# Hurwitz Zeta Functions and Ramanujan’s Identity for Odd Zeta Values

Parth Chavan

###### Abstract.

Inspired by a famous formula of Ramanujan for odd zeta values, we prove an
analogous formula involving the Hurwitz zeta function. We introduce a new
integral kernel related to the Hurwitz zeta function, generalizing the
integral kernel associated to Ramanujan’s identity. We also derive several
infinite families of identities analogous to Ramanujan’s formula.

###### Key words and phrases:

Hurwitz zeta function, Ramanujan’s formula for $\zeta(2n+1)$, Dirichlet
series.

###### 2020 Mathematics Subject Classification:

Primary $11$Mxx, $11$B$68$ and $11$F$03$

##  1\. Introduction

The Riemann zeta function $\zeta(s)$, defined by the series

$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}},$

with $\mathfrak{R}(s)>1$, is one of the most important special functions of
mathematics. The critical strip $0<\mathfrak{R}(s)<1$ is undoubtedly the most
famous region in the complex plane on account of the unsolved problem
regarding the location of non-trivial zeros of $\zeta(s)$, namely, the Riemann
Hypothesis, however the right-half plane $\mathfrak{R}(s)>1$ also has its own
share of interesting unsolved problems, such as the arithmetic nature of odd
zeta values.

One of the identities given by Ramanujan that has attracted the attention of
many mathematicians over the years is the following intriguing identity
involving the odd values of the Riemann zeta function:

######  Theorem 1.1 (Ramanujan’s formula for $\zeta(2n+1)$).

If $\alpha$ and $\beta$ are positive real numbers such that
$\alpha\beta=\pi^{2}$ and if $n\in\mathbb{Z}\setminus{\\{0\\}}$, then we have

$\displaystyle\alpha^{-n}\left\\{\dfrac{1}{2}\,\zeta(2n+1)+\sum_{m=1}^{\infty}\dfrac{m^{-2n-1}}{e^{2\alpha
m}-1}\right\\}$
$\displaystyle-\left(-\beta\right)^{-n}\left\\{\dfrac{1}{2}\,\zeta(2n+1)+\sum_{m=1}^{\infty}\dfrac{m^{-2n-1}}{e^{2\beta
m}-1}\right\\}$
$\displaystyle=2^{2n}\sum_{k=0}^{n+1}\dfrac{\left(-1\right)^{k-1}B_{2k}\,B_{2n-2k+2}}{\left(2k\right)!\left(2n-2k+2\right)!}\,\alpha^{n-k+1}\beta^{k}.$
(1.1)

where $B_{n}$ denotes the $n$-th Bernoulli number.

Theorem 1.1 appears as Entry 21 in Chapter 14 of Ramanujan’s second notebook
[4]. It also appears in a formerly unpublished manuscript of Ramanujan that
was published in its original handwritten form with his lost notebook [1]. For
an elementary proof of Theorem 1.1 we refer the reader to [6]. For history and
developments related to Ramanujan’s formulas we refer the reader to [3]. The
first published proof of Theorem 1.1 is due to S.L. Marulkar [13], although he
was not aware that this formula can be found in Ramanujan’s Notebooks. The
function $\frac{1}{e^{2\pi x}-1}$ appears in several Ramanujan’s identities
and has the integral representation

$\frac{1}{e^{2\pi x}-1}=\int_{(c)}\frac{\zeta(1-s)}{2\cos\left(\frac{\pi
s}{2}\right)}x^{-s}\mathrm{d}s,$

where $(c)$ denoted the vertical line line $\Re(s)=c$ and $c$ is an arbitrary
real number such that $c>1$. We call this function Ramanujan’s kernel. In this
article we generalize introduce a two parameter generalization of Ramanujan’s
kernel and obtain several Ramanujan type identities. One reason to study this
kernel is to obtain more information on the arithmetic nature of
$\zeta(2m+1)$. By including a free parameter $a$, one can derive other results
involving the zeta function by differentiating or integrating against $a$.

Ramanujan’s kernel has simple poles at $x=0$ and $x=\pm in,n\in\mathbb{N}$
with residue $\frac{1}{2\pi}$ at $0,\pm in$ and thus has the partial fraction
expansion

$\frac{1}{e^{2\pi x}-1}=-\frac{1}{2}+\frac{1}{2\pi
x}+\frac{x}{\pi}\sum_{n=1}^{\infty}\frac{1}{n^{2}+x^{2}}.$

We generalize this result to a meromorphic function with simple poles at
$x=0$,$x=e^{\frac{(2j+1)i\pi}{2k}}(n+a)$ where $j\in\\{0,1,\ldots,2k-1\\}$,
$n\in\mathbb{N}$ and $a\in\mathbb{C}$ with residue $\frac{1}{2k\pi}$ at
$e^{\frac{(2j+1)i\pi}{2k}}(n+a)$ defined as follows

######  Definition 1.2.

Let $x\in\mathbb{R}^{+},a\in\mathbb{C}$ and $k\in\mathbb{N}$. Define the
Hurwitz kernel by

$\displaystyle\Psi\left(x,a;k\right)$
$\displaystyle:=\int_{\left(c\right)}\frac{\zeta(1-s,a)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}x^{-s}\mathrm{d}s$
$\displaystyle=\frac{2a-1}{2\pi
x}-\frac{1}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}+\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{x^{2k-1}}{x^{2k}+(n+a)^{2k}},$
(1.2)

where $\zeta(s,a)$ is the Hurwitz zeta function defined in Section 2. Note
that $\Psi(x,1;1)$ is Ramanujan’s kernel. A. Dixit et al ask in [10] whether a
Ramanujan type identity for Hurwitz zeta exists and we answer this question
positively. Ramanujan’s identity gives an expression for the convolution of
Riemann zeta at even arguments. In the same spirit it can be asked whether
such an expression exists for the convolution of Riemann zeta at odd
arguments, which was positively answered in [9]:

######  Theorem 1.3.

Let $\alpha$ and $\beta$ be two complex numbers such that
$\Re(\alpha)>0,\Re(\beta)>0$ and $\alpha\beta=4\pi^{2}$. Let $\psi$ denote the
digamma function. Then for $m\in\mathbb{N}$, we have

$\displaystyle\left(-\beta\right)^{-m}\left\\{2\gamma\zeta(2m+1)+\sum_{n=1}^{\infty}\dfrac{1}{n^{2m+1}}\left(\psi\left(\dfrac{\imath
n\beta}{2\pi}\right)+\psi\left(-\dfrac{\imath
n\beta}{2\pi}\right)\right)\right\\}$
$\displaystyle+\alpha^{-m}\left\\{2\gamma\zeta(2m+1)+\sum_{n=1}^{\infty}\dfrac{1}{n^{2m+1}}\left(\psi\left(\dfrac{\imath
n\alpha}{2\pi}\right)+\psi\left(-\dfrac{\imath
n\alpha}{2\pi}\right)\right)\right\\}$
$\displaystyle=-2\sum_{k=1}^{m-1}\left(-1\right)^{k}\zeta(2k+1)\,\zeta(2m-2k+1)\alpha^{k-m}\beta^{-k}.$

The above identity can also be found in [5].

It can be observed that

$\displaystyle\frac{1}{2i\pi}\int_{(c)}\frac{\zeta(1-s)x^{-s}}{2\sin\left(\frac{\pi
s}{2}\right)}ds$
$\displaystyle=\frac{\log(x)+\gamma}{\pi}-\frac{x^{2}}{\pi}\sum_{n=1}^{\infty}\frac{1}{n(x^{2}+n^{2})}$
$\displaystyle=\frac{1}{\pi}\left(\log(x)-\frac{\psi(ix)+\psi(-ix)}{2}\right),$

where $c>1$. We generalize Theorem 1.3 to its Hurwitz zeta analog involving
another new kernel.

######  Definition 1.4.

Let $x\in\mathbb{R}^{+},a\in\mathbb{C}$ and $k\in\mathbb{N}$. Define the odd
Hurwitz kernel by

$\displaystyle\Phi(x,a;k)=\int_{(c)}\frac{\zeta(1-s,a)}{2\sin\left(\frac{\pi
s}{2k}\right)}x^{-s}ds$ (1.3)
$\displaystyle=\frac{\log(x)+\gamma_{0}(a)}{\pi}-\frac{x^{2k}}{\pi}\sum_{n=0}^{\infty}\frac{1}{(n+a)((n+a)^{2k}+x^{2k})}$
(1.4)

Finally we give a relation between kernels $\Phi(x,a;k)$ and $\Psi(x,a;b)$ in
Theorem 2.8. Throughout the remainder of the paper we assume $k\in\mathbb{N}$
and $a,b\in\mathbb{R}^{+}/\mathbb{N}$ unless specified. For ease of notation
we define

$\Psi\left(\frac{\alpha
x}{\pi},a;k\right)=\Psi_{\alpha}(x,a;k)\,\,\text{and}\,\,\Phi\left(\frac{\alpha
x}{\pi},a;k\right)=\Phi_{\alpha}(x,a;k).$

We prove our main results except Theorem 2.4 and Theorem 2.6 in two ways,
first using complex analysis and then using tools borrowed from [5]. For the
convenience of the reader we include Dirichlet series setup and notations
which can be found in [5].

###  1.1. Notations

For a sequence of non-zero complex numbers $\left\\{x_{n}\right\\}$ and a
sequence of associated complex weights $\left\\{a_{n}\right\\}$, a Dirichlet
series is defined as

$\zeta_{x,a}\left(N\right)=\sum_{n=1}^{\infty}\frac{a_{n}}{x_{n}^{N}}.$ (1.5)

Note that we assume that the Dirichlet series is convergent for $N\geqslant
1.$ If the series diverges at $N=1$, but has a finite abscissa of convergence,
we can obtain analogous results. We call the function

$\psi_{x,a}\left(z\right)=\sum_{N=1}^{\infty}\zeta_{x,a}\left(N\right)z^{N}$
(1.6)

associated to $\zeta_{x,a}$ the zeta generating function. Using geometric
series, we can check that the generating function $\psi_{x,a}\left(z\right)$
can be expressed in terms of the weights $\left\\{a_{n}\right\\}$ and zeros
$\left\\{x_{n}\right\\}$ as follows

$\psi_{x,a}\left(z\right)=\sum_{n=1}^{\infty}a_{n}\,\dfrac{z}{x_{n}-z}.$ (1.7)

Finally, we introduce the following notation: the modified sequence of weights
$\left\\{a.\psi_{y,b}\right\\}_{n\geqslant 1}$ is defined by

$\left(a.\psi_{y,b}\right)_{n}=a_{n}\psi_{y,b}\left(x_{n}\right)$

so that the corresponding Dirichlet series is

$\zeta_{x,a.\psi_{y,b}}\left(N\right)=\sum_{n=1}^{\infty}\frac{a_{n}\psi_{y,b}\left(x_{n}\right)}{x_{n}^{N}}.$

Similarly, we denote as $\left(a.\psi_{y,b}.\psi_{z,c}\right)$ the sequence
defined by

$\left(a.\psi_{y,b}.\psi_{z,c}\right)_{n}=a_{n}\psi_{y,b}\left(x_{n}\right)\psi_{z,c}\left(x_{n}\right)$

with associated Dirichlet series

$\zeta_{x,a.\psi_{y,b}.\psi_{z,c}}\left(N\right)=\sum_{n=1}^{\infty}\frac{a_{n}\psi_{y,b}\left(x_{n}\right)\psi_{z,c}\left(x_{n}\right)}{x_{n}^{N}}.$

Finally the convolution of two Dirichlet series is defined as

$\left(\zeta_{y,b}*\zeta_{x,a}\right)\left(N+1\right)=\sum_{k=1}^{N}\zeta_{y,b}\left(k\right)\zeta_{x,a}\left(N+1-k\right)$

and the $n-$fold convolution of $n$ Dirichlet series
$\zeta_{x^{\left(1\right)},a^{\left(1\right)}},\dots,\zeta_{x^{\left(n\right)},a^{\left(n\right)}}$
as

$\left(\stackrel{{\scriptstyle[}}{{i}}=1]{n}{*}\zeta_{x^{\left(i\right)},a^{\left(i\right)}}\right)\left(N+1\right)=\sum\zeta_{x^{\left(1\right)},a^{\left(1\right)}}\left(k_{1}\right)\dots\zeta_{x^{\left(n\right)},a^{\left(n\right)}}\left(k_{n}\right)$

where the sum is over the set of indices

$\left\\{\left(k_{1},k_{2},\ldots,k_{n}\right):1\leqslant k_{i}\leqslant
N,\sum_{i=1}^{n}k_{i}=N+1\right\\}.$

The proof of Theorem 1.5 below can be found in [5]

######  Theorem 1.5.

For a set of $n\geqslant 2$ Dirichlet series
$\left\\{\zeta_{x^{\left(i\right)},a^{\left(i\right)}}\right\\}{}_{1\leqslant
i\leqslant n}$, we have, evaluated at argument $N+1$ removed for clarity,

$\stackrel{{\scriptstyle[}}{{i}}=1]{n}{*}\zeta_{x^{\left(i\right)},a^{\left(i\right)}}=\sum_{i=1}^{n}\zeta_{x^{\left(i\right)},a^{\left(i\right)}.\prod_{1\leqslant
k\neq i\leqslant n}\psi_{x^{\left(k\right)}}}$

The special case $n=2$ reads

$\displaystyle\zeta_{y,b}*\zeta_{x,a}=\zeta_{x,a.\psi_{y}}+\zeta_{y,b.\psi_{x}}$
(1.8)

For the sake of clarity, let us rephrase this identity (1.8) in a more
explicit way:

$\left(\zeta_{y,b}*\zeta_{x,a}\right)\left(N+1\right)=\sum_{n=1}^{\infty}\left\\{\frac{a_{n}\psi_{y}\left(x_{n}\right)}{x_{n}^{N+1}}+\frac{b_{n}\psi_{x}\left(y_{n}\right)}{y_{n}^{N+1}}\right\\},$

##  2\. Main results

The Bernoulli polynomials $B_{n}(x)$ are defined through their generating
function as

$\frac{te^{xt}}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}(x)\frac{t^{n}}{n!},$ (2.1)

and the Bernoulli numbers $B_{n}$ are defined by $B_{n}=B_{n}(0)$. The Hurwitz
zeta function for complex variable $s$ with $\Re(s)>1$ and
$a\in\mathbb{C}/\mathbb{Z}_{\leq 0}$ is defined by

$\zeta(s,a)=\sum_{n=0}^{\infty}{\frac{1}{(n+a)^{s}}}.$ (2.2)

The Hurwitz zeta function has the integral representation [2]

$\zeta(s,a)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}e^{-ax}}{1-e^{-x}}\mathrm{d}s,$
(2.3)

valid for $\Re(s)>1$. Moreover, it has an analytic continuation represented by
the following contour integral [15]:

$\zeta(s,a)={\frac{-\Gamma(1-s)}{2\pi
i}}\int_{C}{\frac{(-z)^{s-1}e^{-az}}{1-e^{-z}}}dz,$ (2.4)

where $C$ is the Hankel contour counterclockwise around the positive real axis
and the principal branch for exponentiation is used. The Laurent series
expansion of Hurwitz zeta zentered at $s=1$ is given by

$\zeta(s,a)={\frac{1}{s-1}}+\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{n!}}\gamma_{n}(a)(s-1)^{n}$

where $\gamma_{n}(a)$ are generalized Stieltjes constants.

The integral (2.4) defines $\zeta(s,a)$ for all $s\in\mathbb{C}$, with a
single pole at $s=1$ and corresponding residue $1$. Observing the analogy
between Ramanujan’s kernel and $\Psi(x,a;k)$ we derive a Ramanujan type
formula involving $\Psi(x,a;k)$ in the theorem below.

######  Theorem 2.1.

Let $\alpha,\beta\in\mathbb{R}^{+}$ such that $\alpha\beta=\pi^{2}$, let
$a,b\in\mathbb{C}/\mathbb{Z}_{\leq 0}$ and $k,N\in\mathbb{N}$. Then, we have

$\displaystyle\beta^{k\left(N+1\right)-1}\sum_{n=0}^{\infty}\dfrac{\Psi_{\alpha}(n+b,a;k)}{\left(n+b\right)^{2k\left(N+1\right)-1}}-\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\sum_{n=0}^{\infty}\dfrac{\Psi_{\beta}(n+a,b;k)}{\left(n+a\right)^{2k\left(N+1\right)-1}}$
$\displaystyle=\dfrac{\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1,a)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}-\frac{\beta^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1,b)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}$
$\displaystyle+\sum_{p=0}^{N+1}\left(-1\right)^{p+1}\zeta(2kp,a)\,\zeta(2k\left(N-p+1\right),b)\,\alpha^{kp-1}\beta^{k(N+1-p)-1}.$
(2.5)

Substituting the power series expansion of $\Psi_{\alpha}(x,a;k)$ in equation
(2.1) we get

$\displaystyle\beta^{k(N+1)-1}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(N+1)-1}}\sum_{n=0}^{\infty}\frac{\alpha^{k-1}(n+b)^{2k-1}}{\alpha^{k}(n+b)^{2k}+\beta^{k}(n+a)^{2k}}$
(2.6)
$\displaystyle-(-1)^{N}\alpha^{k(N+1)-1}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(N+1)-1}}\sum_{n=0}^{\infty}\frac{\beta^{k-1}(n+a)^{2k-1}}{\beta^{k}(n+a)^{2k}+\alpha^{k}(n+b)^{2k}}$
$\displaystyle=\sum_{p=1}^{N}\left(-1\right)^{p+1}\zeta(2kp,a)\,\zeta(2k\left(N-p+1\right),b)\,\alpha^{kp-1}\beta^{k(N+1-p)-1}$

When $N=2q+1,q\in\mathbb{N}$ and $\alpha=\beta=\pi$ we have

$\displaystyle\sum_{n=0}^{\infty}\frac{\psi_{\pi}(n+b,a;k)}{(n+b)^{4k(q+1)-1}}+\sum_{n=0}^{\infty}\frac{\psi_{\pi}(n+a,b;k)}{(n+a)^{4k(q+1)-1}}$
(2.7)
$\displaystyle=\frac{1}{\pi}\sum_{p=0}^{2q+2}(-1)^{p+1}\zeta(2kp,a)\zeta(2k(2q-p+2),b)$
$\displaystyle-\frac{1}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}\left(\zeta(4k(q+1)-1,a)+\zeta(4k(q+1)-1,b)\right).$

Since

$\Psi_{\pi}(x,a;k)=\Psi(x,a;k)=\frac{2a-1}{2\pi
x}-\frac{1}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}+\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{x^{2k-1}}{x^{2k}+(n+a)^{2k}},$

we have

$\displaystyle\frac{2a-1(\zeta(4k(q+1),a)+\zeta(4k(q+1),b))}{2}+\sum_{n=0}^{\infty}\frac{1}{(n+b)^{4k(q+1)}}\sum_{j=0}^{\infty}\frac{1}{1+\left(\frac{j+a}{n+b}\right)^{2k}}$
(2.8)
$\displaystyle+\sum_{n=0}^{\infty}\frac{1}{(n+a)^{4k(q+1)}}\sum_{j=0}^{\infty}\frac{1}{1+\left(\frac{j+b}{n+a}\right)^{2k}}=\sum_{p=0}^{2q+2}(-1)^{p+1}\zeta(2kp,a)\zeta(2k(2q-p+2),b),$

which can be written as

$\displaystyle\sum_{n=0}^{\infty}\frac{1}{(n+b)^{4k(q+1)}}\sum_{j=0}^{\infty}\frac{1}{1+\left(\frac{j+a}{n+b}\right)^{2k}}+\sum_{n=0}^{\infty}\frac{1}{(n+a)^{4k(q+1)}}\sum_{j=0}^{\infty}\frac{1}{1+\left(\frac{j+b}{n+a}\right)^{2k}}$
(2.9)
$\displaystyle=\sum_{p=1}^{2q+1}(-1)^{p+1}\zeta(2kp,a)\zeta(2k(2q-p+2),b).$

With $a=b$ we have

$\displaystyle\sum_{n=0}^{\infty}\frac{1}{(n+a)^{4k(q+1)}}\sum_{j=0}^{\infty}\frac{1}{1+\left(\frac{j+a}{n+a}\right)^{2k}}=\frac{1}{2}\sum_{p=0}^{2q+1}(-1)^{p+1}\zeta(2kp,a)\zeta(2k(2q-p+2),a)$
(2.10)

Setting $a=b$ in Theorem 2.1 we deduce:

######  Corollary 2.2.

Let $\alpha\beta=\pi^{2}$, $a\in\mathbb{C}/\mathbb{Z}_{\leq 0}$ and
$k,N\in\mathbb{N}$. We have the identity

$\displaystyle\beta^{k\left(N+1\right)-1}\sum_{n=0}^{\infty}\dfrac{\Psi_{\alpha}(n+a,a;k)}{\left(n+a\right)^{2k\left(N+1\right)-1}}-\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\sum_{n=0}^{\infty}\dfrac{\Psi_{\beta}(n+a,a;k)}{\left(n+a\right)^{2k\left(N+1\right)-1}}$
$\displaystyle=\dfrac{\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1,a)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}-\frac{\beta^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1,a)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}$
$\displaystyle+\sum_{p=0}^{N+1}\left(-1\right)^{p}\zeta(2kp,a)\,\zeta(2k\left(N-p+1\right),a)\,\alpha^{kp-1}\beta^{k(N+1-p)-1}.$
(2.11)

As a corollary, when $a=1$ we have the following convolution of Riemann zeta
functions:

######  Corollary 2.3.

Let $\alpha\beta=\pi^{2}$

$\displaystyle\beta^{k\left(N+1\right)-1}\sum_{n=1}^{\infty}\dfrac{\Psi_{\alpha}(n,1;k)}{\left(n\right)^{2k\left(N+1\right)-1}}-\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\sum_{n=1}^{\infty}\dfrac{\Psi_{\beta}(n,1;k)}{\left(n\right)^{2k\left(N+1\right)-1}}$
$\displaystyle=\dfrac{\left(-1\right)^{N}\alpha^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}-\frac{\beta^{k\left(N+1\right)-1}\zeta(2k\left(N+1\right)-1)}{2k\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}$
$\displaystyle+\sum_{p=0}^{N+1}\left(-1\right)^{p}\zeta(2kp)\,\zeta(2k\left(N-p+1\right))\,\alpha^{kp-1}\beta^{k(N+1-p)-1}.$
(2.12)

When $k=1$ in Theorem 2.1 we recover the identity

$\displaystyle\sum_{k=1}^{N}\left(-\alpha\right)^{N+1-k}\beta^{k}\zeta\left(2k,a\right)\zeta\left(2N+2-2k,b\right)$
$\displaystyle=\dfrac{\beta^{N+1}}{2}\,i\sqrt{\frac{\alpha}{\beta}}\sum_{n=0}^{\infty}\frac{1}{\left(a+n\right)^{2N+1}}\left[\psi\left(b+i\sqrt{\frac{\alpha}{\beta}}\left(a+n\right)\right)-\psi\left(b-i\sqrt{\frac{\alpha}{\beta}}\left(a+n\right)\right)\right]$
(2.13)
$\displaystyle+\frac{\left(-\alpha\right)^{N+1}}{2}\,i\sqrt{\frac{\beta}{\alpha}}\sum_{n=0}^{\infty}\frac{1}{\left(b+n\right)^{2N+1}}\left[\psi\left(a+i\sqrt{\frac{\beta}{\alpha}}\left(b+n\right)\right)-\psi\left(a-i\sqrt{\frac{\beta}{\alpha}}\left(b+n\right)\right)\right]$

which has been recently proved in [5] using Theorem 1.5. Their method allows
for straightforward generalizations of these transformations. Conversely, our
method provides rigorous convergence guarantees, which their method does not
provide.

Ramanujan’s identity 1.1 has the following special case, namely, for
$\alpha,\beta>0$ with $\alpha\beta=\pi^{2}$,

$\alpha^{m}\sum_{n=1}^{\infty}\frac{n^{2m-1}}{e^{2\alpha
n}-1}-(-\beta)^{m}\sum_{n=1}^{\infty}\frac{n^{2m-1}}{e^{2\beta
n}-1}=(\alpha^{m}-(-\beta)^{m})\frac{B_{2m}}{4m}\qquad(m>1),$ (2.14)

which can be found in [3, 14]. The following theorem provides a generalization
of equation (2.14) involving the Hurwitz kernel $\Psi(x,a;k)$.

######  Theorem 2.4.

Let $\alpha,\beta\in\mathbb{R}^{+}$ such that $\alpha\beta=\pi^{2}$ and
$B_{j}(x)$ denote Bernoulli polynomials. Then, the following identity holds

$\displaystyle\alpha^{km+1}\sum_{n=0}^{\infty}\left(n+b\right)^{2km+1}\left[\Psi_{\alpha}(n+b,a;k)-\sum_{p=1}^{m}\frac{B_{2kp+1}(a)}{2kp+1}\left(\frac{\pi}{\alpha}\right)^{2kp+1}\right]$
(2.15)
$\displaystyle+\left(-1\right)^{m}\beta^{km+1}\sum_{n=0}^{\infty}\left(n+a\right)^{2km+1}\left[\Psi_{\beta}(n+a,b;k)-\sum_{p=1}^{m}\frac{B_{2kp+1}(b)}{2kp+1}\left(\frac{\pi}{\beta}\right)^{2kp+1}\right]$
$\displaystyle=\frac{\left(-1\right)^{m+1}\beta^{km+1}B_{2km+2}(a)}{4k(km+1)\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}-\frac{\alpha^{km+1}B_{2km+2}(b)}{4k(km+1)\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)}$
$\displaystyle+\sum_{p=0}^{m}(-1)^{p}\frac{B_{2kp+1}(a)\beta^{kp}B_{2k\left(m-p\right)+1}(b)\alpha^{k\left(m-p\right)}}{\left(2kp+1\right)\left(2k\left(m-p\right)+1\right)}.$

As a special case when $a=b=1$ we have:

######  Proposition 2.5.

Let $\alpha,\beta>0$ and $\alpha\beta=\pi^{2}$. The following identity holds

$\displaystyle(\alpha)^{km+1}\sum_{n=1}^{\infty}n^{2km+1}\Psi_{\alpha}(n,1;k)+(-1)^{m}\beta^{km+1}\sum_{n=1}^{\infty}n^{2km+1}\Psi_{\beta}(n,1;k)$
(2.16)
$\displaystyle=\frac{1}{4k(km+1)\cos\left(\frac{\pi(k-1)}{2k}\right)}\left(\alpha^{km+1}B_{2km+2}-(-1)^{m+1}\beta^{km+1}B_{2km+2}\right).$

Moreover, plugging in $\alpha=\beta=\pi$ with $m=2p$ in Proposition 2.5
produces

$\sum_{n=1}^{\infty}\dfrac{\Psi_{\pi}(n,1;k)}{n^{-4kp-1}}=\frac{B_{4kp+2}}{4k\left(2kp+1\right)\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)},$
(2.17)

which is a generalization of Glaisher’s famous identity [12]

$\sum_{n=1}^{\infty}\frac{n^{4m+1}}{e^{2\pi n}-1}=\frac{B_{4m+2}}{2(4m+2)}.$
(2.18)

When we put $k=1$ in Proposition 2.5 we recover equation (2.14).

As a companion of Theorem 2.4 we have the following identity for $\Phi(x,a;k)$

######  Theorem 2.6.

Let $\alpha,\beta>0$ and $\alpha\beta=\pi^{2}$. The following identity holds

$\displaystyle\alpha^{km}\sum_{n=0}^{\infty}\left(n+b\right)^{2km-1}\left[\Phi_{\alpha}(n+b,a;k)-\sum_{p=1}^{m}\frac{B_{2kp}(a)}{2kp}\left(\frac{\pi}{\alpha}\right)^{2kp}\right]$
(2.19)
$\displaystyle+(-1)^{m}\beta^{km}\sum_{n=0}^{\infty}\left(n+a\right)^{2km-1}\left[\Phi_{\beta}(n+a,b;k)-\sum_{p=1}^{m}\frac{B_{2kp}(b)}{2kp}\left(\frac{\pi}{\beta}\right)^{2kp}\right]$
$\displaystyle=\frac{-\alpha^{km}}{\pi}\left(\frac{B_{2km}(b)}{2km}\left(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a)\right)+\frac{\partial}{\partial
s}\zeta(s-2km+1,b)|_{s=0}\right)$
$\displaystyle+\frac{(-1)^{m+1}\beta^{km}}{\pi}\left(\frac{B_{2km}(a)}{2km}\left(\log\left(\frac{\beta}{\pi}\right)+\gamma_{0}(b)\right)+\frac{\partial}{\partial
s}\zeta(s-2km+1,a)\big{|}_{s=2km}\right)$
$\displaystyle+\sum_{p=1}^{m-1}(-1)^{p}\
\frac{\beta^{kp}B_{2kp}(a)\alpha^{k(m-p)}B_{2k\left(m-p\right)}(b)}{\pi\left(2kp\right)\left(2k\left(m-p\right)\right)}$

We now give Hurwitz zeta generalization of Theorem 1.3 involving the kernel
$\Phi(x,a;k)$.

######  Theorem 2.7.

Let $\alpha,\beta>0$ and $\alpha\beta=\pi^{2}$. The following identity holds

$\displaystyle\beta^{km}\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;k)}{(n+b)^{2km+1}}+(-\alpha^{k})^{m}\sum_{n=0}^{\infty}\frac{\Phi_{\beta}(n+a,b;k)}{(n+a)^{2m+1}}$
(2.20)
$\displaystyle=\sum_{i=1}^{m-1}(-1)^{i}\frac{\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i)+1,b)}{\pi}$
$\displaystyle+\frac{\beta^{km}}{\pi}\left(\zeta(2km+1,b)\left(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a)\right)-\frac{\partial}{\partial
s}\zeta(2km+1+s,b)\big{|}_{s=0}\right)$
$\displaystyle+\frac{(-1)^{m}\alpha^{km}}{\pi}\left(\zeta(2km+1,a)\left(\gamma_{0}(b)-\log\left(\frac{\alpha}{\pi}\right)\right)-\frac{\partial}{\partial
s}\zeta(2km+1+s,a)\big{|}_{s=0}\right)$

We recover Theorem 1.3 when we put $a=b=k=1$ in Theorem 2.7.

Substituting the power series expansion of $\Phi_{\alpha}(n+b,a;k)$ and
$\Phi_{\beta}(n+a,b;k)$ we have

$\displaystyle\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;k)}{(n+b)^{2km+1}}$
$\displaystyle=\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2km+1}}\left(\frac{\log\left(\frac{\alpha(n+b)}{\pi}\right)+\gamma_{0}(a)}{\pi}-\frac{\alpha^{k}(n+b)^{2k}}{\pi}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}\right)$
$\displaystyle\sum_{n=0}^{\infty}\frac{\Phi_{\beta}(n+a,b;k)}{(n+a)^{2km+1}}$
$\displaystyle=\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2km+1}}\left(\frac{\log\left(\frac{\beta(n+a)}{\pi}\right)+\gamma_{0}(b)}{\pi}-\frac{\beta^{k}(n+a)^{2k}}{\pi}\sum_{i=0}^{\infty}\frac{1}{(i+b)(\beta^{k}(i+b)^{2k}+\alpha^{k}(n+a)^{2k})}\right)$

Thus the identity can be written as

$\displaystyle-\pi^{2k-1}\beta^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}$
(2.21)
$\displaystyle-\pi^{2k-1}(-\alpha)^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{(i+b)(\beta^{k}(i+b)^{2k}+\alpha^{k}(n+a)^{2k})}$
$\displaystyle=\sum_{i=1}^{m-1}(-1)^{i}\frac{\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i)+1,b)}{\pi}$

In [7] , Dixit et al gave an expression for convolution of riemann zeta
function at odd and even values, namely, for $\alpha,\beta>0$ with
$\alpha\beta=\pi^{2}$ and $m\in\mathbb{N}$

$\beta^{-\left(m-\frac{1}{2}\right)}\left\\{\frac{1}{2}\zeta(2m)+\sum_{n=0}^{\infty}\frac{n^{-2m}}{e^{2n\beta}-1}\right\\}-\sum_{k=0}^{m-1}(-1)^{k+1}\frac{\zeta(2k)\zeta(2m-2k+1)}{\pi^{2k}}\beta^{2k-m-\frac{1}{2}}$
(2.22)
$=(-1)^{m+1}\alpha^{-\left(m-\frac{1}{2}\right)}\left\\{\frac{\gamma}{\pi}\zeta(2m)+\frac{1}{2\pi}\sum_{n=1}^{\infty}n^{-2m}\left(\psi\left(\frac{in\alpha}{\pi}\right)+\psi\left(\frac{-in\alpha}{\pi}\right)\right)\right\\}.$

Equation (2.22) is natural companion to Ramanujan’s identity and can be
interpreted as a relation between $\Phi_{\alpha}(n,1;1)$ and
$\Psi_{\beta}(n,1;1)$. The following t“heorem provides extension of this
relation to Hurwitz zeta function and containing the kernes
$\Phi_{\alpha}(x,a;k)$ and $\Psi_{\beta}(x,a;k)$

######  Theorem 2.8.

Let $\alpha,\beta>0$ and $\alpha\beta=\pi^{2}$. The following identity holds:

$\displaystyle\beta^{km}\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;k)}{(n+b)^{2km}}+\pi(-1)^{m-1}\alpha^{km-1}\sum_{n=0}^{\infty}\frac{\Psi_{\beta}(n+a,b;k)}{(n+a)^{2km}}$
(2.23)
$\displaystyle=\frac{\beta^{km}}{\pi}\left(\zeta(2km,b)\left(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a)\right)-\frac{\partial}{\partial
s}\zeta(2km+s,b)\big{|}_{s=0}\right)+(-1)^{m}\pi\alpha^{km-1}\frac{\zeta(2km,a)}{2k\sin\left(\frac{\pi}{2k}\right)}$
$\displaystyle+\sum_{i=1}^{m}\frac{(-1)^{i}\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i),b)}{\pi}$

When we put $a=b=k=1$ in Theorem 2.8 we recover equation (2.22). Substituting
the power series expansion of $\Phi_{\alpha}(n+b,a;k),\Psi_{\beta}(n+a,b;k)$
we have

$\displaystyle\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;k)}{(n+b)^{2km}}$
$\displaystyle=\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2km}}\left(\frac{\log\left(\frac{\alpha(n+b)}{\pi}\right)+\gamma_{0}(a)}{\pi}-\frac{\alpha^{k}(n+b)^{2k}}{\pi}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}\right)$
$\displaystyle\sum_{n=0}^{\infty}\frac{\Psi_{\beta}(n+a,b;k)}{(n+a)^{2km}}$
$\displaystyle=\frac{2b-1}{2\beta}\zeta(2km+1,a)-\frac{\zeta(2km,a)}{2k\cos\left(\frac{\pi}{2k}(k-1)\right)}+\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{\beta^{k-1}}{\alpha^{k}(n+a)^{2k}+\beta^{k}(i+b)^{2k}}$

Thus the identity can be written as

$\displaystyle\pi^{2k-1}(-1)^{m-1}\alpha^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{\beta^{k}(n+a)^{2k}+\alpha^{k}(i+b)^{2k}}$
(2.24)
$\displaystyle-\pi^{2k-1}\beta^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(m-1)}}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}$
$\displaystyle+\sum_{i=1}^{m-1}\frac{(-1)^{i}\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i),b)}{\pi}$

##  3\. Proofs

Before proving these results we state an important inequality which we will
use throughout the proofs. Stirling’s formula for the gamma function on a
vertical strip states that for $a\leq\sigma\leq b$ and $|t|\geq 1$,

$|\Gamma(\sigma+it)|=(2\pi)^{\frac{1}{2}}|t|^{\sigma-\frac{1}{2}}e^{-\frac{\pi}{2}|t|}\left(1+O\left(\frac{1}{|t|}\right)\right).$
(3.1)

The reflection formula for the gamma function is
$\Gamma(1-z)\Gamma(z)={\frac{\pi}{\sin\pi z}},z\not\in\mathbb{Z}$ Thus as
$\Im(s)\to\infty$ using the reflection formula and (3.1) we have the
inequality

$\frac{1}{\left|\sin\left(\frac{\pi
s}{2k}\right)\right|}=2\exp\left(\frac{-\pi}{2}\left|\frac{\Im(s)}{k}\right|\right)\left(1+O\left(\frac{1}{|\Im(s)|}\right)\right).$
(3.2)

A variant of the reflection formula for the gamma function is
$\Gamma\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}-z\right)=\frac{\pi}{\cos(\pi
z)},z\notin\mathbb{Z}-\frac{1}{2}$. Thus as $\Im(s)\to\infty$ using the
variant of reflection formula and (3.1) we have the inequality

$\frac{1}{\left|\cos\left(\frac{\pi(s+k-1)}{2k}\right)\right|}=2\exp\left(\frac{-\pi}{2}\left|\frac{\Im(s)}{k}\right|\right)\left(1+O\left(\frac{1}{|\Im(s)|}\right)\right)$
(3.3)

###  3.1. First Proof of Theorem 2.1

Let $\Psi_{\alpha}(x)=\Psi\left(\frac{\alpha x}{\pi}\right)$. On account of
absolute convergence of the Hurwitz zeta function for $\Re(s)>1$ we have

$\sum_{n=0}^{\infty}(n+b)^{-2kN-2k+1}\Psi_{\alpha}\left(n+b,a\right)=\int_{(c)}\frac{\zeta(1-s,a)\zeta(2kN+2k-1+s,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}ds.$

We now evaluate this integral by shifting the line of integration. Consider
rectangular the contour determined by the line segments
$[c-iT,c+iT],[c+iT,d+iT],[d+iT,d-iT],[d-iT,c-iT]$ where $d=-c-2kN-2k+2$.
Inside this domain, the integrand has simple poles at $0,-2kN-2k+2$ due to
$\zeta(1-s,a)$ and $\zeta(2kN+2k-1+s,a)$ respectively. It also has simple
poles at the integers $-2kp+1$ where $p\in\\{0,1,\ldots,N+1\\}$ due to the
cosine term in the denominator. The residues at these poles are

$\displaystyle
R_{0}=\frac{-\zeta(2kN+2k-1,b)}{2k\cos\left(\frac{\pi(k-1)}{2k}\right)},$
$\displaystyle
R_{-2kN-2k+2}=\left(\frac{\alpha}{\pi}\right)^{2kN+2k-2}\frac{\zeta(2kN+2k-1,a)}{2k\cos\left(\frac{\pi(-2kN-k+1)}{2k}\right)},$
$\displaystyle
R_{-2kp+1}=(-1)^{p}\left(\frac{\alpha}{\pi}\right)^{2kp-1}\zeta(2kp,a)\zeta(2k(N+1-p),b).$

Thus by Cauchy’s residue formula we have

$\displaystyle\frac{1}{2i\pi}\left[\int_{c-iT}^{c+iT}+\int_{c+iT}^{d+iT}+\int_{d+iT}^{d-iT}+\int_{d-iT}^{c-iT}\right]\frac{\zeta(1-s,a)\zeta(2kN+2k-1+s,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=R_{0}+R_{-2kN-2k+2}+\sum_{p=0}^{N+1}R_{-2kp+1}.$

From elementary bounds on the Hurwitz zeta function and Equation (3.3), it can
be seen that as $T\to\infty$, the integrals along horizontal segments tend to
zero. Under the change of variables $s\to-s-2kN-2k+2$ we have

$\int_{(d)}\frac{\zeta(1-s,a)\zeta(2kN+2k-1+s,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}$
$=(-1)^{N}\left(\frac{\alpha}{\pi}\right)^{2kN+2k-2}\int_{(c)}\frac{\zeta(1-s,b)\zeta(2kN+2k-1+s,a)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s},$

which proves Theorem 2.1.

###  3.2. Second Proof of Theorem 2.1

Choose

$a_{n}=b_{n}=1\,\,,x_{n}=\frac{-(n+a)^{2k}}{\alpha^{k}}\,\,\text{and}\,\,y_{n}=\frac{(n+b)^{2k}}{\beta^{k}}$

which has the corrosponding zeta function and zeta generating function as

$\zeta_{x,a}(q)=(-1)^{q}\alpha^{kq}\zeta(2kq,a)\,\,,\,\,\psi_{x,a}(z)=\sum_{n=0}^{\infty}\frac{\alpha^{k}z}{(n+a)^{2k}-\beta^{k}z}$

Using Theorem 1.5 we have

$\displaystyle\sum_{p=1}^{N}(-1)^{p}\zeta(2kp,a)\alpha^{kp}\zeta(2k(N+1-p),b)\beta^{k(N+1-p)}$
(3.4)
$\displaystyle=(-1)^{N}\alpha^{k(N+1)-1}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(N+1)-1}}\sum_{n=0}^{\infty}\frac{\beta^{k-1}(n+a)^{2k-1}}{\beta^{k}(n+a)^{2k}+\alpha^{k}(n+b)^{2k}}$
$\displaystyle-\beta^{k(N+1)-1}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(N+1)-1}}\sum_{n=0}^{\infty}\frac{\alpha^{k-1}(n+b)^{2k-1}}{\alpha^{k}(n+b)^{2k}+\beta^{k}(n+a)^{2k}}$

which can be seen equivalent to equation (2.6).

###  3.3. Proof of Theorem 2.4

On account of the absolute convergence of $\zeta(s,a)$ for $\Re(s)>1$ we have

$\displaystyle\sum_{n=0}^{\infty}\left(n+b\right)^{2km+1}\left[\Psi_{\alpha}(n+b,a;k)-\sum_{p=1}^{m}\frac{B_{2kp+1}(a)}{2kp+1}\left(\frac{\pi}{\alpha}\right)^{2kp+1}\right]$
$\displaystyle=\frac{1}{2i\pi}\int_{(d)}\frac{\zeta(1-s,a)\zeta(s-2km-1,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s,$

where $2km+2<d<2km+3$ since

$\int_{(c)}\frac{\zeta(1-s,a)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}x^{-s}\mathrm{d}s=\int_{(d)}\frac{\zeta(1-s,a)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}x^{-s}\mathrm{d}s-\sum_{p=1}^{m}\frac{B_{2kp+1}(a)}{2kp+1}x^{-2kp-1}.$

We now evaluate this integral by shifting the line of integration and using
the Cauchy’s Residue Theorem. Consider the rectangular contour determined by
the line segments $[d-iT,d+iT],[d+iT,e+iT],[e+iT,e-iT],[e-iT,d-iT]$ where
$e=2km+2-d$. Inside this domain, the integrand has simple poles at $0,2km+2$
due to $\zeta(1-s,a)$ and $\zeta(s-2km-2,b)$ respectively. It also has simple
poles at the integers $2kp+1$ where $p\in\\{0,1,\ldots,m\\}$ due to cosine
term in the denominator. The residues at these poles are

$\displaystyle
R_{0}=-\frac{B_{2km+2}(b)}{2k(2km+2)\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)},$
$\displaystyle
R_{2km+2}=(-1)^{m+1}\left(\frac{\pi}{\alpha}\right)^{2km+2}\frac{\alpha^{km+1}B_{2km+2}(b)}{2k(2km+2)\cos\left(\frac{\pi\left(k-1\right)}{2k}\right)},$
$\displaystyle
R_{2kp+1}=(-1)^{p}\left(\frac{\pi}{\alpha}\right)^{2kp+1}\frac{B_{2kp+1}(a)B_{2k\left(m-p\right)+1}(b)}{\pi\left(2kp+1\right)\left(2k\left(m-p\right)+1\right)}.$

Thus by Cauchy’s residue formula we have

$\displaystyle\frac{1}{2i\pi}\left[\int_{d-iT}^{d+iT}+\int_{d+iT}^{e+iT}+\int_{e+iT}^{e-iT}+\int_{e-iT}^{d-iT}\right]\frac{\zeta(1-s,a)\zeta(s-2km-1,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=R_{0}+R_{2km+2}+\sum_{p=0}^{m}R_{2kp+1}.$

From elementary bounds on the Hurwitz zeta and Equation 3.3, it can be seen
that as $T\to\infty$, the integrals along horizontal segments tend to zero.
Under the change of variables $s\to 2km+2-s$ we have

$\displaystyle\int_{(e)}\frac{\zeta(1-s,a)\zeta(s-2km-1,b)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=\left(\frac{\alpha}{\pi}\right)^{-2km-2}(-1)^{m}\int_{(d)}\frac{\zeta(1-s,b)\zeta(s-2km-1,a)}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s}\mathrm{d}s$

which proves Theorem 2.4.

###  3.4. Proof of Theorem 2.6

On account of the absolute convergence of $\zeta(s,a)$ for $\Re(s)>1$ we have

$\displaystyle\sum_{n=0}^{\infty}\left(n+b\right)^{2km-1}\left[\Phi_{\alpha}(n+b,a;k)-\sum_{p=1}^{m}\frac{B_{2kp}(a)}{2kp}\left(\frac{\pi}{\alpha}\right)^{2kp}\right]$
$\displaystyle=\frac{1}{2i\pi}\int_{(d)}\frac{\zeta(1-s,a)\zeta(s-2km+1,b)}{2k\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s,$

where $2km<d<2km+1$ since

$\int_{(c)}\frac{\zeta(1-s,a)}{2k\sin\left(\frac{\pi
s}{2k}\right)}x^{-s}\mathrm{d}s=\int_{(d)}\frac{\zeta(1-s,a)}{2k\sin\left(\frac{\pi
s}{2k}\right)}x^{-s}\mathrm{d}s-\sum_{p=1}^{m}\frac{B_{2kp}(a)}{2kp}x^{-2kp}.$

We now evaluate this integral by shifting the line of integration and using
the Residue Theorem. Consider the contour determined by the line segments
$[d-iT,d+iT],[d+iT,e+iT],[e+iT,e-iT],[e-iT,d-iT]$ where $e=2km-d$. The
integrand has poles of order two at $0,2km$. It also has simple poles at the
integers $2kp$ where $p\in\\{1,\ldots,m-1\\}$ due to cosine term in the
denominator. The residues at these poles are

$\displaystyle
R_{0}=\frac{-1}{\pi}\left(\frac{B_{2km}(b)}{2km}\left(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a)\right)+\frac{\partial}{\partial
s}\zeta(s-2km+1,b)|_{s=0}\right)$ $\displaystyle
R_{2km}=\frac{(-1)^{m}}{\pi}\left(\frac{\pi}{\alpha}\right)^{2km}\left(\frac{B_{2km}(a)}{2km}\left(\log\left(\frac{\alpha}{\pi}\right)-\gamma_{0}(b)\right)-\frac{\partial}{\partial
s}\zeta(s-2km+1,a)\big{|}_{s=2km}\right)$ $\displaystyle
R_{2kp}=(-1)^{p}\left(\frac{\pi}{\alpha}\right)^{2kp}\frac{B_{2kp}(a)B_{2k\left(m-p\right)}(b)}{\pi\left(2kp\right)\left(2k\left(m-p\right)\right)}.$

Thus we have

$\displaystyle\frac{1}{2i\pi}\left[\int_{d-iT}^{d+iT}+\int_{d+iT}^{e+iT}+\int_{e+iT}^{e-iT}+\int_{e-iT}^{d-iT}\right]\frac{\zeta(1-s,a)\zeta(s-2km+1,b)}{2k\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=R_{0}+R_{2km}+\sum_{p=1}^{m-1}R_{2kp}.$

From elementary bounds on the Hurwitz zeta and Equation 3.3, it can be seen
that as $T\to\infty$, the integrals along horizontal segments tend to zero.
Under the change of variables $s\to 2km-s$ we have

$\displaystyle\int_{(e)}\frac{\zeta(1-s,a)\zeta(s-2km-1,b)}{2k\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=\left(\frac{\alpha}{\pi}\right)^{-2km}(-1)^{m-1}\int_{(d)}\frac{\zeta(1-s,b)\zeta(s-2km-1,a)}{2k\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s}\mathrm{d}s$

which proves Theorem 2.6.

###  3.5. First Proof of Theorem 2.7

On account of absolute convergence of $\zeta(s,b)$ for $\Re(s)>1$ we have

$\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;k)}{(n+b)^{2km+1}}=\int_{(c)}\frac{\zeta(1-s,a)\zeta(2km+1+s,b)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}ds$ (3.5)

where $1<c<2$. We now evaluate this integral by shifting the line of
integration and using Residue Theorem. Consider the recangular contour
determined by the line segments
$[c-iT,c+iT],[c+iT,d+iT],[d+iT,d-iT],[d-iT,c-iT]$ where $d=-2km-c$. Inside
this domain, the integrand has simple poles at $0,-2km$ due to $\zeta(1-s,a)$
and $\zeta(s-2m-1,b)$ respectively. It also has simple poles at the integers
$-2ki$ where $i\in\\{1,\ldots,m-1\\}$ due to sine term in the denominator. The
residues at these poles are

$\displaystyle
R_{0}=\frac{1}{\pi}\left(\zeta(2km+1,b)\left(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a)\right)-\frac{\partial}{\partial
s}\zeta(2km+1+s,b)\big{|}_{s=0}\right)$ $\displaystyle
R_{-2km}=\frac{(-1)^{m}}{\pi}\left(\frac{\alpha}{\pi}\right)^{2km}\left(\zeta(2km+1,a)\left(\gamma_{0}(b)-\log\left(\frac{\alpha}{\pi}\right)\right)+\frac{\partial}{\partial
s}\zeta(2km+1+s,a)\big{|}_{s=0}\right)$ $\displaystyle
R_{-2i}=(-1)^{i}\left(\frac{\alpha}{\pi}\right)^{2ki}\frac{\zeta(2ki+1,a)\zeta(2k(m-i)+1,b)}{\pi}$

Thus we have

$\displaystyle\frac{1}{2i\pi}\left[\int_{c-iT}^{c+iT}+\int_{c+iT}^{d+iT}+\int_{d+iT}^{d-iT}+\int_{d-iT}^{c-iT}\right]\frac{\zeta(1-s,a)\zeta(s+2km+1,b)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=R_{0}+R_{-2km}+\sum_{i=1}^{m-1}R_{-2ki}$

From elementary bounds on Hurwitz zeta and (3.3), it can be seen that as
$T\to\infty$, the integrals along horizontal segments tend to zero. Under the
change of variables $s\to-2km-s$ we have

$\displaystyle\int_{(d)}\frac{\zeta(1-s,a)\zeta(s+2km+1,b)}{2\sin\left(\frac{\pi
s}{2}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=\left(\frac{\alpha}{\pi}\right)^{2km}(-1)^{m}\int_{(c)}\frac{\zeta(1-s,b)\zeta(s-2km-1,a)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s}\mathrm{d}s$

which proves Theorem 2.7.

###  3.6. Second Proof of Theorem 2.7

Choose

$a_{n}=\frac{1}{n+a}\,\,x_{n}=-\frac{(n+a)^{2k}}{\alpha^{k}}\,\,b_{n}=\frac{1}{n+b}\,\,\text{and}\,\,y_{n}=\frac{(n+b)^{2k}}{\beta^{k}}$

so that the corrosponding zeta functions are

$\zeta_{x,a}(i)=(-1)^{i}\alpha^{ki}\zeta(2ki+1,a)\,\,,\,\,\zeta_{y,b}(m-i)=\beta^{k(m-i)}\zeta(2k(m-i)+1,b)$

and corrosponding zeta generating functions are

$\psi_{x,a}(z)=-\sum_{n=0}^{\infty}\frac{\alpha^{k}z}{(n+a)((n+a)^{2k}+\alpha^{k}z)}\,\,,\,\,\psi_{y,b}(z)=\sum_{n=0}^{\infty}\frac{\beta^{k}z}{(n+b)((n+b)^{2k}-\beta^{k}z)}$

Using Theorem 1.5 we have

$\displaystyle=\sum_{i=1}^{m-1}(-1)^{i}\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i)+1,b)$
$\displaystyle-\pi^{2k}\beta^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}$
$\displaystyle-\pi^{2k}(-\alpha)^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{(i+b)(\beta^{k}(i+b)^{2k}+\alpha^{k}(n+a)^{2k})}$

which is equivalent to equation (2.21).

###  3.7. First Proof of Theorem 2.8

On account of the absolute convergence of $\zeta(s,a)$ for $\Re(s)>1$, $k>1$
we have

$\sum_{n=0}^{\infty}\frac{\Phi_{\alpha}(n+b,a;1)}{(n+b)^{2km}}=\frac{1}{2i\pi}\int_{(c)}\frac{\zeta(1-s,a)\zeta(2km+s,b)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}ds$

We now evaluate this integral by shifting the line of integration and using
Cauchy’s Residue Theorem. Consider the contour determined by the line segments
$[c-iT,c+iT],[c+iT,d+iT],[d+iT,d-iT],[d-iT,c-iT]$ where $d=-c-2km+1$. Inside
this domian, the integrand has a simple pole at $-2km+1$ due to
$\zeta(2km+s,b)$ and a pole of order two at $s=0$. It also has simple poles at
the integers $-2ki$ where $i\in\\{1,\ldots,m\\}$ due to sine term in the
denominator. The residues at these poles are

$\displaystyle
R_{0}=\frac{1}{\pi}\left(\zeta(2km,b)(\log\left(\frac{\alpha}{\pi}\right)+\gamma_{0}(a))-\frac{\partial}{\partial
s}\zeta(2km+s,b)\big{|}_{s=0}\right)$ $\displaystyle
R_{-2km+1}=\left(\frac{\alpha}{\pi}\right)^{2km-1}\frac{(-1)^{m}\zeta(2km,a)}{2\sin\left(\frac{\pi}{2k}\right)}$
$\displaystyle
R_{-2ki}=(-1)^{i}\left(\frac{\alpha}{\pi}\right)^{2ki}\frac{\zeta(2ki+1,a)\zeta(2k(m-i),b)}{\pi}$

Thus by Cauchy’s Residue theorem we have

$\displaystyle\frac{1}{2i\pi}\left[\int_{c-iT}^{c+iT}+\int_{c+iT}^{d+iT}+\int_{d+iT}^{d-iT}+\int_{d-iT}^{c-iT}\right]\frac{\zeta(1-s,a)\zeta(s-2km-1,b)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=R_{0}+R_{-2km+1}+\sum_{i=1}^{m}R_{-2ki}$

From elementary bounds on Hurwitz zeta and 3.2, it can be seen that as
$T\to\infty$, the integrals along horizontal segments tend to zero. Under the
change of variables $s\to-s-2km+1$ we have

$\displaystyle\int_{(d)}\frac{\zeta(1-s,a)\zeta(2km+s,b)}{2\sin\left(\frac{\pi
s}{2k}\right)}\left(\frac{\alpha}{\pi}\right)^{-s}\mathrm{d}s$
$\displaystyle=\left(\frac{\alpha}{\pi}\right)^{2km-1}(-1)^{m}\int_{(c)}\frac{\zeta(2km+s,a)\zeta(1-s,b)}{2\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s}\mathrm{d}s$

However,

$\displaystyle\int_{(c)}\frac{\zeta(2km+s,a)\zeta(1-s,b)}{2\cos\left(\frac{\pi(s+k-1)}{2k}\right)}\left(\frac{\beta}{\pi}\right)^{-s}\mathrm{d}s=\sum_{n=0}^{\infty}\frac{\Psi_{\beta}(n+a,b;k)}{(n+a)^{2m}}$

which proves Theorem 2.8.

###  3.8. Second Proof of Theorem 2.8

Choose

$a_{n}=\frac{1}{n+a}\,\,x_{n}=-\frac{(n+a)^{2k}}{\alpha^{k}}\,\,b_{n}=1\,\,\text{and}\,\,y_{n}=\frac{(n+b)^{2k}}{\beta^{k}}$

so the the corrosponding zeta functions are

$\zeta_{x,a}(i)=(-1)^{i}\alpha^{i}\zeta(2ki+1,a)\,\,,\,\,\zeta_{y,b}(m-i)=\beta^{k(m-i)}\zeta(2k(m-i),b)$

and zeta generating functions are

$\psi_{x,a}(z)=-\sum_{n=0}^{\infty}\frac{1}{n+a}\frac{\alpha^{k}z}{(n+a)^{2k}+\alpha^{k}z}\,\,,\,\,\psi_{y,b}(z)=\sum_{n=0}^{\infty}\frac{\beta^{k}z}{(n+b)^{2k}-\beta^{k}z}$

Using Theorem 1.5 we have

$\displaystyle\sum_{i=1}^{m-1}(-1)^{i}\alpha^{ki}\zeta(2ki+1,a)\beta^{k(m-i)}\zeta(2k(m-i),b)$
$\displaystyle=\pi^{2k}(-\alpha^{k})^{m-1}\sum_{n=0}^{\infty}\frac{1}{(n+a)^{2k(m-1)+1}}\sum_{i=0}^{\infty}\frac{1}{\beta^{k}(n+a)^{2k}+\alpha^{k}(i+b)^{2k}}$
$\displaystyle-\pi^{2k}\beta^{k(m-1)}\sum_{n=0}^{\infty}\frac{1}{(n+b)^{2k(m-1)}}\sum_{i=0}^{\infty}\frac{1}{(i+a)(\beta^{k}(i+a)^{2k}+\alpha^{k}(n+b)^{2k})}$

which can be seen equivalent to equation (2.8).

##  4\. Concluding Remarks and future directions

Integral transforms are useful in number theory as can be witnessed from the
results obtained in this paper. In a recent paper A. Dixit et al [8] found a
Ramanujan Type identity for squares of zeta by considering the kernel

$\Omega(x)=\int_{(c)}\frac{\zeta(1-s,a)^{2}}{2k\cos\left(\frac{\pi
s}{2}\right)}x^{-s}ds,$

which they call Koshliakov kernel first introduced by N.S. Koshliakov in [11]
stated as follows

$\displaystyle\left(-\beta^{2}\right)^{-N}\left\\{\zeta^{2}\left(2N+1\right)\left(\gamma+\log\left(\frac{\beta}{\pi}\right)-\frac{\zeta^{\prime}\left(2N+1\right)}{\zeta\left(2N+1\right)}\right)+\sum_{n=1}^{+\infty}\frac{\tau_{0}(n)}{n^{2N+1}}\,\Omega\left(\dfrac{\beta^{2}n}{\pi^{2}}\right)\right\\}$
$\displaystyle-\left(\alpha^{2}\right)^{-N}\left\\{\zeta^{2}\left(2N+1\right)\left(\gamma+\log\left(\frac{\alpha}{\pi}\right)-\frac{\zeta^{\prime}\left(2N+1\right)}{\zeta\left(2N+1\right)}\right)+\sum_{n=1}^{+\infty}\frac{\tau_{0}(n)}{n^{2N+1}}\,\,\Omega\left(\dfrac{\alpha^{2}n}{\pi^{2}}\right)\right\\}$
$\displaystyle=2^{4N}\pi\sum_{j=0}^{N+1}\frac{\left(-1\right)^{j}\mathcal{B}_{2j}^{2}\mathcal{B}_{2N+2-2j}^{2}}{\left(\left(2j\right)!\right)^{2}\left(\left(2N+2-2j\right)!\right)^{2}}\left(\alpha^{2}\right)^{j}\left(\beta^{2}\right)^{N+1-j}.$

We have restated their transformation in terms of the Koshliakov kernel
$\Omega(x)$, which has equivalent expressions [8].

By considering the generalized Koshaliakov kernel

$\Omega(x,a;k)=\int_{(c)}\frac{\zeta(1-s,a)^{2}}{2k\cos\left(\frac{\pi(s+k-1)}{2k}\right)}x^{-s}ds$

one can obtain two parameter generalization of A. Dixits identity. However
$\zeta(1-s,a)^{2}$ does not have simple coefficients so we consider the
special case $a=\frac{1}{n}$ where $n\in\mathbb{N}$. Similar to the odd zeta
kernel one can define a new kernel

$\Omega\left(x,a;k\right)=\int_{(c)}\frac{\zeta\left(1-s,a\right)^{2}}{2k\sin\left(\frac{\pi
s}{2k}\right)}x^{-s}ds$

to find Generalization of Theorem 2.7. This will be explored in a future
publication.

##  5\. Acknowledgements

The author would like to thank Christophe Vignat and Tanay Wakhare for their
guidance and support throughout the completion of this work and for taking
time to read a rough draft of this manuscript.

## References

  * [1] G. E. Andrews and B.C. Berndt, Ramanujan’s Lost Notebook, Part IV, Springer, New York, 2013.
  * [2] T. M. Apostol. Introduction to analytic number theory. Springer, 1998 .
  * [3] B. C. Berndt and A. Straub, Ramanujan’s formula for $\zeta(2n+1)$, in: Exploring the Riemann Zeta Function: 190 years from Riemann’s Birth, H. Montgomery, A. Nikeghbali and M. Th. Rassias Eds, Springer, 13–34, 2017.
  * [4] R. A Rankin and B. C. Berndt, Ramanujan’s notebooks, part ii (Springer-Verlag, Berlin–heidelberg–new york, 1989), 359 pp., 3 540 96794 x,£ 55. Proceedings of the Edinburgh Mathematical Society, 33(1):166–166, 1990
  * [5] P. Chavan, S. Chavan, C. Vignat, and T. Wakhare. Dirichlet series under standard convolutions: Variations on Ramanujan’s identity for odd zeta values. arXiv preprint arXiv:2107.06457, 2021
  * [6] S. Chavan. An elementary proof of Ramanujan’s identity for odd zeta values, J. Classical Anal., 19 (2), 139-147, 2022.
  * [7] A. Dixit, A. Kesarwani, and R. Kumar. A generalized modified Bessel function and explicit transformations of certain Lambert series. arXiv preprint arXiv:2012.12064, 2020.
  * [8] A. Dixit and R. Gupta, On squares of odd zeta values and analogues of Eisenstein series, Adv. in Appl. Math., 110, 86 – 119, 2019.
  * [9] A. Dixit, R. Gupta, and R. Kumar. Extended higher Herglotz functions I. functional equations. arXiv preprint arXiv:2107.02607
  * [10] A. Dixit and R. Gupta, Koshliakov zeta functions I. Modular relations, Adv. Math. 393 (2021), Paper No.108093.
  * [11] N. S. Koshlyakov. On Voronoi’s sum formula. Mess. Math., 58:30–32, 1929.
  * [12] J. Whitbread and L. Glaisher. On the series which represent the twelve elliptic and the four zeta functions. Mess. Math, 18:1–84, 1889.
  * [13] S. L. Malurkar, On the application of Herr Mellin’s integrals to some series, J. Indian Math. Soc. 16 (1925/26), 130–138
  * [14] S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay, 1957, 2nd edition, 2012.
  * [15] E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambridge University Press, (1963)