and 2.2. As a special case of one of these functional equations, we get (1.6) of Zagier.

For x > 0, one of the beautiful identities that F (x) satisfies [26] is

J(x) = F (2x)− 2F (x) + F(x

2

)+

π2

12x, (1.14)

where

J(x) :=

∫ 1

0

log (1 + tx)

1 + tdt. (1.15)

Radchenko and Zagier [26] have obtained special values of F (x) at positive rationals and quadraticunits with the help of which they are able to evaluate special values of J(x), for example,

J(4 +√

17) = −π2

6+

1

2log2(2) +

1

2log(2) log(4 +

√17),

J

(2

5

)=

11π2

240+

3

4log2(2)− 2 log2

(√5 + 1

2

).

Earlier, Herglotz [16] as well as Muzaffar and Williams [25] had evaluated such integrals but only

of a particular type, that is, J(n +√n2 − 1). For example, Herglotz [16, Equation (70a)] showed

that

J(4 +√

15) = −π2

12(√

15− 2) + log(2) log(√

3 +√

5) + log

(√5 + 1

2

)log(2 +

√3).

Regarding this integral, Chowla [6, p. 372] remarked, ‘A direct evaluation of this definite integral isprobably difficult’. Indeed, these evaluations come from not only employing methods from analyticnumber theory but also those from algebraic number theory, which is why they are all the moreinteresting!

In this paper, a generalization of (1.14) involving Fk,N (x) is obtained in Theorem 2.8. Lastly,we obtain some results showcasing the intimate relationship between generalized Lambert seriesand extended higher Herglotz functions.

2. Main results

Throughout this paper, the notation∑(N−1)

j=−(N−1)′′ will denote a sum over over j = −(N −

1),−(N − 3), · · · , N − 3, N − 1.


2.1. Functional equations for extended higher Herglotz functions Fk,N (x).

Theorem 2.1. Let k,N ∈ N such that 1 < k ≤ N and x ∈ C\(−∞, 0]. Let

B(k,N, x) :=

{(−1)k+N+1x1/Nζ

(1 + k

N

)if k = N

0 if k 6= N. (2.1)

Then

x1−kN Fk,N (x)− (−1)k

N

(N−1)∑j=−(N−1)

′′eiπj(k−1)/NFN+k−1

N, 1N

(e−

iπjN

x1/N

)

=π

N

ζ(1 + k−1

N

)sin(πN (k − 1)

) + x1−kN(− (γ + log x) ζ(k) +Nζ ′(k)

)− 1

xN+k−1N

ζ(k +N) + B(k,N, x). (2.2)

Also,

F1,N (x) +1

N

(N−1)∑j=−(N−1)

′′F1, 1

N

(e−

iπjN

x1/N

)

=1

N

{π2

6− (N − 1)γ log x+

1

2(log x)2 −Nγ2 − (N2 + 1)γ1

}− ζ(N + 1)

x+ B(1, N, x). (2.3)

We note that when N = 1, (2.3) gives Zagier’s (1.6) as a special case.In the next theorem, we obtain a different kind of functional equation for Fk,N (x), more specifi-

cally, for a combination of Fk,N functions. There is a trade-off between the two types of functionalequations obtained in Theorems 2.1 and 2.2, namely, Theorem 2.1 is valid for any natural numbersk and N such that 1 < k ≤ N whereas Theorem 2.2 is valid for any odd natural numbers k andN . These two types are, indeed, equivalent when k and N are odd such that 1 < k ≤ N . This isshown in subsection 3.1.

The idea for deriving the second type of functional equation, considered in Theorem 2.2 below,stemmed from a beautiful result obtained in [9, Theorem 2.2]. This identity, which was instrumentalin deriving the main transformation for a generalized Lambert series in [9], is as follows.

Let u ∈ C be fixed such that Re(u) > 0. Then,

∞∑m=1

∫ ∞0

t cos(t)

t2 +m2u2dt =

1

2

{log( u

2π

)− 1

2

(ψ

(iu

2π

)+ ψ

(−iu2π

))}. (2.4)

Theorem 2.2. Let k and N be odd natural numbers. Let Re(x) > 0. Then,

N−1∑j=−(N−1)

′′exp

(− iπ(k−1)j

2N

){FN+k−1

N, 1N

(i(2πx

) 1N e

iπj2N

)+ FN+k−1

N, 1N

(−i(2πx

) 1N e

iπj2N

)}= (−1)

k+12 N

(2πx

) k−1N

{Fk,N

(ix2π

)+ Fk,N

(− ix

2π

)+ 2

((γ − log

(2πx

))ζ(k)−Nζ ′(k)

)+ C (k,N, x)

},

(2.5)

where, for 1−kN 6= −2

⌊k2N

⌋,

C (k,N, x) :=πζ(N+k−1N

)N sin

(π2N (1− k)

) ( x2π

) k−1N

+ 4π

b k2Nc∑

j=1

(−1)j(2π)−2j−1ζ(k − 2Nj)ζ(2j + 1)x2j ,

(2.6)


and for 1−kN = −2

⌊k2N

⌋,

C (k,N, x) :=2(−1)

k−12N

N

( x2π

) k−1N [(

γ + log(2πx

))ζ(1 + k−1

N

)− ζ ′

(1 + k−1

N

)]+ 4π

b k2Nc−1∑

j=1

(−1)j(2π)−2j−1ζ(k − 2Nj)ζ(2j + 1)x2j . (2.7)

The remaining case not considered in the above theorem, namely, 1−kN = −2

⌊k2N

⌋= 0, that is,

k = 1, is given in the following result.

Theorem 2.3. Let k and N be odd natural numbers and Re(x) > 0. Then,

N−1∑j=−(N−1)

′′eiπ2(j−1)

{F1, 1

N

(i(2πx

) 1N e

iπj2N

)+ F1, 1

N

(−i(2πx

) 1N e

iπj2N

)}= −N

{F1,N

(ix2π

)+ F1,N

(− ix

2π

)− 1

N

(π2

12− 2Nγ2 + 2(N − 1)γ log

(2π

x

)+ log2(2π)− log

(4π2

x

)log x− 2(N2 + 1)γ1

)}.

Ramanujan’s famous formula for ζ(2m + 1) [27, p. 173, Ch. 14, Entry 21(i)], [28, pp. 319-320,formula (28)], [3, pp. 275-276] is given for α, β > 0, Re(α) > 0,Re(β) > 0, αβ = π2 and m ∈ Z\{0},by

α−m

{1

2ζ(2m+ 1) +

∞∑n=1

n−2m−1

e2αn − 1

}= (−β)−m

{1

2ζ(2m+ 1) +

∞∑n=1

n−2m−1

e2βn − 1

}

− 22mm+1∑j=0

(−1)jB2jB2m+2−2j(2j)!(2m+ 2− 2j)!

αm+1−jβj , (2.8)

where Bn is the nth Bernoulli number. The Lambert series in (1.13) enables us to obtain a two-parameter generalization of (2.8) given in [9, Theorem 2.4]. See (4.8) below.

As a special case, Theorem 2.2 gives a beautiful transformation for a combination of the Vlasenko-Zagier higher Herglotz function Fk(x) which is analogous to (2.8).

Corollary 2.4. Let α and β be two complex numbers with Re(α) > 0,Re(β) > 0 and αβ = 4π2.Then for m ∈ N,

α−m

{2γζ(2m+ 1) +

∞∑n=1

1

n2m+1

(ψ

(inα

2π

)+ ψ

(− inα

2π

))}

= −(−β)−m

{2γζ(2m+ 1) +

∞∑n=1

1

n2m+1

(ψ

(inβ

2π

)+ ψ

(− inβ

2π

))}

− 2m−1∑j=1

(−1)jζ(1− 2j + 2m)ζ(2j + 1)αj−mβ−j . (2.9)

It is important to note here that Corollary 2.4 cannot be obtained for any general α, β such thatαβ = 4π2 by applying Vlasenko and Zagier’s (1.8) once with k = 2m + 1 and x = iα/(2π), andagain with k = 2m+ 1 and x = iβ/(2π), and adding the resulting two equations, for, doing so will


result in

∞∑n=1

ψ(inα2π

)n2m+1

+ (−β)mα−m∞∑n=1

ψ(− inα

2π

)n2m+1

+

∞∑n=1

ψ(inβ2π

)n2m+1

+ (−α)mβ−m∞∑n=1

ψ(− inβ

2π

)n2m+1

= −γ(2 + (−β)mα−m + (−α)mβ−m

)ζ(2m+ 1) + iζ(2m+ 2)

(2π

β+

2π

α− (−β)mα−mβ

2π

− (−α)mβ−mα

2π

)−

2m∑r=2

ζ(r)ζ(2m+ 2− r)(− i

2π

)r−1 (αr−1 + βr−1

). (2.10)

It is clear that it is impossible to get Corollary 2.4 from (2.10) for any α, β such that αβ = 4π2 be-cause of the factors of the form (−β)mα−m and (−α)mβ−m in front of the series

∑∞n=1 n

−2m−1ψ(− inα

2π

)and

∑∞n=1 n

−2m−1ψ(− inβ

2π

)respectively in (2.10) whereas we do not have such factors in Corollary

2.4.Only in the case when α = β = 2π and m is replaced by 2m is Corollary 2.4 obtainable from

(2.10). We state below this special case of Corollary 2.4, namely for m ∈ N,

∞∑n=1

1

n4m+1(ψ(in) + ψ(−in)) = −2γζ(4m+ 1)−

2m−1∑j=1

(−1)jζ(2j + 1)ζ(4m+ 1− 2j). (2.11)

Letting m = 1 in Corollary 2.4 gives a beautiful modular relation:

Corollary 2.5. Let α and β be two complex numbers with Re(α) > 0,Re(β) > 0 and αβ = 4π2.Then

1

α

{2γζ(3) +

∞∑n=1

1

n3

(ψ

(inα

2π

)+ ψ

(−inα

2π

))}=

1

β

{2γζ(3) +

∞∑n=1

1

n3

(ψ

(inβ

2π

)+ ψ

(−inβ

2π

))}.

(2.12)

Remark 2.2. Corollary 2.4 is an analogue of Ramanujan’s formula for ζ(2m+ 1), that is, (2.8).Indeed, the combination of Vlasenko-Zagier higher Herglotz functions in Corollary 2.4 seems tomimick the role played by the Lambert series in Ramanujan’s formula.

This phenomenon is seen to be true even when we consider the generalization of Corollary 2.4given in Theorem 2.2. Indeed, the combination of the extended higher Herglotz functions Fk,N (x)in Theorem 2.2 plays a similar role as that played by the generalized Lambert series in Theorem1.2 of [11]. The latter theorem is the special case a = 1 of (4.8).

2.3. Asymptotics of the extended higher Herglotz functions. An immediate application ofthe functional equations given in the above theorems is in obtaining the asymptotic expansionsof the extended higher Herglotz functions Fk,N (x) as x → 0. These asymptotic expansions fordifferent conditions on the parameters involved are collected together in the theorem below.

Theorem 2.6. Let x ∈ C\(−∞, 0] and k,N are positive real numbers such that k + N > 1. LetB(k,N, x) be defined in (2.1). Then as x→∞,

Fk,N (x) ∼ − 1

2xζ(k +N)−

∞∑n=1

B2n

2nζ(k + 2nN)x−2n. (2.13)

Also, as x→ 0,

(i) For k,N ∈ N such that 1 < k ≤ N, we have

Fk,N (x) ∼ −ζ(k +N)

x− (γ + log x) ζ(k) +Nζ ′(k) +

π

N

ζ(1 + k−1

N

)sin(πN (k − 1)

)x k−1N + x

k−1N B(k,N, x)


+ (−1)k∞∑m=1

(−1)m(N+1)ζ(k −mN)ζ (1 +m)xm; (2.14)

(ii) For N ∈ N,

F1,N (x) ∼ −ζ(N + 1)

x+

1

N

{π2

6− (N − 1)γ log x+

1

2log2 x−Nγ2 − (N2 + 1)γ1

}+ B(1, N, x)−

∞∑m=1

(−1)m(N+1)ζ(1−mN)ζ (1 +m)xm. (2.15)

Letting N = 1 gives Radchenko and Zagier’s asymptotic expansions for F (x) as x → ∞ and asx→ 0 as can be seen from [26, Equation (7)].

Theorem 2.7. Let k and N be positive real numbers such that k +N > 1 and let Re(x) > 0. LetC (k,N, x) be defined in (2.6) and (2.7). Then as x→∞,

Fk,N

(ix

2π

)+ Fk,N

(− ix

2π

)∼∞∑n=1

(−1)n+1

nB2nζ(k + 2nN)

(2π

x

)2n

, (2.16)

Also, for odd natural numbers k and N , as x→ 0,

Fk,N

(ix

2π

)+ Fk,N

(− ix

2π

)∼ 2

{(log

(2π

x

)− γ)ζ(k) +Nζ ′(k)

}− C (k,N, x) +

(−1)N+k+2

2

N

( x2π

) k−1N

exp

(− iπ(N + k − 1)

2N

)×∞∑n=1

B2n

n cos(πnN

)ζ (N + k + 2n− 1

N

)( x2π

) 2nN, (2.17)

and for an odd natural number N , as x→ 0,

F1,N

(ix

2π

)+ F1,N

(− ix

2π

)∼ 1

N

(π2

12− 2Nγ2 + 2(N − 1)γ log

(2π

x

)+ log2(2π)− log

(4π2

x

)log x− 2(N2 + 1)γ1

)+

i

N(−1)

N+12

∞∑n=1

B2n

n cos(πnN

)ζ (1 +2n

N

)( x2π

) 2nN. (2.18)

2.4. An interesting relation between a generalized polylogarithm function and the ex-tended higher Herglotz functions. The polylogarithm is a generalization of the dilogarithmfunction of (1.3). It is defined by

Lis(t) :=∞∑n=1

tn

ns.

It is clear that Lis(t) converges for any complex s as long as |t| < 1. It can be analytically continuedfor |t| ≥ 1. There are a number of generalizations of polylogarithm in the literature.

In the following result, we encountered a new generalization of the polylogarithm function. Wedefine it for N ∈ N, s ∈ C and |t| < 1 by

NLis(t) :=

∞∑n=1

tnN

ns. (2.19)

Clearly, 1Lis(t) = Lis(t) for |t| < 1.


Theorem 2.8. Let k and N be any positive integers. Let NLis(t) be defined in (2.19). Define

Jk,N (x) :=

∫ 1

0

(2k−1

u− 1− 2Nu(2

N−1)

u2N − 1− 2k−1 − 1

u log u

)NLik(−ux) du. (2.20)

Then for x > 0, we have

Jk,N (x) = Fk,N (2Nx)−(

2k−1 + 21−k)

Fk,N (x) + Fk,N

( x

2N

)+(

2N + 2−N − 2k−1 − 21−k) ζ(k +N)

x. (2.21)

The integral in (2.20) is a natural generalization of (1.15) in the setting of Fk,N (x) in the sensethat it arises naturally while extending the Radchenko-Zagier relation between J(x) and F (x) in(1.14). Indeed, it can be easily checked that J1,1(x) = J(x) and that (1.14) follows by lettingk = N = 1 in Theorem 2.8.

The above theorem leads to the following relation involving polylogarithm and Vlasenko andZagier’s higher Herglotz function Fk(x) from (1.12), which is valid for a positive integer k > 1:

Corollary 2.9. Let x > 0 and let k > 1 be a positive integer. Let Fk(x) be defined in (1.7). Then∫ 1

0

(2k−1

u− 1− 2u

u2 − 1− 2k−1 − 1

u log u

)Lik(−ux) du

= Fk(2x)−(

2k−1 + 21−k)Fk(x) + Fk

(x2

)+(

2− 2k−1 − 21−k)(

ζ ′(k)− ζ(k) log x+1

xζ(k + 1)

)+

1

2xζ(k + 1).

For k odd, Corollary 2.9, in turn, gives an evaluation of an integral involving polylogarithmfunction in terms of the higher Herglotz function Fk(2), special values of the Riemann zeta functionand its derivatives as stated below.

Corollary 2.10. Let k > 1 be an odd positive integer. Let Fk(x) be defined in (1.7). Then∫ 1

0

(2k−1

u− 1− 2u

u2 − 1− 2k−1 − 1

u log u

)Lik(−u) du

=(

1− 21−k)Fk(2) + (2k−1 − 1)γζ(k) +

(1

2− 2−k

)ζ(k + 1) +

(2− 2k−1 − 21−k

)ζ ′(k)

+k−1∑r=2

(−1)r−1(

2k−2 + 2−k − 2r−k)ζ(r)ζ(k + 1− r). (2.22)

2.5. Transformations involving higher Herglotz functions and generalized Lambert se-ries. In [10, Corollary 2.13], the following companion of Ramanujan’s formula (2.8) was obtained.Let m ∈ N. If α and β are complex numbers such that Re(α) > 0, Re(β) > 0, and αβ = π2, then

α−(m− 12)

{1

2ζ(2m) +

∞∑n=1

n−2m

e2nα − 1

}−m−1∑j=0

22j−1B2j

(2j)!ζ(2m− 2j + 1)α2j−m− 1

2

= (−1)m+1β−(m− 12)

{γ

πζ(2m) +

1

2π

∞∑n=1

n−2m(ψ

(inβ

π

)+ ψ

(− inβ

π

))}. (2.23)

In what follows, we give a one-parameter extension of (2.23).


Theorem 2.11. Let N be an odd positive integer. For αβN = πN+1 with Re(α) > 0, Re(β) > 0,and m ≥ 1,

α−( 2NmN+1

− 12)

(1

2ζ(2Nm+ 1−N) +

∞∑n=1

n−2Nm−1+N

e(2n)Nα − 1

)

−m−1∑j=0

B2j ζ(2Nm+ 1− 2Nj)

(2j)!2N(2j−1)α2j− 2Nm

N+1− 1

2

=22m(N−1)

NπN+1

2

(−1)m+1β−( 2NmN+1

−N2 )

[Nγ

2N−1ζ(2m)

+1

2N

(N−1)2∑

j=− (N−1)2

∞∑n=1

1

n2m

(ψ

(iβ

2π(2n)1/Ne

iπjN

)+ ψ

(− iβ

2π(2n)1/Ne

iπjN

))]. (2.24)

It is straightforward to see that letting N = 1 in the above theorem gives (2.23).

Theorem 2.12. Let N be an odd positive integer. If α, β > 0 such that αβN = πN+1,∞∑n=1

nN−1

exp ((2n)Nα)− 1− 1

2NαN(Nγ − log(2π)− (N − 1)(log 2− γ))

=1

2Nα(N + 1)log

(α

β

)+

2

2NαN

N−12∑

j=−(N−12 )

∞∑n=1

[log

(β

2π(2n)

1N e

iπjN

)

− 1

2

{ψ

(iβ

2π(2n)

1N e

iπjN

)+ ψ

(−i β

2π(2n)

1N e

iπjN

)}]+

{α2 , if, N = 1

0, if, N > 1.(2.25)

As an application of the asymptotic expansions derived in Theorem 2.7, we obtain the completeasymptotic expansions for the Lambert series with the help of Theorem 2.11 above.

Theorem 2.13. Let N be an odd positive integer and let m ≥ 1. As α→ 0,∞∑n=1

n−2Nm−1+N

e(2n)Nα − 1∼ 1

2Nαζ(2Nm+ 1)− 1

2ζ(2Nm+ 1−N) +

(−1)m+1

Nπ2m2(2m−1)(N−1)α2m−1

×∞∑`=1

(−1)`+1B2`N

`ζ (2m+ 2`)

(2N−1α

π

)2`

+ DN (m,α), (2.26)

where

DN (m,α) :=2(2m−1)(N−1)

Nπ2m(−1)m+1α2m−1

{ζ(2m)

(Nγ − log

(α2N−1

π

))− ζ ′(2m)

}

+m−1∑j=1

B2j ζ(2Nm+ 1− 2Nj)

(2j)!2N(2j−1)α2j−1.

Also, in particular, as α→ 0,∞∑n=1

n−2m

e2nα − 1∼ ζ(2m+ 1)

2α− 1

2ζ(2m) +

(−1)m+1α2m−1

π2m

{ζ(2m)

(γ − log

(απ

))− ζ ′(2m)

}

+(−1)m+1α2m−1

π2m

∞∑`=1

(−1)`+1B2`

`ζ (2m+ 2`)

(απ

)2`


+

m−1∑j=1

B2j ζ(2m+ 1− 2j)

(2j)!(2α)(2j−1). (2.27)

We note in passing that the well-known result [12, p. 903, Formula 8.361.8], for Re(x) > 0,namely

ψ(x)− log x = −∫ ∞0

(1

1− e−t− 1

t

)e−xt dt,

gives an integral representation for Fk,N (x):

Fk,N (x) = −∫ ∞0

(1

1− e−t− 1

t

)NLik

(e−xt

)dt, (2.28)

where NLik(t) is defined in (2.19), whereas Binet’s formula [31, p. 251] for Re(x) > 0, namely

ψ(x) +1

2x− log x = −

∫ ∞0

2t

(t2 + x2)(e2πt − 1)dt,

yields a representation for Fk,N (x) in terms of a generalized Lambert series for Re(x) > 0:

Fk,N (x) = −∫ ∞0

( ∞∑n=1

n−k

exp(2πnN t)− 1

)2t

t2 + x2dt− 1

2xζ(k +N).

3. Proofs of functional equations satisfied by Fk,N (x)

Proof of Theorem 2.1. Employing the well-known functional equation

ψ(x+ 1) = ψ(x) +1

x(3.1)

in the definition of Fk,N (x) in (1.10), we get

Fk,N (x) =

∞∑n=1

ψ(nNx+ 1)− log(nNx)

nk− 1

xζ(k +N). (3.2)

We need the following formula due to Kloosterman [29, Equation 2.9.2], which is valid in 0 < c =Re(z) < 1 and x ∈ C\(−∞, 0]:

1

2πi

∫(c)

−πζ(1− z)sin(πz)

x−z dz = ψ(x+ 1)− log x, (3.3)

where here and in the sequel,∫(c) denotes the line integral

∫ c+i∞c−i∞ .

Thus, invoking (3.3) in (3.2), we get for 0 < c = Re(z) < 1,

Fk,N (x) =1

2πi

∞∑n=1

1

nk

∫(c)

−πζ(1− z)sin(πz)

(nNx

)−zdz − 1

xζ(k +N)

=1

2πi

∫(c)

−πζ(1− z)ζ (k +Nz)

sin(πz)x−z dz − 1

xζ(k +N),

by interchanging the order of summation and integration. Employing the change of variable z =− sN −

k−1N in the above equation, we see that for −N − k + 1 < c′ = Re(s) < 1− k,

Fk,N (x) =1

Nxk−1N I − 1

xζ(k +N), (3.4)

where,

I = I(k,N, x) :=1

2πi

∫(c′)

πζ(1 + s+k−1

N

)ζ(1− s)

sin(πN (s+ k − 1)

) xsN ds.


Now construct a rectangular contour formed by the line segments [c′− iT, c′+ iT ], [c′+ iT, c′′+ iT ],[c′′ + iT, c′′ − iT ], [c′′ − iT, c′ − iT ] where 0 < c′′ < N − k + 1. Since we wish to employ (3.3) againafter shifting the line of integration from Re(s) = c′ to Re(s) = c′′, we also need 0 < c′′ < 1. In anycase, we must have N − k + 1 > 0, or, in other words, k ≤ N . Note that in the process of shiftingthe line of integration, we encounter a double pole of the integrand at s = 1− k and a simple poleat s = 0.

Here, and throughout the paper, Ra denotes the residue of the associated integrand at its poleat a. Stirling’s formula in the vertical strip p ≤ σ ≤ q reads [7, p. 224]

|Γ(s)| =√

2π|t|σ−12 e−

12π|t|(

1 +O

(1

|t|

))(3.5)

as |t| → ∞.Then letting T → ∞, noting that the integrals over the horizontal segments thereby go to zero

(due to (3.5)) and employing Cauchy’s residue theorem, we have

I = I1 − (R0 +R1−k), (3.6)

where

I1 = I1(k,N, x) :=1

2πi

∫(c′′)

πζ(1 + s+k−1

N

)ζ(1− s)

sin(πN (s+ k − 1)

) xsN ds,

and the residues R0 and R1−k are easily seen to be

R0 = −πζ(1 + k−1

N

)sin(πN (k − 1)

) , (3.7)

R1−k = −Nx1−kN(− (γ + log x) ζ(k) +Nζ ′(k)

). (3.8)

To evaluate I1, we use the following well known result [9, Lemma 4.3]

1

sin(z)=

1

sin(Nz)

(N−1)∑j=−(N−1)

′′eijz (3.9)

with z = π(s+ k − 1)/N so as to obtain

I1 =

(N−1)∑j=−(N−1)

′′eiπj(k−1)/N

1

2πi

∫(c′′)

πζ(1− s)ζ(1 + s+k−1

N

)sin (π(s+ k − 1))

(e−

iπjN

x1/N

)−sds

= (−1)k(N−1)∑

j=−(N−1)


∞∑n=1

1

n1+k−1N

1

2πi

∫(c′′)

−πζ(1− s)sin (πs)

((nx

) 1Ne−

iπjN

)−sds, (3.10)

where in the last step we used the series representation of ζ(1 + s+k−1

N

)since k ≥ 1 and c′′ > 0

and interchanged the order of summation and integration which can be easily justified by standardarguments. Now invoke (3.3) again in (3.10) to obtain

I1 = (−1)k(N−1)∑

j=−(N−1)


∞∑n=1

1

n1+k−1N

{ψ

((nx

) 1Ne−

iπjN + 1

)− log

((nx

) 1Ne−

iπjN

)}

= (−1)k(N−1)∑

j=−(N−1)


∞∑n=1

1

n1+k−1N

{ψ

((nx

) 1Ne−

iπjN

)− log

((nx

) 1Ne−

iπjN

)}

+ (−1)kx1/Nζ

(1 +

k

N

) (N−1)∑j=−(N−1)

′′eiπjk/N , (3.11)


where we used (3.1) in the last step. From (3.9) and the fact that 1 < k ≤ N , we have

(N−1)∑j=−(N−1)

′′eiπjk/N =

{(−1)N+1N if k = N0 if 1 < k < N.

(3.12)

Using (3.12) in (3.11) and recalling the definition of (1.10), we see that

I1 = (−1)k(N−1)∑

j=−(N−1)


N, 1N

(e−

iπjN

x1/N

)+NB(k,N, x), (3.13)

where B(k,N, x) is defined in (2.1). Substituting (3.7), (3.8) and (3.13) in (3.6), we see that

I = (−1)k(N−1)∑

j=−(N−1)


N, 1N

(e−

iπjN

x1/N

)+NB(k,N, x) + π

ζ(1 + k−1

N

)sin(πN (k − 1)

)+Nx

1−kN(− (γ + log x) ζ(k) +Nζ ′(k)

). (3.14)

The proof now follows from substituting (3.14) in (3.4) and multiplying both sides of the resulting

equation by x1−kN .

Equation (2.3) can be similarly proved. The only difference is that k = 1 renders the pole ats = 0 of the integrand of to be a pole of order 3 because of which the residue at it has to be modifiedaccordingly. �

Proof of Theorem 2.2. The fact that k and N are odd is used several times in the proof withoutmention.

Case 1: 1−kN 6= −2

⌊k2N

⌋.

Let J denote the left-hand side of (2.5). Using (1.10), we rewrite J in the form

J = −2N−1∑

j=−(N−1)

′′ij exp

(− iπ(k +N − 1)j

2N

) ∞∑n=1

1

nk+N−1N

{log

((2πn

x

) 1N

eiπj2N

)

− 1

2

[ψ

(i

(2πn

x

) 1N

eiπj2N

)+ ψ

(−i(

2πn

x

) 1N

eiπj2N

)]}. (3.15)

Define

X`,n :=2π`

(2πn

x

) 1N

(3.16)

X∗`,n,j :=2π`

(2πn

x

) 1N

eiπj2N = X`,ne

iπj2N (3.17)

Employing (2.4) in (3.15), we observe that

J = −4N−1∑

j=−(N−1)

′′ij exp

(− iπ(k +N − 1)j

2N

) ∞∑n=1

1

nk+N−1N

∞∑`=1

∫ ∞0

t cos t

t2 +X∗`,n,j2 dt

= −4N−1∑

j=−(N−1)

′′ij exp

(− iπ(k +N − 1)j

2N

)π

2

∞∑n,`=1

1

nk+N−1N

1

2πi

∫(c1)

Γ(s1)

tan(πs12

) (X∗`,n,j)−s1 ds1,

(3.18)


where 1 < c1 := Re(s1) < 2. The last step follows from the following result [9, Lemma 4.1]

1

2πi

∫(c1)

Γ(s1)

tan(πs12 )u−s1 ds1 =

2

π

∫ ∞0

t cos t

u2 + t2dt, (3.19)

which is valid for 0 < Re(s1) = c1 < 2 and Re(u) > 0 4.From (3.16), (3.17) and (3.18) and interchanging the order of summation and integration, we

have

J =−1

i

∞∑n,`=1

1

nk+N−1N

∫(c1)

Γ(s1)

tan(πs12

) N−1∑j=−(N−1)

′′ij exp

(− iπj

2N(s1 + k +N − 1)

)X−s1`,n ds1. (3.20)

From [9, Equation (4.2)], for N odd,

cos(Nz)

cos(z)= (−1)

N−12

N−1∑j=−(N−1)

′′ij exp(−ijz). (3.21)

Using (3.21) with z = π(s1 + k +N − 1)/(2N) in (3.20), we observe that

J =−1

i

∞∑n,`=1

1

nk+N−1N

∫(c1)

Γ(s1)

tan(πs12

)(−1)N−1

2cos(π2 (s1 + k +N − 1)

)cos(π2

(s1+k+N−1

N

)) X−s1`,n ds1

=(−1)

k−12

2i

∞∑n,`=1

1

nk+N−1N

∫(c1)

Γ(s1) cos(πs12

)cos(π2

(s1+k+N−1

N

))X−s1`,n ds1.

Performing the change of variable s1 = Ns − k − N + 1 in the above equation, we see that fork+NN < c < k+N+1

N ,

J = −(−1)N+1

2 N

i

∞∑n,`=1

1

nk+N−1N

∫(c1)

Γ(Ns− k −N + 1) sin(πNs2

)cos(πs2

) X−Ns+k+N−1`,n ds

=−Ni

(−1)N+1

2

(2π

(2π

x

) 1N

)k+N−1

×∫(c)

Γ(Ns− k −N + 1) sin(πNs2

)cos(πs2

) ζ(s)ζ(Ns− k −N + 1)

(2π

(2π

x

) 1N

)−Nsds, (3.22)

where in the last step, we used the definition of X`,n in (3.16) and the series representations forζ(s) and ζ(Ns− k −N + 1).

Now use the asymmetric form of the functional equation for ζ(s) [29, p. 13, Equation (2.1.1)],namely,

ζ(s) = 2sπs−1Γ(1− s)ζ(1− s) sin(πs

2

), (3.23)

in (3.22) to deduce that

J =−N2πi

(−1)k+12

(2π

x

) k+N−1N

∫(c)ζ(k +N −Ns) tan

(πs2

)Γ(1− s)ζ(1− s)xs ds. (3.24)

4We take 1 < c1 = Re(s1) < 2 in the integral in (3.18) and not 0 < c1 = Re(s1) < 2 so that we are able to use theseries representation of ζ(s) while evaluating the integral in (3.22)


If we replace s by 1− s in the above equation then for −k+1N < c2 < − k

N , we get

J =−N2πi

(−1)k+12

(2π

x

) k+N−1N

∫(c2)

ζ(k +Ns)cos(πs2

)sin(πs2

)Γ(s)ζ(s)x1−s ds. (3.25)

We wish to use the series representation for ζ(s) in (3.25) for which we transform it by shift-ing the line of integration to 1 < d = Re(s) < 2. Consider the contour formed by the lines[c2− iT, c2 + iT ], [c2 + iT, d+ iT ], [d+ iT, d− iT ] and [d− iT, c2− iT ]. Note that inside this contourthe integrand has(1) a simple pole at s = 1−k

N (due to ζ(k +Ns));

(2) a double order pole at s = 0 (due to sin(πs2

)and Γ(s));

(3) simple poles at s = −2j due to the zeros of sin(πs2

), 1 ≤ j ≤

⌊k2N

⌋(because −2j ≥ − k

N which

implies that j ≤⌊k2N

⌋). This is because the poles of Γ(s) at s = −j, 1 ≤ j ≤ k

N are canceled by

the zeros of ζ(s) and cos(πs2

)at negative even and odd integers respectively.

Note that using (3.5), one can see that the integrals over the horizontal segments of the contourtend to zero as the height T →∞. Therefore invoking Cauchy’s theorem, we deduce that

1

2πi

∫(c2)

ζ(k +Ns) cot(πs

2

)Γ(s)ζ(s)x1−s ds

= −R0 −R 1−kN−b k

2N c∑j=1

R−2j +1

2πi

∫(d)ζ(k +Ns) cot

(πs2

)Γ(s)ζ(s)x1−s ds. (3.26)

The residues in (3.26) are calculated to be

R0 =x

π

{(γ − log

(2π

x

))ζ(k)−Nζ ′(k)

},

R−2j = 2(−1)j(2π)−2j−1ζ(k − 2Nj)ζ(1 + 2j)x1+2j ,

R 1−kN

=π

N

( x2π

) k+N−1N ζ

(1 + k−1

N

)sin(π2

(1−kN

)) . (3.27)

Also, the integral on the right-hand side of (3.26) is evaluated to

1

2πi

∫(d)ζ(k +Ns) cot

(πs2


= x∞∑

n,`=1

1

`k1

2πi

∫(d)

Γ(s)

tan(πs2

)(x`Nn)−s ds

=2x

π

∞∑`=1

1

`k

∞∑n=1

∫ ∞0

t cos t

(t2 + (x`Nn)2)dt

=x

π

∞∑`=1

1

`k

{log

(`Nx

2π

)− 1

2

(ψ

(i`Nx

2π

)+

(−i`Nx

2π

))}, (3.28)

where in the second step we used (3.19) and in the last step used (2.4).Now substitute (3.27) and (3.28) in (3.26) so as to derive

1

2πi

∫(c)ζ(k +Ns) cot

(πs2


=x

π

∞∑`=1

1

`k

{log

(`Nx

2π

)− 1

2

(ψ

(i`Nx

2π

)+

(−i`Nx

2π

))}


− x

π

{(γ − log

(2π

x

))ζ(k)−Nζ ′(k)

}

− 2

b k2N c∑j=1

(−1)j(2π)−2j−1ζ(k − 2Nj)ζ(1 + 2j)x1+2j

− π

N

( x2π

) k+N−1N

sec

(π

2

(k +N − 1

N

))ζ

(k +N − 1

N

). (3.29)

Finally substitute (3.29) in (3.25) and use the definitions of Fk,N (x) and C (k,N, x) from (1.10)and (2.6) respectively to arrive at (2.5).

Case 2: 1−kN = −2

⌊k2N

⌋6= 0.

The only change in this case is the contribution of the double order pole of the integrand in (3.25)at s = (1 − k)/N due to ζ(k + Ns) and sin

(πs2

). Due to this, the residue at (1 − k)/N of the

integrand of the integral on the left-hand side of (3.26) now becomes

2(−1)k−12N

N

( x2π

)N+k−1N

[(γ + log

(2π

x

))ζ

(1 +

k − 1

N

)− ζ ′

(1 +

k − 1

N

)].

�

3.1. Proof of the equivalence of Theorems 2.1 and 2.2 for odd natural numbers k andN such that 1 < k ≤ N . Assume k and N to be odd positive integers such that 1 < k ≤ N .Employing (2.2) once, with x replaced by ix

2π and then again with x replaced by − ix2π , and then

adding the resulting two equations, we arrive at

Fk,N

(ix

2π

)+ Fk,N

(− ix

2π

)=( x

2π

) k−1N {E1 + E2 + E3}+ E4, (3.30)

where

E1 : =2π

N

ζ(N+k−1N

)sin(π(k−1)N

) cos( π

2N(k − 1)

)(3.31)

E2 : = ik−1N B

(k,N,

ix

2π

)+ (−i)

k−1N B

(k,N,− ix

2π

)E3 : =

(−1)k

N

N−1∑j=−(N−1)

′′eiπj(k−1)

N

[ik−1N F k+N−1

N, 1N

((ix

2π

)− 1N

e−iπjN

)

+(−i)k−1N F k+N−1

N, 1N

((− ix

2π

)− 1N

e−iπjN

)](3.32)

E4 : = −2γζ(k) + 2Nζ ′(k)− ζ(k)

(log

(ix

2π

)+ log

(− ix

2π

)). (3.33)

Using the definition of B(k,N, x) from (2.1), it is easy to see that E2 = 0 for 1 < k < N , and thatfor k = N ,

E2 = 2(−1)k+N+1( x

2π

) 1Nζ

(1 +

k

N

)cos

(πk

2N

).

Since the cosine function vanishes for k = N , we conclude that E2 = 0 in this case as well. Hence

E2 = 0 for 1 < k ≤ N. (3.34)


We next simplify E3. To that end, use the definition of Fk,N (x) from (1.10) in (3.32) so that

E3 =(−1)k

N

N−1∑j=−(N−1)

′′eiπj(k−1)

N

i k−1N

∞∑n=1

ψ((

ix2πn

)− 1N e−

iπjN

)− log

((ix2πn

)− 1N e−

iπjN

)nk+N−1N

+(−i)k−1N

∞∑n=1

ψ((− ix

2πn

)− 1N e−

iπjN

)− log

((− ix

2πn

)− 1N e−

iπjN

)nk+N−1N

.Now write both the digamma functions in the above equation in the form ψ(x) = ψ(x+ 1)− 1/x

by using the functional equation (3.1) and then use Kloosterman’s formula (3.3) twice in the aboveequation so as to get, for 0 < c = Re(s) < 1,

E3 =(−1)k

N

N−1∑j=−(N−1)

′′eiπj(k−1)

N

[−( x

2π

) 1NeiπjN

(eiπk2N + e−

iπk2N

)ζ

(1 +

k

N

)

+1

2πi

∫(c)

−πζ(1− s)sin(πs)

(( x

2πn

)− 1Ne−

iπjN

)−s (is+k−1N + (−i)

s+k−1N

)ds

]

=2(−1)k+1

N

( x2π

) 1N

cos

(πk

2N

)ζ

(1 +

k

N

) N−1∑j=−(N−1)

′′eiπjkN +

2(−1)k

2πiN

N−1∑j=−(N−1)

′′eiπj(k−1)

N

×∫(c)

−πζ(1− s)sin(πs)

ζ

(s+ k +N − 1

N

)cos

(π(s+ k − 1)

2N

)(( x2π

)− 1Ne−

iπjN

)−sds. (3.35)

We now use (3.9) in (3.35) and employ the change of variable s = s1 − (k +N − 1) in the integralon the right-hand side of (3.35) to arrive at, for k +N − 1 < λ = Re(s1) < k +N ,

E3 =(−1)k+1

N

(xπ

) 1Nζ

(1 +

k

N

)sin(πk)

sin(πk2N

) +2(−1)k

2πiN

( x2π

)− (k+N−1)N

N−1∑j=−(N−1)

′′e−iπj

×∫(λ)

πζ(k +N − s1)(−1)k+N sin(πs1)

ζ(s1N

)sin(πs1

2N

)(( x2π

)− 1Ne−

iπjN

)−sds1.

Note that the first term on the right-hand side of the above equation vanishes as the sine functionis zero since k is an integer. Also as k and N are odd positive integers therefore (−1)k+N = 1 andj runs over even integers hence we have e−iπj = 1. Therefore E3 reduces to

E3 = 2(−1)k

2πiN

( x2π

)− (k+N−1)N

N−1∑j=−(N−1)

′′∫(λ)

πζ(k +N − s1)sin(πs1)

ζ(s1N

)sin(πs1

2N

)

×(( x

2π

)− 1Ne−

iπjN

)−s1ds1. (3.36)


Employing the change of variable s1 = Ns in the integral in (3.36), we have, for k+N−1N < λ∗ =

Re(s) < k+NN ,

E3 =2(−1)k

2πi

( x2π

)− (k+N−1)N

N−1∑j=−(N−1)

′′∫(λ∗)

πζ(k +N −Ns)sin(πNs)

ζ (s) sin(πs

2

)(xeiπj2π

)sds

=2(−1)k

2πi

( x2π

)− (k+N−1)N

∫(λ∗)

ζ(k +N −Ns)sin(πNs)

sin2(πs

2

)Γ(1− s)ζ (1− s)xs

N−1∑j=−(N−1)

′′eiπjsds,

(3.37)

where in the last step we used (3.23). Now use (3.9) in (3.37) to obtain

E3 =(−1)k

2πi

( x2π

)− (k+N−1)N

∫(λ∗)

ζ(k +N −Ns)Γ(1− s)ζ (1− s) tan(πs

2

)xsds. (3.38)

From (3.15) and (3.24), for k+NN < c < k+N+1

N , we have

1

2πi

∫(c)ζ(k +N −Ns)Γ(1− s)ζ (1− s) tan

(πs2

)xsds

=−(−1)

k+12

N

( x2π

) k+N−1N

N−1∑j=−(N−1)

′′e−

iπ(k−1)j2N

{F k+N−1

N, 1N

(i

(2π

x

) 1N

eiπj2N

)

+F k+N−1N

, 1N

(−i(

2π

x

) 1N

eiπj2N

)}. (3.39)

We want to employ (3.39) in (3.38), for which, we need to shift the line of integration of (3.38)to k+N

N < c < k+N+1N . Consider the contour formed by the line segments [λ∗ − iT, c − iT ], [c −

iT, c + iT ], [c + iT, λ∗ + iT ] and [λ∗ + iT, λ∗ − iT ]. Note that the integrand has no poles insidethis contour. The condition 1 < k ≤ N implies −1 ≤ 1 − k+N

N < −1N , ensures that the pole of the

integrand of the left-hand side of (3.39) at s = k+NN does not lie inside the contour. Due to the

same reason, ζ(1 − s) also does not have a pole inside the contour. Therefore applying Cauchy’sresidue theorem to the integral in (3.38) and using the fact that the integral along the horizontalsegments go to zero as the height T of the contour tends to ∞, for k+N

N < c < k+N+1N , we see that

E3 =(−1)k

2πi

( x2π

)− (k+N−1)N

∫(c)ζ(k +N −Ns)Γ(1− s)ζ (1− s) tan

(πs2

)xsds

=(−1)

k+12

N

N−1∑j=−(N−1)

′′e−

iπ(k−1)j2N

{F k+N−1

N, 1N

(i

(2π

x

) 1N

eiπj2N

)

+F k+N−1N

, 1N

(−i(

2π

x

) 1N

eiπj2N

)}, (3.40)

where in the last step we used (3.39).Note that k and N odd and 1 < k ≤ N imply that 1−k

N 6= −2⌊k2N

⌋, which, in turn, implies that

the finite sum in (2.6) is empty. Therefore, the equivalence of Theorems 2.1 and 2.2 follows uponsubstituting (3.31), (3.33), (3.34) and (3.40) in (3.30) to arrive at (2.5).

Proof of Theorem 2.3. Let k = 1 in (3.25). The proof is almost similar to that of Theorem 2.2except that the pole of the integrand in (3.25) at 0 is of order three whence the corresponding


residue becomes

R0 =−x

2Nπ

{−2γ2N +

π2

12+ 2γ(N − 1) log

(2π

x

)+ log2(2π)− log

(4π2

x

)log(x)− 2(N2 + 1)γ1

}.

�

Proof of Corollary 2.4. Let N = 1, k = 2m + 1,m ≥ 1, in Theorem 2.2 and use the definitionof Fk,N (x) in (1.10). Let α = x and β = 4π2/α. Since m ≥ 1, we can separate the expressionsinvolving logarithm. This results in a lot of simplification thereby resulting in (2.9). �

Proof of Corollary 2.5. Let m = 1 in Corollary 2.4 and notice that the finite sum on the right-hand side vanishes thereby giving (2.12). �

3.2. Proof of a relation between Fk,N (x) and a generalized polylogarithm.

Proof of Theorem 2.8. Employing (2.28) and repeatedly using the fact∞∑n=1

a(2n) =1

2

∞∑n=1

(1 +

(−1)n)a(n), we see that

Fk,N (2Nx)−(

2k−1 + 21−k)

Fk,N (x) + Fk,N

( x

2N

)= −

∫ ∞0

(1

1− e−t− 1

t

)[ ∞∑n=1

e−(2n)Nxt

nk−(

2k−1 + 21−k) ∞∑n=1

e−nNxt

nk+∞∑n=1

e−(n/2)Nxt

nk

]dt

= −∫ ∞0

(1

1− e−t− 1

t

)[2k−1

∞∑n=1

(1 + (−1)n)e−n

Nxt

nk− 2k−1

∞∑n=1

e−nNxt

nk

−∞∑n=1

(1 + (−1)n)e−(n/2)

Nxt

nk+

∞∑n=1

e−(n/2)Nxt

nk

]dt

= −∫ ∞0

(1

1− e−t− 1

t

)[2k−1

∞∑n=1

(−1)ne−n

Nxt

nk−∞∑n=1

(−1)ne−(n/2)

Nxt

nk

]dt

= −∫ ∞0

(1

et − 1− 1

t

)[2k−1

∞∑n=1

(−1)ne−n

Nxt

nk−∞∑n=1

(−1)ne−(n/2)

Nxt

nk

]dt

−∫ ∞0

(2k−1

∞∑n=1

(−1)ne−n

Nxt

nk−∞∑n=1

(−1)ne−(n/2)

Nxt

nk

)dt,

where the last step was simplified using1

1− e−t=

1

et − 1+ 1. Since

∫ ∞0

e−at dt =1

afor a > 0, we

have

Fk,N (2Nx)−(

2k−1 + 21−k)

Fk,N (x) + Fk,N

( x

2N

)= −2k−1

∫ ∞0

(1

et − 1− 1

t

) ∞∑n=1

(−1)ne−n

Nxt

nkdt+ 2N

∫ ∞0

(1

e2Ny − 1− 1

2Ny

) ∞∑n=1

(−1)ne−n

Nxy

nkdy

−(2k−1 − 2N

)x

∞∑n=1

(−1)n

nk+N,


where, in the second integral, we employed the change of variable t = 2Ny. Since∑∞

n=1(−1)nn−s =(21−s − 1)ζ(s) for Re(s) > 0, we deduce that

Fk,N (2Nx)− (2k−1 + 21−k)Fk,N (x) + Fk,N

( x

2N

)= −

∫ ∞0

(2k−1

et − 1− 2N

e2N t − 1− (2k−1 − 1)

t

) ∞∑n=1

(−1)ne−n

Nxt

nkdt

− (2N + 2−N − 2k−1 − 21−k)ζ(k +N)

x. (3.41)

Making change of variable e−t = u in the integral on the right-hand side of (3.41), using thedefinitions of NLis(t) and Jk,N (x) given in (2.19) and (2.20) respectively, using the elementary fact

(−1)nN

= (−1)n and simplifying, we arrive at (2.21). This completes the proof.�

The next result which gives a relation between an integral with polylogarithm in its integrandand the Vlasenko-Zagier higher Herglotz function, and is analogous to (1.14), was missing in theliterature.

Proof of Corollary 2.9. Let N = 1 in the Theorem 2.8 and employ (1.11). �

Proof of Corollary 2.10. Let x = 1 in Corollary 2.9. This gives∫ 1

0

(2k−1

u− 1− 2u

u2 − 1− 2k−1 − 1

u log u

)Lik(−u) du

= Fk(2)−(

2k−1 + 21−k)Fk(1) + Fk

(1

2

)+(

2− 2k−1 − 21−k) (ζ ′(k) + ζ(k + 1)

)+

1

2ζ(k + 1).

(3.42)

Now let k > 1 be an odd integer. We first calculate Fk(1). Letting x = 1 in (1.8) and simplifying,we obtain

Fk(1) = −γζ(k)− ζ(k + 1)− 1

2

k−1∑r=2

(−1)r−1ζ(r)ζ(k + 1− r). (3.43)

Also, employing (1.8) again with x = 2 and simplifying, we deduce that

Fk

(1

2

)= −21−kFk(2)− 2(1 + 2−k−1)ζ(k + 1)− (1 + 21−k)γζ(k)− 21−k

k−1∑r=2

(−2)r−1ζ(r)ζ(k + 1− r).

(3.44)

Now, substitute (3.43) and (3.44) in (3.42) so as to obtain (2.22) upon simplification. �

4. Proofs of transformations and asymptotic expansions of the extended higherHerglotz functions and generalized Lambert series

Proof of Theorem 2.6. The formula [1, p. 259, formula 6.3.18]

ψ(x) ∼ log x+

∞∑n=1

ζ(1− n)x−n,


as x→∞, | arg x| < π implies that

Fk,N (x) ∼∞∑n=1

ζ(1− n)ζ(k + nN)x−n (4.1)

= − 1

2xζ(k +N)−

∞∑n=1

B2n

2nζ(k + 2nN)x−2n,

where in the second step we used the facts that ζ(−2j) = 0 for j ∈ N and [2, p. 266, Theorem12.16]

ζ(−n) =

{−1

2 , if n = 0,

−Bn+1

n+1 , if n ∈ N.(4.2)

We now establish (2.14) using (2.2) and (4.1). Upon using (4.1), as x→ 0, we have

(N−1)∑j=−(N−1)


N, 1N

((e−iπj

x

)1/N)

∼(N−1)∑

j=−(N−1)


∞∑n=1

ζ(1− n)ζ(N+k+n−1

N

)((e−iπj

x

)1/N)n=∞∑n=1

ζ(1− n)ζ

(N + k + n− 1

N

)xn/N

(N−1)∑j=−(N−1)

′′eiπj(k+n−1)/N

=∞∑n=1

ζ(1− n)ζ

(N + k + n− 1

N

)xn/N

(N−1)∑j=−(N−1)

′′eiπj(k+n−1)/N . (4.3)

From (3.9), we know that if m ∈ N,

(N−1)∑j=−(N−1)

′′eiπj(k+n−1)/N =

{(−1)m(N+1)N if n = mN − k + 10 if n 6= mN − k + 1.

(4.4)

Employing (4.4) in (4.3), we see that, as x→ 0,

(N−1)∑j=−(N−1)


N, 1N

((e−iπj

x

)1/N)

∼ N∞∑

m≥ kN

(−1)m(N+1)ζ(k −mN)ζ (1 +m)xm+ 1−kN . (4.5)

Now divide (2.2) by x(1−k)/N , then take x→ 0 in the resulting identity to obtain (2.14).Equation (2.15) can be proved by letting k = 1 in (4.5) and putting the resulting asymptotic

expansion in (2.3). �

Proof of Theorem 2.7. From (2.13), it is easy to observe that as x→∞,

Fk,N

(ix

2π

)+ Fk,N

(− ix

2π

)∼∞∑n=1

(−1)n+1

nB2nζ(k + 2nN)

(2π

x

)2n

.


To obtain (2.17) as x→ 0, we first rewrite (2.5) as,

Fk,N

(ix2π

)+ Fk,N

(− ix

2π

)= −

{2((γ − log

(2πx

))ζ(k)−Nζ ′(k)

)+ C (k,N, x)

}+

(−1)k+12

N

( x2π

) k−1N

×N−1∑

j=−(N−1)

′′exp

(iπj2 −

iπ(N+k−1)2N

){FN+k−1

N, 1N

(i(2πx

) 1N e

iπj2N

)+ FN+k−1

N, 1N

(−i(2πx

) 1N e

iπj2N

)}.

(4.6)

Using (4.1), as x→ 0,

N−1∑j=−(N−1)

′′exp

(iπj2

){FN+k−1

N, 1N

(i(2πx

) 1N e

iπj2N

)+ FN+k−1

N, 1N

(−i(2πx

) 1N e

iπj2N

)}

∼ 2N−1∑

j=−(N−1)

′′exp

(iπj2

) ∞∑n=1

ζ(1− n)ζ

(k + n+N − 1

N

)( x2π

) nNe−iπnj2N cos

(πn2

)

= 2∞∑n=1

(−1)nζ(1− 2n)ζ

(k + 2n+N − 1

N

)( x2π

) 2nN

N−1∑j=−(N−1)

′′exp

(iπj

2− iπnj

N

),

where in the last step the fact that cosine function vanishes for odd values of n. Using (3.9) and(4.2) in the above equation, we arrive at

N−1∑j=−(N−1)

′′exp

(iπj2

){FN+k−1

N, 1N

(i(2πx

) 1N e

iπj2N

)+ FN+k−1

N, 1N

(−i(2πx

) 1N e

iπj2N

)}

= −∞∑n=1

B2n

nζ

(k + 2n+N − 1

N

)( x2π

) 2nN sin

(πN2

)cos(πnN

) . (4.7)

Hence employ (4.7) in (4.6) to obtain (2.17).To obtain (2.18), let k = 1 in (4.7) and then substitute the resultant in Theorem 2.3. �

Proof of Theorem 2.11. We need Theorem 2.4 from [9] given below.


Let 0 < a ≤ 1, let N be an odd positive integer and α, β > 0 such that αβN = πN+1. Then forany positive integer m,

α−2NmN+1

((a− 1

2

)ζ(2Nm+ 1) +

m−1∑j=1

B2j+1(a)

(2j + 1)!ζ(2Nm+ 1− 2jN)(2Nα)2j

+∞∑n=1

n−2Nm−1exp(−a(2n)Nα

)1− exp (−(2n)Nα)

)

=(−β

2NN+1

)−m 22m(N−1)

N

[(−1)m+1(2π)2mB2m+1(a)Nγ

(2m+ 1)!+

1

2

∞∑n=1

cos(2πna)

n2m+1

+ (−1)N+3

2

N−12∑

j=−(N−1)

2

(−1)j{ ∞∑n=1

n−2m−1 cos(2πna)

exp(

(2n)1N βe

iπjN

)− 1

+(−1)j+

N+32

2π

∞∑n=1

sin(2πna)

n2m+1

(ψ(iβ2π (2n)

1N e

iπjN

)+ ψ

(−iβ2π (2n)

1N e

iπjN

))}]

+ (−1)m+N+32 22Nm

bN+12N

+mc∑j=0

(−1)jB2j(a)BN+1+2N(m−j)

(2j)!(N + 1 + 2N(m− j))!α

2jN+1βN+

2N2(m−j)N+1 . (4.8)

The main idea is to differentiate both sides of the above identity with respect to a and then let

a = 1 while using the facts ddaB2j(a) = 2jB2j−1(a), Bn(1) = (−1)nBn, B2m = 2(−1)m+1(2m)!ζ(2m)

(2π)2m

and αβN = πN+1. Inducting the ζ(2Nm + 1) from the resulting identity in the finite sum on theresulting left-hand side as its j = 0 term, we see that

α−2NmN+1

m−1∑j=0

B2j+1(a)

(2j + 1)!ζ(2Nm+ 1− 2jN)(2Nα)2j − 2Nα

∞∑n=1

n−2Nm−1+N

exp((2n)Nα)− 1

=(−β

2NN+1

)−m 22m(N−1)

N

[2Nγζ(2m) +

N−12∑

j=−(N−1)

2

∞∑n=1

1

n2m

(ψ

(iβ

2π(2n)

1N e

iπjN

)

+ψ

(−iβ2π

(2n)1N e

iπjN

))]+ 2N−1α1− 2Nm

N+1 ζ(2Nm+ 1−N).

The result now follows by dividing both sides of the above equation by 2N√α, using the fact

αβN = πN+1 and rearranging the resulting equation. �


Proof of Theorem 2.12. Let 0 < a ≤ 1 and N be an odd positive integer. Let α, β > 0 be suchthat αβ = πN+1. Then from [9, Theorem 2.7], we have

∞∑n=1

exp(−a(2n)Nα)

n(1− exp(−(2n)Nα))− 1

N(−1)

N+32

N−12∑

j=− (N−1)2

(−1)j( ∞∑n=1

cos(2πna)

n(

exp(

(2n)1N βe

iπjN

)− 1)

+(−1)j+

N+12

π

∞∑n=1

sin(2πna)

n

{log(β2π (2n)

1N e

iπjN

)− 1

2

(ψ(iβ2π (2n)

1N e

iπjN

)+ ψ

(−iβ2π (2n)

1N e

iπjN

))})=

1

N((a− 1) log(2π) + log Γ(a)) +

(a− 1

2

){(N − 1)(log 2− γ)

N+

log(α/β)

N + 1

}

+ (−1)N+3

2

bN+12N c∑j=0

(−1)jB2j(a)BN+1−2Nj(2j)!(N + 1− 2Nj)!

α2jN+1βN−

2N2jN+1 . (4.9)

Differentiate both sides of (4.9) with respect to a, let a = 1 in the resulting identity, and thendivide both sides by −2Nα to arrive at (2.25) upon simplification.

�

Proof of Theorem 2.13. Using (1.10), we rewrite (2.24) as

∞∑n=1

n−2Nm−1+N

e(2n)Nα − 1= −1

2ζ(2Nm+ 1−N) +

(−1)m+122m(N−1)βN/2

NπN+1

2 2N√α

(β

α

)− 2NmN+1

×

(N−1)2∑

j=− (N−1)2

(F2m, 1

N

(iβ

2π21/Ne

iπjN

)+ F2m, 1

N

(− iβ

2π21/Ne

iπjN

))+ α

2NmN+1

− 12 EN (m,α, β),

(4.10)

where EN (m,α, β) is defined by

EN (m,α, β) :=2(2m−1)(N−1)

NπN+1

2

(−1)m+1β−2NmN+1

+N2

{Nζ(2m)

(γ + log

(β

2π21/N

))− ζ ′(2m)

}

+

m−1∑j=0

B2j ζ(2Nm+ 1− 2Nj)

(2j)!2N(2j−1)α2j− 2Nm

N+1− 1

2 . (4.11)

To study the behavior of the series on the left-hand side of (4.10) as α → 0, we examine theseries on the right-hand side of (4.10) as β →∞ using the relation αβN = πN+1.

We now invoke (2.16) with k = 2m, N replaced by 1/N and x = β21/Neiπj/N in the expressionbelow, then replace j by j/2 to see that as β →∞,

(N−1)2∑

j=− (N−1)2

(F2m, 1

N

(iβ

2π21/Ne

iπjN

)+ F2m, 1

N

(− iβ

2π21/Ne

iπjN

))

∼∞∑n=1

(−1)n+1B2n

nζ

(2m+

2n

N

)(β

2π21/N

)−2n N−1∑j=−(N−1)

′′e−

iπnjN .


From (3.9), for ` ∈ N,

(N−1)∑j=−(N−1)

′′e−

iπnjN =

{N(−1)`(N−1) if n = `N0 if n 6= `N.

Hence,

(N−1)2∑

j=− (N−1)2

(F2m, 1

N

(iβ

2π21/Ne

iπjN

)+ F2m, 1

N

(− iβ

2π21/Ne

iπjN

))

∼ N∞∑`=1

(−1)`+1B2`N

`Nζ (2m+ 2`)

(β

2π21/N

)−2`N. (4.12)

Thus from (4.10), (4.11) and (4.12) and the relation β = π(N+1)/N

α1/N , we arrive at (2.26) upon simpli-fication.

To obtain (2.27), simply put N = 1 in (2.26) and simplify. �

5. Concluding Remarks

The primary goal of this paper was to obtain functional equations for a new generalization ofthe Herglotz function, namely Fk,N (x). We obtained two different kinds of functional equationsrelating Fk,N (x) with FN+k−1

N, 1N

(ξ/x), where ξ is some root of unity, the first of which (Theorem

2.1) reduces to Zagier’s (1.6) when k = N = 1.

(1) Although we were unable to get a three-term functional equation for Fk,N (x) similar to (1.6)and (1.9), we have an idea that might help suggest the form of such an equation, if it exists. Thisis explained below.

Note that when we let x = 1 in (1.5) and (1.14), the corresponding Herglotz functions havethe same arguments 5. Provided this phenomenon persists when we transition from the Herglotzfunction to the extended higher Herglotz function Fk,N (x), the three-term functional relation thatis sought might be involving

Fk,N (xN ), Fk,N ((1 + x)N ), and Fk,N

(xN

(1 + x)N

),

for, not only do they reduce to F (x), F (x+1) and F(

xx+1

)respectively when k = N = 1 which are

the same as the ones occurring in (1.6) but they also reduce to Fk,N (1), Fk,N (2N ) and Fk,N (2−N )when we let x = 1 which are indeed the same as those occurring in (2.21) for x = 1.

Such a three-term functional relation, if/when obtained, would be a first-of-its-kind result sincenowhere in the literature has there been a relation which involves like powers (which are greaterthan 1) of x, x+ 1 and x/(x+ 1) in the arguments of the associated functions.

(2) Equation (4.8) involves the series, namely,

∞∑n=1

sin(2πna)

n2m+1

(ψ(iβ2π (2n)

1N e

iπjN

)+ ψ

(−iβ2π (2n)

1N e

iπjN

)),

5In fact, when we let x = 1 in equations (1.5) and (1.14), they exactly match since J(1) = 12

log2(2).


which, if we differentiate with respect to a and then let a = 1, leads to the following combinationof the extended higher Herglotz functions

∞∑n=1

1

n2m

(ψ(iβ2π (2n)

1N e

iπjN

)+ ψ

(−iβ2π (2n)

1N e

iπjN

)).

This was, in fact, our motivation to study Fk,N (x).This suggests a further question - does there exist a transformation for a more general series

∞∑n=1

cos(2πna)

nk

(ψ

(iα

2π(2n)N

)+ ψ

(−iα2π

(2n)N))

,

where k ≥ 1 and 0 < a ≤ 1. This would then generalize Theorems 2.2 and 2.3 and the generalresult, if obtained, would be analogous to (4.8) which is the corresponding result in the settingof generalized Lambert series. This would then complete the analogy between extended higherHerglotz functions and generalized Lambert series as specified in Remark 2.2.

(3) As explained in the introduction, the closed-form evaluation of the integral J(x) in (1.15) isof interest not only from the point of view of analytic number theory but also of algebraic numbertheory. In the same vein, it would be worthwhile to study for which values of k,N and x canthe integral in (2.20) be evaluated in closed-form. In Corollary 2.10, we have evaluated it in an“almost” closed-form since it has the constant Fk(2) in its evaluation.

(4) Page 220 of Ramanujan’s Lost Notebook [28] contains a beautiful modular relation for α, βpositive such that αβ = 1, namely, if

φ(x) := ψ(x) +1

2x− log x,

and ξ(s) and Ξ(t) denote Riemann’s functions respectively defined by [29, p. 16, Equations (2.1.12),(2.1.14)]

ξ(s) := (s− 1)π−12 sΓ(1 + 1

2s)ζ(s),

Ξ(t) := ξ(12 + it),

then

√α

{γ − log(2πα)

2α+

∞∑n=1

φ(nα)

}=√β

{γ − log(2πβ)

2β+

∞∑n=1

φ(nα)

}

= − 1

π3/2

∫ ∞0

∣∣∣∣Ξ( t2)

Γ

(−1 + it

4

)∣∣∣∣2 cos(t2 logα

)1 + t2

dt.

The first equality in this formula seems to be in the realms of the theory of Herglotz function. Aproof of it can be found, for example, in [4]. It would be certainly of merit to see if another proofof it could be obtained through the theory of Herglotz function.

(5) While the finite sum in Ramanujan’s formula for ζ(2m+1), that is, (2.8), involves only even-indexed Bernoulli numbers, or, in other words, even zeta values, that in (2.23) involves a productof an even zeta value and an odd zeta value. The remaining case of having products of only oddzeta values in the summand of the finite sum is covered by our Corollary 2.4.

In view of the fact that the finite sum in (2.8) has given rise to a nice theory of Ramanujanpolynomials [13], [24], it would be interesting to study the analogous polynomials stemming from(2.9) and (2.23).


(6) Our method for proving Theorem 2.1 limits k to be between 1 and N (N inclusive). Doesthere exist a functional equation for Fk,N (x) where k which are greater than N?

Acknowledgements

The authors sincerely thank Professor Kenneth S. Williams for kindly sending them a copy of [16].The first author’s research was partially supported by SERB-DST CRG grant CRG/2020/002367.The third author is a postdoctoral fellow funded in part by the same grant. Both sincerely thankSERB-DST for the support. The third author also thanks IIT Gandhinagar for financial support.

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Discipline of Mathematics, Indian Institute of Technology, Gandhinagar, Palaj, Gandhinagar382355, Gujarat, India

Email address: [email protected]; rajat [email protected]; [email protected]

extended higher herglotz functions i. functional equations

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