Extended Higher Herglotz Functions I. Functional Equations 2

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K denote a number field and ζK (s) the Dedekind zeta-function of K. If A denotes an ideal class 1 group of K, then we can write ζK(s) = A ζ(s, A), where, for Re(s) > 1, ζ(s, A) = a A N(a)s , with N(a) being the norm of the fractional ideal a. Then the Kronecker limit formula is concerned∈ P P with the evaluation of the constant term ̺(A) in the Laurent series expansion of ζ(s, A). This was done by Kronecker [18] in the case of an imaginary quadratic field K. While Hecke [15] obtained a corresponding formula for a real quadratic field K, it was not explicit as one of the expressions in 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 real quadratic fields. Zagier’s formula gives a new representation for the value log ǫ ̺( ) := lim Ds/2ζ(s, ) , B s 1 B − s 1 → − where is an element of the group of narrow ideal classes of invertible fractional -ideals1 of the B OD quadratic order = Z + Z D+√D of discriminant D > 0, ζ(s, ) is the sum of N(a) s over all OD 2 B − invertible D-ideals a in the class , and ǫ = ǫD is the smallest unit > 1 in D of norm 1. The reason suchO a representation is worthB studying is because the special value at sO= 1 of the Dirichlet s 1/2 series LK(s,χ) = a χ(a)N(a)− equals D− χ( )̺( ) for any character χ on the narrow ideal class group. Zagier’s formula [35, p. 164] readsB B B P P ̺( )= P (w, w′), (1.1) B w Red( ) ∈X B where the function P (x,y) for x>y> 0 is defined as y π2 x 1 1 x P (x,y) := F (x) F (y)+Li + log γ log(x y)+ log , (1.2) − 2 x − 6 y − 2 − 4 y with Red( ) being the set of larger roots w = ( b + √D)/(2a) of all reduced primitive quadratic forms Q(X,YB )= aX2 + bXY + cY 2(a, c > 0, a +−b + c< 0) of discriminant D belonging to the class . Here Li (t) is the dilogarithm function B 2 ∞ tn (0 0. EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 3

1 1 π2 F (x)+ F = 2F (1) + log2(x) (x 1)2, (1.6) x 2 − 6x − where [35, Equation (7.12)] 1 π2 F (1) = γ2 γ , −2 − 12 − 1 and γ and γ1 denote Euler’s constant and the first Stieltjes constant respectively. 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 obtain asymptotic expansions of F (x) as x 0 and x 1. Functional equations of the type in (1.5) occur in 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 to a beautiful article of Zagier [37] on this topic. Several generalizations of the Herglotz function (also known sometimes as the Herglotz-Zagier function) have been studied in the literature. Ishibashi [17] studied the jth order Herglotz function defined by j ∞ R′ (nx) log (nx) Φ (x) := j − (j N), j n ∈ n 1 X− where Rj(x) is a generalization of the function R(x) studied by Deninger [8], and is defined as the solution 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 function Φj(x) appears in his explicit representation [17, Theorem 3] of the Laurent series coefficients of

ZQℓ (s), the zeta function associated to indefinite quadratic forms Qℓ, where Ds/2ζ(s, )= Z (s) (Re(s) > 1). B Qℓ Xℓ Zagier [35] had earlier obtained the constant coefficient and the coefficient of (s 1) 1 in the Laurent − − series expansion of Ds/2ζ(s, ). The Herglotz function F (x) makes its conspicuous presence in Zagier’s expression of the constantB coefficient as can be seen from (1.1) and (1.2). Masri [21] generalized F (x) in a different direction by considering

∞ α(n) (ψ(nx) log(nx)) F (α,s,x) := − (Re(s) > 0,x> 0), ns nX=1 where α : Z C is a periodic function with period M. Vlasenko and Zagier [30] studied the following special→ case of Masri’s F (α,s,x) when α(n) 1 and s > 1 is a natural number but without considering log(nx) term (which is not needed for≡ convergence of the series since s> 1):

∞ ψ(nx) F (x) := (k N,k > 1,x C ( , 0]) . (1.7) k nk ∈ ∈ \ −∞ n=1 X They termed it the higher Herglotz function2 and obtained following beautiful functional relations for it [30, Equations (11), (12)]: k 1 k 1 1 k 1 − r 1 F (x) + ( x) − F = γζ(k) 1 + ( x) − ζ(r)ζ(k + 1 r)( x) − k − k x − − − − − r=2 X 2 ∞ k−1 It is important to note that Vlasenko and Zagier [30] deﬁne their higher Herglotz function by n=1 ψ(nx)/n for k > 2. P EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 4

1 + ζ(k + 1) ( x)k , (1.8) − − x and3

k 1 x + 1 k 1 F (x) F (x + 1) + ( x) − F = ( x) − (ζ(k, 1) + ζ(k + 1) γζ(k)) k − k − k x − − k 1 k − r 1 ( x) 1 ζ(k + 1 r, r)( x) − + ζ(k + 1) − , − − − x + 1 − x r=1 X (1.9) where 1 ζ(m,n) := (m 2,n 1). pmqn ≥ ≥ p>q>X0 The function Fk(x) plays an instrumental role in the higher Kronecker “limit” formulas developed in [30]. The goal of our paper is to study a novel generalization of F (x), which we call the extended higher Herglotz function. We deﬁne it by

∞ ψ(nN x) log(nN x) F (x) := − (x C ( , 0]) , (1.10) k,N nk ∈ \ −∞ Xn=1 where k and N are positive real numbers such that k + N > 1. It is clear that when k> 1, one can write ∞ ψ(nN x) ∞ log(nN x) F (x)= k,N nk − nk n=1 n=1 X X ∞ ψ(nN x) = + Nζ′(k) ζ(k) log x. (1.11) nk − nX=1 However, we keep (1.10) as the definition of the extended higher Herglotz function so as to keep many special cases under one roof. Indeed, from (1.7) and (1.11), we have for k N, ∈ F (x)= F (x)+ ζ′(k) ζ(k) log x. (1.12) k,1 k − 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 Herglotz function turns up after differentiating a transformation for the generalized Lambert series 2Nm 1 N ∞ n− − exp a(2n) α − (1.13) 1 exp ( (2n)N α) n=1 X − − 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 2Nm 1+N − − (N odd) exp ((2n)N α) 1 n=1 X − as α 0+ which can be seen in Theorem 2.13. Obtaining asymptotic expansions for Lambert series→ and its generalizations has received a lot of attention since last century because of their applications 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)). EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 5

nz p. 163, Theorem 7.15]) obtained the asymptotic expansion of n∞=1 1/(e 1) as z 0 which was used by Lukkarinen [20] to not only study the moments of the Riemann− zeta function→ but to P 2 also obtain meromorphic continuation of the modiﬁed Mellin transform of ζ(1/2+ ix) deﬁned 1 2 s N | | for Re(s) > 1 by 1∞ ζ 2 + ix x− dx. For odd N , Wigert himself [33, p. 9-10] obtained the asymptotic expansion of the generalized Lambert series∈ R

∞ 1 ∞ ∞ N = e n jx. nN x − n=1 e 1 n=1 X − Xj=1 X 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 generalized Lambert series [33]. Zagier [36] had conjectured that the asymptotic expansion of the Lambert series associated with 2 nx + the square of the Ramanujan tau function τ(n), that is of n∞=1 τ (n)e− , as x 0 , can be expressed in terms of the non-trivial zeros of the Riemann zeta function. This conjecture→ was proved by Hafner and Stopple [14]. Thus obtaining asymptotiPc expansions of Lambert series is a useful endeavor. In this paper, we obtain two different kinds of functional equations satisfied by Fk,N (x). See Theorems 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 x π2 J(x)= F (2x) 2F (x)+ F + , (1.14) − 2 12x where 1 log (1 + tx) J(x) := dt. (1.15) 1+ t Z0 Radchenko and Zagier [26] have obtained special values of F (x) at positive rationals and quadratic units with the help of which they are able to evaluate special values of J(x), for example, π2 1 1 J(4 + √17) = + log2(2) + log(2) log(4 + √17), − 6 2 2 2 11π2 3 √5 + 1 J = + log2(2) 2 log2 . 5 240 4 − 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 − π2 √5 + 1 J(4 + √15) = (√15 2) + log(2) log(√3+ √5) + log log(2 + √3). − 12 − 2 ! Regarding this integral, Chowla [6, p. 372] remarked, ‘A direct evaluation of this definite integral is probably difficult’. Indeed, these evaluations come from not only employing methods from analytic number theory but also those from algebraic number theory, which is why they are all the more interesting! 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 series and extended higher Herglotz functions.

2. Main results (N 1) Throughout this paper, the notation − ′′ will denote a sum over over j = (N j= (N 1) − − 1), (N 3), ,N 3,N 1. − − − − · · · − − P EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 6

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 ∈ ≤ ∈ \ −∞ ( 1)k+N+1x1/N ζ 1+ k if k = N B(k,N,x) := N . (2.1) 0− if k = N 6 Then

k (N 1) iπj 1−k ( 1) − e− N N F ′′ iπj(k 1)/N F − x k,N (x) − e − N+k 1 , 1 − N N N x1/N j= (N 1) ! −X− k 1 − π ζ 1+ N− 1 k 1 = + x N (γ + log x) ζ(k)+ Nζ′(k) ζ(k + N)+ B(k,N,x). (2.2) N π − − N+k−1 sin N (k 1) x N − Also,

(N 1) iπj 1 − e N F ′′F − 1,N (x)+ 1, 1 N N x1/N j= (N 1) ! −X− 1 π2 1 ζ(N + 1) = (N 1)γ log x + (log x)2 Nγ2 (N 2 + 1)γ + B(1,N,x). (2.3) N 6 − − 2 − − 1 − x 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 functional equations obtained in Theorems 2.1 and 2.2, namely, Theorem 2.1 is valid for any natural numbers k and N such that 1 < k N whereas Theorem 2.2 is valid for any odd natural numbers k and N. These two types are, indeed,≤ equivalent when k and N are odd such that 1 < k N. This is shown 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 instrumental in 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, ∈ ∞ ∞ t cos(t) 1 u 1 iu iu dt = log ψ + ψ − . (2.4) t2 + m2u2 2 2π − 2 2π 2π m=1 0 X Z Theorem 2.2. Let k and N be odd natural numbers. Let Re(x) > 0. Then,

N 1 − iπ(k 1)j 1 iπj 1 iπj ′′ F − 2π N 2N F − 2π N 2N exp 2N− N+k 1 , 1 i x e + N+k 1 , 1 i x e − N N N N − j= (N 1) −X− n o k−1 k+1 2π ix ix 2π = ( 1) 2 N N F + F + 2 γ log ζ(k) Nζ′(k) + C (k,N,x) , − x k,N 2π k,N − 2π − x − (2.5)

1 k k where, for − = 2 , N 6 − 2N k N+k 1 k−1 ⌊ 2N ⌋ πζ N − x N j 2j 1 2j C (k,N,x) := + 4π ( 1) (2π)− − ζ(k 2Nj)ζ(2j + 1)x , N sin π (1 k) 2π − − 2N j=1 − X (2.6) EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 7

1 k k and for − = 2 , N − 2N − k 1 k−1 2( 1) 2N x N 2π k 1 k 1 C (k,N,x) := − γ + log ζ 1+ − ζ′ 1+ − N 2π x N − N k 1 ⌊ 2N ⌋− j 2j 1 2j + 4π ( 1) (2π)− − ζ(k 2Nj)ζ(2j + 1)x . (2.7) − − Xj=1 1 k k The remaining case not considered in the above theorem, namely, N− = 2 2N = 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 − iπ 1 iπj 1 iπj ′′ 2 (j 1) F 2π N 2N F 2π N 2N e − 1, 1 i x e + 1, 1 i x e N N − j= (N 1) −X− n o 1 π2 2π = N F ix + F ix 2Nγ2 + 2(N 1)γ log − 1,N 2π 1,N − 2π − N 12 − − x 4π2 + log2(2π) log log x 2(N 2 + 1)γ . − x − 1 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 ∈ \{ }

2m 1 2m 1 m 1 ∞ n− − m 1 ∞ n− − α− ζ(2m +1)+ = ( β)− ζ(2m +1)+ 2 e2αn 1 − 2 e2βn 1 ( n=1 ) ( n=1 ) X − X − m+1 j 2m ( 1) B2jB2m+2 2j m+1 j j 2 − − α − β , (2.8) − (2j)!(2m + 2 2j)! Xj=0 − th where Bn is the n 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, ∈

m ∞ 1 inα inα α− 2γζ(2m +1)+ ψ + ψ n2m+1 2π − 2π ( n=1 ) X m ∞ 1 inβ inβ = ( β)− 2γζ(2m +1)+ 2m+1 ψ + ψ − − ( n 2π − 2π ) nX=1 m 1 − j j m j 2 ( 1) ζ(1 2j + 2m)ζ(2j + 1)α − β− . (2.9) − − − Xj=1 It is important to note here that Corollary 2.4 cannot be obtained for any general α, β such that αβ = 4π2 by applying Vlasenko and Zagier’s (1.8) once with k = 2m +2 and x = iα/(2π), and again with k = 2m +2 and x = iβ/(2π), and adding the resulting two equations, for, doing so will EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 8 result in inα inα ψ inβ ψ inβ ∞ ψ 2π m m ∞ ψ 2π ∞ 2π m m ∞ 2π + ( β) α− − + + ( α) β− − n2m+1 − n2m+1 n2m+1 − n2m+1 n=1 n=1 n=1 n=1 X X X X m m m m m m 2π 2π ( β) α− β = γ 2 + ( β) α− + ( α) β− ζ(2m +1)+ iζ(2m + 2) + − − − − β α − 2π m m 2m r 1 ( α) β− α i − r 1 r 1 − ζ(r)ζ(2m + 2 r) α − + β − . (2.10) − 2π − − −2π ! r=2 X It is clear that it is impossible to get Corollary 2.4 from (2.10) for any α, β such that αβ = 4π2 be- m m m m 2m 1 inα cause of the factors of the form ( β) α and ( α) β in front of the series ∞ n ψ − − − − n=1 − − − 2π 2m 1 inβ and ∞ n ψ respectively in (2.10) whereas we do not have such factors in Corollary n=1 − − − 2π P 2.4. P 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, ∈ 2m 1 ∞ 1 − (ψ(in)+ ψ( in)) = 2γζ(4m + 1) ( 1)j ζ(2j + 1)ζ(4m + 1 2j). (2.11) n4m+1 − − − − − nX=1 Xj=1 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 ∞ 1 inα inα 1 ∞ 1 inβ inα 2γζ(3) + ψ + ψ − = 2γζ(3) + ψ + ψ − . α n3 2π 2π β n3 2π 2π ( n=1 ) ( n=1 ) X X (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 to mimick 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 2.4 given 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 Theorem 1.2 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 of the functional equations given in the above theorems is in obtaining the asymptotic expansions of the extended higher Herglotz functions Fk,N (x) as x 0. These asymptotic expansions for diﬀerent conditions on the parameters involved are collecte→d together in the theorem below. Theorem 2.6. Let x C ( , 0] and k, N are positive real numbers such that k + N > 1. Let B(k,N,x) be deﬁned in∈ (2.1)\ −∞. Then as x , →∞ 1 ∞ B2n 2n F (x) ζ(k + N) ζ(k + 2nN)x− . (2.13) k,N ∼−2x − 2n nX=1 Also, as x 0, → (i) For k, N N such that 1 < k N, we have ∈ ≤ k 1 − − ζ(k + N) π ζ 1+ N− k 1 k 1 Fk,N (x) (γ + log x) ζ(k)+ Nζ′(k)+ x N + x N B(k,N,x) ∼− x − N sin π (k 1) N − EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 9

∞ + ( 1)k ( 1)m(N+1)ζ(k mN)ζ (1 + m) xm; (2.14) − − − m=1 X (ii) For N N, ∈ ζ(N + 1) 1 π2 1 F (x) + (N 1)γ log x + log2 x Nγ2 (N 2 + 1)γ 1,N ∼− x N 6 − − 2 − − 1 ∞ + B(1,N,x) ( 1)m(N+1)ζ(1 mN)ζ (1+ m) xm. (2.15) − − − m=1 X Letting N = 1 gives Radchenko and Zagier’s asymptotic expansions for F (x) as x and as x 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. Let C (k,N,x) be deﬁned in (2.6) and (2.7). Then as x , →∞ 2n ix ix ∞ ( 1)n+1 2π F + F − B ζ(k + 2nN) , (2.16) k,N 2π k,N −2π ∼ n 2n x n=1 X Also, for odd natural numbers k and N, as x 0, → ix ix F + F k,N 2π k,N −2π N+k+2 k−1 2π ( 1) 2 x N iπ(N + k 1) 2 log γ ζ(k)+ Nζ′(k) C (k,N,x)+ − exp − ∼ x − − N 2π − 2N 2n ∞ B N + k + 2n 1 x N 2n ζ − , (2.17) × n cos πn N 2π n=1 N X and for an odd natural number N, as x 0, → ix ix F + F 1,N 2π 1,N −2π 1 π2 2π 4π2 2Nγ2 + 2(N 1)γ log + log2(2π) log log x 2(N 2 + 1)γ ∼ N 12 − − x − x − 1 2n i N+1 ∞ B2n 2n x N + ( 1) 2 ζ 1+ . (2.18) N − n cos πn N 2π n=1 N X 2.4. An interesting relation between a generalized polylogarithm function and the extended higher Herglotz functions. The polylogarithm is a generalization of the dilogarithm function of (1.3). It is deﬁned by

∞ tn Li (t) := . s ns n=1 X It is clear that Lis(t) converges for any complex s as long as t < 1. It can be analytically continued for 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. We deﬁne it for N N,s C and t < 1 by ∈ ∈ | | N ∞ tn Li (t) := . (2.19) N s ns Xn=1 Clearly, Li (t)=Li (t) for t < 1. 1 s s | | EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 10

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

1 k 1 N (2N 1) k 1 2 − 2 u − 2 − 1 x Jk,N (x) := − N Lik( u ) du. (2.20) u 1 − u2N 1 − u log u − Z0 − − ! Then for x> 0, we have

N k 1 1 k x J (x)= F (2 x) 2 − + 2 − F (x)+ F k,N k,N − k,N k,N 2N N N k 1 1 k ζ(k + N) + 2 + 2− 2 − 2 − . (2.21) − − x The integral in (2.20) is a natural generalization of (1.15) in the setting of Fk,N (x) in the sense that 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 letting k = N = 1 in Theorem 2.8. The above theorem leads to the following relation involving polylogarithm and Vlasenko and Zagier’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 2k 1 2u 2k 1 1 − − − Li ( ux) du u 1 − u2 1 − u log u k − Z0 − − k 1 1 k x k 1 1 k 1 = F (2x) 2 − + 2 − F (x)+ F + 2 2 − 2 − ζ′(k) ζ(k) log x + ζ(k + 1) k − k k 2 − − − x 1 + ζ(k + 1). 2x For k odd, Corollary 2.9, in turn, gives an evaluation of an integral involving polylogarithm function in terms of the higher Herglotz function Fk(2), special values of the Riemann zeta function and 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 2k 1 2u 2k 1 1 − − − Li ( u) du u 1 − u2 1 − u log u k − Z0 − − 1 k k 1 1 k k 1 1 k = 1 2 − F (2) + (2 − 1)γζ(k)+ 2− ζ(k +1)+ 2 2 − 2 − ζ′(k) − k − 2 − − − k 1 − r 1 k 2 k r k + ( 1) − 2 − + 2− 2 − ζ(r)ζ(k + 1 r). (2.22) − − − r=2 X 2.5. Transformations involving higher Herglotz functions and generalized Lambert series. 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 ∈ 2m m 1 2j 1 (m 1 ) 1 ∞ n− − 2 − B2j 2j m 1 α− − 2 ζ(2m)+ ζ(2m 2j + 1)α − − 2 2 e2nα 1 − (2j)! − ( n=1 ) X − Xj=0 m+1 (m 1 ) γ 1 ∞ 2m inβ inβ = ( 1) β− − 2 ζ(2m)+ n− ψ + ψ . (2.23) − π 2π π − π ( n=1 ) X In what follows, we give a one-parameter extension of (2.23). EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 11

Theorem 2.11. Let N be an odd positive integer. For αβN = πN+1 with Re(α) > 0, Re(β) > 0, and m 1, ≥ 2Nm 1+N 2Nm 1 1 ∞ n− − ( N+1 2 ) α− − ζ(2Nm + 1 N)+ N 2 − e(2n) α 1 ! nX=1 − m 1 − B2j ζ(2Nm + 1 2Nj) N(2j 1) 2j 2Nm 1 − 2 − α − N+1 − 2 − (2j)! Xj=0 2m(N 1) 2 − 2Nm N Nγ m+1 ( N+1 2 ) = N+1 ( 1) β− − N 1 ζ(2m) Nπ 2 − "2 − (N−1) 2 1 ∞ 1 iβ iπj iβ iπj 1/N N 1/N N + N 2m ψ (2n) e + ψ (2n) e . (2.24) 2 − n 2π −2π # j= (N 1) n=1 −X2 X 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, ∞ nN 1 1 − (Nγ log(2π) (N 1)(log 2 γ)) exp ((2n)N α) 1 − 2N αN − − − − n=1 − X N−1 2 1 α 2 ∞ β 1 iπj = log + log (2n) N e N 2N α(N + 1) β 2N αN 2π N−1 " j= ( ) n=1 −X2 X α 1 β 1 iπj β 1 iπj , if,N = 1 ψ i (2n) N e N + ψ i (2n) N e N + 2 (2.25) − 2 2π − 2π 0, if, N> 1. # ( As an application of the asymptotic expansions derived in Theorem 2.7, we obtain the complete asymptotic 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, ≥ → 2Nm 1+N m+1 ∞ n− − 1 1 ( 1) (2m 1)(N 1) 2m 1 ζ(2Nm + 1) ζ(2Nm + 1 N)+ − 2 − − α − e(2n)N α 1 ∼ 2N α − 2 − Nπ2m n=1 − X 2ℓ ∞ ( 1)ℓ+1B 2N 1α − 2ℓN ζ (2m + 2ℓ) − + D (m, α), (2.26) × ℓ π N Xℓ=1 where (2m 1)(N 1) N 1 2 − − m+1 2m 1 α2 − D (m, α) := ( 1) α − ζ(2m) Nγ log ζ′(2m) N Nπ2m − − π − ( ) m 1 − B2j ζ(2Nm + 1 2Nj) N(2j 1) 2j 1 + − 2 − α − . (2j)! Xj=1 Also, in particular, as α 0, → 2m m+1 2m 1 ∞ n− ζ(2m + 1) 1 ( 1) α − α ζ(2m)+ − ζ(2m) γ log ζ′(2m) e2nα 1 ∼ 2α − 2 π2m − π − n=1 ( ) X − ( 1)m+1α2m 1 ∞ ( 1)ℓ+1B α 2ℓ + − − − 2ℓ ζ (2m + 2ℓ) π2m ℓ π Xℓ=1 EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 12

m 1 − B2j ζ(2m + 1 2j) (2j 1) + − (2α) − . (2.27) (2j)! Xj=1 We note in passing that the well-known result [12, p. 903, Formula 8.361.8], for Re(x) > 0, namely

∞ 1 1 xt ψ(x) log x = e− dt, − − 1 e t − t Z0 − − gives an integral representation for Fk,N (x):

∞ 1 1 xt F (x)= Li e− dt, (2.28) k,N − 1 e t − t N k Z0 − − where N Lik(t) is deﬁned in (2.19), whereas Binet’s formula [31, p. 251] for Re(x) > 0, namely 1 2t ψ(x)+ log x = ∞ dt, 2x − − (t2 + x2)(e2πt 1) Z0 − yields a representation for Fk,N (x) in terms of a generalized Lambert series for Re(x) > 0: k F ∞ ∞ n− 2t 1 k,N (x)= N 2 2 dt ζ(k + N). − 0 exp(2πn t) 1! t + x − 2x Z nX=1 − 3. Proofs of functional equations satisfied by Fk,N (x) Proof of Theorem 2.1. Employing the well-known functional equation 1 ψ(x +1) = ψ(x)+ (3.1) x in the deﬁnition of Fk,N (x) in (1.10), we get ∞ ψ(nN x + 1) log(nN x) 1 F (x)= − ζ(k + N). (3.2) k,N nk − x n=1 X We need the following formula due to Kloosterman [29, Equation 2.9.2], which is valid in 0 < c = Re(z) < 1:

1 πζ(1 z) z − − x− dz = ψ(x + 1) log x, (3.3) 2πi sin(πz) − Z(c) c+i where here and in the sequel, (c) denotes the line integral c i ∞. Thus, invoking (3.3) in (3.2), we get for 0 < c = Re(z) < 1,− ∞ R R 1 ∞ 1 πζ(1 z) z 1 F (x)= − − nN x − dz ζ(k + N) k,N 2πi nk sin(πz) − x n=1 Z(c) X 1 πζ(1 z)ζ (k + Nz) z 1 = − − x− dz ζ(k + N), 2πi sin(πz) − x Z(c) by interchanging the order of summation and integration. Employing the change of variable z = s k 1 − in the above equation, we see that for N k + 1 < c = Re(s) < 1 k, − N − N − − ′ − 1 k−1 1 F (x)= x N I ζ(k + N), (3.4) k,N N − x where, s+k 1 1 πζ 1+ N− ζ(1 s) s I = I(k,N,x) := − x N ds. 2πi ′ sin π (s + k 1) Z(c ) N − EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 13

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) again ′′ ′′ − ′′ − ′ − ′′ − after shifting the line of integration from Re(s)= c′ to Re(s)= c′′, we also need 0 < c′′ < 1. In any case, we must have N k + 1 > 0, or, in other words, k N. Note that in the process of shifting the line of integration,− we encounter a double pole of the≤ integrand at s = 1 k and a simple pole at s = 0. − Here, and throughout the paper, Ra denotes the residue of the associated integrand at its pole at a. Stirling’s formula in the vertical strip p σ q reads [7, p. 224] ≤ ≤ σ 1 1 π t 1 Γ(s) = √2π t − 2 e− 2 | | 1+ O (3.5) | | | | t | | 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 s+k 1 1 πζ 1+ N− ζ(1 s) s I1 = I1(k,N,x) := − x N ds, 2πi ′′ sin π (s + k 1) Z(c ) N − and the residues R0 and R1 k are easily seen to be − k 1 ζ 1+ N− R0 = π , (3.7) − sin π (k 1) N − 1−k R1 k = Nx N (γ + log x) ζ(k)+ Nζ′(k) . (3.8) − − − To evaluate I1, we use the following well known result [9, Lemma 4.3] (N 1) 1 1 − = ′′eijz (3.9) sin(z) sin(Nz) j= (N 1) −X− with z = π(s + k 1)/N so as to obtain − (N 1) s+k 1 iπj s − πζ(1 s)ζ 1+ − N − ′′ iπj(k 1)/N 1 N e− I1 = e − − ds 2πi ′′ sin (π(s + k 1)) x1/N j= (N 1) Z(c ) ! −X− − (N 1) 1 s − ∞ 1 1 πζ(1 s) n N iπj − k ′′ iπj(k 1)/N N = ( 1) e − k−1 − − e− ds, (3.10) − 1+ 2πi ′′ sin (πs) x j= (N 1) n=1 n N Z(c ) −X− X s+k 1 where in the last step we used the series representation of ζ 1+ N− since k 1 and c′′ > 0 and interchanged the order of summation and integration which can be easily justiﬁed≥ by standard arguments. Now invoke (3.3) again in (3.10) to obtain

(N 1) 1 1 − ∞ iπj iπj k ′′ iπj(k 1)/N 1 n N n N I1 = ( 1) e − − ψ e− N + 1 log e− N − 1+ k 1 x − x j= (N 1) n=1 n N −X− X (N 1) 1 1 − ∞ iπj iπj k ′′ iπj(k 1)/N 1 n N n N = ( 1) e − − ψ e− N log e− N − 1+ k 1 x − x j= (N 1) n=1 n N −X− X (N 1) k − + ( 1)kx1/N ζ 1+ ′′eiπjk/N , (3.11) − N j= (N 1) −X− EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 14 where we used (3.1) in the last step. From (3.9) and the fact that 1 < k N, we have ≤ (N 1) − ( 1)N+1N if k = N ′′eiπjk/N = (3.12) 0− if 1

(N 1) iπj − e− N k ′′ iπj(k 1)/N F − B I1 = ( 1) e − N+k 1 , 1 + N (k,N,x), (3.13) − N N x1/N j= (N 1) ! −X− where B(k,N,x) is deﬁned in (2.1). Substituting (3.7), (3.8) and (3.13) in (3.6), we see that

(N 1) iπj k 1 − e− N ζ 1+ − k ′′ iπj(k 1)/N F − B N I = ( 1) e − N+k 1 , 1 + N (k,N,x)+ π π − N N x1/N sin (k 1) j= (N 1) ! N −X− − 1−k + Nx N (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 1−k equation by x N . Proof of Theorem 2.2. The fact that k and N are odd is used several times in the proof without mention. 1 k k Case 1: N− = 2 2N . Let J denote the6 left-hand− side of (2.5). Using (1.10), we rewrite J in the form N 1 1 − ∞ N iπj ′′ j iπ(k + N 1)j 1 2πn J = 2 i exp − − log e 2N − − 2N k+N 1 x j= (N 1) n=1 n N ( ! −X− X 1 1 1 2πn N iπj 2πn N iπj ψ i e 2N + ψ i e 2N . (3.15) − 2 x − x " ! !# ) Deﬁne 1 2πn N X :=2πℓ (3.16) ℓ,n x 1 2πn N iπj iπj X∗ :=2πℓ e 2N = X e 2N (3.17) ℓ,n,j x ℓ,n Employing (2.4) in (3.15), we observe that

N 1 − ∞ ∞ ′′ j iπ(k + N 1)j 1 ∞ t cos t J = 4 i exp − − dt k+N 1 2 2 − − 2N N t + X j= (N 1) n=1 n ℓ=1 Z0 ℓ,n,j∗ −X− X X N 1 − ∞ ′′ j iπ(k + N 1)j π 1 1 Γ(s1) s1 = 4 i exp − − X∗ − ds1, − − 2N 2 k+N 1 2πi tan πs1 ℓ,n,j j= (N 1) n,ℓ=1 n N Z(c1) 2 −X− X (3.18) where 1 < c1 := Re(s1) < 2. The last step follows from the following result [9, Lemma 4.1]

1 Γ(s1) s1 2 ∞ t cos t u− ds = dt, (3.19) 2πi tan( πs1 ) 1 π u2 + t2 Z(c1) 2 Z0 EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 15

4 which is valid for 0 < Re(s1)= c1 < 2 and Re(u) > 0 . From (3.16), (3.17) and (3.18) and interchanging the order of summation and integration, we have N 1 1 ∞ 1 Γ(s ) − iπj 1 ′′ j s1 J = − − i exp (s1 + k + N 1) X− ds1. (3.20) i k+N 1 tan πs1 −2N − ℓ,n n,ℓ=1 n N Z(c1) 2 j= (N 1) X −X− From [9, Equation (4.2)], for N odd, N 1 − cos(Nz) N 1 − j = ( 1) 2 ′′i exp( ijz). (3.21) cos(z) − − j= (N 1) −X− Using (3.21) with z = π(s + k + N 1)/(2N) in (3.20), we observe that 1 − π ∞ N−1 cos (s + k + N 1) 1 1 Γ(s1) 2 1 s1 J = − − ( 1) 2 − X− ds k+N 1 πs1 ℓ,n 1 i tan − π s1+k+N 1 n N (c1) 2 cos − n,ℓX=1 Z 2 N k−1 πs 2 ∞ Γ(s ) cos 1 ( 1) 1 1 2 s1 = − k+N−1 Xℓ,n− ds1. 2i π s1+k+N 1 n,ℓ=1 n N (c1) cos − X Z 2 N Performing the change of variable s1 = Ns k N + 1 in the above equation, we see that for k+N k+N+1 − − N

1 πs 1 s ζ(k + Ns) cot Γ(s)ζ(s)x − ds 2πi 2 Z(c2) k 2N ⌊ ⌋ 1 πs 1 s = R0 R 1−k R 2j + ζ(k + Ns) cot Γ(s)ζ(s)x − ds. (3.26) − − N − − 2πi (d) 2 Xj=1 Z The residues in (3.26) are calculated to be x 2π R = γ log ζ(k) Nζ′(k) , 0 π − x − j 2j 1 1+2j R 2j = 2( 1) (2π)− − ζ(k 2Nj)ζ(1 + 2j)x , − − − k+N−1 k 1 π x N ζ 1+ − − N R 1 k = π 1 k . (3.27) N N 2π sin − 2 N Also, the integral on the right-hand side of (3.26) is evalua ted to

1 πs 1 s ζ(k + Ns) cot Γ(s)ζ(s)x − ds 2πi (d) 2 Z ∞ 1 1 Γ(s) N s = x k πs (xℓ n)− ds ℓ 2πi (d) tan 2 n,ℓX=1 Z 2x ∞ 1 ∞ t cos t = ∞ dt π ℓk (t2 + (xℓN n)2) n=1 0 Xℓ=1 X Z x ∞ 1 ℓN x 1 iℓN x iℓN x = log ψ + − , (3.28) π ℓk 2π − 2 2π 2π Xℓ=1 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 πs 1 s ζ(k + Ns) cot Γ(s)ζ(s)x − ds 2πi (c) 2 Z x ∞ 1 ℓN x 1 iℓN x iℓN x = log ψ + − π ℓk 2π − 2 2π 2π Xℓ=1 x 2π γ log ζ(k) Nζ′(k) − π − x − k 2N ⌊ ⌋ j 2j 1 1+2j 2 ( 1) (2π)− − ζ(k 2Nj)ζ(1 + 2j)x − − − Xj=1 EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 17

k+N−1 π x N π k + N 1 k + N 1 sec − ζ − . (3.29) − N 2π 2 N N 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). 1 k k Case 2: N− = 2 2N = 0. The only change in− this case6 is the contribution of the double order pole of the integrand in (3.25) πs at s = (1 k)/N due to ζ(k + Ns) and sin 2 . 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 − − k 1 N+k−1 2( 1) 2N x N 2π k 1 k 1 − γ + log ζ 1+ − ζ′ 1+ − . N 2π x N − N

3.1. Proof of the equivalence of Theorems 2.1 and 2.2 for odd natural numbers k and N such that 1 < k N. Assume k and N to be odd positive integers such that 1 < k N. ≤ ix ix ≤ Employing (2.2) once, with x replaced by 2π and then again with x replaced by 2π , and then adding the resulting two equations, we arrive at −

k−1 ix ix x N F + F = E + E + E + E , (3.30) k,N 2π k,N −2π 2π { 1 2 3} 4 where N+k 1 2π ζ N − π E1 := cos (k 1) (3.31) π(k 1) N sin − 2N − N k−1 ix k−1 ix E := i N B k,N, + ( i) N B k,N, 2 2π − −2π 1 k N 1 ( 1) − iπj(k−1) k−1 ix − N iπj ′′ N N F − N E3 := − e i k+N 1 , 1 e− N N N 2π j= (N 1) " ! −X− 1 k−1 ix − N iπj N F − N +( i) k+N 1 , 1 e− (3.32) − N N −2π !# ix ix E := 2γζ(k) + 2Nζ′(k) ζ(k) log + log . (3.33) 4 − − 2π −2π Using the deﬁnition of B(k,N,x) from (2.1), it is easy to see that E2 = 0 for 1 < k < N, and that for k = N,

1 x N k πk E = 2( 1)k+N+1 ζ 1+ cos . 2 − 2π N 2N Since the cosine function vanishes for k = N, we conclude that E2 = 0 in this case as well. Hence E = 0 for 1 < k N. (3.34) 2 ≤ We next simplify E3. To that end, use the deﬁnition of Fk,N (x) from (1.10) in (3.32) so that

1 1 ix iπj ix iπj k N 1 − N N − N N − iπj(k−1) k−1 ∞ ψ 2πn e− log 2πn e− ( 1) ′′ − E = − e N i N − 3 N  k+N 1 j= (N 1) n=1 n N −X−  X  EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 18

1 1 ix iπj ix iπj − N N − N N k−1 ∞ ψ 2πn e− log 2πn e− +( i) N − − − . − k+N−1  n=1 n N X   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− above equation so as to get, for 0 < c = Re(s) < 1,

N 1 k − 1 ( 1) − iπj(k 1) x N iπj iπk iπk k E = − ′′e N e N e 2N + e− 2N ζ 1+ 3 N − 2π N j= (N 1) −X− 1 s − − 1 πζ(1 s) x − N iπj − s+k 1 s+k 1 + − − e− N i N + ( i) N ds 2πi (c) sin(πs) 2πn − # Z N 1 N 1 k+1 1 k − 2( 1) x N πk k − iπjk 2( 1) − iπj(k 1) = − cos ζ 1+ ′′e N + − ′′e N N 2π 2N N 2πiN j= (N 1) j= (N 1) −X− −X− 1 s πζ(1 s) s + k + N 1 π(s + k 1) x − N iπj − − − ζ − cos − e− N ds. (3.35) × (c) sin(πs) N 2N 2π Z We now use (3.9) in (3.35) and employ the change of variable s = s1 (k + N 1) in the integral on the right-hand side of (3.35) to arrive at, for k + N 1 < λ = Re(−s ) < k +−N, − 1

1 (k+N−1) N 1 k+1 k − ( 1) x N k sin(πk) 2( 1) x − N ′′ iπj E3 = − ζ 1+ + − e− N π N sin πk 2πiN 2π 2N j= (N 1) −X− 1 s πζ(k + N s1) s1 πs1 x N iπj − − N k+N − ζ sin e− ds1. × (λ) ( 1) sin(πs1) N 2N 2π Z − Note that the ﬁrst term on the right-hand side of the above equation vanishes as the sine function is zero since k is an integer. Also as k and N are odd positive integers therefore ( 1)k+N = 1 and iπj − j runs over even integers hence we have e− = 1. Therefore E3 reduces to

(k+N−1) N 1 k − ( 1) x − N ′′ πζ(k + N s1) s1 πs1 E3 = 2 − − ζ sin 2πiN 2π sin(πs1) N 2N j= (N 1) Z(λ) −X− 1 s1 x − N iπj − e− N ds . (3.36) × 2π 1 k+N 1 Employing the change of variable s1 = Ns in the integral in (3.36), we have, for N − < λ∗ = k+N Re(s) < N ,

(k+N−1) N 1 s k − iπj 2( 1) x − N ′′ πζ(k + N Ns) πs xe E3 = − − ζ (s)sin ds 2πi 2π ∗ sin(πNs) 2 2π j= (N 1) Z(λ ) −X− k (k+N−1) N 1 2( 1) x N ζ(k + N Ns) πs − = − − − sin2 Γ(1 s)ζ (1 s) xs ′′eiπjsds, 2πi 2π ∗ sin(πNs) 2 − − Z(λ ) j= (N 1) −X− (3.37) EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 19 where in the last step we used (3.23). Now use (3.9) in (3.37) to obtain

k (k+N−1) ( 1) x − N πs s E3 = − ζ(k + N Ns)Γ(1 s)ζ (1 s) tan x ds. (3.38) 2πi 2π (λ∗) − − − 2 Z k+N k+N+1 From (3.15) and (3.24), for N

Proof of Corollary 2.4. Let N = 1 in Theorem 2.2 and use the definition of Fk,N (x) in (1.10). Since m 1, we can separate the expressions involving logarithm. This results in a lot of simplifi- cation 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). EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 20

3.2. Proof of a relation between Fk,N (x) and a generalized polylogarithm. ∞ 1 ∞ Proof of Theorem 2.8. Employing (2.28) and repeatedly using the fact a(2n) = (1 + 2 n=1 n=1 ( 1)n)a(n), we see that X X − N k 1 1 k x F (2 x) 2 − + 2 − F (x)+ F k,N − k,N k,N 2N (2n)N xt nN xt (n/2)N xt ∞ 1 1 ∞ e− k 1 1 k ∞ e− ∞ e− = 2 − + 2 − + dt − 1 e t − t nk − nk nk 0 − "n=1 n=1 n=1 # Z − X X X nN xt nN xt ∞ 1 1 k 1 ∞ n e− k 1 ∞ e− = 2 − (1 + ( 1) ) 2 − − 1 e t − t − nk − nk Z0 − − " n=1 n=1 N X N X ∞ e (n/2) xt ∞ e (n/2) xt (1 + ( 1)n) − + − dt − − nk nk n=1 n=1 # X X nN xt (n/2)N xt ∞ 1 1 k 1 ∞ n e− ∞ n e− = 2 − ( 1) ( 1) dt − 1 e t − t − nk − − nk 0 − " n=1 n=1 # Z − X X nN xt (n/2)N xt ∞ 1 1 k 1 ∞ n e− ∞ n e− = 2 − ( 1) ( 1) dt − et 1 − t − nk − − nk 0 " n=1 n=1 # Z − X X nN xt (n/2)N xt ∞ k 1 ∞ n e− ∞ n e− 2 − ( 1) ( 1) dt, − − nk − − nk 0 n=1 n=1 ! Z X X 1 1 ∞ at 1 where the last step was simpliﬁed using t = t + 1. Since e− dt = for a> 0, we 1 e− e 1 0 a have − − Z

N k 1 1 k x F (2 x) 2 − + 2 − F (x)+ F k,N − k,N k,N 2N nN xt nN xy k 1 ∞ 1 1 ∞ n e− N ∞ 1 1 ∞ n e− = 2 − ( 1) dt + 2 ( 1) dy − et 1 − t − nk 2N y − 2N y − nk 0 n=1 0 e 1 n=1 Z − X Z − X k 1 N n 2 − 2 ∞ ( 1) − − , − x nk+N n=1 X N n s where, in the second integral, we employed the change of variable t = 2 y. Since n∞=1( 1) n− = 1 s − (2 − 1)ζ(s) for Re(s) > 0, we deduce that − P N k 1 1 k x F (2 x) (2 − + 2 − )F (x)+ F k,N − k,N k,N 2N k 1 N k 1 nN xt ∞ 2 2 (2 1) ∞ e = − − − ( 1)n − dt − et 1 − 2N t − t − nk 0 e 1 n=1 Z − − X N N k 1 1 k ζ(k + N) (2 + 2− 2 − 2 − ) . (3.41) − − − x t Making change of variable e− = u in the integral on the right-hand side of (3.41), using the deﬁnitions of N Lis(t) and Jk,N (x) given in (2.19) and (2.20) respectively, using the elementary fact N ( 1)n = ( 1)n and simplifying, we arrive at (2.21). This completes the proof. − − EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 21

The next result which gives a relation between an integral with polylogarithm in its integrand and the Vlasenko-Zagier higher Herglotz function, and is analogous to (1.14), was missing in the literature. 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 2k 1 2u 2k 1 1 − − − Li ( u) du u 1 − u2 1 − u log u k − Z0 − − k 1 1 k 1 k 1 1 k 1 = F (2) 2 − + 2 − F (1) + F + 2 2 − 2 − ζ′(k)+ ζ(k + 1) + ζ(k + 1). k − k k 2 − − 2 (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 k 1 1 − r 1 F (1) = γζ(k) ζ(k + 1) ( 1) − ζ(r)ζ(k + 1 r). (3.43) k − − − 2 − − r=2 X Also, employing (1.8) again with x = 2 and simplifying, we deduce that

k 1 1 1 k k 1 1 k 1 k − r 1 F = 2 − F (2) 2(1 + 2− − )ζ(k + 1) (1 + 2 − )γζ(k) 2 − ( 2) − ζ(r)ζ(k + 1 r). k 2 − k − − − − − r=2 X (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 higher Herglotz functions and generalized Lambert series Proof of Theorem 2.6. The formula [1, p. 259, formula 6.3.18]

∞ n ψ(x) log x + ζ(1 n)x− , ∼ − n=1 X as x , arg x <π implies that →∞ | |

∞ n F (x) ζ(1 n)ζ(k + nN)x− (4.1) k,N ∼ − n=1 X 1 ∞ B2n 2n = ζ(k + N) ζ(k + 2nN)x− , −2x − 2n nX=1 where in the second step we used the facts that ζ( 2j) = 0 for j N and [2, p. 266, Theorem 12.16] − ∈ 1 , if n = 0, ζ( n)= − 2 (4.2) − Bn+1 , if n N. (− n+1 ∈ We now establish (2.14) using (2.2) and (4.1). Upon using (4.1), as x 0, we have → (N 1) iπj 1/N − e− ′′ iπj(k 1)/N F − e − N+k 1 , 1 N N x j= (N 1) ! −X− EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 22

(N 1) N+k+n 1 − ∞ ζ(1 n)ζ − ′′ iπj(k 1)/N N e − − n ∼ −iπj 1/N j= (N 1) n=1 e −X− X x (N 1) − ∞ N + k + n 1 n/N iπj(k+n 1)/N = ζ(1 n)ζ − x ′′e − − N n=1 j= (N 1) X −X− (N 1) − ∞ N + k + n 1 n/N iπj(k+n 1)/N = ζ(1 n)ζ − x ′′e − . (4.3) − N n=1 j= (N 1) X −X− From (3.9), we know that if m N, ∈ (N 1) − ( 1)m(N+1)N if n = mN k + 1 ′′eiπj(k+n 1)/N = (4.4) − 0− if n = mN − k + 1. j= (N 1) 6 − −X− Employing (4.4) in (4.3), we see that, as x 0, → (N 1) iπj 1/N − e− ′′ iπj(k 1)/N F − e − N+k 1 , 1 N N x j= (N 1) ! −X− − ∞ m(N+1) m+ 1 k N ( 1) ζ(k mN)ζ (1 + m) x N . (4.5) ∼ − − m k X≥ N (1 k)/N Now divide (2.2) by x − , 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 , →∞ 2n ix ix ∞ ( 1)n+1 2π F + F − B ζ(k + 2nN) . k,N 2π k,N −2π ∼ n 2n x n=1 X To obtain (2.17) as x 0, we ﬁrst rewrite (2.5) as, → k+1 k−1 ix ix 2π ( 1) 2 x N F + F = 2 γ log ζ(k) Nζ′(k) + C (k,N,x) + − k,N 2π k,N − 2π − − x − N 2π N 1 1 1 − iπj iπ(N+k 1) iπj iπj ′′ F − 2π N 2N F − 2π N 2N exp 2 2N − N+k 1 , 1 i x e + N+k 1 , 1 i x e . × − N N N N − j= (N 1) −X− n o (4.6) Using (4.1), as x 0, → N 1 1 1 − iπj iπj iπj ′′ F − 2π N 2N F − 2π N 2N exp 2 N+k 1 , 1 i x e + N+k 1 , 1 i x e N N N N − j= (N 1) −X− n o N 1 n − − iπj ∞ k + n + N 1 x N iπnj πn 2 ′′ exp ζ(1 n)ζ − e 2N cos ∼ 2 − N 2π 2 j= (N 1) n=1 −X− X 2n N 1 ∞ k + 2n + N 1 x N − iπj iπnj = 2 ( 1)nζ(1 2n)ζ − ′′ exp , − − N 2π 2 − N n=1 j= (N 1) X −X− EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 23 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 1 1 − iπj iπj iπj ′′ F − 2π N 2N F − 2π N 2N exp 2 N+k 1 , 1 i x e + N+k 1 , 1 i x e N N N N − j= (N 1) −X− n o 2n πN ∞ B k + 2n + N 1 x N sin = 2n ζ − 2 . (4.7) − n N 2π cos πn n=1 N X 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 for any positive integer≤ m,

m 1 2Nm 1 − B2j+1(a) N 2j α− N+1 a ζ(2Nm +1)+ ζ(2Nm + 1 2jN)(2 α) − 2 (2j + 1)! − Xj=1 2Nm 1 N ∞ n− − exp a(2n) α + − 1 exp ( (2n)N α) n=1 X − − 2m(N 1) m+1 2m 2N m 2 − ( 1) (2π) B2m+1(a)Nγ 1 ∞ cos(2πna) = β N+1 − − + − N (2m + 1)! 2 n2m+1 nX=1 N−1 2 2m 1 N+3 j ∞ n− − cos(2πna) + ( 1) 2 ( 1) − − 1 iπj −(N−1) exp (2n) N βe N 1 j= n=1 X2 X − j+ N+3 ( 1) 2 ∞ sin(2πna) iβ 1 iπj iβ 1 iπj + − ψ (2n) N e N + ψ − (2n) N e N 2π n2m+1 2π 2π nX=1 N+1 2N +m j 2N2(m−j) m+ N+3 2Nm ⌊ ⌋ ( 1) B2j(a)BN+1+2N(m j) 2j N+ + ( 1) 2 2 − − α N+1 β N+1 . (4.8) − (2j)!(N +1+2N(m j))! Xj=0 − The main idea is to diﬀerentiate both sides of the above identity with respect to a and then let d n 2( 1)m+1(2m)!ζ(2m) a = 1 while using the facts B2j(a) = 2jB2j 1(a), Bn(1) = ( 1) Bn, B2m = − 2m da − − (2π) and αβN = πN+1. Inducting the ζ(2Nm + 1) from the resulting identity in the ﬁnite sum on the resulting left-hand side as its j = 0 term, we see that

m 1 2Nm 1+N 2Nm − B2j+1(a) N 2j N ∞ n− − α− N+1 ζ(2Nm + 1 2jN)(2 α) 2 α  (2j + 1)! − − exp((2n)N α) 1 n=1 Xj=0 X −  N−1  2m(N 1) 2 2N m 2 − ∞ 1 iβ 1 iπj N+1 − N N = β 2Nγζ(2m)+ 2m ψ (2n) e − N − − n 2π j= (N 1) n=1 X2 X

iβ 1 iπj N 1 1 2Nm +ψ − (2n) N e N + 2 − α − N+1 ζ(2Nm + 1 N). 2π − EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 24

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 such that αβ = πN+1. Then from [9, Theorem≤ 2.7], we have

N−1 N 2 ∞ exp( a(2n) α) 1 N+3 j ∞ cos(2πna) − ( 1) 2 ( 1) n(1 exp( (2n)N α)) − N − − 1 iπj (N−1) n exp (2n) N βe N 1 n=1 − − j= n=1 X −X2 X − j+ N+1 ( 1) 2 ∞ sin(2πna) β 1 iπj 1 iβ 1 iπj iβ 1 iπj + − log (2n) N e N ψ (2n) N e N + ψ − (2n) N e N π n 2π − 2 2π 2π n=1 X 1 1 (N 1)(log 2 γ) log(α/β) = ((a 1) log(2π) + log Γ(a)) + a − − + N − − 2 N N + 1 N+1 2N j 2 N+3 ⌊ ⌋ ( 1) B2j(a)BN+1 2Nj 2j N 2N j + ( 1) 2 − − α N+1 β − N+1 . (4.9) − (2j)!(N + 1 2Nj)! Xj=0 − Diﬀerentiate both sides of (4.9) with respect to a, let a = 1 in the resulting identity, and then divide both sides by 2N α to arrive at (2.25) upon simpliﬁcation. − Proof of Theorem 2.13. Using (1.10), we rewrite (2.24) as

2Nm 2Nm 1+N m+1 2m(N 1) N/2 ∞ n 1 ( 1) 2 β β − N+1 − − = ζ(2Nm + 1 N)+ − − (2n)N α N+1 e 1 −2 − Nπ 2 2N √α α nX=1 − (N−1) 2 iβ 1/N iπj iβ 1/N iπj 2Nm 1 F 1 2 e N + F 1 2 e N + α N+1 − 2 EN (m, α, β), 2m, N 2m, N × − 2π −2π j= (N 1) −X2 (4.10) where EN (m, α, β) is deﬁned by

(2m 1)(N 1) 2 − − m+1 2Nm + N β 1/N E (m, α, β) := ( 1) β− N+1 2 Nζ(2m) γ + log 2 ζ′(2m) N N+1 − 2π − Nπ 2 ( ) m 1 − B2j ζ(2Nm + 1 2Nj) N(2j 1) 2j 2Nm 1 + − 2 − α − N+1 − 2 . (4.11) (2j)! Xj=0 To study the behavior of the series on the left-hand side of (4.10) as α 0, we examine the series on the right-hand side of (4.10) as β using the relation αβN = π→N+1. →∞ We now invoke (2.16) with k = 2m, N replaced by 1/N and x = β21/N eiπj/N in the expression below, then replace j by j/2 to see that as β , →∞ (N−1) 2 iβ 1/N iπj iβ 1/N iπj F 1 2 e N + F 1 2 e N 2m, N 2m, N − 2π −2π j= (N 1) −X2 2n N 1 ∞ n+1 B2n 2n β 1/N − − iπnj ( 1) ζ 2m + 2 ′′e− N . ∼ − n N 2π n=1 j= (N 1) X −X− EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 25

From (3.9), for ℓ N, ∈ (N 1) − ℓ(N 1) iπnj N( 1) − if n = ℓN ′′e N = − 0− if n = ℓN. j= (N 1) 6 −X− Hence,

(N−1) 2 iβ 1/N iπj iβ 1/N iπj F 1 2 e N + F 1 2 e N 2m, N 2m, N − 2π −2π j= (N 1) −X2 2ℓN ∞ B β − N ( 1)ℓ+1 2ℓN ζ (2m + 2ℓ) 21/N . (4.12) ∼ − ℓN 2π Xℓ=1 (N+1)/N Thus from (4.10), (4.11) and (4.12) and the relation β = π , we arrive at (2.26) upon simpli- α1/N 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 of the Herglotz function, namely Fk,N (x). We obtained two diﬀerent kinds of functional equations relating Fk,N (x) with F N+k−1 1 (ξ/x), where ξ is some root of unity, the ﬁrst of which (Theorem N , N 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. This is explained below. Note that when we let x = 1 in (1.5) and (1.14), the corresponding Herglotz functions have the same arguments 5. Provided this phenomenon persists when we transition from the Herglotz function to the extended higher Herglotz function Fk,N (x), the three-term functional relation that is sought might be involving xN F (xN ), F ((1 + x)N ), and F , k,N k,N k,N (1 + x)N x for, not only do they reduce to F (x), F (x+1) and F x+1 respectively when k = N = 1 which are N N the same as the ones occurring in (1.6) but they also reduce to Fk,N (1), Fk,N (2 ) and Fk,N (2− ) 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 since nowhere in the literature has there been a relation which involves like powers (which are greater than 1) of x, x +1 and x/(x + 1) in the arguments of the associated functions.

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

∞ sin(2πna) iβ 1 iπj iβ 1 iπj ψ (2n) N e N + ψ − (2n) N e N , n2m+1 2π 2π n=1 X

5 1 2 In fact, when we let x = 1 in equations (1.5) and (1.14), they exactly match since J(1) = 2 log (2). EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 26 which, if we diﬀerentiate with respect to a and then let a = 1, leads to the following combination of the extended higher Herglotz functions

∞ 1 iβ 1 iπj iβ 1 iπj ψ (2n) N e N + ψ − (2n) N e N . n2m 2π 2π Xn=1 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

∞ cos(2πna) iα iα ψ (2n)N + ψ − (2n)N , nk 2π 2π n=1 X where k 1 and 0 < a 1. This would then generalize Theorems 2.2 and 2.3 and the general result, if≥ obtained, would≤ be analogous to (4.8) which is the corresponding result in the setting of generalized Lambert series. This would then complete the analogy between extended higher Herglotz 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) is of interest not only from the point of view of analytic number theory but also of algebraic number theory. In the same vein, it would be worthwhile to study for which values of k, N and x can the integral in (2.20) be evaluated in closed-form. In Corollary 2.9, 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, β positive such that αβ = 1, namely, if 1 φ(x) := ψ(x)+ log x, 2x − and ξ(s) and Ξ(t) denote Riemann’s functions respectively deﬁned by [29, p. 16, Equations (2.1.12), (2.1.14)] 1 s 1 ξ(s) := (s 1)π− 2 Γ(1 + s)ζ(s), − 2 1 Ξ(t) := ξ( 2 + it), then γ log(2πα) ∞ γ log(2πβ) ∞ √α − + φ(nα) = β − + φ(nα) 2α 2β ( n=1 ) ( n=1 ) X p 2 t X 1 ∞ t 1+ it cos log α = Ξ Γ − 2 dt. −π3/2 2 4 1+ t2 Z0

The ﬁrst equality in this formula seems to be in the realms of the theory of Herglotz function. A proof of it can be found, for example, in [4]. It would be certainly of merit to see if another proof of 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 product of an even zeta value and an odd zeta value. The remaining case of having products of only odd zeta 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 Ramanujan polynomials [13], [24], it would be interesting to study the analogous polynomials stemming from (2.9) and (2.23). EXTENDED HIGHER HERGLOTZ FUNCTIONS I. FUNCTIONAL EQUATIONS 27

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Discipline of Mathematics, Indian Institute of Technology, Gandhinagar, Palaj, Gandhinagar 382355, Gujarat, India Email address: [email protected]; rajat [email protected]; [email protected]