# A new Ramanujan-type identity for $L(2k+1,\chi_{1})$

Shashi Chourasiya  Shashi Chourasiya
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS>,  Md
Kashif Jamal  Md Kashif Jamal
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Bibekananda Maji  Bibekananda Maji
Department of Mathematics
Indian Institute of Technology Indore
Simrol, Indore, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

One of the celebrated formulas of Ramanujan is about odd zeta values, which
has been studied by many mathematicians over the years. A notable extension
was given by Grosswald in 1972. Following Ramanujan’s idea, we rediscovered a
Ramanujan-type identity for $\zeta(2k+1)$ that was first established by
Malurkar and later by Berndt using different techniques. In the current paper,
we extend the aforementioned identity of Malurkar and Berndt to derive a new
Ramanujan-type identity for $L(2k+1,\chi_{1})$, where $\chi_{1}$ is the
principal character modulo prime $p$. In the process, we encounter a new
family of Ramanujan-type polynomials and we notice that a particular case of
these polynomials has been studied by Lalín and Rogers in 2013. Furthermore,
we establish a character analogue of Grosswald’s identity and a few more
interesting results inspired from the work of Gun, Murty and Rath.

2020 Mathematics Subject Classification. Primary 11M06; Secondary 26C10 .
Keywords and phrases. Riemann zeta function, odd zeta values, Ramanujan
polynomials, reciprocal polynomials

Dedicated to Srinivasa Ramanujan on his $134$th birth anniversary

##  1\. introduction

The Riemann zeta function $\zeta(s)$ is one of the most important special
functions in number theory and its theory plays a crucial role for the
development of analytic number theory. In 1734, Euler established an exact
formula for $\zeta(2k)$ in terms of powers of $\pi$ and Bernoulli number
$B_{2k}$. Mainly, he showed that, for any natural number $k$,

$\zeta(2k)=(-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!}.$ (1.1)

This formula instantly tells us that all even zeta values are transcendental
since $\pi$ is transcendental and Bernoulli numbers are rational. However, in
the literature no such simple explicit formula for $\zeta(2k+1)$ exists and
not much is known about the algebraic nature of odd zeta values. Stunningly,
Roger Apréy [1, 2], in 1959, proved that $\zeta(3)$ is an irrational number,
but the arithmetic nature of $\zeta(3)$ is not known yet. In 2001, Zudilin
[39] proved that atleast one of $\zeta(5)$, $\zeta(7)$, $\zeta(9)$, and
$\zeta(11)$ is irrational. Around the same time, Rivoal [35], Ball and Rivoal
[3] proved that infinitely many odd zeta values are irrational. These are the
current best results in this direction.

Ramanujan’s notebook and lost-notebook contain many intriguing identities and
one of the most celebrated identities is about $\zeta(2k+1)$. The identity is
the following:

$\displaystyle(4\alpha)^{-k}\left\\{\frac{1}{2}\zeta(2k+1)+\sum_{n=1}^{\infty}\frac{n^{-2k-1}}{e^{2\alpha
n}-1}\right\\}$
$\displaystyle-(-4\beta)^{-k}\left\\{\frac{1}{2}\zeta(2k+1)+\sum_{n=1}^{\infty}\frac{n^{-2k-1}}{e^{2\beta
n}-1}\right\\}$
$\displaystyle=\sum_{j=0}^{k+1}(-1)^{j-1}\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{k+1-j}\beta^{j},$
(1.2)

where $\alpha\beta=\pi^{2}$ with $\alpha,\beta>0$, and
$k\in\mathbb{Z}\backslash\\{0\\}$. This identity can be found in Ramanujan’s
second notebook [33, p. 173, Ch. 14, Entry 21(i)] as well as in the lost
notebook [34, p. 319, Entry (28)]. In the course of time, this identity took
attention of many mathematicians. To know more about this formula readers are
encouraged to see the paper of Berndt and Straub [11]. Recently Dixit and the
third author [19] established a beautiful generalization of (1.2) while
extending an identity of Kanemitsu et al. [24] and another generalization to
the Hurwitz zeta function can be found in Dixit et. al. [18]. Character
analogues of Ramanujan’s formula have been studied by Berndt [5, 6] and
Bradley [13]. Further, various generalizations of (1.2) have been made by many
mathematicians, readers are refer to see [4, 7, 11, 14, 20].

Now setting $\alpha=\beta=\pi$ and replacing $k$ by $2k+1$ in (1.2), one can
obtain the following Lerch’s formula for $\zeta(4k+3)$,

$\displaystyle\zeta(4k+3)=\pi^{4k+3}2^{4k+2}\sum_{j=0}^{2k+2}(-1)^{j+1}\frac{B_{2j}}{(2j)!}\frac{B_{4k+4-2j}}{(4k+4-2j)!}-2\sum_{n=1}^{\infty}\frac{1}{n^{4k+3}(e^{2\pi
n}-1)}.$ (1.3)

To obtain (1.2), Ramanujan used the partial fraction expansion of
$\cot(\sqrt{w\alpha})\coth(\sqrt{w\beta})$, unfortunately he made an error. He
[33, p. 171, Ch. 14, Entry 19(i)], [34, p. 318, Entry (21)] offered the below
partial fraction decomposition

$\pi^{2}xy\cot(\pi x)\coth(\pi y)=1+2\pi
xy\sum_{n=1}^{\infty}{\frac{n\coth{\big{(}\frac{\pi
nx}{y}\big{)}}}{n^{2}+y^{2}}-2\pi
xy\sum_{n=1}^{\infty}{\frac{n\coth{\big{(}\frac{\pi
ny}{x}\big{)}}}{n^{2}-x^{2}}}},$ (1.4)

in which the two infinite series on the right hand side diverge individually.
The corrected version of the above partial fraction formula was later
established by R. Sitaramchandrarao [36] as below:

$\displaystyle\pi^{2}xy\cot(\pi x)\coth{(\pi y)}$
$\displaystyle=1+\frac{\pi^{2}}{3}(y^{2}-x^{2})$ $\displaystyle-2\pi
xy\sum_{m=1}^{\infty}\left(\frac{y^{2}\coth{\big{(}\frac{\pi
mx}{y}\big{)}}}{m(m^{2}+y^{2})}+\frac{x^{2}\coth{\big{(}\frac{\pi
my}{x}\big{)}}}{m(m^{2}-x^{2})}\right).$ (1.5)

On replacing $\pi x$ by $\sqrt{w\alpha}$ and $\pi y$ by $\sqrt{w\beta}$,
followed by some elementary calculations and comparing the coefficients of
$w^{k}$, for $k\geq 1$, identity $\eqref{RamaF}$ can be easily obtained. The
first published proof of (1.2) was given by Malurkar [25] in 1925. One of the
notable extensions of the Ramanujan’s formula (1.2) was given by Emil
Grosswald [22] in 1972. For any $z$ lying in the upper half plane
$\mathbb{H}$,

$\displaystyle F_{2k+1}(z)-z^{2k}F_{2k+1}\left(-\frac{1}{z}\right)$
$\displaystyle=\frac{1}{2}\zeta(2k+1)(z^{2k}-1)$ $\displaystyle+\frac{(2\pi
i)^{2k+1}}{2z}\sum_{j=0}^{k+1}z^{2k+2-2j}\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!},$
(1.6)

where $F_{k}(z)=\sum_{n=1}^{\infty}\sigma_{-k}(n)e^{2\pi inz}$ with
$\sigma_{-k}(n)=\sum\limits_{d|n}d^{-k}$. Setting $z=i\beta/\pi$,
$\alpha\beta=\pi^{2}$, with $\alpha,\beta>0$, Grosswald’s identity immediately
gives Ramanujan’s identity (1.2) for odd zeta values. More importantly, the
above identity establishes a connection with the Eisenstein series $E_{2k}(z)$
for the full modular group $SL_{2}(\mathbb{Z})$ since the Fourier series
expansion suggests that, for any $k\geq 2$,

$\displaystyle E_{2k}(z)=1-\frac{4k}{B_{2k}}F_{1-2k}(z).$

Again, for $k\geq 3$ odd, we can think $F_{k}(z)$ as an Eichler integral of
the first kind. To know more about this connection, readers are encouraged to
see [12, 21, 23]. The polynomial on the right hand side of (1.6) has been
studied by Gun, Murty and Rath [23] and they termed it as _Ramanujan’s
polynomial_ , namely,

$R_{2k+1}(z)=\sum_{j=0}^{k+1}z^{2k+2-2j}\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}.$
(1.7)

Murty, Smith and Wang [26] proved that $R_{2k+1}(z)$ has only unimodular roots
with multiplicity $1$ apart from four real roots, for $k\geq 4$. More
specifically, they have shown that the only zeros of $R_{2k+1}(z)$ that are
roots of unity at $\pm i$ which occur only if $k$ is an even natural number,
and $\pm\rho,\pm\rho^{2}$ with $\rho=e^{2\pi i/3}$ if and only if $3$ divides
$k$. This indicates that there are zeros of $R_{2k+1}(z)$ that are lying in
$\mathbb{H}$ and not $2k$-th roots of unity. Therefore, Grosswald’s identity
$\eqref{grosswald_eq}$ leads to

$\displaystyle\zeta(2k+1)=\frac{2}{z^{2k}-1}\left(F_{2k+1}(z)-z^{2k}F_{2k+1}\left(-\frac{1}{z}\right)\right)$
(1.8)

for some $z\in\mathbb{H}\cap\overline{\mathbb{Q}}$. The above observation
inspired Gun, Murty and Rath [23] to study the nature of the special values of
the function $G_{2k+1}(z)$ defined as

$\displaystyle
G_{2k+1}(z):=\frac{1}{z^{2k}-1}\left(F_{2k+1}(z)-z^{2k}F_{2k+1}\left(-\frac{1}{z}\right)\right).$
(1.9)

They obtained the following result.

Let $k\in\mathbb{N}\cup\\{0\\}$. Define $\delta_{k}=0,1,2,3$ accordingly as
the gcd$(k,6)$ equals $1,2,3$ or $6$, respectively. For any algebraic
$z\in\mathbb{H}$, the quantity

$F_{2k+1}(z)-z^{2k}F_{2k+1}\left(-\frac{1}{z}\right)$

is transcendental with at most $2k+2+\delta_{k}$ exceptions.

Ramanujan’s [33, p. 171, Ch. 14, Entry 19(i)], [34, p. 318, Entry (21)] main
idea was to use the partial fraction decomposition (1.4) to derive the formula
(1.2) for $\zeta(2m+1)$. Subsequent to that on the same page, he [33, p. 171,
Ch. 14, Entry 19(iii)] offered the following partial fraction decomposition
for product of two tangent functions:
_Let $x$ and $y$ be complex numbers such that $y/x$ is not purely imaginary.
Then_

$\displaystyle\frac{\pi}{4}\tan\left(\frac{\pi
x}{2}\right)\tanh\left(\frac{\pi
y}{2}\right)=y^{2}\sum_{n=0}^{\infty}{\frac{\tanh\left((2n+1)\frac{\pi
x}{2y}\right)}{(2n+1)\\{(2n+1)^{2}+y^{2}\\}}+x^{2}\sum_{n=0}^{\infty}{\frac{\tanh\left((2n+1)\frac{\pi
y}{2x}\right)}{(2n+1)\\{(2n+1)^{2}-x^{2}\\}}}}.$ (1.10)

In the next section, we state all the main results on this paper.

##  2\. Main results

Quite surprisingly, the above formula (1.10) gives us a Ramanujan-type
identity for $\zeta(2k+1)$.

######  Theorem 2.1.

Let $\alpha$ and $\beta$ be two positive real numbers such that
$\alpha\beta=\frac{\pi^{2}}{4}$. For any integer $k\geq 1$, we have

$\displaystyle(4\alpha)^{-k}\left(\frac{1}{2}\zeta(2k+1)\left(1-2^{-2k-1}\right)-\sum_{n=0}^{\infty}\frac{(2n+1)^{-2k-1}}{(e^{2(2n+1)\alpha}+1)}\right)$
$\displaystyle-(-4\beta)^{-k}\left(\frac{1}{2}\zeta(2k+1)\left(1-2^{-2k-1}\right)-\sum_{n=0}^{\infty}\frac{(2n+1)^{-2k-1}}{(e^{2(2n+1)\beta}+1)}\right)$
$\displaystyle=\sum_{j=1}^{k}(-1)^{j-1}(2^{2j}-1)(2^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{k+1-j}\beta^{j}.$
(2.1)

######  Remark 1.

This identity was first proved by Malurkar [25], although his derivation was
completely based on complex analysis technique. In 1978, Berndt [8, Theorem
4.7] also recovered the same identity while studying a generalized Eisenstien
series. Recently, Dixit and Gupta [17, Corollary 4.2] have also obtained the
above identity, while studying Koshliakov zeta function. Here we would like to
emphasise that the above theorem is not only true for positive integers $k$,
but also true for negative integers $k$. In the later case, one must consider
the finite sum involving Bernoulli numbers as an empty sum.

######  Remark 2.

An alternative form of the above result is the following identity:

$\displaystyle(4\alpha)^{-k}\sum_{n=0}^{\infty}\frac{\tanh{((2n+1)\alpha})}{(2n+1)^{2k+1}}-(-4\beta)^{-k}\sum_{n=0}^{\infty}\frac{\tanh{((2n+1)\beta})}{(2n+1)^{2k+1}}$
$\displaystyle=2\sum_{j=1}^{k}(-1)^{j-1}(2^{2j}-1)(2^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{k+1-j}\beta^{j}.$
(2.2)

As an immediate implication of (2.2), we obtain an exact evaluation of an
infinite series associated to tan-hyperbolic function.

######  Corollary 2.2.

For $k\geq 0$, we have

$\displaystyle\sum_{n=0}^{\infty}\frac{\tanh\left((2n+1)\frac{\pi}{2}\right)}{(2n+1)^{4k+3}}$
$\displaystyle=\frac{\pi^{4k+3}}{2}\sum_{j=1}^{2k+1}(-1)^{j-1}(2^{2j}-1)(2^{4k+4-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{4k+4-2j}}{(4k+4-2j)!}.$
(2.3)

Special cases $k=0$ and $k=1$ of (2.3) are noted down by Ramanujan as Entry
$25$ $(iii)$, $(iv)$ in Chapter $14$ of his second notebook [33, p. 180]. The
above identity has been also proved by many mathematicians and to know more
about this identity we refer to [8]. This identity immediately implies that
the above infinite series converges to a transcendental number. More
generally, Berndt [8, Theorem 4.11] proved that
$\sum_{n=0}^{\infty}\frac{\tanh\left((2n+1)\frac{\pi\theta}{2}\right)}{(2n+1)^{2k+1}}$
is a transcendental number, when $\theta$ is some certain quadratic irrational
number. Similar behaviour for other trigonometric Dirichlet series have been
studied by Lalín et al. [27] and Straub [38]. The identity (2.3) also provides
a beautiful formula for $\zeta(4k+3)$ analogous to Lerch’s famous identity
(1.3). For any $k\geq 0$,

$\displaystyle\zeta(4k+3)\left(1-\frac{1}{2^{4k+3}}\right)$
$\displaystyle=\frac{\pi^{4k+3}}{2}\sum_{n=1}^{2k+1}(-1)^{n+1}(2^{2n}-1)(2^{4k+4-2n}-1)\frac{B_{2n}}{(2n)!}\frac{B_{4k+4-2n}}{(4k+4-2n)!}$
$\displaystyle+2\sum_{n=0}^{\infty}\frac{(2n+1)^{-4k-3}}{(e^{(2n+1)\pi}+1)}.$
(2.4)

Now we would like to highlight that Theorem 2.1 can also be derived using
contour integration technique and while doing so we realized that a more
general version of it is indeed true. Here we state a one-variable
generalization of Theorem 2.1.

######  Theorem 2.3.

Let $p$ be a prime number and $\alpha$, $\beta>0$ such that
$\alpha\beta=\frac{\pi^{2}}{p^{2}}$. Let $\chi_{1}$ denote the principal
Dirichlet character modulo $p$. Then for any
$k\in\mathbb{Z}\backslash\\{0\\}$, we have

$\displaystyle(4\alpha)^{-k}\left(\frac{p-1}{2}L(2k+1,\chi_{1})-\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}e^{-2n\alpha}\right)$
$\displaystyle-(-4\beta)^{-k}\left(\frac{p-1}{2}L(2k+1,\chi_{1})-\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}e^{-2n\beta}\right)$
$\displaystyle=\sum_{j=1}^{k}{(-1)^{j-1}(p^{2j}-1)(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{k+1-j}\beta^{j}},$
(2.5)

where $L(s,\chi)$ is the Dirichlet $L$-function and

$\displaystyle a_{n}=\begin{cases}1,&\text{if}\,\,\gcd(n,p)=1,\\\
1-p,&\text{if}\,\,\gcd(n,p)=p.\end{cases}$ (2.6)

Letting $\alpha=\beta=\frac{\pi}{p}$ and replacing $k\to 2k+1$ as an odd
positive integer, Theorem 2.3 immediately provides an interesting identity for
$L(4k+3,\chi_{1})$ analogous to Lerch’s formula (1.3) for $\zeta(4k+3)$.

######  Corollary 2.4.

Let $p$ be a prime number and $\chi_{1}$ be the principal Dirichlet character
modulo $p$. For any integer $k\geq 0$, we have

$\displaystyle L(4k+3,\chi_{1})$
$\displaystyle=\frac{2}{p-1}\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}{\frac{\chi_{1}(d)}{d^{4k+3}}}e^{-\frac{2n\pi}{p}}\right)}$
$\displaystyle+\frac{2^{4k}}{2(p-1)}\left(\frac{\pi}{p}\right)^{4k+3}\sum_{j=1}^{k}{(-1)^{j-1}(p^{2j}-1)(p^{4k+4-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{4k+4-2j}}{(4k+4-2j)!}}.$
(2.7)

######  Corollary 2.5.

Let $\alpha,\beta>0$ such that $\alpha\beta=\pi^{2}/p^{2}$. For any integer
$k\geq 0$, we have

$\displaystyle\alpha^{k+1}\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\chi_{1}(d)d^{2k+1}}\bigg{)}}e^{-2n\alpha}$
$\displaystyle-(-\beta)^{k+1}\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\chi_{1}(d)d^{2k+1}}\bigg{)}}e^{-2n\beta}$
$\displaystyle=(p-1)(p^{2k+1}-1)\left(\alpha^{k+1}-(-\beta)^{k+1}\right)\frac{B_{2k+2}}{4k+4}.$

One can observe that the above identity is a resemblant of the following
identity of Ramanujan [32, Vol. 1, p. 259].

_Let $k\geq 0$ be a positive integer. For $\alpha,\beta>0$ with
$\alpha\beta=\pi^{2}$, we have_

$\displaystyle\alpha^{k+1}\sum_{n=1}^{\infty}\frac{n^{2k+1}}{e^{2n\alpha}-1}-(-\beta)^{k+1}\sum_{n=1}^{\infty}\frac{n^{2k+1}}{e^{2n\beta}-1}=(\alpha^{k+1}-(-\beta)^{k+1})\frac{B_{2k+2}}{4k+4}.$
(2.8)

The point to be noted that the Theorem 2.3 is not true for $k=0$ since
$L(s,\chi_{1})$ has a simple pole at $s=1$. Corresponding to $k=0$, we obtain
the below identity.

######  Theorem 2.6.

Let us consider the variables as we defined in Theorem 2.3. Then

$\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\frac{\chi_{1}(d)}{d}}}e^{-2n\alpha}\bigg{)}-\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\frac{\chi_{1}(d)}{d}}}e^{-2n\beta}\bigg{)}=\frac{(p-1)^{2}}{2p}\log\left(\frac{\pi}{p\alpha}\right).$
(2.9)

In particular, for $p=2$, we derive an interesting identity.

######  Corollary 2.7.

For $\alpha,\beta>0$ with $\alpha\beta=\pi^{2}/4$, we get

$\sum_{n=0}^{\infty}{\frac{(2n+1)^{-1}}{e^{2(2n+1)\alpha}+1}}-\sum_{n=0}^{\infty}{\frac{(2n+1)^{-1}}{e^{2(2n+1)\beta}+1}}=\frac{\log(\pi)-\log(2\alpha)}{4}.$
(2.10)

######  Remark 3.

The identity (2.10) is clearly analogous to the following well-known identity
of Ramanujan [32, Ch. 16, Entry 27(iii)] [34, p. 320, Formula (29)]. For
$\alpha,\beta>0$ and $\alpha\beta=\pi^{2}$,

$\sum_{n=1}^{\infty}\frac{1}{n(e^{2n\alpha}-1)}-\sum_{n=1}^{\infty}\frac{1}{n(e^{2n\beta}-1)}=\frac{\beta-\alpha}{12}+\frac{1}{4}\log\left(\frac{\alpha}{\beta}\right),$
(2.11)

which is equivalent to the transformation formula for the logarithm of the
Dedekind eta function $\eta(z)$, a half-integral weight modular form. To know
more about it, readers can see [9, p. 256], [10, p. 43]. It would be
interesting to find such an equivalent form for the identity (2.10).

###  2.1. Analogue of Grosswald’s identity and Ramanujan-type polynomials

In the introduction, we have already seen that the Grosswald’s identity (1.6)
is a natural generalization of Ramanujan’s formula (1.2). The generalized
divisor function $\sigma_{-k}(n)=\sum_{d|n}d^{-k}$ appeared in Grosswald’s
identity. A natural character analogue of the generalized divisor function is
defined by $\sigma_{-k,\chi}(n)=\sum_{d|n}\chi(d)d^{-k}$, where $\chi$ is a
Dirichlet character. This character analogue of the divisor function has been
studied by many mathematicians. Now we state a generalization Theorem 2.3,
namely, a Grosswald-type identity for complex variable $z$.

######  Theorem 2.8.

Let $p$ be a prime number and $\chi_{1}$ be the principal Dirichlet character
modulo $p$. For any $k\in\mathbb{N}$ and $z\in\mathbb{H}$, we have

$\displaystyle\left(zp\right)^{2k}$
$\displaystyle\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)=\frac{p-1}{2}L(2k+1,\chi_{1})\left\\{\left(zp\right)^{2k}-1\right\\}$
$\displaystyle+$ $\displaystyle\frac{(2\pi
i)^{2k+1}}{2z\,p^{2k+2}}\sum_{j=1}^{k}{(p^{2j}-1)(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}}(pz)^{2k+2-2j},$
(2.12)

where

$\displaystyle\mathfrak{F}_{k,\chi_{1}}(z):=\sum_{n=1}^{\infty}a_{n}\sigma_{-k,\chi_{1}}(n)\exp(2\pi
inz).$ (2.13)

Here we can clearly observe that the polynomial present on the right hand side
of the above theorem is an analogue of the Ramanujan’s polynomial (1.7). For
any prime number $p$ and $k\in\mathbb{N}$, we define this Ramanujan-type
polynomial as

$\displaystyle\mathfrak{R}_{2k+1,p}(z):=\sum_{j=1}^{k}{(p^{2j}-1)(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}}(pz)^{2k+2-2j}.$
(2.14)

Interestingly, the above polynomial for $p=2$ has been studied by Lalín and
Rogers [28, Theorem 2]. Utilizing their result, we can immediately say that
all the complex zeros of the polynomial $\mathfrak{R}_{2k+1,2}(z)$ lie on the
circle $|z|=1/2$ and roots are simple. Next, we state a more general
conjecture, which suggests that a more general version of Lalín and Rogers’s
result might be true.

######  Conjecture 2.9.

Let $p$ be a prime number and $k$ be a natural number. The Ramanujan-type
polynomial $\mathfrak{R}_{2k+1,p}(z)$ has only real zero $z=0$ of multiplicity
$2$, and rest are the non real zeros on the circle $|z|=1/p$ and are simple.

We state another conjecture, which gives information about the zeros of the
Ramanujan-type polynomial $\mathfrak{R}_{2k+1,p}(z)$ that are $2k$th roots of
unity.

######  Conjecture 2.10.

If $2|k$, then the polynomials $\mathfrak{R}_{2k+1,p}(z)$ and $(pz)^{2k}-1$
have only common roots $z=\pm i/p$, and if $2\nmid k$, they do have any common
roots.

In the final section, we provide numerical evidences for these two
conjectures. Now assuming these conjectures, inspired from the work of Gun,
Murty and Rath [23], we obtain the below mentioned results. First, let us
define a quantity, extracted from Theorem 2.8, for any $z\in\mathbb{H}$ with
$(pz)^{2k}\neq 1$, as

$\displaystyle\mathfrak{G}_{2k+1,\chi_{1}}(z):=\frac{\left(zp\right)^{2k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)}{(pz)^{2k}-1}.$
(2.15)

######  Proposition 2.11.

The set

$\\{\mathfrak{G}_{2k+1,\chi_{1}}(z)|\hskip
2.84544ptz\in\mathbb{H}\cap\overline{\mathbb{Q}},\hskip 5.69046pt(pz)^{2k}\neq
1\\}$

contains at most one algebraic number.

######  Theorem 2.12.

Let $k\geq 0$ be an integer and $p$ be a prime. Define $\delta_{k}=0,1$
respectively depending on if $\gcd(k,2)$ equals to $1$, or $2$. Then, with at
most $2k+\delta_{k}$ exceptions, the quantity

$\displaystyle\left(zp\right)^{2k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)$
(2.16)

is transcendental for every algebraic number $z\in\mathbb{H}$. This suggests
that, there are at most $2k+\delta_{k}$ algebraic numbers $z$ on the upper
half plane such that $\mathfrak{F}_{2k+1,\chi_{1}}(z)$ and
$\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)$ are both
algebraic.

##  3\. Preliminaries

The Riemann zeta function $\zeta(s)$ obeys the following asymmetric form of
functional equation, namely,

$\zeta(s)=2^{s}\pi^{s-1}\zeta(1-s)\Gamma(1-s)\sin\left(\frac{\pi
s}{2}\right).$ (3.1)

The Laurent series expansion of $\Gamma(s)$ around the point $s=0$ is the
following

$\displaystyle\Gamma(s)=\frac{1}{s}-\gamma+\frac{1}{2}\left(\gamma^{2}+\frac{\pi^{2}}{6}\right)s-\frac{1}{6}\left(\gamma^{3}+\gamma\frac{\pi^{2}}{2}+2\zeta(3)\right)s^{2}+\cdots,$
(3.2)

where $\gamma$ is the well-known Euler-Mascheroni constant. The gamma function
$\Gamma(s)$ satisfies the below reflection formula, namely,

$\displaystyle\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)},\hskip
14.22636pt\forall s\in\mathbb{C}\backslash\mathbb{Z}.$ (3.3)

The next result provides an important information about the asymptotic growth
of $\Gamma(s)$. For $s=\sigma+iT$, in a vertical strip
$\alpha\leq\sigma\leq\beta$,

$|\Gamma(\sigma+iT)|=\sqrt{2\pi}|T|^{\sigma-1/2}e^{-\frac{1}{2}|T|}\left(1+\mathcal{O}\left(\frac{1}{|T|}\right)\right),\quad\textrm{as}\,\,|T|\to\infty.$
(3.4)

Now we will state a few results which will be essential in proving our main
results.

######  Lemma 3.1.

For any $k\in\mathbb{Z}$, one has

$\Gamma(s)\zeta(s)\zeta(s+2k+1)=(-1)^{k}(2\pi)^{2s+2k}\Gamma(-s-2k)\zeta(1-s)\zeta(-s-2k).$
(3.5)

  * Proof.

Employing the functional equation (3.1) of $\zeta(s)$ and the reflection
formula (3.3) for $\Gamma(s)$, one can obtain this identity. ∎

######  Lemma 3.2.

Let $p$ be a prime number and $a_{n}$ be the function as we defined in (2.6).
Let $\chi_{1}$ be the principal Dirichlet character $mod$ $p$. Then for
$\operatorname{Re}(s)>\max\\{1,-2k\\}$,

$\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}\frac{1}{n^{s}}=L(s+2k+1,\chi_{1})\zeta(s)(1-p^{1-s}).$
(3.6)

  * Proof.

First, we note that $\zeta(s)(1-p^{1-s})=\sum_{n=1}^{\infty}a_{n}n^{-s}$ for
$\operatorname{Re}(s)>1$. Using Dirichlet convolution, for
$\operatorname{Re}(s)>\max\\{1,-2k\\}$, one can verify that

$L(s+2k+1,\chi_{1})\zeta(s)(1-p^{1-s})=\sum_{n=1}^{\infty}\left(\frac{a_{\frac{n}{d}}\chi_{1}(d)}{d^{2k+1}}\right)\frac{1}{n^{s}}.$

Note that as $\chi_{1}(d)$ survives only when $\gcd(d,p)=1$, and in that case,
we can check that $a_{\frac{n}{d}}=a_{n}$. This completes the proof of this
lemma. ∎

##  4\. Proof of main results

  * Proof of Theorem 2.1 .

Let $\alpha$ and $\beta$ be positive number satisfying
$\alpha\beta=\frac{\pi^{2}}{4}$. Substituting $\frac{\pi
x}{2}=\sqrt{w\alpha}$, $\frac{\pi y}{2}=\sqrt{w\beta}$ in the partial fraction
decomposition (1.10), we see that

$\displaystyle\frac{\pi}{4}\tan\left(\sqrt{w\alpha}\right)\tanh\left(\sqrt{w\beta}\right)$
$\displaystyle=w\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{(2n+1)\\{\alpha(2n+1)^{2}+w\\}}$
$\displaystyle+w\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\beta\\}}{(2n+1)\\{\beta(2n+1)^{2}-w\\}}.$
(4.1)

Now we can write

$\displaystyle
w\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{(2n+1)\\{\alpha(2n+1)^{2}+w\\}}$
$\displaystyle=w\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{\alpha(2n+1)^{3}}\left(1+\frac{w}{\alpha(2n+1)^{2}}\right)^{-1}$
$\displaystyle=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{(2n+1)^{2k+1}}\frac{(-1)^{k+1}w^{k}}{\alpha^{k}}.$
(4.2)

Similarly, one can show that

$\displaystyle
w\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\beta\\}}{(2n+1)\\{\beta(2n+1)^{2}-w\\}}=\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\beta\\}}{(2n+1)^{2k+1}}\frac{w^{k}}{\beta^{k}}.$
(4.3)

In view of (4.2) and (4.3), the coefficient of $w^{k}$ of the right hand side
expression of (4.1) is

$\displaystyle\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{(2n+1)^{2k+1}}\frac{(-1)^{k+1}}{\alpha^{k}}+\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\beta\\}}{(2n+1)^{2k+1}}\frac{1}{\beta^{k}}.$
(4.4)

Now we mention the Laurent series expansions [16, p. 5] for $\tan z$ and
$\tanh z$ around $z=0$, i.e., for $0<|z|<\frac{\pi}{2}$,

$\displaystyle\tan z$
$\displaystyle=\sum_{j=1}^{\infty}(-1)^{j-1}2^{2j}(2^{2j}-1)\frac{B_{2j}}{(2j)!}z^{2j-1},\quad\tanh
z=\sum_{n=1}^{\infty}2^{2n}(2^{2n}-1)\frac{B_{2n}}{(2n)!}z^{2n-1}.$

On substituting the above series expansions, the left hand side of (4.1)
becomes

$\displaystyle\frac{\pi}{4}\left(\sum_{j=1}^{\infty}(-1)^{j-1}2^{2j}(2^{2j}-1)\frac{B_{2j}}{(2j)!}(w\alpha)^{j-\frac{1}{2}}\right)\left(\sum_{n=1}^{\infty}2^{2n}(2^{2n}-1)\frac{B_{2n}}{(2n)!}(w\beta)^{n-\frac{1}{2}}\right).$

Replacing $n+j-1$ by $k$, the above product can be re-written as

$\displaystyle
2^{2k+1}\sum_{k=1}^{\infty}\sum_{j=1}^{k+1}(-1)^{j-1}(2^{2j}-1)(2^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{j}\beta^{k+1-j}w^{k}.$
(4.5)

Now collecting the coefficient of $w^{k}$ from (4.5) and together with (4.4),
we arrive at

$\displaystyle\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\alpha\\}}{(2n+1)^{2k+1}}\frac{(-1)^{k+1}}{\alpha^{k}}+\sum_{n=0}^{\infty}\frac{\tanh\\{(2n+1)\beta\\}}{(2n+1)^{2k+1}}\frac{1}{\beta^{k}}$
$\displaystyle=2^{2k+1}\sum_{j=1}^{k+1}(-1)^{j-1}(2^{2j}-1)(2^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{j}\beta^{k+1-j}.$
(4.6)

This identity is nothing but the equation (2.2). Here we point out that the
term corresponding to $j=k+1$ vanishes due to the presence of the factor
$(2^{2k+2-2j}-1)$. Now plugging $\tanh z=1-\frac{2}{e^{2z}+1}$ and multiply by
$(-1)^{k+1}$ in ( Proof of Theorem 2.1 .), we deduce that

$\displaystyle\frac{1}{\alpha^{k}}\sum_{n=0}^{\infty}\left(\frac{1}{(2n+1)^{2k+1}}-\frac{2(2n+1)^{-2k-1}}{e^{2(2n+1)\alpha}+1}\right)+\frac{(-1)^{k+1}}{\beta^{k}}\sum_{n=0}^{\infty}\left(\frac{1}{(2n+1)^{2k+1}}-\frac{2(2n+1)^{-2k-1}}{e^{2(2n+1)\beta}+1}\right)$
$\displaystyle=2^{2k+1}\sum_{j=1}^{k}(-1)^{k+j}(2^{2j}-1)(2^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{j}\beta^{k+1-j}.$
(4.7)

Finally, observe that

$\displaystyle\zeta(2k+1)\left(1-2^{-2k-1}\right)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2k+1}}.$
(4.8)

Now making use of (4.8) and replacing $j$ by $k+1-j$ in ( Proof of Theorem 2.1
.), one can obtain (2.1). ∎

  * Proof of Corollary 2.2.

Letting $\alpha=\beta=\pi/2$, replacing $k$ by $2k+1$ in (2.2) yields (2.3).
To obtain (2.4), we have to use the identity $\tanh z=1-\frac{2}{e^{2z}+1}$ in
(2.3). ∎

  * Proof of Theorem 2.3 .

First, we point out that the infinite series

$\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}e^{-2nx}$

is convergent for any positive real number $x$. Making use of the inverse
Mellin transform for $\Gamma(s)$ and employing Lemma 3.2, for
$\operatorname{Re}(s)=c>\max\\{1,-2k\\}$, we can readily show that

$\displaystyle\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}e^{-2nx}=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)(1-p^{1-s})L(s+2k+1,\chi_{1})(2x)^{-s}\mathrm{d}s,$
(4.9)

where $\chi_{1}$ is the principal Dirichlet character modulo $p$. Note that
$L(s+2k+1,\chi_{1})=\zeta(s+2k+1)(1-p^{-s-2k-1})$. Thus, we must try to
evaluate the following line integration

$\displaystyle I_{p,k}(x):=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}\mathrm{d}s.$
(4.10)

Let us define the integrand function

$f(s):=\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}.$

Now we shall try to investigate the poles of the integrand function. We know
that $\Gamma(s)$ has simple poles at negative integers including zero and
$\zeta(s)$ has a simple pole at $s=1$. Again, $\zeta(s)$ has trivial zeros at
negative even integers. Note that $(1-p^{1-s})(1-p^{-s-2k-1})$ has trivial
zeros at $s=1$ and $s=-(2k+1)$.
Case I: For $k>0$, one can check that the only simple poles of the integrand
function $f(s)$ are at $0,-1,-3,-5,\cdots,-2k+1$, and $-2k$.
Case II: When $k$ is a negative integer, $\zeta(s+2k+1)$ has trivial zeros at
all negative odd integers. Therefore, in that case, the integrand function
$f(s)$ has poles only at $0$ and $-2k$.

Now we construct a suitable rectangular contour $\mathfrak{C}$ consisting of
the vertices $c-iT,c+iT,d+iT,d-iT$ with counter-clockwise direction. Here $T$
is some large positive quantity with $c>\max\\{1,-2k\\}$ and $d$ is cleverly
chosen to be less than $\min\\{0,-2k-1\\}$ so that all the poles of the
integrand function $f(s)$ lie inside the contour $\mathfrak{C}$. Now appealing
to Cacuchy’s residue theorem, we see that

$\displaystyle\frac{1}{2\pi
i}\left(\int_{c-iT}^{c+iT}+\int_{c+iT}^{d+iT}+\int_{d+iT}^{d-iT}+\int_{d-iT}^{c-iT}\right)f(s)\mathrm{d}s=R_{0}+R_{-2k}+\mathfrak{R}(x),$
(4.11)

where

$\displaystyle\mathfrak{R}(x)=\begin{cases}\displaystyle\sum_{j=1}^{k}R_{-(2j-1)},&{\rm
if}\,\,k>0,\\\ 0,&{\rm if}\,\,k<0.\end{cases}$ (4.12)

and $R_{\rho}$ denotes the residual term corresponding to the pole at
$s=\rho$. Now we evaluate the residual terms, i.e.,

$\displaystyle R_{0}=\lim_{s\to 0}sf(s)$ $\displaystyle=\lim_{s\to
0}s\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}$
$\displaystyle=\zeta(0)\zeta(2k+1)(1-p)(1-p^{-2k-1})=\frac{(p-1)L(2k+1,\chi_{1})}{2},$
(4.13) $\displaystyle R_{-2k}$
$\displaystyle=\lim_{s\to-2k}(s+2k)\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}$
$\displaystyle=(1-p^{1+2k})(1-p^{-1})\frac{\zeta^{{}^{\prime}}(-2k)(2x)^{2k}}{(2k)!}$
$\displaystyle=(-1)^{k+1}(p-1)\left(\frac{px}{\pi}\right)^{2k}\frac{L(2k+1,\chi_{1})}{2},$
(4.14)

to obtain the final step we have used the identities
$\zeta^{{}^{\prime}}(-2k)=(-1)^{k}\frac{(2k)!}{2(2\pi)^{2k}}\zeta(2k+1)$ and
$L(2k+1,\chi_{1})=\zeta(2k+1)(1-p^{-2k-1})$. Again, we calculate

$\displaystyle\sum_{j=1}^{k}R_{-(2j-1)}$
$\displaystyle=\sum_{j=1}^{k}\lim_{s\to-(2j-1)}(s+2j-1)\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}$
$\displaystyle=\sum_{j=1}^{k}\frac{(-1)^{2j-1}}{(2j-1)!}\zeta(-2j+1)\zeta(2k-2j+2)(1-p^{2j})(1-p^{2j-2k-2})(2x)^{2j-1}$
$\displaystyle=\sum_{j=1}^{k}\frac{B_{2j}}{(2j)!}(1-p^{2j})(1-p^{2j-2k-2})\zeta(2k-2j+2)(2x)^{2j-1}$
$\displaystyle=\sum_{j=1}^{k}\frac{(-1)^{k+j}}{2}(1-p^{2j})(1-p^{2j-2k-2})\frac{B_{2j}}{(2j)!}\frac{B_{2k-2j+2}}{(2k-2j+2)!}(2\pi)^{2k-2j+2}(2x)^{2j-1}$
$\displaystyle=\sum_{j=1}^{k}\frac{(-1)^{j}}{2}\left(p^{2j}-1\right)\left(p^{2k+2-2j}-1\right)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}(2x)^{2k+1}\left(\frac{\pi}{px}\right)^{2j}.$
(4.15)

To obtain the final expression, in the penultimate step we have used Euler’s
formula (1.1) and in the ultimate step we replaced the variable $j$ by
$k+1-j$. Now with the help of Stirling’s formula (3.4) for $\Gamma(s)$ and
estimate for $\zeta(s)$, one can show that the contribution of the horizontal
integrals vanish as $T\rightarrow\infty$. Thus, letting $T\to\infty$ in
(4.11), we arrive at

$\displaystyle\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}f(s)\mathrm{d}s=\frac{1}{2\pi
i}\int_{d-i\infty}^{d+i\infty}f(s)\mathrm{d}s+R_{0}+R_{-2k}+\mathfrak{R}(x).$
(4.16)

At this moment, one of our main goals is to simplify the following left
vertical integral

$\displaystyle J_{p,k}(x):=\frac{1}{2\pi
i}\int_{d-i\infty}^{d+i\infty}\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})(2x)^{-s}\mathrm{d}s,$
(4.17)

where $d<\min\\{0,-2k\\}$. In order to write this integral as an infinite
series we will make a sutiable change of variable, as discussed below. Replace
$s$ by $-s-2k$ to obtain

$\displaystyle J_{p,k}(x)$ $\displaystyle=\frac{1}{2\pi
i}\int_{(-d-2k)}\Gamma(-s-2k)\zeta(-s-2k)\zeta(1-s)(1-p^{1+s+2k})(1-p^{s-1})(2x)^{s+2k}\mathrm{d}s.$
(4.18)

Here we note that the new line of integration
$\operatorname{Re}(s)=-d-2k>\max\\{1,-2k\\}$ as we have considered
$d<\min\\{0,-2k\\}$. Now employing Lemma 3.1 in (4.18) the left vertical
integral $J_{p,k}(x)$ takes the shape as

$\displaystyle\left(\frac{px}{\pi}\right)^{2k}\frac{(-1)^{k}}{2\pi
i}\int_{(-d-2k)}\Gamma(s)\zeta(s)\zeta(s+2k+1)(1-p^{1-s})(1-p^{-s-2k-1})\left(\frac{2\pi^{2}}{p^{2}x}\right)^{-s}\mathrm{d}s.$

In view of (4.9) and (4.10), we can clearly see that the above integral is
equals to

$\displaystyle J_{p,k}(x)$
$\displaystyle=(-1)^{k}\left(\frac{px}{\pi}\right)^{2k}I_{p,k}\left(\frac{\pi^{2}}{p^{2}x}\right)$
$\displaystyle=(-1)^{k}\left(\frac{px}{\pi}\right)^{2k}\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d^{2k+1}}\right)}e^{-\frac{2n\pi^{2}}{p^{2}x}}.$
(4.19)

Eventually, combining (4.9), (4.16) and (4.19) and collecting all the residual
terms (4.13), (4.14), (4.15), and substituting $x=\alpha$ and
$\alpha\beta=\pi^{2}/p^{2}$ and simplifying further, one can complete the
proof of Theorem 2.3. ∎

  * Proof of Corollary 2.4.

We substitute $\alpha=\beta=\frac{\pi}{p}$ and replace $k$ by $2k+1$ as a
positive odd integer in Theorem 2.3 to complete the proof of this corollary. ∎

  * Proof of Corollary 2.5.

Considering $k$ as a negative integer i.e., replacing $k$ by $-(k+1)$ in
Theorem 2.3, for $k\geq 0$, we obtain

$\displaystyle(4\alpha)^{k+1}\left(\frac{p-1}{2}L(-2k-1,\chi_{1})-\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\chi_{1}(d)d^{2k+1}}\bigg{)}}e^{-2n\alpha}\right)$
$\displaystyle=(-4\beta)^{k+1}\left(\frac{p-1}{2}L(-2k-1,\chi_{1})-\sum_{n=1}^{\infty}{a_{n}\bigg{(}\sum_{d|n}{\chi_{1}(d)d^{2k+1}}\bigg{)}}e^{-2n\beta}\right).$

Now using the fact that $L(-2k-1,\chi_{1})=(p^{2k+1}-1)\frac{B_{2k+2}}{2k+2}$
and simplifying, one can finish the proof. ∎

  * Proof of Theorem 2.6.

The proof of this identity goes along the same line as in Theorem 2.3, so we
will mention those important steps where it differs from the previous one. In
view of (4.9) and (4.10), with $k=0$, we can show that

$\displaystyle\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d}\right)}e^{-2nx}=\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)(1-p^{1-s})L(s+1,\chi_{1})(2x)^{-s}\mathrm{d}s,$
(4.20)

where $c>1$ and $L(s+1,\chi_{1})=)\zeta(s+1)(1-p^{-s-1})$. Thus, in this case,
our integrand function is

$f(s)=\Gamma(s)\zeta(s)\zeta(s+1)(1-p^{1-s})(1-p^{-s-1})(2x)^{-s}.$

One can notice that the poles of $\Gamma(s)$ at negative even integers are
getting neutralized by the trivial zeros of $\zeta(s)$, and the poles of
$\Gamma(s)$ at negative odd integers will get cancelled by the trivial zeros
of $\zeta(s+1)(1-p^{-s-1})$. Hence, the only pole of the integrand function
$f(s)$ is at $s=0$ of order $2$. Now we would like to shift the line of
integration to the left, on line $\operatorname{Re}(s)=d$ with $d<0$. Invoking
Cauchy’s residue theorem, one can show that

$\displaystyle\frac{1}{2\pi
i}\int_{c-i\infty}^{c+i\infty}f(s)\mathrm{d}s=\frac{1}{2\pi
i}\int_{d-i\infty}^{d+i\infty}f(s)\mathrm{d}s+R_{0}+R_{1},$ (4.21)

where $R_{0}$ and $R_{1}$ denote the residual terms corresponding to the poles
at $s=0$ and $s=1$ respectively. Now we jot down the Laurent series expansions
of each term present in $f(s)$ around $s=0$, namely,

$\displaystyle\Gamma(s)$
$\displaystyle=\frac{1}{s}-\gamma+\frac{1}{2}\left(\gamma^{2}+\frac{\pi^{2}}{6}\right)s+\cdots,$
$\displaystyle\zeta(s)$
$\displaystyle=-\frac{1}{2}+\zeta^{{}^{\prime}}(0)s+\cdots,$
$\displaystyle\zeta(s+1)$
$\displaystyle=\frac{1}{s}+\gamma-\gamma_{1}s+\cdots,$
$\displaystyle(1-p^{1-s})$ $\displaystyle=1-p+(p\log p)s-\frac{1}{2}p(\log
p)^{2}s^{2}+\cdots,$ $\displaystyle(1-p^{-s-1})$
$\displaystyle=\left(1-\frac{1}{p}\right)+\left(\frac{\log
p}{p}\right)s+\cdots,$ $\displaystyle(2x)^{-s}$
$\displaystyle=1+\left(\log\frac{1}{2x}\right)s+\frac{1}{2}\left(\log\frac{1}{2x}\right)^{2}s^{2}+\cdots,$

where $\gamma$ is the Euler-Mascheroni constant, and $\gamma_{n}$ are the
Stieltjes constants. Utilizing the above Laurent series expansions, one can
see that the following Laurent series expansion of $f(s)$ around $s=0$ holds,

$f(s)=\frac{(p-1)^{2}}{2p}\frac{1}{s^{2}}+\frac{(p-1)^{2}}{2p}\log\left(\frac{\pi}{px}\right)\frac{1}{s}+\cdots.$
(4.22)

This indicates that

$R_{0}=\lim_{s\rightarrow 0}\frac{{\rm d}}{{\rm
d}s}s^{2}f(s)=\frac{(p-1)^{2}}{2p}\log\left(\frac{\pi}{px}\right).$ (4.23)

The simplification of the left vertical integral goes along the same vein as
in Theorem 2.3, so we won’t repeat the calculation here. Mainly, in view of
(4.19), we reach at

$\displaystyle J_{p,0}(x)=\frac{1}{2\pi
i}\int_{d-i\infty}^{d+i\infty}f(s)\mathrm{d}s=\sum_{n=1}^{\infty}{a_{n}\left(\sum_{d|n}\frac{\chi_{1}(d)}{d}\right)}e^{-\frac{2n\pi^{2}}{p^{2}x}}.$
(4.24)

Ultimately, combining (4.20), (4.21), (4.24), and together with the residual
term $R_{0}$ in (4.23), and substituting $x=\alpha$ and
$\alpha\beta=\pi^{2}/p^{2}$, we settle the proof of (2.9). ∎

  * Proof of Theorem 2.8.

First, we note that Theorem 2.3 can be analytically continued for
$\operatorname{Re}(\alpha)>0,\operatorname{Re}(\beta)>0$. Multiplying by
$(4\alpha)^{k}$ and substituting $\alpha=-i\pi z$ and
$\beta=\frac{i\pi}{p^{2}z}$ on both sides of Theorem 2.3 yields

$\displaystyle\frac{p-1}{2}L(2k+1,\chi_{1})$
$\displaystyle\left\\{1+(-1)^{k+1}\left(\frac{\alpha}{\beta}\right)^{k}\right\\}=\mathfrak{F}_{2k+1,\chi_{1}}(z)+(-1)^{k+1}\left(\frac{\alpha}{\beta}\right)^{k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)$
$\displaystyle+2^{2k}\sum_{j=1}^{k}{(-1)^{j-1}(p^{2j}-1)(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{2k+1-j}\beta^{j}},$

where
$\mathfrak{F}_{k,\chi_{1}}(z)=\sum_{n=1}^{\infty}a_{n}\sigma_{-k,\chi_{1}}(n)\exp(2\pi
inz).$ Simplifying further, it turns out that

$\displaystyle\frac{p-1}{2}L(2k+1,\chi_{1})$
$\displaystyle\left\\{1-\left(zp\right)^{2k}\right\\}=\mathfrak{F}_{2k+1,\chi_{1}}(z)-\left(zp\right)^{2k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)$
$\displaystyle+$ $\displaystyle\frac{(2\pi
i)^{2k+1}}{2z}\sum_{j=1}^{k}{(1-p^{-2j})(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}}z^{2k+2-2j}.$

Upon simplification, one can complete the proof. ∎

  * Proof of Proposition 2.11.

This result can be proved by the contradiction method. Let us suppose that
there are at least two distinct complex algebraic numbers $z_{1}$ and $z_{2}$,
satisfying $(zp)^{2k}\neq 1$, such that $\mathfrak{G}_{2k+1,\chi_{1}}(z_{1})$
and $\mathfrak{G}_{2k+1,\chi_{1}}(z_{1})$ are both algebraic. Now, Theorem 2.8
indicates that

$\displaystyle\mathfrak{G}_{2k+1,\chi_{1}}(z_{1})$
$\displaystyle=\frac{p-1}{2}L(2k+1,\chi_{1})+\frac{(2\pi
i)^{2k+1}}{2z_{1}\,p^{2k+2}}\frac{\mathfrak{R}_{2k+1,p}(z_{1})}{\left((pz_{1})^{2k}-1\right)},$
(4.25) $\displaystyle\mathfrak{G}_{2k+1,\chi_{1}}(z_{2})$
$\displaystyle=\frac{p-1}{2}L(2k+1,\chi_{1})+\frac{(2\pi
i)^{2k+1}}{2z_{2}\,p^{2k+2}}\frac{\mathfrak{R}_{2k+1,p}(z_{2})}{\left((pz_{2})^{2k}-1\right)}.$
(4.26)

Subtracting the second equation from the first, we get

$\displaystyle\mathfrak{G}_{2k+1,\chi_{1}}(z_{1})-\mathfrak{G}_{2k+1,\chi_{1}}(z_{2})=\frac{(2\pi
i)^{2k+1}}{2p^{2k+2}}\left(\frac{\mathfrak{R}_{2k+1,p}(z_{1})}{z_{1}\left((pz_{1})^{2k}-1\right)}-\frac{\mathfrak{R}_{2k+1,p}(z_{2})}{z_{2}\left((pz_{2})^{2k}-1\right)}\right).$
(4.27)

Note that the above right side expression is transcendental, since $\pi$ is
transcendental and $\mathfrak{R}_{2k+1}(z)$ is algebraic for $z$ being an
algebraic number, whereas the left side expression is an algebraic number,
which arises a contradiction. Thus, no such distinct $z_{1},z_{2}$ exists.
This finishes the proof.

∎

  * Proof of Theorem 2.12.

Let $z\in\mathbb{H}\cup\overline{\mathbb{Q}}$ with $(pz)^{2k}\neq 1$.
Proposition 2.11 implies that the following set

$\\{\mathfrak{G}_{2k+1,\chi_{1}}(z)|\hskip
2.84544ptz\in\mathbb{H}\cap\overline{\mathbb{Q}},\hskip 5.69046pt(pz)^{2k}\neq
1\\}$

contains at most one algebraic number. Let us call that algebraic number as
$A$. Now we define a set

$\displaystyle S:=\left\\{z\in\mathbb{H}\cap\overline{\mathbb{Q}},\hskip
5.69046pt(pz)^{2k}\neq 1\big{|}\mathfrak{G}_{2k+1,\chi_{1}}(z)=A\right\\}.$
(4.28)

Theorem 2.8 indicates that the elements of the set $S$ satisfy the following
polynomial:

$\displaystyle
A\left\\{(zp)^{2k}-1\right\\}=\frac{p-1}{2}L(2k+1,\chi_{1})\left\\{\left(zp\right)^{2k}-1\right\\}+$
$\displaystyle\frac{(2\pi i)^{2k+1}}{2z\,p^{2k+2}}\mathfrak{R}_{2k+1,p}(z),$
(4.29)

which is a polynomial of degree $2k$. Therefore, we have at most $2k$
algebraic numbers $z\in\mathbb{H}$ that are satisfying $(zp)^{2k}\neq 1$ for
which $\mathfrak{G}_{2k+1,\chi_{1}}(z)$ is algebraic, that is, the quantity
$\left(zp\right)^{2k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)$
is algebraic.

On the other hand, we consider
$z\in\mathbb{H}\cup\mathbb{\overline{\mathbb{Q}}}$ that are satisfying
$(zp)^{2k}=1$. Now Theorem 2.8 gives us that

$\displaystyle\left(zp\right)^{2k}$
$\displaystyle\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)=\frac{(2\pi
i)^{2k+1}}{2z\,p^{2k+2}}\mathfrak{R}_{2k+1,p}(z).$

Note that the left side expression will be an algebraic number only when
$\mathfrak{R}_{2k+1,p}(z)=0$, otherwise it will arise a contradiction to the
fact that $\pi$ is a transcendental number. Here we employ Conjecture 2.10,
that says, if $2|k$, the polynomials $\mathfrak{R}_{2k+1,p}(z)$ and
$(pz)^{2k}-1$ have only one common root, namely, $i/p$, lying in the upper
half plane. Again, if $2\nmid k$, they do not have any common roots. This
gives the clarification of defining $\delta_{k}$. Therefore, combining both of
the cases, we can clearly see that the expression
$\left(zp\right)^{2k}\mathfrak{F}_{2k+1,\chi_{1}}\left(-\frac{1}{p^{2}z}\right)-\mathfrak{F}_{2k+1,\chi_{1}}(z)$
is transcendental, for any $z\in\mathbb{H}\cup\mathbb{\overline{\mathbb{Q}}}$
with at most $2k+\delta_{k}$ exceptions. This ends the proof of this theorem.

∎

##  5\. concluding remarks

Inspired from Ramanujan’s idea of proving his famous formula for
$\zeta(2k+1)$, in the current paper, we have established a new Ramanujan-type
formula for $L(2k+1,\chi_{1})$, namely, Theorem 2.3. And then motivated from
Grosswald’s generalization of Ramanujan’s formula, we have extended Theorem
2.3 in the upper half-plane and obtained Theorem 2.8, which influenced us to
define a new Ramanujan-type polynomial (2.14) as

$\displaystyle\mathfrak{R}_{2k+1,p}(z)=\sum_{j=1}^{k}{(p^{2j}-1)(p^{2k+2-2j}-1)\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}}(pz)^{2k+2-2j}.$
(5.1)

Replacing $z$ by $z/p$, one can check that the above polynomial
$\mathfrak{R}_{2k+1,p}(z/p)$ is a reciprocal polynomial. It is clear that
Conjecture 2.9 is equivalent of proving that all the non-real roots
$\mathfrak{R}_{2k+1,p}(z/p)$ will lie on $|z|=1$. To know more about the
location of zeros of self-reciprocal polynomials, readers are encouraged to
see [15, 29, 30, 37].

Table 1. Numerical evidence for Conjecture 2.9 and 2.10. In the below table, we have evaluated zeros of a few Ramanujan-type polynomial $\mathfrak{R}_{2k+1,p}(z/p)$.  $p$ | $k$ | $\mathfrak{R}_{2k+1,p}(z/p)$ | Real roots | Non-real roots
---|---|---|---|---
$2$ | $1$ | $\frac{z^{2}}{16}$ | $0,0$ | -
$2$ | $2$ | $-\frac{z^{2}}{192}-\frac{z^{4}}{192}$ | $0,0$ | $\pm i$
$3$ | $4$ | $-\frac{41z^{2}}{11340}-\frac{13z^{4}}{4860}-\frac{13z^{6}}{4860}-\frac{41z^{8}}{11340}$ | $0,0$ | $\pm i,\pm\sqrt{\frac{16}{123}\pm\frac{i\sqrt{14873}}{123}}$
$5$ | $3$ | $\frac{31z^{2}}{30}+\frac{169z^{4}}{225}+\frac{31z^{6}}{30}$ | $0,0$ | $\pm\sqrt{-\frac{169}{465}\pm\frac{4i\sqrt{11729}}{465}}$
$5$ | $4$ | $-\frac{4069z^{2}}{6300}-\frac{403z^{4}}{900}-\frac{403z^{6}}{900}-\frac{4069z^{8}}{6300}$ | $0,0$ | $\pm i,\pm\sqrt{\frac{48}{313}\pm\frac{19\,i\sqrt{265}}{313}}$
$7$ | $3$ | $\frac{1634z^{2}}{105}+\frac{100z^{4}}{9}+\frac{1634z^{6}}{105}$ | $0,0$ | $\pm\sqrt{-\frac{875}{2451}\pm\frac{4i\sqrt{327611}}{2451}}$
$7$ | $4$ | $-\frac{1201z^{2}}{63}-\frac{817z^{4}}{63}-\frac{817z^{6}}{63}-\frac{1201z^{8}}{63}$ | $0,0$ | $\pm i,\pm\sqrt{\frac{192}{1201}\pm\frac{i\sqrt{1405537}}{1201}}$

One can easily verify that all complex zeros of the polynomials, mentioned in
the Table 1, are lying on $|z|=1$. While collecting these numerical data we
have observed that Conjecture 2.9 might be true, even if we replace the prime
number $p$ by any natural number $n\geq 2$.

The Table 1 also indicates that, $\pm i$ are the only zeros of
$\mathfrak{R}_{2k+1,p}(z/p)$ that are $2k$th roots of unity when $k$ is
divisible by $2$. This suggests that Conjecture 2.10 is also valid for the
polynomials mentioned in Table 1.

Table 2. Verification of Theorem 2.3: The left-hand side sum over $n$ considered only first $1000$ terms with specific values of $p$, $k$ and $\alpha$, $\beta$. We have used Mathematica software to evaluate this numerical data. $p$ | $k$ | $\alpha$ | $\beta$ | Left-hand side | Right-hand side
---|---|---|---|---|---
$2$ | $1$ | $\frac{\pi}{2}$ | $\frac{\pi}{2}$ | $0.154212568767021...$ | $0.154212568767021...$
$2$ | $2$ | $1$ | $\frac{\pi^{2}}{4}$ | $0.018857641090379...$ | $0.018857641090379...$
$2$ | $2$ | $\pi$ | $\frac{\pi}{4}$ | $-0.030279567070605...$ | $-0.030279567070605...$
$2$ | $3$ | $\frac{\pi}{2}$ | $\frac{\pi}{2}$ | $0.003699346990223...$ | $0.003699346990223...$
$2$ | $3$ | $\pi$ | $\frac{\pi}{4}$ | $0.010833801899940...$ | $0.010833801899940...$
$3$ | $2$ | $\pi$ | $\frac{\pi}{9}$ | $-0.226840615859532...$ | $-0.226840615859532...$
$3$ | $3$ | $\pi$ | $\frac{\pi}{9}$ | $0.161004035376020...$ | $0.161004035376020...$
$5$ | $1$ | $\frac{\pi}{5}$ | $\frac{\pi}{5}$ | $1.579136704174297...$ | $1.579136704174297...$
$5$ | $3$ | $\pi$ | $\frac{\pi}{25}$ | $3.915620336334279...$ | $3.915620336334286...$

Acknowledgement: The authors would like to thank Prof. Bruce Berndt and Prof.
Atul Dixit for giving valuable suggestions. We are also thankful to the
Computational Number Theory (CNT) Lab, IIT Indore for providing conductive
research environment. The third author wants to thank SERB for the Start-Up
Research Grant SRG/2020/000144.

## References

  * [1] R. Apéry, _Irrationalité de $\zeta{(2)}$ et $\zeta(3)$_, Astéerisque, 61 (1979), 11–13.
  * [2] R. Apéry, _Interpolation de fractions continues et irrationalité de certaines constantes_ , Bull. Section des Sci., Tome III, Bibliothéque Nationale, Paris, 1981, 37—63.
  * [3] K. Ball and T. Rivoal, _Irrationalité d’une infinité de valeurs de la fonction zêta aux entiers impairs_ (French), Invent. Math. 146 no. 1 (2001), 193–207.
  * [4] S. Banerjee and R. Kumar, _Explicit identities on zeta values over imaginary quadratic field_ , submitted for publication, arXiv:2105.04141, 2021.
  * [5] B. C. Berndt, _Character transformation formulae similar to those for the Dedekind eta-function_ , Analytic Number Theory, Proc. Sym. Pure Math., No. 24, Amer. Math. Soc., Providence, 1973, pp. 9–30.
  * [6] B. C. Berndt, _On Eisenstein series with characters and the values of Dirichlet $L$-functions_, Acta Arith. 28 (1975), 299–320.
  * [7] B. C. Berndt, _M_ odular transformations and generalizations of several formulae of Ramanujan, Rocky Mountain J. Math. 7 (1977), 147–189.
  * [8] B. C. Berndt, _Analytic Eisenstein series, theta-functions, and series relations in the spirit of Ramanujan_ , J. Reine Angew. Math. , 303(304) (1978), 332–365.
  * [9] B. C. Berndt, Ramanujan’s Notebooks, Part II, Springer-Verlag, New York, 1989.
  * [10] B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991
  * [11] B. C. Berndt and A. Straub, _Ramanujan’s formula for $\zeta(2n+1)$_, Exploring the Riemann zeta function, Eds. H. Montgomery, A. Nikeghbali, and M. Rassias, pp. 13–34, Springer, 2017.
  * [12] B.C. Berndt and A. Straub, _On a secant Dirichlet series and Eichler integrals of Eisenstein series_ , Math. Z. 284 (2016), no. 3, 827–852.
  * [13] D. M. Bradley, _S_ eries acceleration formulas for Dirichlet series with periodic coefficients, Ramanujan J. 6 (2002), 331–346.
  * [14] P. Chavan, S. Chavan, C. Vignat, and T. Wakhare, _Dirichlet series under standard convolutions: Variations on Ramanujan’s identity for odd zeta values_ , arXiv:2107.06457.
  * [15] W. Chen, _On the polynomials with all their zeros on the unit circle_. J. Math. Anal. Appl. 190 (1995), no. 3, 714–724.
  * [16] H. Cohen, Number Theory Volume II: Analytic and Modern Tolls, Springer, New York, 2007.
  * [17] A. Dixit and R. Gupta, _Koshliakov zeta functions I: Modular relations_ , accepted, Adv. Math. 2021.
  * [18] A. Dixit, R. Gupta, R. Kumar, and B. Maji, _Generalized Lambert series, Raabe’s cosine transform and a generalization of Ramanujan’s formula for $\zeta(2m+1)$_, Nagoya Math. J. 239 (2020), 232–293.
  * [19] A. Dixit and B. Maji, _G_ eneralized Lambert series and arithmetic nature of odd zeta values, Proc. Roy. Soc. Edinburg Sect. A: Mathematics 150 (2020), 741–769.
  * [20] A. Gupta and B. Maji, On Ramanujan’s formula for $\zeta(1/2)$ and $\zeta(2m+1)$, J. Math. Anal. Appl. 507 (2022), 125738.
  * [21] L. Goldstein, _Zeta functions and Eichler integrals_ , Acta Arith. 36 (1980) 229–256.
  * [22] E. Grosswald, _‘Remarks concerning the values of the Riemann zeta function at integral, odd arguments’_ , J. Number Theory 4 (1972) 225–235.
  * [23] S. Gun, M. R. Murty, and P. Rath, _Transcendental values of certain Eichler integrals,_ Bull. London Math. Soc., $2011$.
  * [24] S. Kanemitsu, Y. Tanigawa and M. Yoshimoto, _On the values of the Riemann zeta-function at rational arguments._ Hardy-Ramanujan J. 24 (2001), 11–19.
  * [25] S. L. Malurkar, _O_ n the application of Herr Mellin’s integrals to some series, J. Indian Math. Soc. 16 (1925/26), 130–138.
  * [26] R. Murty, C. Smyth and R. Wang, _Zeros of Ramanujan polynomials_ , J . Ramanujan Math. Soc. 26 (2011) 107–125.
  * [27] M. Lalín, F. Rodrigue and M. Rogers, _Secant zeta functions_ , J. Math. Anal. Appl., 409 (2014), 197–204.
  * [28] M. Lalín and M. Rogers, _Variations of the Ramanujan polynomials and remarks on $\zeta(2j+1)/\pi^{2j+1}$_, Functiones et Approximatio 48.1 (2013), 91–111.
  * [29] P. Lakatos, _On zeros of reciprocal polynomials_ , C. R. Math. Rep. Acad. Sci. Canada 24 (2002), 91–96.
  * [30] P. Lakatos and L. Losonczi, _Polynomials with all zeros on the unit circle_ , Acta Math. Hungar. 125, no. 4, (2009), 341–356.
  * [31] S. Ramanujan, Collected Papers, Cambridge University Press, Cambridge, $1927$; reprinted by Chelsea, New York, $1962$; reprinted by the American Mathematical Society, Providence, RI, $2000$.
  * [32] S. Ramanujan, _Notebooks of Srinivasa Ramanujan_ , Volume 1, Tata Institute of Fundamental Research, Bombay, Second edition 2012.
  * [33] S. Ramanujan, _Notebooks of Srinivasa Ramanujan_ , Volume 2, Tata Institute of Fundamental Research, Bombay, Second edition 2012.
  * [34] S. Ramanujan, _The Lost Notebook and Other Unpublished Papers_ , Narosa, New Delhi, 1988.
  * [35] T. Rivoal, _La fonction zêta de Riemann prend une infinité de valeurs irrationnelles aux entiers impairs_ , C. R. Acad. Sci. Paris Sér. I Math. 331 no. 4 (2000), 267–270.
  * [36] R. Sitaramchandrarao, _Ramanujan’s formula for $\zeta(2n+1)$,_ Madurai Kamaraj University Technical Report 4, pp. 70–117, 1997.
  * [37] A. Schinzel, _Self-inversive polynomials with all zeros on the unit circle_ , Ramanujan J., 9 (2005), 19–23.
  * [38] A. Straub, _Special values of trigonometric Dirichlet series and Eichler integrals_ , Ramanujan J. 41 (2016), 269–285.
  * [39] W. W. Zudilin, _One of the numbers $\zeta(5),\zeta(7),\zeta(9)$ and $\zeta(11)$ is irrational_ (Russian), Uspekhi Mat. Nauk 56 No. 4 (2001), 149–150; translation in Russian Math. Surveys 56 No. 4 (2001), 774–776.