Applications of the Lipschitz Summation Formula and a Generalization of Raabe’s Cosine Transform

Abstract

General summation formulas have been proved to be very useful in analysis, number theory and other branches of mathematics. The Lipschitz summation formula is one of them. In this paper, we give its application by providing a new transformation formula which generalizes that of Ramanujan. Ramanujan’s result, in turn, is a generalization of the modular transformation of Eisenstein series $E_k(z)$ on SL$_2({\mathbb {Z}})$, where $z\rightarrow -1/z, z\in {\mathbb {H}}$. The proof of our result involves delicate analysis containing Cauchy Principal Value integrals. A simpler proof of a recent result of ours with Kesarwani giving a non-modular transformation for $\sum _{n=1}^{\infty }\sigma _{2m}(n)e^{-ny}$ is also derived using the Lipschitz summation formula. In the pursuit of obtaining this transformation, we naturally encounter a new generalization of Raabe’s cosine transform whose several properties are also demonstrated. As an application of our results, we get a generalization of Wright’s asymptotic estimate for the generating function of the number of plane partitions of a positive integer n.

1 Introduction

Let $\alpha , \beta >0$ with $\alpha \beta =\pi ^2$ and $m\in {\mathbb {Z}}\backslash \{0\}$. Ramanujan’s famous formula for $\zeta (2m+1)$ is given by¹ [45, p. 173, Ch. 14, Entry 21(i)], [44, pp. 319–320, formula (28)], [4, pp. 275–276]

$$\begin{aligned}&\alpha ^{-m}\left\{ \frac{1}{2}\zeta (2m+1)+\sum _{n=1}^\infty \frac{n^{-2m-1}}{e^{2n\alpha }-1}\right\} =(-\beta )^{-m}\left\{ \frac{1}{2}\zeta (2m+1)+\sum _{n=1}^\infty \frac{n^{-2m-1}}{e^{2n\beta }-1}\right\} \nonumber \\&\qquad -2^{2m}\sum _{k=0}^m\frac{(-1)^{k}B_{2k}B_{2m+2-2k}}{(2k)!(2m+2-2k)!}\alpha ^{m+1-k}\beta ^k, \end{aligned}$$

(1.1)

where, as customary, $\zeta (s)$ denotes the Riemann zeta function and $B_n$ denotes the $n^{th }$ Bernoulli number defined by

$$\begin{aligned} \sum _{n=0}^{\infty }\frac{B_n z^n}{n!}=\frac{z}{e^{z}-1} \hspace{5mm}(|z|<2\pi ). \end{aligned}$$

The above formula has received enormous attention from numerous mathematicians over the years and has been rediscovered many times, for example, see [24, Theorem 9] and [35]. It is an impressive result, for it encapsulates not only the transformation formulas of the Eisenstein series on SL$_2({\mathbb {Z}})$ and the corresponding ones for their Eichler integrals but also the transformation property of the logarithm of the Dedekind eta function. For a delightful historical account on it, we refer the reader to the excellent survey [6]. There are several generalizations of (1.1) in the literature, for example, [10, 13, 14, 16, 17, 25] and [31]. In his second notebook [45, p. 269], Ramanujan himself provided the following generalization of (1.1).

Let $\alpha$ and $\beta$ be two positive real numbers such that $\alpha \beta =4\pi ^2$. Then for $\textrm{Re}(s)>2$, we have

$$\begin{aligned}&\alpha ^{s/2}\left\{ \frac{\Gamma (s)\zeta (s)}{(2\pi )^s}+\cos \left( \frac{\pi s}{2}\right) \sum _{n=1}^\infty \frac{n^{s-1}}{e^{n\alpha }-1}\right\} \nonumber \\&\quad =\beta ^{s/2}\left\{ \cos \left( \frac{\pi s}{2}\right) \frac{\Gamma (s)\zeta (s)}{(2\pi )^s}+\sum _{n=1}^\infty \frac{n^{s-1}}{e^{n\beta }-1}\right. \nonumber \\&\quad \left. \quad -\sin \left( \frac{\pi s}{2}\right) \textrm{PV}\int _0^\infty \frac{x^{s-1}}{e^{2\pi x}-1}\cot \left( \frac{1}{2}\beta x\right) dx\right\} , \end{aligned}$$

(1.2)

where $\textrm{PV}$ denotes the Cauchy principal value integral. The above formula has been proved in [5, p. 416]. Also see [7, Section 9] for a recent generalization of (1.2).

Unfortunately, Ramanujan’s formula (1.2) has not received as much attention as (1.1). But it is also a noteworthy result because it not only gives the transformation formula for the Eisenstein series on SL$_2({\mathbb {Z}})$ in the special case $s=2m, m\in {\mathbb {N}}, m>1$, but also reveals the obstruction to modularity for other values of s, which is evident due to the appearance of the integral on its right-hand side. Note that the last term involving the integral disappears for $s=2m$.

One of the goals of this paper is to derive a generalization of (1.2):

Theorem 1.1

Let $\textrm{Re}(\alpha ),\textrm{Re}(\beta )>0$ such that $\alpha \beta =4\pi ^2$. Let $0\le a<1$. Then, for $\textrm{Re}(s)>2$, the following transformation holds

$$\begin{aligned}&\alpha ^{s/2}\left\{ \frac{\Gamma (s)\zeta (s)}{(2\pi )^s}+\frac{1}{2}\sum _{n=1}^\infty n^{s-1}\left( \frac{e^{\pi is/2}}{e^{n\alpha -2\pi ia}-1}+\frac{e^{-\pi is/2}}{e^{n\alpha +2\pi ia}-1}\right) \right\} \nonumber \\&=\beta ^{s/2}\left\{ \frac{\Gamma (s)}{(2\pi )^{s}}\sum _{k=1}^\infty \frac{\cos \left( \frac{\pi s}{2}+2\pi ak\right) }{k^s}+\sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)\beta }-1}\right. \nonumber \\&\quad \left. -\frac{1}{2i}\textrm{PV}\int _0^\infty x^{s-1} \left( \frac{e^{\pi is/2}}{e^{2\pi x-2\pi ia}-1}-\frac{e^{-\pi is/2}}{e^{2\pi x+2\pi ia}-1}\right) \cot \left( \frac{1}{2}\beta x\right) dx\right\} . \end{aligned}$$

(1.3)

The above theorem reduces to Ramanujan’s formula (1.2) for $a=0$. Also for $s=2m, m\in {\mathbb {N}}, m>1$, and $a=0$, it gives (1.1).

A special case of Theorem 1.1 is the new transformation given below.

Corollary 1.2

Let $m\in {\mathbb {N}}$ and $m>1$. For $\textrm{Re}(\alpha ),\textrm{Re}(\beta )>0$ such that $\alpha \beta =4\pi ^2$, we have

$$\begin{aligned} \alpha ^m\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{n\alpha }+1}+(-\beta )^m\sum _{n=1}^\infty \frac{(n-1/2)^{2m-1}}{e^{(n-1/2)\beta }-1}=-\left\{ \alpha ^m-(2^{1-2m}-1)(-\beta )^m\right\} \frac{B_{2m}}{4m}. \end{aligned}$$

(1.4)

Equation (1.4) is a “hybrid” analogue of the following transformation formula for the Eisenstein series over SL$_2({\mathbb {Z}})$ in that the role of n in the first series is played by $n-1/2$ in the second.

$$\begin{aligned} \alpha ^m\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{n\alpha }-1}-(-\beta )^m\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{n\beta }-1}=\left\{ \alpha ^m-(-\beta )^m\right\} \frac{B_{2m}}{4m}. \end{aligned}$$

The above formula can be obtained from (1.1) by replacing m by $-m$. As an application of Corollary 1.2, we obtain the following results, which, to the best of our knowledge, are new.

Corollary 1.3

For any odd positive integer m greater than 1,

$$\begin{aligned} \sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\pi }+1}-\frac{1}{2^{2m-1}}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{n\pi }-1} =-(1+2^{1-2m})\frac{B_{2m}}{4m}. \end{aligned}$$

(1.5)

Moreover,

$$\begin{aligned} \sum _{n=1}^\infty \frac{n}{e^{2n\pi }+1}-\frac{1}{2}\sum _{n=1}^\infty \frac{n}{e^{n\pi }-1}=-\frac{1}{16}+\frac{1}{8\pi }, \end{aligned}$$

(1.6)

and hence, at least one of the two series is transcendental.

We next describe another consequence of Theorem 1.1. First, let $s=2\,m$ and $a=1/4$ in (1.3), then let $s=2\,m$ and $a=3/4$ in (1.3), and add the corresponding sides of the resulting identities. This leads to the transformation between just the infinite series which we record below in (1.7). Similarly subtracting the corresponding sides of the two resulting identities expresses a principal value integral in terms of a Lambert series, which is given in (1.8).

Corollary 1.4

Let $\textrm{Re}(\alpha ),\textrm{Re}(\beta )>0$ such that $\alpha \beta =4\pi ^2$. For $m\in {\mathbb {N}}, m>1$,

$$\begin{aligned}&\alpha ^m2^{4m-1}\left\{ \frac{\Gamma (2m)\zeta (2m)}{(2\pi )^{2m}}+(-1)^{m+1}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\alpha }+1}\right\} \nonumber \\&\quad =\beta ^m\left\{ (-1)^{m+1}\frac{\Gamma (2m)\zeta (2m)(2^{2m-1}-1)}{(2\pi )^{2m}}+\sum _{n=1}^\infty \frac{(4n-1)^{2m-1}}{e^{(4n-1)\beta /4}-1}\right. \nonumber \\&\quad \quad \left. +\sum _{n=1}^\infty \frac{(4n-3)^{2m-1}}{e^{(4n-3)\beta /4}-1}\right\} , \end{aligned}$$

(1.7)

and,

$$\begin{aligned} \textrm{PV}\int _0^\infty \textrm{sech}(2\pi x)\cot \left( 2\beta x\right) x^{2m-1}dx=(-1)^{m+1}4^{1-2m}\sum _{n=1}^\infty \frac{\chi (n)n^{2m-1}}{e^{n\beta }-1}, \end{aligned}$$

(1.8)

where $\chi (n)$ is a Dirichlet character modulo 4 given by

$$\begin{aligned} \chi (n)= {\left\{ \begin{array}{ll} 1, \qquad {\textrm{ if }}\ n\equiv 1\pmod {4},\\ -1, \quad {\textrm{ if }}\ n\equiv 3\pmod {4},\\ 0, \qquad {\textrm{ if }}\ n\equiv 0, 2\pmod {4}. \end{array}\right. } \end{aligned}$$

(1.9)

We note that the series on the right-hand side of (1.8) cannot be treated using [10, Theorem 1].

We now transition towards the second goal of our paper. Recently, the current authors, along with Kesarwani [16], extensively studied a more general Lambert series

$$\begin{aligned} \sum _{n=1}^\infty \frac{n^s}{e^{ny}-1}=\sum _{n=1}^\infty \sigma _s(n)e^{-ny} \quad (s\in {\mathbb {C}},\ \textrm{Re}(y)>0) \end{aligned}$$

(1.10)

than the ones appearing in (1.1). Here $\sigma _s(n):=\sum _{d|n}d^s$ is the generalized divisor function. Observe that the series in (1.10) is the same one that appears in Ramanujan’s (1.2). Among other results, they obtained [16, Theorem 2.5] an explicit transformation for any $\textrm{Re}(s)>-1$ and Re$(y)>0$, which is given next.

$$\begin{aligned}&\sum _{n=1}^\infty \sigma _s(n)e^{-ny}+\frac{1}{2}\left( \left( \frac{2\pi }{y}\right) ^{1+s}\textrm{cosec}\left( \frac{\pi s}{2}\right) +1\right) \zeta (-s)-\frac{1}{y}\zeta (1-s)\nonumber \\&\quad =\frac{2\pi }{y\sin \left( \frac{\pi s}{2}\right) }\sum _{n=1}^\infty \sigma _{s}(n)\Bigg (\frac{(2\pi n)^{-s}}{\Gamma (1-s)} {}_1F_2\left( 1;\frac{1-s}{2},1-\frac{s}{2};\frac{4\pi ^4n^2}{y^2} \right) \nonumber \\&\quad -\left( \frac{2\pi }{y}\right) ^{s}\cosh \left( \frac{4\pi ^2n}{y}\right) \Bigg ), \end{aligned}$$

(1.11)

where ${}_1F_2(a;b,c;z):=\sum _{n=0}^\infty \frac{(a)_nz^n}{(b)_n(c)_nn!},\ z\in {\mathbb {C}},\ (a)_n=\frac{\Gamma (a+n)}{\Gamma (a)}$, is the generalized hypergeometric function.

The explicit transformations of the type (1.11) are always desirable due to their possible applications in analytic number theory, especially in the theory of zeta functions. See the recent paper [2] for an application of (1.11) in the theory of $\zeta (s)$ by applying the operator $\frac{d}{ds}\big |_{s=0}$ on both sides, thereby resulting in a transformation of the Lambert series of the logarithm, that is, $\displaystyle \sum _{n=1}^\infty \frac{\log (n)}{e^{ny}-1}$.

The authors of [16, Theorem 2.5] also analytically continued (1.11) to Re$(s)>-2m-3,\ m\in {\mathbb {N}}\cup \{0\}$. Then, as a special case, they not only obtained Ramanujan’s formula (1.1) and the transformation formula of the logarithm of the Dedekind eta function but also new transformations when s is an even integer. For example, they established an explicit transformation [16, Theorem 2.11] for the series in (1.10) when $s=2m,\ m>0$. We record it below in (1.13). It comprises two special functions Shi(z) and Chi(z), known as the hyperbolic sine and hyperbolic cosine integrals, respectively defined by [38, p. 150, Equation (6.2.15), (6.2.16)]

$$\begin{aligned} \textrm{Shi}(z):=\int _0^z\frac{\sinh (t)}{t}\ dt,\hspace{4mm} \textrm{Chi}(z):=\gamma +\log (z)+\int _0^z\frac{\cosh (t)-1}{t}\ dt, \end{aligned}$$

(1.12)

where $\gamma$ is Euler’s constant. Let $m\in {\mathbb {N}}$. Then for $\textrm{Re}(y)>0$, we have [16, Theorem 2.11]

$$\begin{aligned} \sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}&-\frac{(2m)!}{y^{2m+1}}\zeta (2m+1)+\frac{B_{2m}}{2my}\nonumber \\&=(-1)^m\frac{2}{\pi }\left( \frac{2\pi }{y}\right) ^{2m+1} \sum _{n=1}^\infty \sigma _{2m}(n) \Bigg \{\sinh \left( \frac{4\pi ^2n}{y}\right) \textrm{Shi}\left( \frac{4\pi ^2n}{y}\right) \nonumber \\&\qquad -\cosh \left( \frac{4\pi ^2n}{y}\right) \textrm{Chi}\left( \frac{4\pi ^2n}{y}\right) +\sum _{j=1}^m(2j-1)!\left( \frac{4\pi ^2n}{y}\right) ^{-2j}\Bigg \}. \end{aligned}$$

(1.13)

The modular transformation for $\sum _{n=1}^{\infty }\sigma _{2m+1}(n)e^{-ny}$ transforms it into

$$\begin{aligned} \sum _{n=1}^\infty \sigma _{2m+1}(n)e^{-4\pi ^2n/y}=-\sum _{n=1}^\infty \sigma _{2m+1}(n)\Bigg \{\sinh \left( \frac{4\pi ^2n}{y}\right) -\cosh \left( \frac{4\pi ^2n}{y}\right) \Bigg \}.\nonumber \\ \end{aligned}$$

(1.14)

In view of this, it is important to note that while going from $s=2m+1$ to $s=2m$ in $\sum _{n=1}^\infty \sigma _s(n)e^{-ny}$, the expression $\sinh \left( \frac{4\pi ^2n}{y}\right) -\cosh \left( \frac{4\pi ^2n}{y}\right)$ in (1.14) is to be replaced by the corresponding one on the right-hand side of (1.13). (Note that the finite sum $\sum _{j=1}^m(2j-1)!\left( \frac{4\pi ^2n}{y}\right) ^{-2j}$ in the summand of the series on the right-hand side of (1.14) is essential for its convergence; for details, see the proof of (1.13) in Sect. 4.)

Equation (1.13) readily gives the following asymptotic estimate for $\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}$:

Corollary 1.5

Let $m\in {\mathbb {N}}$. As $y\rightarrow 0$ in $|\arg (y)|<\pi /2$,

$$\begin{aligned} \sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}&=\frac{(2m)!}{y^{2m+1}}\zeta (2m+1)-\frac{B_{2m}}{2my}\nonumber \\&\quad -\!\frac{2(-1)^m}{\pi (2\pi )^{2m-1}}\!\sum _{j=1}^{r+1}\!\frac{\Gamma (2m+2j)\zeta (2m\!+2j)\zeta (2j)}{(2\pi )^{4j}}y^{2j-1}\!\!+\!\!O\left( y^{2r+3}\right) .\nonumber \\ \end{aligned}$$

(1.15)

In his first letter to Hardy [9, p. 28], Ramanujan gave the special case $m=1$ of the above result. Watson [49] used the Abel–Plana summation formula to prove this result of Ramanujan.

The case $m=1$ of the series on the left-hand side of (1.13) (or (1.5)) has the following interesting connection with the generating function for plane partitions studied by MacMahon [1, p. 184]:

$$\begin{aligned} \sum _{n=1}^\infty \sigma _{2}(n)e^{-ny}=x\frac{d}{dx}\log (F(x)), \end{aligned}$$

(1.16)

where $F(x):=\prod _{n=1}^\infty \frac{1}{(1-x^n)^n}$, with $y=\log (1/x)$, where $|x|<1$. In his work on finding the asymptotic estimate of q(n), the number of plane partitions of a positive integer n, Wright [51, Lemma 1] first found the asymptotic estimate of F(x) as $x\rightarrow 1^-$. His result on F(x) readily follows from our Corollary 1.5 and is rephrased in the following corollary.

Corollary 1.6

As $x\rightarrow 1^-$, we have

$$\begin{aligned} F(x)=e^c(\log x)^{1/12}\exp \left( \frac{\zeta (3)}{\log ^2x}\right) \exp \left( -\sum _{j=1}^{r+1}\delta _j(\log x)^{2j}\right) \left( 1+O_r\left( (\log x)^{2r+4}\right) \right) , \end{aligned}$$

(1.17)

where, c is a constant, and

$$\begin{aligned} \delta _j:=\frac{\Gamma (2j+2)\zeta (2j+2)\zeta (2j)}{2\pi ^2j(2\pi )^{4j}}. \end{aligned}$$

Wright’s proof (see [51, pp. 180–184]) also gives a representation of the constant c in terms of an integral, namely, $c=\displaystyle 2\int _{0}^{\infty }\frac{y\log (y)}{e^{2\pi y}-1}\, dy=\zeta '(-1)\approx -0.165421\cdots$.

The Lambert series (1.10), whose special cases were considered in (1.1) and (1.13), has been studied by many mathematicians over the years. For a detailed survey, see [16]. One of the earliest mathematicians to study it was Wigert, who wrote several papers on this subject. In [50], Wigert examined the Lambert series when $0<s<1$. Later, Kuylenstierna [32] provided a simple proof of Wigert’s result using double zeta function $\displaystyle \zeta _2(s,\tau ):=\sum _{\begin{array}{c} m,n=0\\ (m,n)\ne (0,0) \end{array}}^\infty \frac{1}{(m+n\tau )^s}$, Re$(s)>2$, $\tau \in {\mathbb {C}}\backslash (-\infty ,0]$. However, both of them were interested only in the asymptotics of the series in (1.10), not in explicit transformations. In his work, Kuylenstierna essentially uses the Lipschitz summation formula [33], namely, $0\le a<1$, $\textrm{Re}(s)>1$ and $\tau \in {\mathbb {H}}$ (the upper half plane),

$$\begin{aligned} \sum _{n=1}^\infty \frac{e^{2\pi i\tau (n-a)}}{(n-a)^{1-s}}=\frac{\Gamma (s)}{(-2\pi i)^s}\sum _{k\in {\mathbb {Z}}}\frac{e^{2\pi iak}}{(k+\tau )^s}. \end{aligned}$$

(1.18)

The Lipschitz summation formula has several nice applications and generalizations, for example, see [3, 30, 40]. Equation (1.18) is usually proved using Poisson summation formula, for example, see [43, pp. 77–79]. For other proofs, one can look at the paper of Vági [48]. Another (rather simple) proof can be found in [41].

It does not seem to be easy to get (1.13) as a special case of Ramanujan’s formula (1.2) or our generalization (1.3), because one has to transform the Lambert series and the principal value integral on the right-hand side of (1.2) into the series in (1.13) involving the special functions Shi(z) and Chi(z).

In this paper, we prove Theorem 1.1 and (1.13) using the Lipschitz summation formula (1.18). The proof of (1.13) through this approach involves a nice generalization of the following identity [14, Theorem 2.2]

$$\begin{aligned} \sum _{n=1}^\infty \int _0^\infty \frac{t\cos (t)}{t^2+n^2u^2}\ dt=\frac{1}{2}\left\{ \log \left( \frac{u}{2\pi }\right) -\frac{1}{2}\left( \psi \left( \frac{iu}{2\pi }\right) +\psi \left( -\frac{iu}{2\pi }\right) \right) \right\} , \end{aligned}$$

(1.19)

where $Re (u)>0$, and $\psi (z):=\Gamma '(z)/\Gamma (z)$ is the digamma function. In [14, Theorem 2.4], the above identity was employed to obtain a two-parameter generalization of (1.1). Various applications of (1.19) can be found in [15, 16].

Observe that the summand of the left-hand side of (1.19) is the Raabe cosine transform defined for Re$(w)>0$ and $y>0$ by [19, p. 144]

$$\begin{aligned} {\mathfrak {R}}(y,w):=\int _0^\infty \frac{t\cos (yt)}{t^2+w^2}\ dt. \end{aligned}$$

Before stating the generalization of (1.19), that is sought for, we first introduce a new generalization of Raabe’s cosine transform, valid for $Re (w)>0, Re (z)>0$ or $z=0$, and $y>0$, by

$$\begin{aligned} {\mathfrak {R}}_z(y,w):=\frac{1}{2}\Gamma (2z+1)\int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \cos (yt)\ dt. \end{aligned}$$

(1.20)

It is easy to see that ${\mathfrak {R}}_0(y,w)={\mathfrak {R}}(y,w)$. Also for $w>0$, ${\mathfrak {R}}_z(y,w)$ satisfies a nice identity, namely,

$$\begin{aligned} w^{2z}{\mathfrak {R}}_z(y,w)=y^{2z}{\mathfrak {R}}_z(w, y), \end{aligned}$$

(1.21)

which is easily seen by making the change of variable $t=xw/y$ in (1.20).

Our first result on ${\mathfrak {R}}_z(y,w)$ gives a closed-form evaluation of an infinite series containing ${\mathfrak {R}}_z(y,w)$.

Theorem 1.7

Let $\zeta (z,a)$ be the Hurwitz zeta function. For $\textrm{Re}(w)>0$ and $\textrm{Re}(z)>0$, we have

$$\begin{aligned}&\frac{2}{\Gamma (2z+1)}\sum _{n=1}^\infty {\mathfrak {R}}_z(2\pi n,w)\nonumber \\&\quad =\sum _{n=1}^\infty \int _0^\infty \left( \frac{1}{(v-iw)^{2z+1}}+\frac{1}{(v+iw)^{2z+1}}\right) \cos (2\pi nv)\ dv\nonumber \\&\quad =\frac{1}{2}\left\{ \zeta (1+2z,iw)+\zeta (1+2z,-iw)-\frac{\cos (\pi z)}{zw^{2z}}\right\} . \end{aligned}$$

(1.22)

Note that this result is not straightforward to obtain as one cannot interchange the order of the summation and integration: doing so leads to a divergent integral. The primary tool to prove this result is Guinand’s generalization of Poisson’s summation formula [23, Theorem 1]; see Theorem 2.2.

The generalized Raabe cosine transform ${\mathfrak {R}}_z(y,w)$ itself can be evaluated in terms of exponential integral functions and incomplete gamma functions which are not so popular. However, the utility of Theorem 1.7 is that the infinite sum of ${\mathfrak {R}}_z(y,w)$ can be evaluated in terms of the well-known functions such as the Hurwitz zeta function $\zeta (z,a)$ and $\cos (z)$.

An immediate consequence of Theorem 1.7 is

Corollary 1.8

Equation (1.19) holds true.

Our next result gives an evaluation of a double integral which is imperative to prove Theorem 1.7.

Theorem 1.9

Let $\textrm{Re}(w)>0$ and $\textrm{Re}(z)>0$. Then

$$\begin{aligned} \int _0^\infty \int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \cos (2\pi vt)\ dtdv=-\frac{1}{2w^{2z+1}}\sin (\pi z). \end{aligned}$$

(1.23)

Equivalently, in the notation of (1.20),

$$\begin{aligned} \int _0^\infty {\mathfrak {R}}_z(2\pi v,w)\ dv=-\frac{1}{4w^{2z+1}}\Gamma (2z+1)\sin (\pi z). \end{aligned}$$

It is effortless to see that for $z\in {\mathbb {N}}\cup \{0\}$, the above integral evaluates to zero. The particular case $z=0$ is already obtained in [14, Lemma 3.4].

We now provide a new equivalent representation for ${\mathfrak {R}}_m(1,w)$, where $m\in {\mathbb {N}}\cup \{0\}$. This representation appears in the transformation of $\sum _{n=1}^{\infty }\sigma _{2m}(n)e^{-ny}$ given in (1.13).

Theorem 1.10

Let $\textrm{Shi}(z)$, $\textrm{Chi}(z)$ and ${\mathfrak {R}}_z(y,w)$ be defined in (1.12) and (1.20) respectively. Let $m\in {\mathbb {N}}\cup \{0\}$ and $\textrm{Re}(w)>0$. Then

$$\begin{aligned} {\mathfrak {R}}_m(1,w)&=\frac{(2m)!}{2}\int _0^\infty \left( \frac{1}{(t-iw)^{2m+1}}+\frac{1}{(t+iw)^{2m+1}}\right) \cos (t)\ dt\nonumber \\&=(-1)^m\left( \sinh (w)\textrm{Shi}(w)-\cosh (w)\textrm{Chi}(w)+\sum _{j=1}^m(2j-1)!w^{-2j}\right) . \end{aligned}$$

(1.24)

The special case $m=0$ of this result was derived in [16, Lemma 9.1].

As mentioned earlier, we provide a new proof of (1.13) in this paper. It is done by employing the Lipschitz summation formula and Theorems 1.7 and 1.10. Deriving it this way is simpler than obtaining it as a special case of (1.11). The latter was done in [16, Section 9].

This paper is organised as follows. In Sect. 2, the proofs of Theorems 1.7, 1.9 and 1.10 are given. Sections 3 and 4 are devoted to proving Theorem 1.1 and (1.13) respectively. In Sect. 5, we give proofs of Corollaries 1.5 and 1.6.

2 The Generalized Raabe Cosine Transform ${\mathfrak {R}}_z(y,w)$

This section is devoted to obtaining the results associated with ${\mathfrak {R}}_z(y,w)$ and which are crucial to proving (1.13). The first result below gives the asymptotic expansion of ${\mathfrak {R}}_z(y,w)$ as $y\rightarrow \infty$.

Lemma 2.1

Let ${\mathfrak {R}}_z(y,w)$ be defined in (1.20). Let $\textrm{Re}(w)>0$ and $\textrm{Re}(z)>0$. Then as $y\rightarrow \infty$,

$$\begin{aligned} {\mathfrak {R}}_z(y,w)\sim -\frac{\cos (\pi z)}{w^{2z}}\sum _{n=1}^\infty \frac{\Gamma (2z+2n)}{(yw)^{2n}}. \end{aligned}$$

Proof

We use the analogue of Watson’s lemma for Laplace transform in the setting of Fourier transforms [37, 12, Equations (1.3), (1.4)]. It states that if the form of h(t) near $t=0$ is given as a series of algebraic powers, that is,

$$\begin{aligned} h(t)\sim \sum _{n=0}^{\infty }b_nt^{n+\lambda -1}\qquad (\textrm{Re}(\lambda )>0) \end{aligned}$$

(2.1)

as $t\rightarrow 0^{+}$, then under certain restrictions on h (see [37, 12, Section 2] for the same),

$$\begin{aligned} \int _{0}^{\infty }e^{ist}h(t)\, \textrm{d}t\sim \sum _{n=0}^{\infty }b_ne^{i(n+\lambda )\pi /2}\Gamma (n+\lambda )s^{-n-\lambda } \end{aligned}$$

(2.2)

as $s\rightarrow \infty$.

Let

$$\begin{aligned} h(t):=\displaystyle \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}. \end{aligned}$$

Then, near $t=0$, it is easy to see that

$$\begin{aligned} h(t)&= (-iw)^{-(2z+1)}\sum _{n=0}^\infty \frac{(2z+1)_n}{n!}\left( \frac{t}{iw}\right) ^n+(iw)^{-(2z+1)}\sum _{n=0}^\infty \frac{(2z+1)_n}{n!}\left( -\frac{t}{iw}\right) ^n\\&=2w^{-(2z+1)}\sum _{n=0}^\infty \frac{(2z+1)_n}{n!w^n}\sin \left( \frac{\pi n}{2}-\pi z\right) t^n. \end{aligned}$$

Therefore, it is clear that our function h(t) satisfies (2.1) with $\lambda =1$ and

$$\begin{aligned} b_n=\frac{2w^{-(2z+1)}(2z+1)_n}{n!w^n}\sin \left( \frac{\pi n}{2}-\pi z\right) . \end{aligned}$$

(2.3)

From (2.2) and (2.3), as $y\rightarrow \infty$,

$$\begin{aligned} \int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) e^{ iyt}\ dt\sim \sum _{n=0}^\infty b_ne^{i(n+1)\frac{\pi }{2}}\Gamma (n+1)y^{-n-1}, \end{aligned}$$

(2.4)

where $b_n$ is given in (2.3). Similarly, as $y\rightarrow \infty$,

$$\begin{aligned} \int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) e^{-iyt}\ dt\sim \sum _{n=0}^\infty b_ne^{i(n+1)\frac{\pi }{2}}\Gamma (n+1)(-y)^{-n-1}. \end{aligned}$$

(2.5)

From (2.4) and (2.5), we see that as $y\rightarrow \infty$,

$$\begin{aligned}&\int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \cos (yt)\ dt\nonumber \\&\sim w^{-(2z+1)}\sum _{n=0}^\infty \frac{(2z+1)_n}{w^ny^{n+1}}e^{i(n+1)\frac{\pi }{2}}\sin \left( \frac{\pi n}{2}-\pi z\right) (1+(-1)^{-n-1})\nonumber \\&=2w^{-(2z+1)}\sum _{n=1}^\infty \frac{(2z+1)_{(2n-1)}}{w^{2n-1}y^{2n}}e^{n\pi i}\sin \left( \frac{\pi (2n-1)}{2}-\pi z\right) \nonumber \\&=-\frac{2w^{-2z}\cos (\pi z)}{\Gamma (2z+1)}\sum _{n=1}^\infty \frac{\Gamma (2z+2n)}{(yw)^{2n}}. \end{aligned}$$

(2.6)

Lemma 2.1 follows upon multiplying both sides of (2.6) by $\frac{1}{2}\Gamma (2z+1)$ and then using the definition of ${\mathfrak {R}}_z(y,w)$ from (1.20). $\square$

Remark 1

The special case $z=0$ of Lemma 2.1 was obtained in [14, Lemma 3.3].

Our next task is to evaluate the double integral in (1.23).

Proof of Theorem 1.9

Note that double integral in (1.23) is not absolutely convergent which means we cannot interchange the order of integration. Securing convergence of the integral over v near $v=0$ is straightforward. Along with this, Lemma 2.1 implies that the double integral in (1.23) is convergent.

We first evaluate a more general integral by introducing the exponential factor $e^{-\frac{v^2}{N}}$ inside the integrand and then take limit $N\rightarrow \infty$. Let N be a positive integer and consider the integral

$$\begin{aligned} I(w,z,N):&=\int _0^\infty \int _0^\infty e^{-\frac{v^2}{N}}\left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \nonumber \\&\quad \times \cos (2\pi vt)\ dtdv \quad (\textrm{Re}(w)>0,\ \textrm{Re}(z)>0). \end{aligned}$$

(2.7)

By invoking Fubini’s theorem we can interchange the order of integration in the above equation to see that

$$\begin{aligned} I(w,z,N)&=\int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \int _0^\infty e^{-\frac{v^2}{N}}\cos (2\pi vt)\ dvdt\nonumber \\&=\frac{\sqrt{\pi N}}{2}\int _0^\infty e^{-N\pi ^2t^2}\left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) dt, \end{aligned}$$

(2.8)

where we used the fact that $e^{-v^2/N}$ is self-reciprocal (up to some factor) with respect to the cosine kernel (See [22, p. 488, Formula 3.896.4]). Next invoke the identity [18, p. 88, Section 2.5.5]

$$\begin{aligned} (1-\sqrt{\xi })^{-2s}+(1+\sqrt{\xi })^{-2s}=2\ {}_2F_1\left( s,s+\frac{1}{2};\frac{1}{2};\xi \right) , \end{aligned}$$

with $\xi =-w^2/t^2$ and $s=z+1/2$ in (2.8) to deduce that

$$\begin{aligned} I(w,z,N)&=\sqrt{\pi N}\int _0^\infty e^{-N\pi ^2t^2}t^{-2z-1}{}_2F_1\left( z+\frac{1}{2},z+1;\frac{1}{2};-\frac{w^2}{t^2}\right) dt\nonumber \\&=\frac{\sqrt{\pi N}}{2}\int _0^\infty e^{-N\pi ^2/x}x^{z-1}{}_2F_1\left( z+\frac{1}{2},z+1;\frac{1}{2};-w^2x\right) dx, \end{aligned}$$

(2.9)

where we made the change of variable $t=1/\sqrt{x}$. From [42, p. 319, Formula 2.21.2.6], for $\textrm{Re}(p)>0, \textrm{Re}(a-\alpha )>0,\ \textrm{Re}(b-\alpha )>0$ and $|\arg (\omega )|<\pi$, we have

$$\begin{aligned}&\int _0^\infty x^{\alpha -1}e^{-p/x}{}_2F_1(a,b;c;-\omega x)dx\\&\quad =\omega ^{-\alpha }\frac{\Gamma (c)\Gamma (\alpha )\Gamma (a-\alpha ) \Gamma (b-\alpha )}{\Gamma (a)\Gamma (b)\Gamma (c-\alpha )}\\&\quad \times {}_2F_2(a-\alpha ,b-\alpha ;1-\alpha ,c-\alpha ;\omega p)\\&\quad +p^\alpha \Gamma (-\alpha ){}_2F_2(a,b;c,\alpha +1;\omega p). \end{aligned}$$

Let $p=N\pi ^2,\ a=z+1/2,\ b=z+1,\ c=1/2,\ \alpha =z$ and $\omega =w^2$ in the above integral evaluation, use the reflection formula for the gamma function $\Gamma (1/2+s)\Gamma (1/2-s)=\pi /\cos (\pi s)$ and substitute the resultant in (2.9) so that for $|\arg (w)|<\pi /2$,

$$\begin{aligned} I(w,z,N)&=\frac{\sqrt{\pi N}}{2}\left\{ \frac{\cos (\pi z)}{zw^{2z}}{}_2F_2\left( \frac{1}{2},1;1-z,\frac{1}{2}-z;N\pi ^2w^2\right) \right. \nonumber \\&\quad \left. +(N\pi ^2)^z\Gamma (-z){}_1F_1\left( z+\frac{1}{2};\frac{1}{2};N\pi ^2w^2\right) \right\} . \end{aligned}$$

(2.10)

We now wish to take limit $N\rightarrow \infty$ on both sides of the above equation. To that end, we need to find the behavior of the functions on the right-hand side as $N\rightarrow \infty$. The following asymptotic is given by Kim [29]: as $x\rightarrow \infty$ in $-\frac{3\pi }{2}<\mathrm {\arg }(x)<\frac{\pi }{2}$, for $\alpha \ne {\mathbb {Z}}\cup \{0\}$,

$$\begin{aligned} {}_2F_2(1,\alpha ;\rho _1,\rho _2;x)\sim \frac{\Gamma (\rho _1)\Gamma (\rho _2)}{\Gamma (\alpha )}\left( K_{22}(x)+L_{22}(-x)\right) , \end{aligned}$$

where, with $\nu =1+\alpha -\rho _1-\rho _2$,

$$\begin{aligned} K_{22}(x)=x^\nu e^x{}_2F_0\left( \rho _1-\alpha ,\rho _2-\alpha ;-;\frac{1}{x}\right) , \end{aligned}$$

and

$$\begin{aligned} L_{22}(x)&=x^{-1}\frac{\Gamma (\alpha -1)}{\Gamma (\rho _1-1)\Gamma (\rho _2-1)}{}_3F_1\left( 1,2-\rho _1,2-\rho _2;2-\alpha ;\frac{1}{x}\right) \\&\quad +x^{-\alpha }\frac{\Gamma (\alpha )\Gamma (1-\alpha )}{\Gamma (\rho _1-\alpha )\Gamma (\rho _1-\alpha )}{}_2F_0\left( 1+\alpha -\rho _1,1+\alpha -\rho _2;-;\frac{1}{x}\right) . \end{aligned}$$

We let $\alpha =1/2,\ \rho _1=1-z,\ \rho _2=1/2-z$ and $x=N\pi ^2w^2$ in the above expression to get, for $-\frac{3\pi }{4}<\mathrm {\arg }(w)<\frac{\pi }{4}$,

$$\begin{aligned}&{}_2F_2\left( \frac{1}{2},1;1-z,\frac{1}{2}-z;N\pi ^2w^2\right) \nonumber \\&\quad \sim \frac{\Gamma (1/2-z)\Gamma (1-z)}{\sqrt{\pi }}\left\{ (N\pi ^2w^2)^{2z}e^{N\pi ^2w^2}{}_2F_0\left( \frac{1}{2}-z,-z;-;\frac{1}{N\pi ^2w^2}\right) \right. \nonumber \\&\left. \quad +\frac{2}{N\pi ^{3/2}w^2\Gamma (-1/2-z)\Gamma (-z)}{}_3F_1\left( 1,1+z,\frac{3}{2}+z;\frac{3}{2};-\frac{1}{N\pi ^2w^2}\right) \right. \nonumber \\&\left. \quad +\frac{\pi (-N\pi ^2w^2)^{-1/2}}{\Gamma (1/2-z)\Gamma (-z)}{}_2F_0\left( \frac{1}{2}+z,1+z;-;-\frac{1}{N\pi ^2w^2}\right) \right\} \end{aligned}$$

(2.11)

as $N\rightarrow \infty$. Also, from [47, p. 189, Exercise 7.7],

$$\begin{aligned}&{}_1F_1(a;c;x)\sim \frac{e^xx^{a-c}\Gamma (c)}{\Gamma (a)}\sum _{n=0}^\infty \frac{(c-a)_n(1-a)_n}{n!}x^{-n}\\&\quad +\frac{e^{-\pi ia}x^{-a}}{\Gamma (c-a)}\sum _{n=0}^\infty \frac{(a)_n(1+a-c)_n}{n!}(-x)^{-n}, \quad x\rightarrow \infty , \end{aligned}$$

where $-\frac{3\pi }{2}<\textrm{arg}(x)<\frac{\pi }{2}$. Upon letting $a=z+1/2,\ c=1/2$ and $x=N\pi ^2w^2$ in the above formula and using the series definition of ${}_2F_0$, for $-\frac{3\pi }{4}<\textrm{arg}(w)<\frac{\pi }{4}$, we see that

$$\begin{aligned}&{}_1F_1\left( z+\frac{1}{2};\frac{1}{2};N\pi ^2w^2\right) \nonumber \\&\quad \sim e^{N\pi ^2w^2}\frac{\sqrt{\pi }(N\pi ^2w^2)^z}{\Gamma (z+1/2)}{}_2F_0\left( -z,\frac{1}{2}-z;-;\frac{1}{N\pi ^2w^2}\right) \nonumber \\&\quad +e^{-\pi i(z+1/2)}\frac{\sqrt{\pi }(N\pi ^2w^2)^{-(z+1/2)}}{\Gamma (-z)}{}_2F_0\left( z+\frac{1}{2},1+z;-;-\frac{1}{N\pi ^2w^2}\right) \end{aligned}$$

(2.12)

as $N\rightarrow \infty$. Substitute (2.11) and (2.12) in (2.10) and observe that the terms involving ${}_2F_0\left( -z,\frac{1}{2}-z;-;\frac{1}{N\pi ^2w^2}\right)$ cancel each other out. Also note that ${}_pF_q(a_1,\cdots , a_p;b_1,\cdots ,b_q;1/N)=1+O(1/N)$, as $N\rightarrow \infty$. Hence, for $-\frac{\pi }{2}<\arg (w)<\frac{\pi }{4}$, as $N\rightarrow \infty$,

$$\begin{aligned} I(w,z,N)&=\frac{\sqrt{\pi }}{2}\left\{ \frac{2^{2z}\Gamma (1-2z)\cos (\pi z)}{zw^{2z}}\left[ \frac{1}{\sqrt{N}}\frac{2^{-2z-1}}{\pi ^2w^2\Gamma (-1-2z)}\left( 1+O\left( \frac{1}{N}\right) \right) \right. \right. \\&\left. \left. \quad +\frac{i2^{-2z-1}}{\sqrt{\pi }w\Gamma (-2z)}\left( 1+O\left( \frac{1}{N}\right) \right) \right] -i\frac{1}{\sqrt{ \pi }} e^{-\pi iz}w^{-2z-1}\left( 1+O\left( \frac{1}{N}\right) \right) \right\} . \end{aligned}$$

We next let $N\rightarrow \infty$ on the both sides of the above equation. By using the dominated convergence theorem, we can take the limit $N\rightarrow \infty$ inside the integral sign in (2.7). Thus,

$$\begin{aligned}&\int _0^\infty \int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \cos (2\pi vt)\ dtdv\nonumber \\&\quad =-\frac{i}{2w^{2z+1}}\left\{ e^{-\pi iz}-\cos (\pi z)\right\} \nonumber \\&\quad =-\frac{i}{2w^{2z+1}}\left\{ e^{-\pi iz}-\frac{e^{i\pi z}+e^{-i\pi z}}{2}\right\} \nonumber \\&\quad =\frac{i}{2w^{2z+1}}\left\{ \frac{e^{i\pi z}-e^{-i\pi z}}{2}\right\} , \end{aligned}$$

(2.13)

which proves our theorem for $-\frac{\pi }{2}<\mathrm {\arg }(w)<\frac{\pi }{4}$. We next prove the result in the remaining region $\frac{\pi }{4}\le \mathrm {\arg }(w)<\frac{\pi }{2}$.

To do so, observe that as $z\rightarrow \infty$, we have [38, p. 411, Formula 16.11.7].

$$\begin{aligned} {}_qF_q\left( a_1,\cdots ,a_q;b_1,\cdots ,b_q;z\right) \sim \frac{\prod _{\ell =1}^q\Gamma (b_\ell )}{\prod _{\ell =1}^q\Gamma (a_\ell )}\left\{ H_{q,q}(ze^{\mp \pi i})+E_{q,q}(z)\right\} , \end{aligned}$$

where $E_{p,q}(z)$ and $H_{p,q}(z)$ are formal infinite series defined by

$$\begin{aligned} E_{p,q}(z)&=(2\pi )^{(p-q)/2}\kappa ^{-\nu -1/2}e^{\kappa z^{1/\kappa }}\sum _{k=0}^\infty c_k(\kappa z^{1/\kappa })^{\nu -k},\qquad p<q+1,\\ H_{p,q}(z)&=\sum _{m=1}^p\sum _{k=0}^\infty \frac{(-1)^k}{k!}\Gamma (a_m+k)\frac{\prod \limits _{\begin{array}{c} {\ell =1}\\ \ell \ne m \end{array}}^p\Gamma (a_\ell -a_m-k)}{\prod \limits _{\begin{array}{c} \ell =1\\ \ell \ne m \end{array}}^q\Gamma (b_\ell -a_m-k)}z^{-a_m-k}, \end{aligned}$$

where $\kappa =q-p+1$, $\nu =a_1+\cdots +a_p-b_1-\cdots -b_q+\frac{1}{2}(q-p)$,

$$\begin{aligned} c_0=1,\quad c_k=-\frac{1}{k\kappa ^\kappa }\sum _{m=0}^{k-1}c_me_{k,m},\quad k\ge 1, \end{aligned}$$

with

$$\begin{aligned} e_{k,m}=\sum _{j=1}^{q+1}(1-\nu -\kappa b_j+m)_{\kappa +k-m}\frac{\prod _{\ell =1}^p(a_\ell -b_j)}{\prod \limits _{\begin{array}{c} \ell =1\\ \ell \ne j \end{array}}^{q+1}(b_\ell -b_j)}, \quad b_{q+1}=1. \end{aligned}$$

By invoking the above asymptotic twice, for $\frac{\pi }{4}\le \mathrm {\arg }(w)<\frac{\pi }{2}$, as $N\rightarrow \infty$,

$$\begin{aligned}&{}_2F_2\left( \frac{1}{2},1;1-z,\frac{1}{2}-z;N\pi ^2w^2\right) \nonumber \\&\quad \sim \frac{\Gamma (1-z)\Gamma \left( \frac{1}{2}-z\right) }{\Gamma (1/2)}\left\{ \frac{e^{\pi i/2}}{\sqrt{N}\pi w}\sum _{k=0}^\infty \frac{(-1)^k\Gamma \left( \frac{1}{2}+k\right) \Gamma \left( \frac{1}{2}-k\right) }{k!\Gamma \left( \frac{1}{2}-z-k\right) \Gamma \left( -z-k\right) }\left( N\pi ^2w^2e^{-\pi i}\right) ^{-k}\right. \nonumber \\&\quad \left. +\frac{e^{\pi i}}{N\pi ^2 w^2}\sum _{k=0}^\infty \frac{(-1)^k\Gamma \left( -\frac{1}{2}-k\right) }{k!\Gamma \left( -z-k\right) \Gamma \left( -\frac{1}{2}-z-k\right) }\left( N\pi ^2w^2e^{-\pi i}\right) ^{-k}\right. \nonumber \\&\quad \left. +\left( N\pi ^2w^2\right) ^{2z}e^{N\pi ^2w^2}\sum _{k=0}^\infty C_k(N\pi ^2w^2)^{-k}\right\} , \end{aligned}$$

(2.14)

and

$$\begin{aligned}&{}_1F_1\left( z+\frac{1}{2};\frac{1}{2};N\pi ^2w^2\right) \nonumber \\&\quad \sim \frac{\Gamma (1/2)}{\Gamma \left( z+\frac{1}{2}\right) }\left\{ (N\pi ^2w^2e^{-\pi i})^{-(z+1/2)}\sum _{k=0}^\infty \frac{(-1)^k\Gamma \left( \frac{1}{2}+z+k\right) }{k!\Gamma (k-z)}(N\pi ^2w^2e^{-\pi i})^{-k}\right. \nonumber \\&\qquad \left. +(N\pi ^2w^2)^ze^{N\pi ^2w^2}\sum _{k=0}^\infty C_k(N\pi ^2w^2)^{-k}\right\} , \end{aligned}$$

(2.15)

where

$$\begin{aligned} C_k=-\frac{1}{k}\sum _{m=0}^{k-1}C_me_{k,m}, \end{aligned}$$

with $C_0=1$ and

$$\begin{aligned} e_{k,m}=2(m-z)_{(k+1-m)}\left( z-\frac{1}{2}\right) -2z\left( \frac{1}{2}-z+m\right) _{(k+1-m)}. \end{aligned}$$

Upon simplifying (2.10), (2.14) and (2.15), and observing that the terms containing $e^{N\pi ^2w^2}$ cancel each other out, for $\frac{\pi }{4}\le \mathrm {\arg }(w)<\frac{\pi }{2}$,

$$\begin{aligned} I(w,z,N)=\frac{e^{\pi i/2}}{2w^{2z+1}}\left( e^{\pi iz}-\cos (\pi z)\right) +O\left( \frac{1}{\sqrt{N}}\right) \end{aligned}$$

as $N\rightarrow \infty$. Employing the dominated convergence theorem to take limit $N\rightarrow \infty$ inside the double integral, we deduce that

$$\begin{aligned} \int _0^\infty \int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \cos (2\pi vt)\ dtdv&=-\frac{1}{2w^{2z+1}}\sin (\pi z). \end{aligned}$$

This along with (2.13) completes the proof of the theorem for $-\frac{\pi }{2}<\arg (w)<\frac{\pi }{2}$. $\square$

As discussed in the introduction, Guinand’s generalization of Poisson’s summation formula [23, Theorem 1] is critical to prove Theorem 1.7. We record Guinand’s result in the following theorem.

Theorem 2.2

If f(x) is an integral, f(x) tends to zero as $x\rightarrow \infty$, and $xf'(x)$ belongs to $L^p(0,\infty )$, for some p, $1<p\le 2$, then

$$\begin{aligned} \lim _{M\rightarrow \infty }\left( \sum _{m=1}^M f(m)-\int _0^M f(v)\, \textrm{d}v\right) =\lim _{M\rightarrow \infty }\left( \sum _{m=1}^M g(m)-\int _0^M g(v)\, \textrm{d}v\right) , \end{aligned}$$

where

$$\begin{aligned} g( x )=2\int _0^{\infty }f(t)\cos (2\pi x t)\, \textrm{d}t. \end{aligned}$$

Proof of Theorem 1.7

Let

$$\begin{aligned} f(v)&:=\frac{1}{(v-iw)^{2z+1}}+\frac{1}{(v+iw)^{2z+1}},\quad \left( \textrm{Re}(w)>0, \textrm{Re}(z)>0\right) ,\nonumber \\ g(x)&:=2\int _0^\infty \left( \frac{1}{(v-iw)^{2z+1}}+\frac{1}{(v+iw)^{2z+1}}\right) \cos (2\pi xv)\ dv. \end{aligned}$$

(2.16)

Now employ Theorem 2.2 with f(x) and g(x) as above. It is easy to see that f satisfies the hypotheses of Theorem 2.2. Invoking Theorem 1.9, we see that

$$\begin{aligned} \sum _{n=1}^\infty g(n)&=\lim _{M\rightarrow \infty }\Bigg \{\sum _{n=1}^M\left( \frac{1}{(n-iw)^{2z+1}}+\frac{1}{(n+iw)^{2z+1}}\right) \\&\quad -\int _0^M\left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \ dt\Bigg \}-\frac{1}{w^{2z+1}}\sin (\pi z). \end{aligned}$$

Note that series and integral on the right-hand side of the above equation exist individually in the limit $M\rightarrow \infty$. Therefore,

$$\begin{aligned}&\sum _{n=1}^\infty g(n)=\sum _{n=1}^\infty \left( \frac{1}{(n-iw)^{2z+1}}+\frac{1}{(n+iw)^{2z+1}}\right) \nonumber \\&\quad -\int _0^\infty \left( \frac{1}{(t-iw)^{2z+1}}+\frac{1}{(t+iw)^{2z+1}}\right) \ dt-\frac{1}{w^{2z+1}}\sin (\pi z). \end{aligned}$$

(2.17)

It is easy to see that for Re$(z)>0$,

$$\begin{aligned} \sum _{n=1}^\infty \frac{1}{(n\mp iw)^{2z+1}}=\zeta (1+2z,1\mp iw). \end{aligned}$$

(2.18)

Also,

$$\begin{aligned} \int _0^\infty \frac{dt}{(t\mp iw)^{2z+1}}=\frac{(\mp i)^{-2z}w^{-2z}}{2z}. \end{aligned}$$

(2.19)

Substitute (2.18) and (2.19) in (2.17) to deduce that

$$\begin{aligned} \sum _{n=1}^\infty g(n)&=\zeta (1+2z,1+iw)+\zeta (1+2z,1-iw)-\frac{\cos (\pi z)}{zw^{2z}}-\frac{1}{w^{2z+1}}\sin (\pi z)\nonumber \\&=\zeta (1+2z,iw)+\zeta (1+2z,-iw)-\frac{\cos (\pi z)}{zw^{2z}}, \end{aligned}$$

(2.20)

which follows using the fact

$$\begin{aligned} \zeta (s,a+1)=\zeta (s,a)-a^{-s}. \end{aligned}$$

(2.21)

Therefore, (2.16) and (2.20) yield Theorem 1.7. $\square$

Proof of Corollary 1.8

We wish to take limit $z\rightarrow 0$ in (1.22). To that end, we use expansions of the functions involved around $z=0$. As $s\rightarrow 1$, we have [22, p. 1038, Formula 9.533.2]

$$\begin{aligned} \zeta (s,a)=\frac{1}{s-1}-\psi (a)+O(|s-1|). \end{aligned}$$

The above equation implies that, as $z\rightarrow 0$,

$$\begin{aligned} \zeta (1+2z,\pm iw)&=\frac{1}{2z}-\psi (\pm iw)+O(|z|). \end{aligned}$$

(2.22)

It is easy to see that

$$\begin{aligned} \frac{\cos (\pi z)}{zw^{2z}}=\frac{1}{z}-2\log (w)+O(|z|), \end{aligned}$$

(2.23)

as $z\rightarrow 0$. Using (2.22) and (2.23), we deduce that

$$\begin{aligned} \lim _{z\rightarrow 0}\left( \zeta (1+2z,iw)\!+\!\zeta (1+2z,-iw)\!-\!\frac{\cos (\pi z)}{zw^{2z}}\right)&=-\!\psi (iw)-\psi (-iw)\!+\!2\log (w). \end{aligned}$$

(2.24)

Let $z\rightarrow 0$ on both sides of (1.22) and use (2.24) so that

$$\begin{aligned} 2\sum _{n=1}^\infty \int _0^\infty \frac{v\cos (2\pi n v)}{v^2+w^2}\ dv =\frac{1}{2}\left\{ 2\log (w)-\left( \psi (iw)+\psi (-iw)\right) \right\} . \end{aligned}$$

(2.25)

Make the change of variable $2\pi nv=t$ on the left-hand side of (2.25) to arrive at

$$\begin{aligned} 2\sum _{n=1}^\infty \int _0^\infty \frac{t\cos (t)}{t^2+(2\pi w)^2n^2}\ dt=\log (w)-\frac{1}{2}\left( \psi (iw)+\psi (-iw)\right) . \end{aligned}$$

Finally let $w=u/(2\pi )$ in the above equation to conclude the proof of the corollary. $\square$

Theorem 1.10 is proved next.

Proof of Theorem 1.10

From [16, Lemma 9.1], for Re$(w)>0$, we have

$$\begin{aligned} \int _{0}^{\infty }\frac{t\cos t\, dt}{t^2+w^2}=\sinh (w)\textrm{Shi}(w)-\cosh (w)\textrm{Chi}(w). \end{aligned}$$

(2.26)

Now (1.24) follows by expanding $\frac{t}{(t^2+w^2)}$ in partial fractions, that is, by writing $\frac{t}{t^2+w^2}=\frac{1}{2}\left( \frac{1}{t-iw}+\frac{1}{t+iw}\right)$, and then by performing integration by parts 2m times the left-hand side of (2.26). $\square$

3 Proof of Our Generalization of a Formula of Ramanujan

We begin with the following result of Hardy [27, pp. 56–57]. This result helps us justify the interchange of the order of the summation and integration having principal values, and will be employed in the proof of Theorem 1.1.

Proposition 3.1

Let

$$\begin{aligned} S(x)=\sum _{k=0}^\infty u_k(x) \end{aligned}$$

be a series whose terms are functions of x and is convergent with the possible exception of a closed enumerable set of points for values of x in a finite interval (a, A). Let $\alpha$ denote one such point in this set. If

• (1) the series S(x) is integrable term by term over any part of (a, A) which does not include $\alpha$,

• (2) the function

  $$\begin{aligned} F(x)=\sum _{k=0}^\infty \textrm{PV}\int _a^x u_k(t) dt \end{aligned}$$

  is a continuous function of x except at $\alpha$, and

• (3)
  $$\begin{aligned} \lim _{\epsilon \rightarrow 0}\left\{ F(\alpha -\epsilon )-F(\alpha +\epsilon )\right\} =0. \end{aligned}$$

  (3.1)

Then, one can interchange the order of summation and integration, namely,

$$\begin{aligned} \textrm{PV}\int _a^A\sum _{k=0}^\infty u_k(t) dt=\sum _{k=0}^\infty \textrm{PV}\int _a^Au_k(t)dt. \end{aligned}$$

Remark 2

We note that [27, pp. 58–59, Section 7] (also see [26, p. 27]) if

$$\begin{aligned} u_k(x)=\frac{v_k(x)}{x-\alpha }, \end{aligned}$$

(3.2)

where $v_k(x)$ is a function of x and has a continuous derivative for all $x\in [a,A]$, then

$$\begin{aligned} \textrm{PV}\int _{\alpha -\epsilon }^{\alpha +\epsilon }u_k(x)dx=2\epsilon v_k'(\alpha +\mu ),\ \mathrm {for\ some}\ \mu \in [-\epsilon ,\epsilon ]. \end{aligned}$$

Also if $|v_k'(x)|<V_k$ for all values $x\in [a,A]$, $V_k$ being independent of x and $\sum _{k=0}^\infty V_k$ is convergent, then the condition (3.1) holds true for $u_k(x)$ given in (3.2).

In the next lemma, we justify the interchange of the order of the summation and principal value integral.

Lemma 3.2

Let $k\in {\mathbb {N}}$, $0<a\le 1$ and $\textrm{Re}(y)>0$. For $\textrm{Re}(s)>2$, we have

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s-1}e^{-4\pi ^2 kx/y}\cot (\pi x)dx\nonumber \\&\quad = \textrm{PV}\int _0^\infty \left( \sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 kx/y}\right) x^{s-1}\cot (\pi x)dx. \end{aligned}$$

(3.3)

Proof

Note that the presence of $\cot (\pi x)$ implies infinitely many singularities of the integrand on the left side of (3.3). To handle this integral efficiently, we use

$$\begin{aligned} \pi \cot (\pi x)=\frac{1}{x}+\sum _{n=1}^\infty \frac{2x}{x^2-n^2}, \end{aligned}$$

(3.4)

so that

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s-1}e^{-4\pi ^2 kx/y}\cot (\pi x)dx\nonumber \\&\quad =\frac{1}{\pi }\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \int _0^\infty x^{s-2}e^{-4\pi ^2 kx/y}dx\nonumber \\&\quad +\frac{2}{\pi }\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx. \end{aligned}$$

(3.5)

We can interchange the order of summation and integration in the first expression on the right-hand side of (3.5) by easily employing [47, p. 30, Theorem 2.1]. The delicate part is to show the same for the second expression on the right, which is done next. We first show that

$$\begin{aligned} \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx=\sum _{n=1}^\infty \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx. \end{aligned}$$

(3.6)

The ingenious argument given in [8, pp. 909–911] can be adapted here as well to prove the above claim. We give the complete details though to make the paper self-contained.

Let $w(t)\in C_0^\infty$ be a smooth function such that $0\le w(t)\le 1$, $\forall \ t\in {\mathbb {R}}$, w(t) has compact support in $\left( -\frac{1}{3},\frac{1}{3}\right)$, and $w(t)=1,\ t\in \left( -\frac{1}{4},\frac{1}{4}\right)$. Note that the right-hand side of (3.6) can be rewritten as

$$\begin{aligned}&\sum _{n=1}^\infty \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx=\sum _{n=1}^\infty \int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\frac{(1-w(x-n))}{x^2-n^2} dx\nonumber \\&\quad +\sum _{n=1}^\infty \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\frac{w(x-n)}{x^2-n^2} dx. \end{aligned}$$

(3.7)

Again, an easy application of [47, p. 30, Theorem 2.1] allows us to interchange the order of summation and integration in the first expression of (3.7). If m is a positive integer such that $m-\frac{1}{2}\le x\le m+\frac{1}{2}$, then

$$\begin{aligned} \sum _{n=1}^\infty \frac{w(x-n)}{x^2-n^2}=\frac{w(x-m)}{x^2-m^2}. \end{aligned}$$

(3.8)

Hence, using (3.8) in the second step below, we have

$$\begin{aligned}&\textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{w(x-n)}{x^2-n^2} dx \\&\quad =\sum _{m=1}^\infty \textrm{PV}\int _{m-1/2}^{m+1/2}x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{w(x-n)}{x^2-n^2} dx \\&\quad =\sum _{m=1}^\infty \textrm{PV}\int _{m-1/2}^{m+1/2}x^{s}e^{-4\pi ^2 kx/y}\frac{w(x-m)}{x^2-m^2} dx \\&\quad =\sum _{m=1}^\infty \textrm{PV}\int _{0}^{\infty }x^{s}e^{-4\pi ^2 kx/y}\frac{w(x-m)}{x^2-m^2} dx. \end{aligned}$$

The above fact along with (3.7) gives

$$\begin{aligned}&\sum _{n=1}^\infty \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx\\&\quad =\textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{(1-w(x-n))}{x^2-n^2} dx+\textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{w(x-n)}{x^2-n^2} dx\\&\quad =\textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx. \end{aligned}$$

This proves the claim in (3.6). Therefore, we can write

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx\\&\quad =\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \sum _{n=1}^\infty \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx. \end{aligned}$$

Fubini’s theorem allows us to interchange the order of the double sum on the right-hand side of the above expression so as to obtain

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx\nonumber \\&\quad =\sum _{n=1}^\infty \sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx. \end{aligned}$$

(3.9)

Now

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx\nonumber \\&\quad =\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \left\{ \left( \int _{0}^\delta +\int _{n+1}^\infty \right) \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx\right. \nonumber \\&\qquad \left. +\frac{1}{2}\int _\delta ^{n+1} \frac{x^{s-1}e^{-4\pi ^2 kx/y}}{x+n} dx+\frac{1}{2}\textrm{PV}\int _\delta ^{n+1} \frac{x^{s-1}e^{-4\pi ^2 kx/y}}{x-n} dx\right\} , \end{aligned}$$

(3.10)

where $0<\delta <1$. Note that there is no need to take the principal value for the first three integrals on the right-hand side of (3.10). Therefore, it is easy to take the summation inside these integrals using the standard techniques, for example, [47, p. 30, theorem 2.1]. To interchange the order of summation and the last integral in (3.10), we now show that the hypotheses of Proposition 3.1 are satisfied. Let us define

$$\begin{aligned} u_k(x):=\frac{v_k(x)}{x-n}\quad \textrm{and}\quad v_k(x):=x^{s-1}e^{-4\pi ^2 kx/y}\sin \left( 2\pi ak\right) . \end{aligned}$$

(3.11)

It is easy to see that the conditions (1) and (2) of Proposition 3.1 are satisfied with $u_k(x)$ being defined in (3.11). To fulfill (3.1), we show that the equivalent condition discussed in Remark 2 is satisfied. To that end, observe that $x\in [\delta ,n+1]$ and use $e^{-x}<3!/x^3, x>0$, so that

$$\begin{aligned} |v_k'(x)|&<\left| x^{s-2}e^{-4\pi ^2 kx/y}\left( s-1-\frac{4\pi ^2kx}{y}\right) \right|<\frac{x^{\textrm{Re}(s)-5}}{(4\pi ^2/y)^3}\frac{3!}{k^3}\left( |s-1|+\frac{4\pi ^2kx}{y}\right) \\&<\frac{M}{(4\pi ^2/y)^3}\frac{3!}{k^3}\left( |s-1|+\frac{4\pi ^2k(n+1)}{y}\right) \\&=:V_k, \end{aligned}$$

where we used the fact that the function $x^{\textrm{Re}(s)-5}$ is continuous on the compact interval $[\delta ,n+1]$, and hence bounded by some constant $M>0$ (which may depend on $\delta$ and n). Since the series $\sum _{k=1}^\infty V_k$ converges, all conditions of Proposition 3.1 are satisfied. Hence we can interchange the order of summation and integration even in the case of the last integral of (3.10). This fact along with the discussion following (3.10) implies that

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx\nonumber \\&\quad =\textrm{PV}\int _0^\infty \left( \sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 kx/y}\right) \frac{x^{s}}{x^2-n^2} dx. \end{aligned}$$

(3.12)

Using the fact $\sin (\theta )=(e^{i\theta }-e^{-i\theta })/(2i)$, we find

$$\begin{aligned} \sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 xk/y}&=\frac{1}{2i}\sum _{k=1}^\infty e^{-\left( \frac{4\pi ^2 x}{y}-2\pi ia\right) k}-\frac{1}{2i}\sum _{k=1}^\infty e^{-\left( \frac{4\pi ^2 x}{y}+2\pi ia\right) k}\nonumber \\&=\frac{1}{2i}\left( \frac{1}{e^{\frac{4\pi ^2 x}{y}-2\pi ia}-1}-\frac{1}{e^{\frac{4\pi ^2 x}{y}+2\pi ia}-1}\right) . \end{aligned}$$

(3.13)

Substitute the above value in (3.12) to arrive at

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty \frac{x^{s}e^{-4\pi ^2 kx/y}}{x^2-n^2} dx\nonumber \\&\quad =\frac{1}{2i}\textrm{PV}\int _0^\infty \left( \frac{1}{e^{\frac{4\pi ^2 x}{y}-2\pi ia}-1}-\frac{1}{e^{\frac{4\pi ^2 x}{y}+2\pi ia}-1}\right) \frac{x^{s}}{x^2-n^2} dx. \end{aligned}$$

(3.14)

Equations (3.9) and (3.14) yield

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx\nonumber \\&=\frac{1}{2i}\sum _{n=1}^\infty \textrm{PV}\int _0^\infty \left( \frac{1}{e^{\frac{4\pi ^2 x}{y}-2\pi ia}-1}-\frac{1}{e^{\frac{4\pi ^2 x}{y}+2\pi ia}-1}\right) \frac{x^{s}}{x^2-n^2} dx. \end{aligned}$$

Again employing the trick that we used after (3.6) to interchange the order of the summation and integration, one can take the sum over n inside the integral on the right-hand side of the above equation to deduce that

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s}e^{-4\pi ^2 kx/y}\sum _{n=1}^\infty \frac{1}{x^2-n^2} dx\nonumber \\&=\frac{1}{2i}\textrm{PV}\int _0^\infty \left( \frac{1}{e^{\frac{4\pi ^2 x}{y}-2\pi ia}-1}-\frac{1}{e^{\frac{4\pi ^2 x}{y}+2\pi ia}-1}\right) \sum _{n=1}^\infty \frac{1}{x^2-n^2} x^sdx. \end{aligned}$$

(3.15)

Substituting (3.15) in (3.5), we obtain

$$\begin{aligned}&\sum _{k=1}^\infty \sin \left( 2\pi ak\right) \textrm{PV}\int _0^\infty x^{s-1}e^{-4\pi ^2 kx/y}\cot (\pi x)dx\\&\quad =\frac{1}{\pi }\int _0^\infty x^{s-2}\sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 kx/y}dx\\&\qquad +\frac{1}{i\pi }\textrm{PV}\int _0^\infty \left( \frac{1}{e^{\frac{4\pi ^2 x}{y}-2\pi ia}-1}-\frac{1}{e^{\frac{4\pi ^2 x}{y}+2\pi ia}-1}\right) \sum _{n=1}^\infty \frac{1}{x^2-n^2} x^sdx\\&\quad =\frac{1}{\pi }\int _0^\infty \left( \sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 kx/y}\right) x^{s-2}dx\\&\qquad +\frac{2}{\pi }\textrm{PV} \int _0^\infty \left( \sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-4\pi ^2 kx/y}\right) \sum _{n=1}^\infty \frac{1}{x^2-n^2} x^sdx, \end{aligned}$$

where in the ultimate step we again used (3.13). Finally employing (3.4) in the above equation, we arrive at (3.3). $\square$

We have now collected all ingredients to give a proof of our generalization of Ramanujan’s formula.

Proof of Theorem 1.1

Letting $\tau = iyj/(2\pi )$, Re$(y)>0$, in (1.18), then taking summation over $j\ge 1$, and then employing the series definition of the Hurwitz zeta function for Re$(s)>1$, we obtain.²

$$\begin{aligned} \sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}&=\frac{\Gamma (s)}{(-2\pi i)^s}\sum _{k\in {\mathbb {Z}}}e^{2\pi iak}\sum _{j=1}^\infty \frac{1}{\left( k+\frac{ijy}{2\pi }\right) ^s}\nonumber \\&=\frac{\Gamma (s)}{y^s}\sum _{k\in {\mathbb {Z}}}e^{2\pi iak}\zeta \left( s,1-\frac{2\pi ik}{y}\right) \nonumber \\&=\frac{\Gamma (s)\zeta (s)}{y^s}+\frac{\Gamma (s)}{y^s}\nonumber \\&\times \sum _{k=1}^\infty \left\{ e^{2\pi iak}\zeta \left( s,1-\frac{2\pi i k}{y}\right) +e^{-2\pi iak}\zeta \left( s,1+\frac{2\pi i k}{y}\right) \right\} . \end{aligned}$$

(3.16)

Invoking the well-known formula [38, p. 609, Formula 25.11.25]

$$\begin{aligned} \Gamma (z)\zeta (z,a)=\int _0^\infty \frac{e^{-ax}}{1-e^{-x}}x^{z-1}dx,\qquad (\textrm{Re}(z)>1,\ \textrm{Re}(a)>0) \end{aligned}$$

(3.17)

in (3.16), we obtain

$$\begin{aligned} \sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}&=\frac{\Gamma (s)\zeta (s)}{y^s}+\frac{1}{y^s}\sum _{k=1}^\infty \int _0^\infty \left( e^{i\left( 2\pi ak+\frac{2\pi kt}{y}\right) }+e^{-i\left( 2\pi ak+\frac{2\pi kt}{y}\right) }\right) \frac{t^{s-1}}{e^{t}-1}dt\nonumber \\&=\frac{\Gamma (s)\zeta (s)}{y^s}+\frac{2}{y^s}\sum _{k=1}^\infty \int _0^\infty \cos \left( 2\pi a k+\frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^{t}-1}dt\nonumber \\&=\frac{\Gamma (s)\zeta (s)}{y^s}+\frac{2}{y^s}\sum _{k=1}^\infty \cos (2\pi a k)\int _0^\infty \cos \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^{t}-1}dt\nonumber \\&\hspace{1.8cm}-\frac{2}{y^s}\sum _{k=1}^\infty \sin (2\pi a k)\int _0^\infty \sin \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^{t}-1}dt. \end{aligned}$$

(3.18)

Our next goal is to evaluate the integrals in (3.18). From [36, p. 42, Formula 1.5.2], for $0<\textrm{Re}(z)<1$, we have

$$\begin{aligned} \int _0^\infty \cos (x)x^{z-1}dx=\Gamma (z)\cos \left( \frac{\pi z}{2}\right) . \end{aligned}$$

Making the change of variable $x=\frac{2\pi k t}{y}$ and replacing z by $s-1+z$ in the above result, we get, for $1-\textrm{Re}(s)<\textrm{Re}(z)<2-\textrm{Re}(s)$,

$$\begin{aligned} \int _0^\infty \cos \left( \frac{2\pi kt}{y}\right) t^{s-1}t^{z-1}dt=\left( \frac{2\pi k}{y}\right) ^{1-s-z}\Gamma (s-1+z)\sin \left( \frac{\pi }{2}(s+z)\right) . \end{aligned}$$

(3.19)

Equation (3.17) with $a=1$, (3.19), and an application of Parseval’s formula [39, p. 83, Equation (3.1.14)] gives, for $1-\textrm{Re}(s)<c=\textrm{Re}(z)<min \left( 0, 2-\textrm{Re}(s)\right)$,

$$\begin{aligned}&\int _0^\infty \cos \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^t-1}dt =\left( \frac{2\pi k}{y}\right) ^{1-s}\frac{1}{2\pi i}\nonumber \\&\qquad \times \int _{(c)} \Gamma (s-1+z)\sin \left( \frac{\pi }{2}(s+z)\right) \Gamma (1-z)\zeta (1-z)\left( \frac{2\pi k}{y}\right) ^{-z}dz\nonumber \\&\quad =\left( \frac{2\pi k}{y}\right) ^{1-s}\left\{ \cos \left( \frac{\pi s}{2}\right) I_1(y,s)+\sin \left( \frac{\pi s}{2}\right) I_2(y,s)\right\} , \end{aligned}$$

(3.20)

where

$$\begin{aligned} I_1(y,s)&:=\frac{1}{2\pi i}\int _{(c)}\Gamma (s-1+z)\sin \left( \frac{\pi z}{2}\right) \Gamma (1-z)\zeta (1-z)\left( \frac{2\pi k}{y}\right) ^{-z}dz,\\ I_2(y,s)&:=\frac{1}{2\pi i}\int _{(c)}\Gamma (s-1+z)\cos \left( \frac{\pi z}{2}\right) \Gamma (1-z)\zeta (1-z)\left( \frac{2\pi k}{y}\right) ^{-z}dz. \end{aligned}$$

(3.21)

Similarly, using the formula [36, p. 42, Formula 1.5.1]

$$\begin{aligned} \int _0^\infty \sin (x)x^{z-1}dx=\Gamma (z)\sin \left( \frac{\pi z}{2}\right) ,\quad (-1<\textrm{Re}(z)<1), \end{aligned}$$

it can be seen that for $-\textrm{Re}(s)<c=\textrm{Re}(z)<min \left( 0, 2-\textrm{Re}(s)\right)$,

$$\begin{aligned} \int _0^\infty \sin \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^t-1}dt =\left( \frac{2\pi k}{y}\right) ^{1-s}\left\{ \sin \left( \frac{\pi s}{2}\right) I_1(y,s)-\cos \left( \frac{\pi s}{2}\right) I_2(y,s)\right\} . \end{aligned}$$

(3.22)

We first evaluate $I_1(y,s)$. Apply the functional equation of the Riemann zeta function [38, p. 603, Formula 25.4.2]

$$\begin{aligned} \zeta (s)=2^{s}\pi ^{s-1}\Gamma (1-s)\zeta (1-s)\sin \left( \frac{\pi s}{2}\right) , \end{aligned}$$

(3.23)

in (3.21) to see that

$$\begin{aligned} I_1(y,s)=\frac{\pi }{2\pi i}\int _{(c)}\Gamma (s-1+z)\zeta (z)\left( \frac{4\pi ^2 k}{y}\right) ^{-z}dz.\nonumber \end{aligned}$$

(3.24)

We want to use the series definition of $\zeta (z)$ to further simplify the above integral. Therefore we shift the line of integration to $d=\textrm{Re}(z)>1$ and use residue theorem thereby obtaining

$$\begin{aligned} I_1(y,s)&=\frac{\pi }{2\pi i}\int _{(d)}\Gamma (s-1+z)\zeta (z)\left( \frac{4\pi ^2 k}{y}\right) ^{-z}dz-\frac{y\Gamma (s)}{4\pi k}\nonumber \\&=\pi \sum _{n=1}^\infty \frac{1}{2\pi i}\int _{(d)}\Gamma (s-1+z)\left( \frac{4\pi ^2 nk}{y}\right) ^{-z}dz-\frac{y\Gamma (s)}{4\pi k}\nonumber \\&=\pi \left( \frac{4\pi ^2 k}{y}\right) ^{s-1}\sum _{n=1}^\infty n^{s-1}e^{-\frac{4\pi ^2 nk}{y}}-\frac{y\Gamma (s)}{4\pi k}, \end{aligned}$$

(3.25)

where in the last step, we used

$$\begin{aligned} e^{-x}=\frac{1}{2\pi i}\int _{(\lambda )}\Gamma (z) x^{-z}dz\hspace{8mm}(\lambda >0). \end{aligned}$$

(3.26)

We now focus on representing the other integral $I_2(y,s)$ in terms of an equivalent integral; see (3.36) below. Again an application of (3.24) in (3.22) yields

$$\begin{aligned} I_2(y,s)=\frac{\pi }{2\pi i}\int _{(c)}\Gamma (s-1+z)\zeta (z)\cot \left( \frac{\pi z}{2}\right) \left( \frac{4\pi ^2 k}{y}\right) ^{-z}dz. \end{aligned}$$

(3.27)

If we shift the line of integration from Re$(z)=c$, where $1-\textrm{Re}(s)<c<\min (0,2-\textrm{Re}(s))$, to $1<\textrm{Re}(z)=d<2$, we encounter a simple pole at $z=0$ of the integrand in the integral of (3.27) due to $\cot (\pi z/2)$. (Note that the pole of $\zeta (z)$ at $z=1$ is annihilated by the zero of $\cot (\pi z/2)$ at $z=1$.) Note that the integrals along the horizontal segments vanish using Stirling’s formula in the vertical strip $p\le \sigma \le q$ [11, p. 224]:

$$\begin{aligned} |\Gamma (s)|=\sqrt{2\pi }|t|^{\sigma -\frac{1}{2}}e^{-\frac{1}{2}\pi |t|}\left( 1+O\left( \frac{1}{|t|}\right) \right) \end{aligned}$$

(3.28)

as $|t|\rightarrow \infty$. Therefore, by the residue theorem and (3.27), we have

$$\begin{aligned} I_2(y,s)=\Gamma (s-1)+\frac{\pi }{2\pi i}\int _{(d)}\Gamma (s-1+z)\zeta (z)\cot \left( \frac{\pi z}{2}\right) \left( \frac{4\pi ^2 k}{y}\right) ^{-z}dz. \end{aligned}$$

(3.29)

Note that we can use the series definition of $\zeta (z)$ in (3.29) and then interchange the order of the summation and integration so as to obtain

$$\begin{aligned} I_2(y,s)&=\Gamma (s-1)+\pi \sum _{n=1}^\infty \frac{1}{2\pi i}\int _{(d)}\Gamma (s-1+z)\cot \left( \frac{\pi z}{2}\right) \left( \frac{4\pi ^2 nk}{y}\right) ^{-z}dz. \end{aligned}$$

(3.30)

From [36, p. 182, Formula 2.4.4], for $-1<c_2<1$, we have

$$\begin{aligned} \frac{1}{2\pi i}\int _{(c_2)}\tan \left( \frac{\pi z}{2}\right) x^{-z}dz=\frac{2}{\pi }\frac{x}{x^2-1},\quad (x\ne \pm 1). \end{aligned}$$

Replacing x by x/n in the above result gives

$$\begin{aligned} \frac{1}{2\pi i}\int _{(c_2)}\tan \left( \frac{\pi z}{2}\right) n^{z-1}x^{-z}dz=\frac{2}{\pi }\frac{x}{x^2-n^2},\quad (x\ne \pm n). \end{aligned}$$

(3.31)

For $c_1>1-\textrm{Re}(s)$, equation (3.26) implies that

$$\begin{aligned} \frac{1}{2\pi i}\int _{(c_1)}\Gamma (s-1+z)\left( \frac{4\pi ^2k}{y}\right) ^{-z}x^{-z}dz=e^{-\frac{4\pi ^2 xk}{y}}\left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}. \end{aligned}$$

(3.32)

We next want to invoke Parseval’s formula [39, p. 83, Equation (3.1.11)] for the functions in (3.31) and (3.32). For that we need to justify the following³

$$\begin{aligned}&\frac{1}{2\pi i}\int _{(d)}\textrm{PV}\int _0^\infty \Gamma (s-1+z)\frac{x^{z-1}}{1-n^2x^2}dxdz\nonumber \\&\quad =\textrm{PV}\int _0^\infty \frac{1}{2\pi i}\int _{(d)}\Gamma (s-1+z)\frac{x^{z-1}}{1-n^2x^2}dzdx, \end{aligned}$$

(3.33)

where $d>0$. Making the change of variable $z=d+it$, we see that

$$\begin{aligned}&\frac{1}{2\pi i}\int _{(d)}\textrm{PV}\int _0^\infty \Gamma (s-1+z)\frac{x^{z-1}}{1-n^2x^2}dxdz\\&\quad =\frac{1}{2\pi }\int _{-\infty }^\infty \textrm{PV}\int _0^\infty \Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt\\&\quad =\frac{1}{2\pi }\int _{-\infty }^\infty \textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt\\&\quad +\frac{1}{2\pi }\int _{-\infty }^\infty \int _{\frac{1}{n}+\epsilon }^\infty \Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt. \end{aligned}$$

Note that the inner integral in the second expression on the right-hand side is a usual improper integral. Therefore, we can interchange the order of the integration by standard methods [47, p. 30, Theorem 2.2]. To justify the same in the first double integral on the right, we proceed as follows. Observe that

$$\begin{aligned}&\frac{1}{2\pi }\int _{-\infty }^\infty \textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt\nonumber \\&\quad =\frac{1}{2\pi }\int _{0}^\infty \textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt\nonumber \\&\quad +\frac{1}{2\pi }\int _{0}^\infty \textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\Gamma (s-1+d-it)\frac{x^{d-it-1}}{1-n^2x^2}dxdt. \end{aligned}$$

(3.34)

We justify the interchange of the order of integration only for the first double integral. That for the second one can be similarly justified. To that end, for $B>0$,

$$\begin{aligned}&\int _{0}^B\textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dxdt\\&\quad =\textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\int _0^B\Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dtdx, \end{aligned}$$

using Hardy’s result [28, p. 94, Theorem 6]. Moreover,

$$\begin{aligned} \lim _{B\rightarrow \infty }\textrm{PV}\int _0^{\frac{1}{n}+\epsilon }\int _B^\infty \Gamma (s-1+d+it)\frac{x^{d+it-1}}{1-n^2x^2}dtdx=0, \end{aligned}$$

which follows from Stirling’s formula (3.28). This shows that the conditions mentioned in [28, p. 94, Section 18] are satisfied. Thus we can interchange the order of integration in the first double integral in (3.34). This finally proves the validity of (3.33).

Hence we can employ Parseval’s formula [39, p. 83, Equation (3.1.11)] for the functions in (3.31) and (3.32) which, for $0<c<2$, gives

$$\begin{aligned}&\frac{1}{2\pi i}\int _{(c)}\Gamma (s-1+z)\cot \left( \frac{\pi z}{2}\right) \left( \frac{4\pi ^2 nk}{y}\right) ^{-z}dz\nonumber \\&\quad =\frac{2}{\pi }\textrm{PV}\int _0^\infty e^{-\frac{4\pi ^2 xk}{y}}\left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}\frac{x}{x^2-n^2}dx. \end{aligned}$$

(3.35)

Substituting the value from (3.35) in (3.30) so that

$$\begin{aligned} I_2(y,s)=\Gamma (s-1)+\sum _{n=1}^\infty \textrm{PV}\int _0^\infty e^{-\frac{4\pi ^2 xk}{y}}\left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}\frac{2x}{x^2-n^2}dx. \end{aligned}$$

By appealing to (3.6) we can take the sum inside the integral in the above equation. Also note that $\Gamma (s-1)=\int _0^\infty e^{-\frac{4\pi ^2 xk}{y}}\left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}\frac{dx}{x}$ for Re$(s)>2, k>0$. Hence

$$\begin{aligned} I_2(y,s)&=\textrm{PV}\int _0^\infty e^{-\frac{4\pi ^2 xk}{y}}\left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}\left\{ \frac{1}{x}+ \sum _{n=1}^\infty \frac{2x}{x^2-n^2}\right\} dx\nonumber \\&=\pi \ \textrm{PV}\int _0^\infty \left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}e^{-\frac{4\pi ^2 xk}{y}}\cot (\pi x)dx, \end{aligned}$$

(3.36)

which follows upon using (3.4). The existence of the principal value integral appearing on the right-hand side of (3.36) is shown by Hardy [26, p. 31]. Substituting (3.25) and (3.36) in (3.20) as well as in (3.23), we get

$$\begin{aligned}&\int _0^\infty \cos \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^t-1}dt\nonumber \\&\quad =\left( \frac{2\pi k}{y}\right) ^{1-s}\left\{ \cos \left( \frac{\pi s}{2}\right) \left( \pi \left( \frac{4\pi ^2 k}{y}\right) ^{s-1}\sum _{n=1}^\infty n^{s-1}e^{-\frac{4\pi ^2 nk}{y}}-\frac{y\Gamma (s)}{4\pi k}\right) \right. \nonumber \\&\qquad \left. +\sin \left( \frac{\pi s}{2}\right) \pi \ \textrm{PV}\int _0^\infty \left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}e^{-\frac{4\pi ^2 xk}{y}}\cot (\pi x)dx\right\} , \end{aligned}$$

(3.37)

and

$$\begin{aligned}&\int _0^\infty \sin \left( \frac{2\pi kt}{y}\right) \frac{t^{s-1}}{e^t-1}dt\nonumber \\&\quad =\left( \frac{2\pi k}{y}\right) ^{1-s}\left\{ \sin \left( \frac{\pi s}{2}\right) \left( \pi \left( \frac{4\pi ^2 k}{y}\right) ^{s-1}\sum _{n=1}^\infty n^{s-1}e^{-\frac{4\pi ^2 nk}{y}}-\frac{y\Gamma (s)}{4\pi k}\right) \right. \nonumber \\&\quad \left. -\cos \left( \frac{\pi s}{2}\right) \pi \ \textrm{PV}\int _0^\infty \left( \frac{4\pi ^2 xk}{y}\right) ^{s-1}e^{-\frac{4\pi ^2 xk}{y}}\cot (\pi x)dx\right\} . \end{aligned}$$

(3.38)

Substituting (3.37) and (3.38) in (3.18) and simplifying, we are led to

$$\begin{aligned}&\sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}\nonumber \\&\quad =\frac{\Gamma (s)\zeta (s)}{y^s}+\left( \frac{2\pi }{y}\right) ^s\cos \left( \frac{\pi s}{2}\right) \sum _{n=1}^\infty n^{s-1}\nonumber \\&\quad \times \sum _{k=1}^\infty \cos \left( 2\pi ak\right) e^{-\frac{4\pi ^2 nk}{y}}-\frac{\Gamma (s)}{(2\pi )^{s}}\cos \left( \frac{\pi s}{2}\right) \sum _{k=1}^\infty \frac{\cos \left( 2\pi ak\right) }{k^s}\nonumber \\&\quad -\left( \frac{2\pi }{y}\right) ^s\sin \left( \frac{\pi s}{2}\right) \sum _{n=1}^\infty n^{s-1}\sum _{k=1}^\infty \sin \left( 2\pi ak\right) e^{-\frac{4\pi ^2 nk}{y}}\nonumber \\&\quad +\frac{\Gamma (s)}{(2\pi )^{s}}\sin \left( \frac{\pi s}{2}\right) \sum _{k=1}^\infty \frac{\sin \left( 2\pi ak\right) }{k^s}\nonumber \\&\quad +\left( \frac{2\pi }{y}\right) ^s\sin \left( \frac{\pi s}{2}\right) \sum _{k=1}^\infty \cos (2\pi ak) \textrm{PV}\int _0^\infty x^{s-1}e^{-\frac{4\pi ^2 xk}{y}}\cot (\pi x)dx\nonumber \\&\quad +\left( \frac{2\pi }{y}\right) ^s\cos \left( \frac{\pi s}{2}\right) \sum _{k=1}^\infty \sin (2\pi ak) \textrm{PV}\int _0^\infty x^{s-1}e^{-\frac{4\pi ^2 xk}{y}}\cot (\pi x)dx. \end{aligned}$$

(3.39)

Invoking Lemma 3.2 we can interchange the summation and integration for the last expression on the right-hand side of (3.39). Also note that one can justify the same for the series involving $\cos (2\pi a k)$. Hence, after simplification, (3.39) becomes

$$\begin{aligned}&\sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}\\&\quad =\frac{\Gamma (s)\zeta (s)}{y^s}+\left( \frac{2\pi }{y}\right) ^s\sum _{n=1}^\infty n^{s-1}\sum _{k=1}^\infty \cos \left( \frac{\pi s}{2}+2\pi ak\right) e^{-4\pi ^2 nk/y}\\&\qquad -\frac{\Gamma (s)}{(2\pi )^{s}}\sum _{k=1}^\infty \frac{\cos \left( \frac{\pi s}{2}+2\pi ak\right) }{k^s}\\&\qquad +\left( \frac{2\pi }{y}\right) ^s \textrm{PV}\int _0^\infty \left( \sum _{k=1}^\infty \sin \left( \frac{\pi s}{2}+2\pi ak\right) e^{-4\pi ^2 kx/y}\right) x^{s-1}\cot (\pi x)dx. \end{aligned}$$

Making the change of variable $x\rightarrow xy/(2\pi )$ in the integral and rearranging the terms in the above expression, we get

$$\begin{aligned}&\frac{\Gamma (s)\zeta (s)}{y^s}+\left( \frac{2\pi }{y}\right) ^s\sum _{n=1}^\infty n^{s-1}\sum _{k=1}^\infty \cos \left( \frac{\pi s}{2}+2\pi ak\right) e^{-4\pi ^2 nk/y}\nonumber \\&\quad =\frac{\Gamma (s)}{(2\pi )^{s}}\sum _{k=1}^\infty \frac{\cos \left( \frac{\pi s}{2}+2\pi ak\right) }{k^s}+\sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}\nonumber \\&\qquad -\textrm{PV} \int _0^\infty \left( \sum _{k=1}^\infty \sin \left( \frac{\pi s}{2}+2\pi ak\right) e^{-2\pi kx}\right) x^{s-1} \cot \left( \frac{1}{2}y x\right) dx. \end{aligned}$$

(3.40)

Now, using the fact $\cos (\theta )=(e^{i\theta }+e^{-i\theta })/2$, we find

$$\begin{aligned}&\sum _{k=1}^\infty \cos \left( \frac{\pi s}{2}+2\pi ak\right) e^{-4\pi ^2 nk/y}\nonumber \\&\quad =\frac{1}{2}e^{\pi is/2}\sum _{k=1}^\infty e^{-\left( \frac{4\pi ^2 n}{y}-2\pi ia\right) k}+\frac{1}{2}e^{-\pi is/2}\sum _{k=1}^\infty e^{-\left( \frac{4\pi ^2 n}{y}+2\pi ia\right) k}\nonumber \\&\quad =\frac{1}{2}\left( \frac{e^{\pi is/2}}{e^{\frac{4\pi ^2 n}{y}-2\pi ia}-1}+\frac{e^{-\pi is/2}}{e^{\frac{4\pi ^2 n}{y}+2\pi ia}-1}\right) . \end{aligned}$$

(3.41)

Similarly,

$$\begin{aligned} \sum _{k=1}^\infty \sin \left( \frac{\pi s}{2}+2\pi ak\right) e^{-2\pi kx}=\frac{1}{2i}\left( \frac{e^{\pi is/2}}{e^{2\pi x-2\pi ia}-1}-\frac{e^{-\pi is/2}}{e^{2\pi x+2\pi ia}-1}\right) . \end{aligned}$$

(3.42)

Substituting (3.41) and (3.42) in (3.40), we get

$$\begin{aligned}&\frac{\Gamma (s)\zeta (s)}{y^s}+\left( \frac{2\pi }{y}\right) ^s\frac{1}{2}\sum _{n=1}^\infty n^{s-1}\left( \frac{e^{\pi is/2}}{e^{\frac{4\pi ^2 n}{y}-2\pi ia}-1}+\frac{e^{-\pi is/2}}{e^{\frac{4\pi ^2 n}{y}+2\pi ia}-1}\right) \nonumber \\&=\frac{\Gamma (s)}{(2\pi )^{s}}\sum _{k=1}^\infty \frac{\cos \left( \frac{\pi s}{2}+2\pi ak\right) }{k^s}+\sum _{n=1}^\infty \frac{(n-a)^{s-1}}{e^{(n-a)y}-1}\\&\quad -\frac{1}{2i}\textrm{PV}\int _0^\infty x^{s-1}\left( \frac{e^{\pi is/2}}{e^{2\pi x-2\pi ia}-1}-\frac{e^{-\pi is/2}}{e^{2\pi x+2\pi ia}-1}\right) \cot \left( \frac{1}{2}y x\right) dx. \end{aligned}$$

Finally, we arrive at (1.3) after multiplying by $\left( 2\pi /y\right) ^{-s}\alpha ^{s/2}$ on the both sides of the above equation and then letting $4\pi ^2/y=\alpha$ with $\alpha \beta =4\pi ^2$. $\square$

Proof of Corollary 1.2

Let $a=1/2$ and $s=2\,m, m\in {\mathbb {N}}$, $m>1$ in Theorem 1.1 and observe that the principal value integral vanishes. The result then follows upon using Euler’s formula [47, p. 5, Equation (1.14)]

$$\begin{aligned} \zeta (2 m ) = (-1)^{m +1} \frac{(2\pi )^{2 m}B_{2 m }}{2 (2 m)!}. \end{aligned}$$

(3.43)

$\square$

Proof of Corollary 1.3

Let $\alpha =\beta =2\pi$ and m be an odd positive integer in Corollary 1.2 so as to get

$$\begin{aligned} \sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\pi }+1}-\frac{1}{2^{2m-1}}\sum _{n=1}^\infty \frac{(2n-1)^{2m-1}}{e^{(2n-1)\pi }-1}=-2^{1-2m}\frac{B_{2m}}{4m}. \end{aligned}$$

(3.44)

Now use the fact

$$\begin{aligned} \sum _{n=1}^\infty \frac{(2n-1)^{2m-1}}{e^{(2n-1)\pi }-1}=\sum _{\begin{array}{c} n=1\\ n\hspace{0.5mm}odd \end{array}}^\infty \frac{n^{2m-1}}{e^{n\pi }-1} = \sum _{n=1}^\infty \frac{n^{2m-1}}{e^{n\pi }-1}-2^{2m-1}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\pi }-1}, \end{aligned}$$

(3.45)

and Glaisher’s evaluation [21]

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{n^{2m-1}}{e^{2n\pi }-1}=\frac{B_{2m}}{4m} \hspace{5mm}(m\hspace{1mm}odd >1) \end{aligned}$$

to arrive at (1.5).

As far as the proof of (1.6) is concerned, we let $m=1$ in (3.44) and use (3.45) and use Schlömilch’s result [46]

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{n}{e^{2n\pi }-1}=\frac{1}{24}-\frac{1}{8\pi }. \end{aligned}$$

$\square$

Proof of Corollary 1.4

Letting $s=2\,m$ and $a=1/4$ in Theorem 1.1 and simplifying, we get

$$\begin{aligned}&\alpha ^m\left\{ \frac{\Gamma (2m)\zeta (2m)}{(2\pi )^{2m}}+(-1)^{m+1}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\alpha }+1}\right\} \nonumber \\&\quad =\beta ^m\left\{ \frac{(-1)^m\Gamma (2m)}{(2\pi )^{2m}}\sum _{k=1}^\infty \frac{\cos (\pi k/2)}{k^{2m}}\right. \nonumber \\&\qquad \left. +\sum _{n=1}^\infty \frac{(n-1/4)^{2m-1}}{e^{(n-1/4)\beta }-1}-\frac{(-1)^m}{2}PV \int _0^\infty \frac{x^{2m-1}\cot (\beta x/2)}{\cosh (2\pi x)}dx\right\} . \end{aligned}$$

(3.46)

Now take $s=2\,m$ and $a=3/4$ in Theorem 1.1 to obtain

$$\begin{aligned}&\alpha ^m\left\{ \frac{\Gamma (2m)\zeta (2m)}{(2\pi )^{2m}}+(-1)^{m+1}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\alpha }+1}\right\} \nonumber \\&\quad =\beta ^m\left\{ \frac{(-1)^m\Gamma (2m)}{(2\pi )^{2m}}\sum _{k=1}^\infty \frac{(-1)^k\cos (\pi k/2)}{k^{2m}}\right. \nonumber \\&\qquad \left. +\sum _{n=1}^\infty \frac{(n-3/4)^{2m-1}}{e^{(n-3/4)\beta }-1}+\frac{(-1)^m}{2}PV \int _0^\infty \frac{x^{2m-1}\cot (\beta x/2)}{\cosh (2\pi x)}dx\right\} . \end{aligned}$$

(3.47)

Now add (3.46) and (3.47) so that

$$\begin{aligned}&\alpha ^m\left\{ 2\frac{\Gamma (2m)\zeta (2m)}{(2\pi )^{2m}}+2(-1)^{m+1}\sum _{n=1}^\infty \frac{n^{2m-1}}{e^{2n\alpha }+1}\right\} \\&\quad =\beta ^m\left\{ \frac{(-1)^m\Gamma (2m)}{(2\pi )^{2m}}\sum _{k=1}^\infty \frac{(1+(-1)^k)\cos (\pi k/2)}{k^{2m}}\right. \\&\qquad \left. +\sum _{n=1}^\infty \frac{(n-1/4)^{2m-1}}{e^{(n-1/4)\beta }-1}+\sum _{n=1}^\infty \frac{(n-3/4)^{2m-1}}{e^{(n-3/4)\beta }-1} \right\} . \end{aligned}$$

Using the fact $\sum _{n=1}^\infty (-1)^k/k^{2m}=2^{-2m}(2-2^{2m})\zeta (2m)$ in the above equation, we arrive at (1.7).

Next, subtracting (3.47) from (3.46) yields

$$\begin{aligned}&\frac{(-1)^m\Gamma (2m)}{(2\pi )^{2m}}\sum _{k=1}^\infty \frac{(1-(-1)^k)\cos (\pi k/2)}{k^{2m}}+\sum _{n=1}^\infty \frac{(n-1/4)^{2m-1}}{e^{(n-1/4)\beta }-1}-\sum _{n=1}^\infty \frac{(n-3/4)^{2m-1}}{e^{(n-3/4)\beta }-1} \\&=(-1)^mPV \int _0^\infty \frac{x^{2m-1}\cot (\beta x/2)}{\cosh (2\pi x)}dx. \end{aligned}$$

Note that the first sum on the left-hand side of the above equation vanishes. Now replacing $\beta$ by $4\beta$ and rewriting the left-hand side in terms of the Dirichlet character $\chi$ defined in (1.9), we are led to (1.8). $\square$

4 A Simpler Proof of the Transformation for $\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}$

In [16], this theorem was obtained for the first time as a corollary of a more general result, namely, (1.11). Hence the absolute convergence of the series on the right-hand side of (1.13) resulted automatically. In what follows, we not only give a direct proof of this result, but also prove from scratch the convergence of the series.

To that end, we first prove the identity for $y>0$ and later extend it to Re$(y)>0$ by analytic continuation. We begin by showing the absolute convergence of the series on the right-hand side of (1.13). Note that for $w>0$, (1.21) and Theorem 1.10 imply

$$\begin{aligned}&\sinh (w)\textrm{Shi}(w)-\cosh (w)\textrm{Chi}(w)+\sum _{j=1}^m(2j-1)!w^{-2j}=(-1)^m{\mathfrak {R}}_m(1,w)\nonumber \\&\quad =\frac{(-1)^m}{w^{2m}}{\mathfrak {R}}_m(w,1). \end{aligned}$$

(4.1)

Now employ Lemma 2.1 for ${\mathfrak {R}}_m(w,1)$, and then let $w=4\pi ^2n/y$, where $y>0$ (as assumed at the beginning of the proof), so that as $n\rightarrow \infty$, we have

$$\begin{aligned}{} & {} \sinh \left( \frac{4\pi ^2n}{y}\right) \textrm{Shi}\left( \frac{4\pi ^2n}{y}\right) -\cosh \left( \frac{4\pi ^2n}{y}\right) \textrm{Chi}\left( \frac{4\pi ^2n}{y}\right) \nonumber \\{} & {} \quad +\sum _{j=1}^m(2j-1)!\left( \frac{4\pi ^2n}{y}\right) ^{-2j}=O_{m, y}\left( \frac{1}{n^{2m+2}}\right) . \end{aligned}$$

(4.2)

The absolute convergence of $\sum _{n=1}^{\infty }\sigma _{2m}(n)/n^{2m+2}$ then implies that of the series on the right-hand side of (1.13) with the help of the above estimate.

We now prove (1.13). Let $a=0$ and $s=2\,m+1$ in (3.16) so that

$$\begin{aligned}&\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}=\frac{(2m)!}{y^{2m+1}}\left\{ \zeta (2m+1)+\sum _{k=1}^\infty \left( \zeta \left( 1+2m,1+\frac{2\pi ik}{y}\right) \right. \right. \\&\quad \left. \left. +\zeta \left( 1+2m,1-\frac{2\pi ik}{y}\right) \right) \right\} . \end{aligned}$$

Using (2.21), we can see that for $m\in {\mathbb {N}}$,

$$\begin{aligned}{} & {} \zeta \left( 1+2m,1+\frac{2\pi ik}{y}\right) +\zeta \left( 1+2m,1-\frac{2\pi ik}{y}\right) \\{} & {} \quad =\zeta \left( 1+2m,\frac{2\pi ik}{y}\right) +\zeta \left( 1+2m,-\frac{2\pi ik}{y}\right) . \end{aligned}$$

Now employ Theorem 1.7 with $z=m$ and $w=2\pi k/y$ in the above equation to see that

$$\begin{aligned}&\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}\\&\quad =\frac{(2m)!}{y^{2m+1}}\Bigg \{\zeta (2m+1)+\frac{\cos (\pi m)}{m}\left( \frac{2\pi }{y}\right) ^{-2m}\sum _{k=1}^\infty \frac{1}{k^{2m}}\\&\quad +2\sum _{k=1}^\infty \sum _{n=1}^\infty \int _0^\infty \left( \frac{1}{(v-2\pi ik/y)^{2m+1}}+\frac{1}{(v+2\pi ik/y)^{2m+1}}\right) \cos (2\pi nv)\ dv\Bigg \}\\&=\frac{(2m)!}{y^{2m+1}}\Bigg \{\zeta (2m+1)+\frac{(-1)^m}{m}\left( \frac{2\pi }{y}\right) ^{-2m}\zeta (2m)+2(2\pi )^{2m}\sum _{k=1}^\infty \sum _{n=1}^\infty n^{2m}\\&\quad \times \int _0^\infty \left( \frac{1}{(t-4\pi ^2 ink/y)^{2m+1}}+\frac{1}{(t+4\pi ^2 ink/y)^{2m+1}}\right) \cos (t)\ dt\Bigg \}, \end{aligned}$$

where in the last step we made change of variable $v=t/(2\pi n)$. Letting $nk=\ell$, we see that

$$\begin{aligned} \sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}&=\frac{(2m)!}{y^{2m+1}}\Bigg \{\zeta (2m+1)+\frac{(-1)^m}{m}\left( \frac{2\pi }{y}\right) ^{-2m}\zeta (2m)+2(2\pi )^{2m}\sum _{\ell =1}^\infty \sum _{n|\ell } n^{2m}\nonumber \\&\quad \times \int _0^\infty \left( \frac{1}{(t-4\pi ^2 i\ell /y)^{2m+1}}+\frac{1}{(t+4\pi ^2 i\ell /y)^{2m+1}}\right) \cos (t)\ dt\Bigg \}. \end{aligned}$$

(4.3)

Next, invoke Theorem 1.10 with $w=4\pi ^2\ell /y$ in (4.3) to arrive at

$$\begin{aligned}&\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny}\nonumber \\&\quad =\frac{(2m)!}{y^{2m+1}}\Bigg \{\zeta (2m+1)+\frac{(-1)^m}{m}\left( \frac{2\pi }{y}\right) ^{-2m}\zeta (2m)\nonumber \\&\qquad +\frac{4(-1)^{m}(2\pi )^{2m}}{(2m)!}\sum _{\ell =1}^\infty \sigma _{2m}(\ell )\Bigg \{\sinh \left( \frac{4\pi ^2\ell }{y}\right) \textrm{Shi}\left( \frac{4\pi ^2\ell }{y}\right) \nonumber \\&\qquad -\cosh \left( \frac{4\pi ^2\ell }{y}\right) \textrm{Chi}\left( \frac{4\pi ^2\ell }{y}\right) +\sum _{j=1}^m(2j-1)!\left( \frac{4\pi ^2\ell }{y}\right) ^{-2j}\Bigg \}. \end{aligned}$$

(4.4)

Using (3.43) in (4.4) and rearranging the terms leads to (1.13) for $y>0$. The result can be extended by analytic continuation to Re$(y)>0$. This is seen as follows. Clearly, the left-hand side of (1.13) is analytic in this region. We now show that the series on the right is also analytic. In order to prove this using Weierstrass’ theorem on analytic functions, we need only show that (4.2) holds for Re$(y)>0$ as well. To that end, employing $(1-\sqrt{\xi })^{-(2\,m+1)}-(1+\sqrt{\xi })^{-(2\,m+1)}=2(2\,m+1)\sqrt{\xi }\ {}_2F_1\left( m+1,m+\frac{3}{2};\frac{3}{2};\xi \right)$, we find that

$$\begin{aligned}&\int _0^\infty \left( \frac{1}{(t-iw)^{2m+1}}+\frac{1}{(t+iw)^{2m+1}}\right) \cos (t) dt=\frac{2(2m+1)}{(-1)^mw^{2m+2}}\nonumber \\&\quad \times \int _{0}^{\infty }t\cos (t){}_2F_{1}\left( m+1,m+\frac{3}{2};\frac{3}{2};-\frac{t^2}{w^2}\right) dt. \end{aligned}$$

(4.5)

The integral on the right can be evaluated in terms of the Meijer G-function ${G^{3,1}_{2,4}}\left( {\begin{matrix}{\begin{array}{l} \quad \quad \quad \quad 1,\frac{3}{2}\\ {1,m+1,m+\frac{3}{2},\frac{3}{2}}\end{array}}\bigg |{\frac{w^2}{4}}\end{matrix}}\right)$ employing [20, p. 81, Formula 8.17.6]. That

$$\begin{aligned}&{G^{3,1}_{2,4}}\left( {\begin{matrix}{\begin{array}{l} \quad \quad \quad \quad 1,\frac{3}{2}\\ {1,m+1,m+\frac{3}{2},\frac{3}{2}}\end{array}}\bigg |{\frac{w^2}{4}}\end{matrix}}\right) \nonumber \\&\quad =-\frac{w^2}{\sqrt{\pi }2^{2m+1}}\sum _{j=1}^{r+1}\frac{\Gamma (2m+2j)}{w^{2j}}+O(w^{-2r-4}),\quad as \ w\rightarrow \infty ,\ Re (w)>0 \end{aligned}$$

(4.6)

can then be obtained using the asymptotic of this Meijer G-function given in [34, p. 179, Theorem 2]. With $w=4\pi ^2n/y, \text {Re}(y)>0$, the first equality of (4.1) and (4.5) finally prove (4.2). This completes the proof of (1.13) for Re$(y)>0$.

$\square$

5 Asymptotics of the Plane Partitions Generating Function

Proof of Corollary 1.5

The following estimate can be directly shown to hold as $y\rightarrow 0$ in Re$(y)>0$ as is done later. However, we first prove it separately for real $y\rightarrow 0^+$ owing to its simplicity.

Indeed, for real $y\rightarrow 0^+$, Lemma 2.1 along with (4.1) imply that

$$\begin{aligned}&\sinh \left( \frac{4\pi ^2n}{y}\right) \textrm{Shi}\left( \frac{4\pi ^2n}{y}\right) -\cosh \left( \frac{4\pi ^2n}{y}\right) \textrm{Chi}\left( \frac{4\pi ^2n}{y}\right) +\sum _{j=1}^m(2j-1)!\left( \frac{4\pi ^2n}{y}\right) ^{-2j}\nonumber \\&\quad =\frac{-1}{(4\pi ^2n)^{2m}}y^{2m}\sum _{j=1}^{r+1}\frac{\Gamma (2m+2j)}{(4\pi ^2n)^{2j}}y^{2j}+O\left( \frac{y^{2r+2m+4}}{n^{2r+2m+4}}\right) . \end{aligned}$$

(5.1)

For complex y in Re$(y)>0$ such that $y\rightarrow 0$, (5.1) is seen to hold from the first equality of (4.1), (4.5) and (4.6).

Substituting (5.1) in (1.13), we deduce that

$$\begin{aligned}&\sum _{n=1}^\infty \sigma _{2m}(n)e^{-ny} \\&\quad =\frac{(2m)!}{y^{2m+1}}\zeta (2m+1)-\frac{B_{2m}}{2my}-(-1)^m\frac{2}{\pi }\left( \frac{2\pi }{y}\right) ^{2m+1}\frac{y^{2m}}{(2\pi )^{4m}} \\&\quad \times \sum _{n=1}^\infty \frac{\sigma _{2m}(n)}{n^{2m}}\sum _{j=1}^{r+1}\frac{\Gamma (2m+2j)}{(4\pi ^2n)^{2j}}y^{2j} +O\left( y^{2r+3}\right) \\&=\frac{(2m)!}{y^{2m+1}}\zeta (2m+1)-\frac{B_{2m}}{2my}-(-1)^m\frac{2}{y\pi (2\pi )^{2m-1}} \\&\quad \times \sum _{j=1}^{r+1}\frac{\Gamma (2m+2j)}{(4\pi ^2)^{2j}}y^{2j}\sum _{n=1}^\infty \frac{\sigma _{2m}(n)}{n^{2m+2j}}+O\left( y^{2r+3}\right) . \end{aligned}$$

Using the well-known identity $\sum _{n=1}^\infty \sigma _a(n)n^{-s}=\zeta (s)\zeta (s-a)$, where $Re (s)>1, Re (s-a)>1,$ in the above expression, we arrive at (1.15). $\square$

We are now ready to derive Wright’s result from [51] as a special case of Corollary 1.5.

Proof of Corollary 1.6

Letting $m=1$ and $y=\log (1/x), |x|<1,$ in Corollary 1.5 and using (1.16), as $x\rightarrow 1^-$, we have

$$\begin{aligned} x\frac{d}{dx}\log F(x)&=-\frac{2\zeta (3)}{(\log x)^3}+\frac{1}{12\log x}\\&-\frac{1}{\pi ^2}\sum _{j=1}^{r+1}\frac{\Gamma (2j+2)\zeta (2j+2)\zeta (2j)}{(2\pi )^{4j}}(\log x)^{2j-1}+O\left( -(\log x)^{2r+3}\right) . \end{aligned}$$

Now divide both sides by x and then integrate with respect to x to get

$$\begin{aligned}&\log F(x)=c+\frac{\zeta (3)}{(\log x)^2}+\frac{1}{12}\log \log x-\frac{1}{\pi ^2}\\&\quad \times \sum _{j=1}^{r+1}\frac{\Gamma (2j+2)\zeta (2j+2)\zeta (2j)}{2j(2\pi )^{4j}}(\log x)^{2j}+O\left( (\log x)^{2r+4}\right) , \end{aligned}$$

where c is an integrating constant. Exponentiating both sides of the above equation, we arrive at (1.17). $\square$

6 Concluding Remarks

For general a with $0\le a<1$, the generalized Lambert series

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{(n-a)^{s-1}}{e^{(n-a)z}-1}\hspace{8mm}(s\in {\mathbb {C}}, Re (z)>0) \end{aligned}$$

does not seem to have been studied before. It makes its appearance for the first time in Theorem 1.1 of our paper. It may be interesting to undertake a further study of this series.

In [10, Theorem 1], Bradley obtains a generalization of Ramanujan’s formula (1.1) for periodic functions g with period $m\in {\mathbb {N}}$. When g is even, his transformation involves the series of the type $\displaystyle \sum _{n=1}^\infty \frac{g(n)n^{-2m-1}}{e^{n\beta }-1}$ whereas for g odd, it involves $\displaystyle \sum _{n=1}^\infty \frac{g(n)n^{-2m}}{e^{n\beta }-1}$. Observe that the series in our (1.8) involves $\displaystyle \sum _{n=1}^\infty \frac{\chi (n)n^{2m-1}}{e^{n\beta }-1}$, where $\chi (n)$ defined in (1.9) is an odd Dirichlet character and $m\in {\mathbb {N}}, m>1$, and hence does not fall under the purview of Bradley’s transformation. Thus it may be worthwhile to see if a more general transformation encompassing our series exists. We note that another series which is not covered by Bradley’s transformation is $\displaystyle \sum _{n=1}^\infty \frac{n^{-2m}}{e^{n\beta }-1}$, for which a transformation was recently obtained in [16, Theorem 2.12].

In [16], (1.13) was derived as a special case of (1.11), whereas in the present paper, this has been accomplished directly. One can then ask if a direct proof of Theorem 2.12 of [16], which is a transformation for $\displaystyle \sum _{n=1}^\infty \frac{n^{-2m}}{e^{n\beta }-1}$, can be derived without resorting to (1.11).