# A Number Field Analogue of Ramanujan’s identity for $\zeta(2m+1)$

Diksha Rani Bansal  Diksha Rani
Department of Mathematics
Indian Institute of Technology Indore
Indore, Simrol, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Bibekananda Maji  Bibekananda Maji
Discipline of Mathematics
Indian Institute of Technology Indore
Indore, Simrol, Madhya Pradesh 453552, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Ramanujan’s famous formula for $\zeta(2m+1)$ has captivated the attention of
numerous mathematicians over the years. Grosswald, in 1972, found a simple
extension of Ramanujan’s formula which in turn gives transformation formula
for Eisenstein series over the full modular group. Recently, Banerjee, Gupta
and Kumar found a number field analogue of Ramanujan’s formula. In this paper,
we present a new number field analogue of the Ramanujan-Grosswald formula for
$\zeta(2m+1)$ by obtaining a formula for Dedekind zeta function at odd
arguments. We also obtain a number field analogue of an identity of
Chandrasekharan and Narasimhan, which played a crucial role in proving our
main identity. As an application, we generalize transformation formula for
Eisenstein series $G_{2k}(z)$ and Dedekind eta function $\eta(z)$. A new
formula for the class number of a totally real number field is also obtained,
which provides a connection with the Kronceker’s limit formula for the
Dedekind zeta function.

2020 Mathematics Subject Classification. Primary 11M06, 11R42; Secondary
11R29.
Keywords and phrases. Eisenstein series, Riemann zeta function, Odd zeta
values, Dedekind zeta function, Ramanujan’s formula, Class number.

_Dedicated to Professor Bruce Berndt on the occasion of his 85th birthday_

##  1\. Introduction

The theory of the Riemann zeta function $\zeta(s)$ is a central object of
study in number theory that holds a substantial place in the mathematical
landscape. The nature of special values of $\zeta(s)$ has a rich history.
Euler gave an exact evaluation for $\zeta(2k)$ in 1734, which establishes a
relation between Bernoulli numbers and even zeta values. More precisely, for
every positive integer $k$,

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

where $B_{2k}$ denotes $2k$th Bernoulli number. The above formula instantly
implies that even zeta values are transcendental. However, we have very little
information about algebraic nature of positive odd zeta values. In 1979, Apery
[1, 2] achieved a breakthrough by proving the irrationality of $\zeta(3)$. In
2001, Rivoal [31], Ball and Rivoal [3] proved that there exist infinitely many
odd zeta values which are irrational. A result due to Zudilin [34], states
that at least one of $\zeta(5),\zeta(7),\zeta(9)$ and $\zeta(11)$ are
irrational, which is the most notable achievement in this area as of now.
Prior to all these results, Ramanujan in his Notebook [28, p. 173, Ch. 14,
Entry 21(i)] as well as in Lost Notebook [29, p. 319, Entry (28)] gave an
intriguing formula involving odd zeta values, that has drawn the attention of
many mathematicians. For any $\alpha,\beta>0$ with
$\alpha\beta=\pi^{2},k\in\mathbb{Z}\setminus\\{0\\}$, we have

$\displaystyle
H_{2k+1}(\alpha)+(-1)^{k+1}H_{2k+1}(\beta)=\sum^{k+1}_{j=0}(-1)^{j-1}\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}\alpha^{k+1-j}\beta^{j},$
(1.2)

where

$H_{2k+1}(x)=(4x)^{-k}\left(\frac{1}{2}\zeta(2k+1)+\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{-2xn}\right),$

and the generalized divisor function
$\sigma_{z}(n)=\sum_{d|n}d^{z},z\in\mathbb{C}.$

Over the years, numerous mathematicians have generalized Ramanujan’s formula
in a variety of ways. Ramanujan [28, Ch. 14, Entry 8(iii)] himself gave a huge
generalization of (1.2). An analogue of Ramanujan’s formula (1.2) for
$L$-functions associated to modular forms was discussed by Razar [30, Theorem
2] and Weil [35] independently. An extension for Dirichlet $L$-function, Lerch
zeta function, and more generally for any Dirichlet series with periodic
coefficients was given by Bradley [10] in 2002. Quite surprisingly, in 1977,
Berndt [8] showed that Euler’s formula (1.1) and Ramanujan’s formula (1.2) are
branches of a bigger tree, that is, they can be obtained from a single
transformation formula for a generalized Eisenstien series. Recently, Dixit
and the second author [16] found an interesting one variable generalization of
(1.2), and later Dixit et. al. [18] and Chavan [12] established different
generalizations of (1.2) for the Hurwitz zeta function. To know more detailed
information about the Ramanujan’s formula (1.2), we refer to [5, p. 276] and a
survey article by Berndt and Straub [9], and an expository paper by Dixit [15]
where one can find recent developments. Readers are also encouraged to see [6,
7, 13, 14, 20, 21, 22].

Ramanujan’s formula (1.2) exhibits a profound correlation with Eisenstein
series. Let $\mathbb{H}$ be the upper half plane. For $z\in\mathbb{H}$ and an
integer $k\geq 2$, we define the holomorphic Eisenstein series $G_{2k}(z)$ of
weight $2k$ for the full modular group $\textrm{SL}_{2}(\mathbb{Z})$,

$\displaystyle
G_{2k}(z)=\sum_{(m,n)\in\mathbb{Z}^{2}\setminus\\{(0,0)\\}}\frac{1}{(m+nz)^{2k}}.$

For $a,b,c,d\in\mathbb{Z}$ with $ad-bc=1$, it satisfies the following modular
transformation:

$\displaystyle G_{2k}\left(\frac{az+b}{cz+d}\right)=(cz+d)^{2k}G_{2k}(z).$
(1.3)

Then the Fourier series expansion of $G_{2k}(z)$ is given by

$\displaystyle
G_{2k}(z)=2\zeta(2k)\left(1+\frac{2}{\zeta(1-2k)}\sum_{n=1}^{\infty}\sigma_{2k-1}(n)e^{2\pi
inz}\right).$

In particular, $G_{2k}(z)$ satisfies the following two main transformation
formulae,

$\displaystyle G_{2k}(z+1)$ $\displaystyle=G_{2k}(z),$ (1.4) $\displaystyle
G_{2k}\left(-\frac{1}{z}\right)$ $\displaystyle=z^{2k}G_{2k}(z),$ (1.5)

which yield (1.3). We now introduce a simple extension of (1.2) given by
Grosswald [19], in 1972, which shows how the above transformation formula
(1.5) of $G_{2k}(z)$ is connected to Ramanujan’s identity. It states that for
$z\in\mathbb{H}$, $k\in\mathbb{Z}\setminus\\{0\\}$,

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

where

$\displaystyle F_{k}(z)=\sum_{n=1}^{\infty}\sigma_{-k}(n)e^{2\pi inz}$ (1.7)

and $R_{2k+1}(z)$ is the Ramanujan polynomial, introduced by Gun, Murty and
Rath [20], defined as

$\displaystyle
R_{2k+1}(z)=\sum_{j=0}^{k+1}z^{2k+2-2j}\frac{B_{2j}}{(2j)!}\frac{B_{2k+2-2j}}{(2k+2-2j)!}.$
(1.8)

More about the above polynomial (1.8) can be seen in the paper by Murty, Smyth
and Wang [26]. Setting $z=i\beta/\pi$, $\alpha\beta=\pi^{2},$ with
$\alpha,\beta>0$, Grosswald’s identity immediately gives Ramanujan’s formula
(1.2). If $k<-1$ in (1.6), then

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

One can easily check that the above formula is nothing but the transformation
formula (1.5) of the Eisenstien series $G_{2k}(z)$. We further note that the
above formula is equivalent to an identity of Ramanujan [5, p. 261, Entry 13].

In the literature, there are several generalizations of $\zeta(s)$, however,
the Dedekind zeta function is considered as one of the most noteworthy
generalizations of $\zeta(s)$ to the number fields. In this paper, our main
aim is to establish a new generalization of Ramanujan’s formula (1.2) for
Dedekind zeta function. Before delving deeper, we first define the Dedekind
zeta function.

####  1.0.1. Dedekind zeta function

Let $\mathbb{F}$ be a number field with the degree
$[\mathbb{F}:\mathbb{Q}]=d=r_{1}+2r_{2}$, where $r_{1}$ and $r_{2}$ denote the
number of real and complex embeddings (upto conjugates) of $\mathbb{F}$. Let
$D$ be the absolute value of the discriminant $D_{\mathbb{F}}$ of
$\mathbb{F}$. Let $\mathcal{O}_{\mathbb{F}}$ be the ring of integers and
$\mathfrak{N}$ be the norm map of $\mathbb{F}$ over $\mathbb{Q}$. Let
$\mathtt{a}_{\mathbb{F}}(n)$ be the number of ideals in
$\mathcal{O}_{\mathbb{F}}$ with norm $n$. Then the Dedekind zeta function
associated to $\mathbb{F}$ is defined by the following Dirichlet series:

$\displaystyle\zeta_{\mathbb{F}}(s)=\sum_{\mathfrak{a}\subset\mathcal{O}_{\mathbb{F}}}\frac{1}{\mathfrak{N}(\mathfrak{a})^{s}}=\sum_{n=1}^{\infty}\frac{\mathtt{a}_{\mathbb{F}}(n)}{n^{s}},\quad\operatorname{Re}(s)>1,$
(1.10)

where $\mathfrak{a}$ runs over the non-zero integral ideals of
$\mathcal{O}_{\mathbb{F}}$. It is well known that $\zeta_{\mathbb{F}}(s)$ has
a simple pole at $s=1$ and the residue is given by the class number formula:

$\displaystyle\lim_{s\rightarrow
1}(s-1)\zeta_{\mathbb{F}}(s)=\frac{2^{r_{1}}(2\pi)^{r_{2}}}{\sqrt{D}}\frac{R_{\mathbb{F}}h_{\mathbb{F}}}{w_{\mathbb{F}}}:=H_{\mathbb{F}},$
(1.11)

where $w_{\mathbb{F}}$ denotes the number of roots of unity contained in
$\mathcal{O}_{\mathbb{F}},h_{\mathbb{F}}$ is the class number, and
$R_{\mathbb{F}}$ denotes the regulator of $\mathbb{F}$. Throughout the paper,
for simplicity, we denote the above residue as $H_{\mathbb{F}}$. Further, we
know that $r=r_{1}+r_{2}-1$ is the rank of the unit group of $\mathbb{F}$ and
$\zeta_{\mathbb{F}}(s)$ has a zero of order $r$ at $s=0$ with

$\displaystyle\lim_{s\rightarrow
0}\frac{\zeta_{\mathbb{F}}(s)}{s^{r}}=-\frac{R_{\mathbb{F}}h_{\mathbb{F}}}{w_{\mathbb{F}}}:=C_{\mathbb{F}}.$
(1.12)

Note that the constants $H_{\mathbb{F}}$ and $C_{\mathbb{F}}$ are related by
the relation $\sqrt{D}H_{\mathbb{F}}=-2^{r_{1}}(2\pi)^{r_{2}}C_{\mathbb{F}}$.
Now we define a _generalized divisor function_ attached to a given number
field $\mathbb{F}$, which was recently studied by Gupta and Pandit [23,
Equation (1.5)]:

$\displaystyle\sigma_{\mathbb{F},\ell}(n)=\sum_{d|n}\mathtt{a}_{\mathbb{F}}(d)\mathtt{a}_{\mathbb{F}}\left(\frac{n}{d}\right)d^{\ell}.$
(1.13)

In their paper, they investigated Riesz sum associated to
$\sigma_{\mathbb{F},\ell}(n)$. One can check that the Dirichlet series
associated to $\sigma_{\mathbb{F},\ell}(n)$ is given by

$\displaystyle\sum_{n=1}^{\infty}\frac{\sigma_{\mathbb{F},\ell}(n)}{n^{s}}=\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s-\ell),\quad\operatorname{Re}(s)>\max\\{1,1+\operatorname{Re}(\ell)\\}.$
(1.14)

A number field analogue of Euler’s identity (1.1) has been given by Klingen
[25] and Siegel [32]. It states that for any totally real number field
$\mathbb{F}$ of degree $n$, we have

$\displaystyle\zeta_{\mathbb{F}}(2m)=\frac{q_{m}\pi^{2mn}}{\sqrt{D}},\quad
m\in\mathbb{N},$ (1.15)

where $q_{m}$ is some fixed non-zero rational number. From the above identity,
we notice that like $\zeta(2m)$, even zeta values over totally real number
fields are also transcendental. Recently, Murty and Pathak [27] have also
studied the arithmetic nature of Dedekind zeta function at odd positive
integers. Their result states that for a given integer $n\geq 1$, at most one
of $\zeta_{\mathbb{F}}(2n+1)$ is rational when $\mathbb{F}$ runs over all
imaginary quadratic fields. Recently, Banerjee, Gupta and Kumar [4] have
studied transformation formula for Dedekind zeta function at odd integers
which gives a generalization of (1.2). However, our generalization is
different from their result.

Now we discuss the Steen function before mentioning the main result. The Steen
function $V(z|a_{1},a_{2},\ldots,a_{n})$ is defined by

$\displaystyle V(z|a_{1},a_{2},\ldots,a_{n})=\frac{1}{2\pi
i}\int_{(c)}\prod_{j=1}^{n}\Gamma(s+a_{j})z^{-s}ds.$ (1.16)

Here and throughout the article the symbol $(c)$ denotes the vertical line
from $c-i\infty$ to $c+i\infty$. We assume that all the poles of
$\Gamma(s+a_{j})$ lie on one side of the vertical line $(c)$. One can check
that this function is a particular case of Meijer $G$-function. Special cases
of Steen function are associated to many other well-known functions such as,

$\displaystyle V(z|0)$ $\displaystyle=e^{-z},~{}~{}\textrm{if}~{}~{}c>0,$
(1.17) $\displaystyle V(z|a,b)$
$\displaystyle=2z^{\frac{1}{2}(a+b)}K_{a-b}(2z^{\frac{1}{2}}),~{}~{}\textrm{if}~{}~{}c>\max\\{-a,-b\\},$
(1.18)

where $K_{\nu}$ denotes the modified Bessel function of the second kind.
Further information about Steen function can be found in [33], [24, p. 63].

Recall that Ramanujan’s formula (1.2) as well as Grosswald’s identity (1.6)
has an infinite series (1.7) containing the divisor function $\sigma_{z}(n)$
and the exponential function, that is, for $k\in\mathbb{Z},z\in\mathbb{H}$,

$\displaystyle F_{k}(z)=\sum_{n=1}^{\infty}\sigma_{-k}(n)e^{2\pi inz}.$ (1.19)

In the current paper, we are interested to study transformation formula for
the following infinite series:

$\displaystyle\mathfrak{F}_{\mathbb{F},k}(z):=\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-k}(n)V\left(-\frac{(2\pi)^{d}niz}{D}\bigg{|}\bar{0}_{d}\right).$
(1.20)

One can easily check that the above series (1.20) reduces to (1.19) when
$\mathbb{F}=\mathbb{Q}$. Now we are ready to state the main results of this
paper in the next section.

##  2\. Main Results

######  Theorem 2.1.

Let $\mathbb{F}$ be a number field of degree $d=r_{1}+2r_{2}$ and
$\zeta_{\mathbb{F}}(s)$ be the Dedekind zeta function defined in (1.10). We
consider $r=r_{1}+r_{2}-1$. For any non-zero integer $k$, we define

$\displaystyle\Lambda_{\mathbb{F},k}(s):=\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}}{D}\right)^{-s}.$
(2.1)

Let $\mathfrak{F}_{\mathbb{F},k}(z)$ be the infinite series defined as in
(1.20). Then for any $z\in\mathbb{H}$ and $k>0$, we have

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k(r_{1}+1)+r_{2}}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)+\sum_{j=1}^{k-1}\mathfrak{R}_{-2j}(z)},$
(2.2)

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z):=\mathfrak{F}_{\mathbb{F},2k+1}(z)-\mathfrak{R}_{0}(z)-\mathfrak{R}_{1}(z),$

and the residual terms are defined as

$\displaystyle\mathfrak{R}_{0}(z)$
$\displaystyle=\frac{1}{(r_{2})!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left(s^{r_{2}+1}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{iD}{(2\pi)^{d}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\frac{1}{(r)!}\lim_{s\rightarrow-(2j-1)}\frac{{\rm d}^{r}}{{\rm
d}s^{r}}\left((s+2j-1)^{r+1}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{-2j}(z)$
$\displaystyle=\frac{1}{(r_{2}-1)!}\lim_{s\rightarrow-2j}\frac{{\rm
d}^{r_{2}-1}}{{\rm
d}s^{r_{2}-1}}\left((s+2j)^{r_{2}}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

Again, for $k<0$, we have

$\displaystyle\mathfrak{U}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k(r_{1}+1)+r_{2}}z^{2k}\mathfrak{U}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+\mathfrak{R}(z),$
(2.3)

where

$\displaystyle\mathfrak{U}_{\mathbb{F},2k+1}(z):=\mathfrak{F}_{\mathbb{F},2k+1}(z)-\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!},$

and

$\displaystyle\mathfrak{R}(z)=\begin{cases}-\frac{i}{4\pi
z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,1,0),\\\
\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)Di}{(2\pi)^{2}z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,0,1),\\\
0,&\quad\text{otherwise.}\end{cases}$ (2.4)

######  Remark 1.

We note that an explicit evaluation of the terms
$\mathfrak{R}_{0}(z),\mathfrak{R}_{-(2j-1)}(z)$ and $\mathfrak{R}_{-2j}(z)$ is
not easy as it involves higher derivatives. However, we can say that
$\mathfrak{R}_{0}(z)$ is a polynomial in $\mathbb{C}[\log(z)]$ of degree
$r_{2}$, whereas $\mathfrak{R}_{-(2j-1)}(z)$ and $\mathfrak{R}_{-2j}(z)$ are
polynomials of the form $z^{2j-1}g(\log(z))$ and $z^{2j}h(\log(z))$, where
$g(z)$ and $h(z)$ are some polynomials of degree $r=r_{1}+r_{2}-1$ and
$r_{2}-1$, respectively.

######  Corollary 2.2.

For $\mathbb{F}=\mathbb{Q}$, Theorem 2.1 reduces to the Ramanujan-Grosswald
formula (1.6).

In the next subsection, we highlight some of the intriguing identities that
are by product of our main result in the case of totally real number field.

###  2.1. Transformation formulae for totally real number fields

The next implication of Theorem 2.1 is stated as a separate theorem since it
can be regarded as a formula for $\zeta_{\mathbb{F}}(2k+1)$ over a totally
real number field.

######  Theorem 2.3.

Let $\mathbb{F}$ be a totally real number field of degree $r_{1}$ and $k$ be a
positive integer. Then we have

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k(r_{1}+1)}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)},$
(2.5)

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)=\mathfrak{F}_{\mathbb{F},2k+1}(z)-C_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+1)-H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{iD}{(2\pi)^{r_{1}}z},$

and the residual term is given by

$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\frac{1}{(r_{1}-1)!}\lim_{s\rightarrow-(2j-1)}\frac{{\rm
d}^{r_{1}-1}}{{\rm
d}s^{r_{1}-1}}\left((s+2j-1)^{r_{1}}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

In particular, for the real quadratic fields we obtain the following identity.

######  Corollary 2.4.

Let $k>0,$ and $\mathbb{F}$ be a real quadratic field i.e.
$\mathbb{F}=\mathbb{Q}(\sqrt{m})$ where m is a positive square free integer.
Then we have

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)},$
(2.6)

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=2\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)K_{0}\left(2\pi\sqrt{\frac{nz}{m}}e^{-\frac{i\pi}{4}}\right)-\zeta_{\mathbb{F}}^{{}^{\prime}}(0)\zeta_{\mathbb{F}}(2k+1)-\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)im}{\pi^{2}z}.$

The term $\mathfrak{R}_{-(2j-1)}(z)$ in (2.6) is given by

$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\lim_{s\rightarrow-(2j-1)}\frac{{\rm d}}{{\rm
d}s}\left((s+2j-1)^{2}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

Next, we present identities for Dedekind zeta function over totally real
number fields at negative odd integers.

######  Theorem 2.5.

Let $k$ be a positive integer and $\mathbb{F}$ be a totally real number field
of degree $r_{1}.$ Then, we have

$\displaystyle
z^{2k}\left\\{\mathfrak{F}_{\mathbb{F},-2k+1}(z)-C_{\mathbb{F}}\zeta_{\mathbb{F}}(1-2k)\right\\}$
$\displaystyle=(-1)^{k(r_{1}+1)}\left\\{\mathfrak{F}_{\mathbb{F},-2k+1}\left(-\frac{1}{z}\right)-C_{\mathbb{F}}\zeta_{\mathbb{F}}(1-2k)\right\\}+\begin{cases}\frac{z^{2k}}{4\pi
zi},&\text{if}\,\,(k,r_{1},r_{2})=(1,1,0),\\\ 0,&\text{otherwise}.\end{cases}$

######  Remark 2.

When $(r_{1},r_{2})=(1,0)$ and $k>1$, then the above formula immediately
transforms into the Grosswald’s identity (1.9). In particular, when
$(k,r_{1},r_{2})=(1,1,0)$, the above formula yields the well known
transformation formula for the Eisenstein series of weight $2$, that is,

$\displaystyle E_{2}\left(-\frac{1}{z}\right)$
$\displaystyle=z^{2}\left(E_{2}(z)+\frac{6}{\pi iz}\right).$

The next result gives an explicit evaluation of the infinite series
$\mathfrak{F}_{\mathbb{F},-2k+1}(z)$ at $z=i$ that involves the class number
and value of the Dedekind zeta function at negative odd integers.

######  Corollary 2.6.

Let $k\geq 3$ and $r_{1}\geq 1$ be odd integers. Then for a totally real field
$\mathbb{F}$ of degree $r_{1}$, we have

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},2k-1}(n)V\left(\frac{(2\pi)^{r_{1}}n}{D}\bigg{|}\bar{0}_{r_{1}}\right)=-\frac{h_{\mathbb{F}}R_{\mathbb{F}}\zeta_{\mathbb{F}}(1-2k)}{w_{\mathbb{F}}}.$
(2.7)

This gives a new identity for the class number of a totally real field. Using
functional equation (3.4) and Siegel’s identity (1.15), one can check that
$\zeta_{\mathbb{F}}(1-2k)$ is a non-zero rational number.

######  Remark 3.

In particular, when $\mathbb{F}=\mathbb{Q}$, Corollary 2.6 gives an exact
evaluation of the following well-known Lambert series

$\displaystyle\sum_{n=1}^{\infty}\sigma_{2k-1}(n)e^{-2\pi n}$
$\displaystyle=-\frac{\zeta(1-2k)}{2}=\frac{B_{2k}}{4k}.$

This identity was first obtained by Glaisher and later rediscovered by
Ramanujan [5, p. 262, Equation (13.1)].

Next, we provide an interesting analogue of the transformation formula (1.5)
for Eisenstien series $G_{2k}(z)$, namely, a transformation formula that sends
$z\rightarrow-\frac{1}{z}$ for real quadratic fields.

######  Corollary 2.7.

Let $m$ be a positive square free integer and
$\mathbb{F}=\mathbb{Q}(\sqrt{m})$. Then for $z\in\mathbb{H}$ and
$k\in\mathbb{N}$, we have

$\displaystyle(iz)^{2k}G_{\mathbb{F},2k}(z)=G_{\mathbb{F},2k}\left(-\frac{1}{z}\right),$

where

$\displaystyle
G_{\mathbb{F},2k}(z):=1-\frac{2}{\zeta_{\mathbb{F}}^{\prime}(0)\zeta_{\mathbb{F}}(1-2k)}\sum_{n=1}^{\infty}\sigma_{\mathbb{F},2k-1}(n)K_{0}\left(2\pi\sqrt{\frac{nz}{m}}e^{-\frac{i\pi}{4}}\right).$

Further, as an application of Theorem 2.1, we give number field analogue of
Ramanujan-Grosswald identity for imaginary number fields.

###  2.2. Transformation formulae for imaginary number fields

######  Theorem 2.8.

Let $\mathbb{F}$ be a purely imaginary number field with degree of extension
$2r_{2}$ over $\mathbb{Q}.$ Then for $k>0,z\in\mathbb{H}$, we have

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k+r_{2}}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)+\sum_{j=1}^{k-1}\mathfrak{R}_{-2j}(z)},$
(2.8)

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)=\mathfrak{F}_{\mathbb{F},2k+1}(z)-\mathfrak{R}_{0}(z)-\mathfrak{R}_{1}(z),$

and the residual terms are defined as

$\displaystyle\mathfrak{R}_{0}(z)$
$\displaystyle=\frac{1}{(r_{2})!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left(s^{r_{2}+1}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{iD}{(2\pi)^{2r_{2}}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\frac{1}{(r_{2}-1)!}\lim_{s\rightarrow-(2j-1)}\frac{{\rm
d}^{r_{2}-1}}{{\rm
d}s^{r_{2}-1}}\left((s+2j-1)^{r_{2}}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{-2j}(z)$
$\displaystyle=\frac{1}{(r_{2}-1)!}\lim_{s\rightarrow-2j}\frac{{\rm
d}^{r_{2}-1}}{{\rm
d}s^{r_{2}-1}}\left((s+2j)^{r_{2}}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

Again, for $k<0$, we have

$\displaystyle\mathfrak{U}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k+r_{2}}z^{2k}\mathfrak{U}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+\mathfrak{R}(z),$
(2.9)

where

$\displaystyle\mathfrak{U}_{\mathbb{F},2k+1}(z):=\mathfrak{F}_{\mathbb{F},2k+1}(z)-\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!},$
(2.10)

and

$\displaystyle\mathfrak{R}(z)=\begin{cases}\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)Di}{(2\pi)^{2}z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,0,1),\\\
0,&\quad\text{otherwise.}\end{cases}$ (2.11)

In particular, for quadratic imaginary fields we obtain the following
transformation formula.

######  Corollary 2.9.

Let $m$ be a positive square free integer and
$\mathbb{F}=\mathbb{Q}(\sqrt{-m})$ be a quadratic imaginary number field. Then
for $k>0$, we have

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k+1}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)+\sum_{j=1}^{k-1}\mathfrak{R}_{-2j}(z)},$
(2.12)

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=2\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)K_{0}\left(2\pi\sqrt{\frac{nz}{m}}e^{-\frac{i\pi}{4}}\right)-\mathfrak{R}_{0}(z)-\mathfrak{R}_{1}(z),$

and the residual terms are defined as

$\displaystyle\mathfrak{R}_{0}(z)$ $\displaystyle=\lim_{s\rightarrow
0}\frac{{\rm d}^{2}}{{\rm
d}s^{2}}\left(s^{2}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{im}{\pi^{2}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=-\frac{\zeta_{\mathbb{F}}^{{}^{\prime}}(1-2j)}{[(2j-1)!]^{2}}\zeta_{\mathbb{F}}(2k-2j+2)\left(\frac{\pi^{2}iz}{m}\right)^{2j-1},$
$\displaystyle\mathfrak{R}_{-2j}(z)$
$\displaystyle=\frac{\zeta_{\mathbb{F}}^{{}^{\prime}}(-2j)}{[(2j)!]^{2}}\zeta_{\mathbb{F}}(2k-2j+1)\left(\frac{\pi^{2}iz}{m}\right)^{2j}.$

The next result also gives another modular transformation property associated
to quadratic imaginary fields.

######  Corollary 2.10.

Let $m$ be a square free positive integer and
$\mathbb{F}=\mathbb{Q}(\sqrt{-m})$. Then for $z\in\mathbb{H}$, we have

$\displaystyle z^{2}\mathfrak{U}_{\mathbb{F},-1}(z)$
$\displaystyle=\mathfrak{U}_{\mathbb{F},-1}\left(-\frac{1}{z}\right)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)izm}{\pi^{2}},$
(2.13)

where

$\displaystyle\mathfrak{U}_{\mathbb{F},-1}(z)=\mathfrak{F}_{\mathbb{F},-1}(z)-\zeta_{\mathbb{F}}(0)\zeta_{\mathbb{F}}^{{}^{\prime}}(-1).$

Moreover, letting $z=i$ in (2.13), we obtain the following exact evaluation:

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},1}(n)K_{0}\left(2\pi\sqrt{\frac{n}{m}}\right)=\frac{1}{2}\zeta_{\mathbb{F}}(0)\zeta^{\prime}_{\mathbb{F}}(-1)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)m}{4\pi^{2}}.$
(2.14)

The below identity holds for any purely imaginary number field.

######  Corollary 2.11.

Let $\mathbb{F}$ be a purely imaginary field of degree $2r_{2}$ and $k$ be a
positive integer. If $(k,r_{2})\neq(1,1)$, then we have

$\displaystyle
z^{2k}\mathfrak{U}_{\mathbb{F},1-2k}(z)=(-1)^{k+r_{2}}\mathfrak{U}_{\mathbb{F},1-2k}\left(-\frac{1}{z}\right),$
(2.15)

where

$\mathfrak{U}_{\mathbb{F},1-2k}(z)=\mathfrak{F}_{\mathbb{F},1-2k}(z)-\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(1-2k)}{(r_{2})!}.$

######  Remark 4.

Moreover, substituting $z=i$ in (2.15) and considering $r_{2}$ as an odd
positive integer, we obtain the following interesting exact evaluation:

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},2k-1}(n)V\left(\frac{(2\pi)^{2r_{2}}n}{D}\bigg{|}\bar{0}_{2r_{2}}\right)=-\frac{h_{\mathbb{F}}R_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(1-2k)}{w_{\mathbb{F}}(r_{2})!}.$
(2.16)

This result is true for purely imaginary number fields and it can be
considered as an analogue of Corollary 2.6, which is true only for real number
fields. In particular, when $\mathbb{F}=\mathbb{Q}(\sqrt{-m})$, then for
$k\geq 3$, we have

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},2k-1}(n)K_{0}\left(2\pi\sqrt{\frac{n}{m}}\right)=\frac{1}{2}\zeta_{\mathbb{F}}(0)\zeta^{\prime}_{\mathbb{F}}(1-2k).$
(2.17)

###  2.3. A number field analogue of transformation formula for Dedekind eta
function

One of the key observations is that our main result, i.e., Theorem 2.1 loses
its validity for $k=0$ as $\zeta_{\mathbb{F}}(2k+1)$ exhibits a simple pole at
$1$. Therefore, corresponding $k=0$, we must handle it separately. Quite
interestingly, in this case, we will derive a number field analogue of
transformation formula for Dedekind eta function $\eta(z)$, which is a half
integral weight modular form.

######  Theorem 2.12.

Let $\mathbb{F}$ be a number field of degree $d$ and
$\mathfrak{F}_{\mathbb{F},1}(z)$ be the infinite series defined as in (1.20).
Then we have

$\displaystyle\mathfrak{T}_{\mathbb{F}}(z)$
$\displaystyle=(-1)^{r_{2}}\mathfrak{T}_{\mathbb{F}}\left(-\frac{1}{z}\right)+\mathfrak{R}_{0}(z),$

where

$\displaystyle\mathfrak{T}_{\mathbb{F}}(z):=\mathfrak{F}_{\mathbb{F},1}(z)-\mathfrak{R}_{1}(z),$

with

$\displaystyle\mathfrak{R}_{0}(z)=\frac{1}{(r_{2}+1)!}{\displaystyle\lim_{s\rightarrow
0}}\frac{{\rm d}^{r_{2}+1}}{{\rm
d}s^{r_{2}+1}}\left(s^{r_{2}+2}\Lambda_{\mathbb{F},0}(s)(-iz)^{-s}\right),\quad\mathfrak{R}_{1}(z)=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)\frac{iD}{(2\pi)^{d}z}.$

In particular, we have the following modular transformation property for
totally real number fields.

######  Corollary 2.13.

Let $\mathbb{F}$ be a totally real number field of degree $r_{1}$ and
$\mathfrak{F}_{\mathbb{F},1}(z)$ be the infinite series defined in (1.20). Let
$H_{\mathbb{F}},\gamma_{\mathbb{F}}$ be the constants defined in (3.5). Then
we have

$\displaystyle\mathfrak{F}_{\mathbb{F},1}(z)-\mathfrak{F}_{\mathbb{F},1}\left(-\frac{1}{z}\right)$
$\displaystyle=a_{0}\gamma_{\mathbb{F}}-a_{0}H_{\mathbb{F}}\left\\{r_{1}\gamma+\log\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)\right\\}+a_{1}H_{\mathbb{F}}$
$\displaystyle+\mathfrak{R}_{1}(z)-\mathfrak{R}_{1}\left(-\frac{1}{z}\right),$
(2.18)

where

$\mathfrak{R}_{1}(z)=\frac{iDH_{\mathbb{F}}\zeta_{\mathbb{F}}(2)}{(2\pi)^{r_{1}}z},\quad
a_{0}=\frac{\zeta_{\mathbb{F}}^{(r_{1}-1)}(0)}{(r_{1}-1)!}=C_{\mathbb{F}},\quad
a_{1}=\frac{\zeta_{\mathbb{F}}^{(r_{1})}(0)}{(r_{1})!}.$

######  Remark 5.

Quite surprisingly, substituting $z=i$ in (2.18) and using the relation
between $H_{\mathbb{F}}$ and $C_{\mathbb{F}}$, we obtain a connection between
class number of a totally real number field of degree $r_{1}$ and the constant
term $\gamma_{\mathbb{F}}$ of the Laurent series expansion (3.5) of the
Dedekind zeta function at $s=1$. Mainly, we obtain

$\displaystyle
C_{\mathbb{F}}=\frac{a_{1}-A\gamma_{\mathbb{F}}}{r_{1}\gamma+\log\left(\frac{(2\pi)^{r_{1}}}{D}\right)},$
(2.19)

where $A=\frac{\sqrt{D}}{2^{r_{1}}(2\pi)^{r_{2}}}$. This suggests that finding
a fomula for class number is also connected with the Kronceker’s limit formula
for the Dedekind zeta function.

Further, letting $\mathbb{F}=\mathbb{Q}$ in Corollary 2.13, we have the
following result.

######  Corollary 2.14.

For any complex number $z\in\mathbb{H}$, we have

$\displaystyle\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}e^{2\pi
inz}-\sum_{n=1}^{\infty}\frac{\sigma(n)}{n}e^{-\frac{2\pi in}{z}}=\frac{i\pi
z^{2}+i\pi}{12z}+\frac{1}{2}\log(-iz),$ (2.20)

which is equivalent to the logarithm of the transformation formula for the
Dedekind eta function $\eta(z)$, namely,

$\displaystyle\eta\left(-\frac{1}{z}\right)=\sqrt{-iz}\,\eta(z).$

##  3\. Required Tools

In this section, we state a few essential results which will be frequently
used in the proof of the main results. The gamma function $\Gamma(z)$
satisfies the following functional equation and reflection formula:

$\displaystyle\Gamma(z+1)$ $\displaystyle=z\Gamma(z),\quad z\in\mathbb{C},$
(3.1) $\displaystyle\Gamma(z)\Gamma(1-z)$ $\displaystyle=\frac{\pi}{\sin\pi
z},\quad z\in\mathbb{C}\backslash\mathbb{Z}$ (3.2)
$\displaystyle\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)$
$\displaystyle=2^{1-2z}\sqrt{\pi}\Gamma(2z).$ (3.3)

We now discuss analytic properties of the Dedekind zeta function
$\zeta_{\mathbb{F}}(s)$.

######  Lemma 3.1.

The Dedekind zeta function $\zeta_{\mathbb{F}}(s)$ has an analytic
continuation in the whole complex plane except for a simple pole at $s=1$. It
also satisfies the following functional equation relating its values at $s$
and $1-s$,

$\displaystyle\Omega_{\mathbb{F}}(s)=\Omega_{\mathbb{F}}(1-s),$ (3.4)

where
$\Omega_{\mathbb{F}}(s)=\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)^{r_{1}}\Gamma(s)^{r_{2}}\zeta_{\mathbb{F}}(s)$.

The Dedekind zeta function $\zeta_{\mathbb{F}}(s)$ satisfies the following
Laurent series expansion at $s=1$:

$\displaystyle\zeta_{\mathbb{F}}(s)=\frac{H_{\mathbb{F}}}{s-1}+\gamma_{\mathbb{F}}+O(s-1).$
(3.5)

Utilizing functional equation (3.4) of $\zeta_{\mathbb{F}}(s)$, one can verify
that it has a zero at $s=0$ of order $r=r_{1}+r_{2}-1$. Therefore, the
following Laurent series expansion at $s=0$ holds:

$\displaystyle\zeta_{\mathbb{F}}(s)=a_{0}s^{r_{1}+r_{2}-1}+a_{1}s^{r_{1}+r_{2}}+O\left(s^{r_{1}+r_{2}+1}\right),$
(3.6)

where $a_{0}=\frac{\zeta^{(r_{1}+r_{2}-1)}(0)}{(r_{1}+r_{2}-1)!}$ and
$a_{1}=\frac{\zeta^{(r_{1}+r_{2})}(0)}{(r_{1}+r_{2})!}$. From (1.12), one can
observe that $a_{0}$ is nothing but the constant $C_{\mathbb{F}}$.

Now we mention an interesting identity due to Chandrasekharan and Narasimhan
[11, Equation (57)]. For any integer $k$, one can prove that

$\displaystyle\Lambda_{k}(s)=(-1)^{k}\Lambda_{k}(-s-2k),$

where $\Lambda_{k}(s)=(2\pi)^{-s}\Gamma(s)\zeta(s)\zeta(s+2k+1).$ We
generalize this identity for Dedekind zeta function $\zeta_{\mathbb{F}}(s)$.
Mainly, we prove the below result, which will play a crucial role in proving
our main identity.

######  Lemma 3.2.

Let $\mathbb{F}$ be a number field of degree $d$ and $\zeta_{\mathbb{F}}(s)$
be the Dededkind zeta function defined in (1.10). For any $k\in\mathbb{Z}$, we
have

$\displaystyle\Lambda_{\mathbb{F},k}(s)=(-1)^{kr_{1}+r_{2}}\Lambda_{\mathbb{F},k}(-s-2k),$

where
$\Lambda_{\mathbb{F},k}(s)=D^{s}(2\pi)^{-ds}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1).$

  * Proof.

We first prove this identity for any non-negative integer $k$. From the
functional equation (3.4) of $\zeta_{\mathbb{F}}(s)$, we have

$\displaystyle\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)^{r_{1}}\Gamma(s)^{r_{2}}\zeta_{\mathbb{F}}(s)=\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)^{r_{1}}\Gamma(1-s)^{r_{2}}\zeta_{\mathbb{F}}(1-s).$
(3.7)

Replace $s$ by $s+2k+1$ in (3.7) to see that

$\displaystyle\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{\frac{s+2k+1}{2}}\Gamma\left(\frac{s+2k+1}{2}\right)^{r_{1}}\Gamma(s+2k+1)^{r_{2}}\zeta_{\mathbb{F}}(s+2k+1)$
$\displaystyle=\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{\frac{-s-2k}{2}}\Gamma\left(-\frac{s}{2}-k\right)^{r_{1}}\Gamma(-s-2k)^{r_{2}}\zeta_{\mathbb{F}}(-s-2k).$
(3.8)

Now we multiply (3.7) and (Proof.) to get

$\displaystyle\left(\frac{D}{\pi^{d}4^{r_{2}}}\right)^{2s+2k}\left\\{\Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{s+1}{2}+k\right)\right\\}^{r_{1}}\left\\{\Gamma(s)\Gamma(s+2k+1)\right\\}^{r_{2}}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)$
$\displaystyle=\left\\{\Gamma\left(\frac{1}{2}-\frac{s}{2}\right)\Gamma\left(-\frac{s}{2}-k\right)\right\\}^{r_{1}}\left\\{\Gamma(1-s)\Gamma(-s-2k)\right\\}^{r_{2}}\zeta_{\mathbb{F}}(1-s)\zeta_{\mathbb{F}}(-s-2k).$
(3.9)

To simplify the above equation further, we use the functional equation (3.1)
of $\Gamma(s)$ repeatedly and find that, for any non-negative integer $k$,

$\displaystyle\Gamma\left(\frac{s+1}{2}+k\right)=\frac{1}{2^{k}}(s+2k-1)(s+2k-3)\cdots(s+1)\Gamma\left(\frac{s}{2}+\frac{1}{2}\right).$
(3.10)

Again, using (3.1) repeatedly, one can show that

$\displaystyle\Gamma\left(-\frac{s}{2}-k\right)=\frac{(-2)^{k}\Gamma\left(-\frac{s}{2}\right)}{(s+2k)(s+2k-2)\cdots(s+2)}.$
(3.11)

Utilize (3.10), (3.11) and duplication formula (3.3) for $\Gamma(s)$ to derive

$\displaystyle\Gamma\left(\frac{s}{2}\right)\Gamma\left(\frac{s+1}{2}+k\right)$
$\displaystyle=\frac{2}{2^{s+k}}\sqrt{\pi}(s+2k-1)(s+2k-3)\cdots(s+1)\Gamma(s),$
(3.12)
$\displaystyle\Gamma\left(\frac{1}{2}-\frac{s}{2}\right)\Gamma\left(-\frac{s}{2}-k\right)$
$\displaystyle=\frac{(-1)^{k}2^{1+s+k}\sqrt{\pi}\Gamma(-s)}{(s+2k)(s+2k-2)\cdots(s+2)}.$
(3.13)

We again employ (3.1) to see that, for any non-negative integer $k$,

$\displaystyle\Gamma(s)\Gamma(s+2k+1)$
$\displaystyle=s(s+1)\cdots(s+2k)\Gamma(s)^{2},$ (3.14)
$\displaystyle\Gamma(-s-2k)\Gamma(1-s)$
$\displaystyle=\frac{-s(\Gamma(-s))^{2}}{(s+2k)(s+2k-1)\cdots(s+1)}.$ (3.15)

Now substituting expressions from (3.12)-(3.15) in (Proof.) and upon
simplification, we have

$\displaystyle\left(\frac{D}{(2\pi)^{d}}\right)^{2s+2k}\left\\{(s+1)(s+2)\cdots(s+2k)\right\\}^{d}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}(\Gamma(-s))^{d}\zeta_{\mathbb{F}}(1-s)\zeta_{\mathbb{F}}(-s-2k).$
(3.16)

Here we have used the fact that $r_{1}+2r_{2}=d$. Applying (3.1) again, one
can check that

$\displaystyle\frac{\Gamma(-s)}{(s+2k)(s+2k-1)\ldots(s+2)(s+1)}=\Gamma(-s-2k).$
(3.17)

Using (3.17) in (3.16), we obtain

$\displaystyle\left(\frac{D}{(2\pi)^{d}}\right)^{2s+2k}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}\Gamma(-s-2k)^{d}\zeta_{\mathbb{F}}(1-s)\zeta_{\mathbb{F}}(-s-2k).$

Now letting
$\Lambda_{\mathbb{F},k}(s)=D^{s}(2\pi)^{-ds}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)$,
the above identity reduces to

$\Lambda_{\mathbb{F},k}(s)=(-1)^{kr_{1}+r_{2}}\Lambda_{\mathbb{F},k}(-s-2k).$

This completes the proof of this lemma for any non-negative integer $k$. In a
similar fashion, one can prove that this identity also holds for any negative
integer as well. ∎

In the next section, we present proofs of our main results.

##  4\. Proof of Main Results

  * Proof of Theorem 2.1.

Using the definition (1.16) of the Steen function, for $y>0$, we write

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)^{d}\left(\frac{(2\pi)^{d}ny}{D}\right)^{-s}{\rm d}s$
$\displaystyle=\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)^{d}\sum_{n=1}^{\infty}\frac{\sigma_{\mathbb{F},-2k-1}(n)}{n^{s}}\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}{\rm
d}s,$ (4.1)

where we choose the line of integration as
$\max\\{1,-2k\\}<\operatorname{Re}(s)=c<\max\\{1,-2k\\}+\epsilon$, with
$0<\epsilon<1$, so that the above Dirichlet series converges absolutely and
uniformly. Hence, the interchange of summation and integration is justifiable.
Now utilize the definition of the Dirichlet series (1.14) in ( Proof of
Theorem 2.1.), to see that

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}{\rm
d}s.$ (4.2)

Now our aim is to study the following vertical line integral,

$\displaystyle J_{\mathbb{F},k}(y)=\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}{\rm
d}s.$

From the definition (2.1) of $\Lambda_{\mathbb{F},k}(s)$, it is clear that the
above integral can be rewritten as

$\displaystyle J_{\mathbb{F},k}(y)=\frac{1}{2\pi
i}\int_{(c)}\Lambda_{\mathbb{F},k}(s)y^{-s}{\rm d}s.$

We now set up a rectangular contour $\mathcal{C}$ made up of the vertices
$c-iT,c+iT,\alpha+iT$ and $\alpha-iT$ taken in counter-clockwise direction. We
already have $c>\max\\{1,-2k\\}$ with $T$ being some large positive quantity
and we wisely choose

$\min\\{-1,-2k-2\\}<\alpha<\min\\{0,-2k-1\\}.$

Before applying Cauchy’s residue theorem, we need to examine all the poles of
$\Lambda_{\mathbb{F},k}(s)$ inside the contour $\mathcal{C}$. We know that
$\Gamma(s)^{d}$ has poles of order $d$ at non-positive integers. Other poles
will depend upon the value of $k$ which will affect location of vertices of
$\mathcal{C}$. Here we divide in two cases.
Case I) Suppose $k>0$.

$c-iT$$c+iT$$\alpha+iT$$\alpha-
iT$0-1-2-3$-2k$1$-2k-1$$-2k-2$$\mathrm{Re}(s)$$\mathrm{Im}(s)$

Now inside the contour $\mathcal{C}$, for $\zeta_{\mathbb{F}}(s),\,s=0$ is a
zero of order $r_{1}+r_{2}-1,\,s\in\\{-2,-4,\ldots,-2k\\}$ are zeros of order
$r_{1}+r_{2},\,s\in\\{-1,-3,\ldots,-2k-1\\}$ are zeros of order $r_{2}$ and
$s=1$ is a simple pole. For $\zeta_{\mathbb{F}}(s+2k+1),s=-2k-1$ is a zero of
order $r_{1}+r_{2}-1$ and $s=-2k$ is a simple pole. Hence final poles of
$\Lambda_{\mathbb{F},k}(s)$ are at $s=0,-2k$ of order $r_{2}+1,s=1,\,-2k-1$
are simple poles, $s\in\\{-2,-4,\ldots,-2k+2\\}$ are poles of order $r_{2}$
and $s\in\\{-1,-3,\ldots,-2k+1\\}$ are poles of order $r_{1}+r_{2}$.
Case II) When $k<0$.

$c-iT$$c+iT$$\alpha+iT$$\alpha-
iT$0123$-2k$$-2k-1$$-1$$\mathrm{Re}(s)$$\mathrm{Im}(s)$

We see that $s=0$ is a zero of order $r_{1}+r_{2}-1$ for
$\zeta_{\mathbb{F}}(s)$ and a zero of order $r_{2}$ for
$\zeta_{\mathbb{F}}(s+2k+1)$. Note that $s=1$ is a simple pole for
$\zeta_{\mathbb{F}}(s)$ and $s=-2k$ is a simple pole for
$\zeta_{\mathbb{F}}(s+2k+1)$. Hence final poles for
$\Lambda_{\mathbb{F},k}(s)$, when $k<0$, are at $s=0,\,1,\,-2k$, which are all
simple poles.

Now, considering the above two cases and applying Cauchy’s residue theorem, we
have

$\displaystyle\frac{1}{2\pi
i}\int_{\mathcal{C}}\Lambda_{\mathbb{F},k}(s)y^{-s}{\rm d}s=\mathcal{R},$
(4.3)

where
$\Lambda_{\mathbb{F},k}(s)=\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)(2\pi)^{-s}D^{s}$
and the residual term is given by

$\displaystyle\mathcal{R}=\begin{cases}R_{0}^{(r_{2}+1)}(y)+R_{1}^{(1)}(y)+R_{-2k}^{(r_{2}+1)}(y)+R_{-2k-1}^{(1)}(y)+{\displaystyle\sum_{j=1}^{k}R_{-(2j-1)}^{(r_{1}+r_{2})}(y)}\\\
\hskip
233.3125pt+{\displaystyle\sum_{j=1}^{k-1}R_{-2j}^{(r_{2})}(y)},&\text{for}\,\,\,k>0,\\\
R_{0}^{(1)}(y)+R_{1}^{(1)}(y)+R_{-2k}^{(1)}(y),&\text{for}\,\,\,k<0,\\\
\end{cases}$ (4.4)

where $R_{\gamma}^{(l)}(y)$ denotes the residue of
$\Lambda_{\mathbb{F},k}(s)y^{-s}$ of order $l$ at $s=\gamma$. Letting
$T\rightarrow\infty$ and making use of Stirling’s formula for $\Gamma(s)$, we
can show that both the horizontal integrals vanish. Now, after taking the left
vertical integral to the right side and using (4.2) in (4.3), we are left with

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}y}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=\mathcal{R}+\frac{1}{2\pi
i}\int_{(\alpha)}\Lambda_{\mathbb{F},k}(s)y^{-s}{\rm d}s.$ (4.5)

Now our main aim is to simplify the above vertical integral, which we defined
as

$\displaystyle I_{\mathbb{F},k}(y):=\frac{1}{2\pi
i}\int_{(\alpha)}\Lambda_{\mathbb{F},k}(s)y^{-s}{\rm d}s.$ (4.6)

Replace $s$ by $-t-2k$, so that
$\operatorname{Re}(t)=-\operatorname{Re}(s)-2k=-\alpha-2k>\max\\{1,-2k\\}.$
Let $\beta=\operatorname{Re}(t)=-\alpha-2k.$ After this change of variable,
the above integral becomes

$\displaystyle I_{\mathbb{F},k}(y)$ $\displaystyle=\frac{1}{2\pi
i}\int_{(\beta)}\Lambda_{\mathbb{F},k}(-t-2k)y^{t+2k}\rm{d}t.$

Now we bring in use of Lemma 3.2 to write as

$\displaystyle I_{\mathbb{F},k}(y)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}y^{2k}\frac{1}{2\pi
i}\int_{(\beta)}\Lambda_{\mathbb{F},k}(t)y^{t}{\rm d}t$
$\displaystyle=(-1)^{(kr_{1}+r_{2})}y^{2k}\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}n}{Dy}\bigg{|}\bar{0}_{d}\right),$
(4.7)

where we have used (4.2) in the last step. Now substituting (4.7) in (4.5), we
obtain

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}y^{2k}\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}n}{Dy}\bigg{|}\bar{0}_{d}\right)+\mathcal{R}.$
(4.8)

Now we are left with calculating the term $\mathcal{R}$ containing all the
residual terms.

Case 1: Let $k>0$. In this case we need to calculate the following terms:

$\displaystyle\mathcal{R}=R_{0}^{(r_{2}+1)}(y)+R_{1}^{(1)}(y)+R_{-2k}^{(r_{2}+1)}(y)+R_{-2k-1}^{(1)}(y)+\sum_{j=1}^{k}R_{-(2j-1)}^{(r_{1}+r_{2})}(y)+\sum_{j=1}^{k-1}R_{-2j}^{(r_{2})}(y).$

Using the definition of residue calculation, one can check that

$\displaystyle R_{0}^{(r_{2}+1)}(y)$
$\displaystyle=\frac{1}{(r_{2})!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left(s^{r_{2}+1}\Lambda_{\mathbb{F},k}(s)y^{-s}\right),$ (4.9)
$\displaystyle R_{1}^{(1)}(y)$ $\displaystyle=\lim_{s\rightarrow
1}(s-1)\Lambda_{\mathbb{F},k}(s)y^{-s}$ $\displaystyle=\lim_{s\rightarrow
1}(s-1)\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{D}{(2\pi)^{d}y},$
(4.10)

in the last step we have used the class number formula (1.11). Further, we
have

$\displaystyle R_{-2k}^{(r_{2}+1)}(y)$
$\displaystyle=\frac{1}{(r_{2})!}\lim_{s\rightarrow-2k}\frac{{\rm
d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left((s+2k)^{r_{2}+1}\Lambda_{\mathbb{F},k}(s)y^{-s}\right).$
(4.11)

Now using Lemma 3.2, the above residual term can be rewritten as follows

$\displaystyle R_{-2k}^{(r_{2}+1)}(y)$
$\displaystyle=\frac{(-1)^{kr_{1}+r_{2}}}{(r_{2})!}\lim_{s\rightarrow-2k}\frac{{\rm
d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left((s+2k)^{r_{2}+1}\Lambda_{\mathbb{F},k}(-s-2k)y^{-s}\right),$
$\displaystyle=\frac{(-1)^{kr_{1}+r_{2}+1}y^{2k}}{(r_{2})!}\lim_{s\rightarrow
0}\frac{{\rm d}^{r_{2}}}{{\rm
d}s^{r_{2}}}\left(s^{r_{2}+1}\Lambda_{\mathbb{F},k}(s)y^{s}\right),$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}R_{0}^{(r_{2}+1)}\left(\frac{1}{y}\right).$
(4.12)

This shows that the residue at $s=0$ and $s=-2k$ are linked by the above
relation. Next, the residue at $s=-2k-1$ is given by

$\displaystyle R_{-2k-1}^{(1)}(y)$
$\displaystyle=\lim_{s\rightarrow-2k-1}(s+2k+1)\Lambda_{\mathbb{F},k}(s)y^{-s}.$

Further, use Lemma 3.2 to see that

$\displaystyle R_{-2k-1}^{(1)}(y)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}\lim_{s\rightarrow-2k-1}(s+2k+1)\Lambda_{\mathbb{F},k}(-s-2k)y^{-s}$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}\lim_{s\rightarrow
1}(s-1)\Lambda_{\mathbb{F},k}(s)y^{s}$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}R^{(1)}_{1}\left(\frac{1}{y}\right)$
(4.13)
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{Dy}{(2\pi)^{d}}.$
(4.14)

The above relation indicates the connection between the residues at $s=1$ and
$s=-2k-1$. The remaining residues at $s=-(2j-1)$ and $s=-2j$ can be evaluated
as follows:

$\displaystyle R_{-(2j-1)}^{(r_{1}+r_{2})}(y)$
$\displaystyle=\frac{1}{(r_{1}+r_{2}-1)!}\lim_{s\rightarrow-(2j-1)}\frac{{\rm
d}^{r_{1}+r_{2}-1}}{{\rm
d}s^{r_{1}+r_{2}-1}}\left((s+2j-1)^{r_{1}+r_{2}}\Lambda_{\mathbb{F},k}(s)y^{-s}\right),$
$\displaystyle=\frac{1}{(r)!}\lim_{s\rightarrow-(2j-1)}\frac{{\rm d}^{r}}{{\rm
d}s^{r}}\left((s+2j-1)^{r+1}\Lambda_{\mathbb{F},k}(s)y^{-s}\right),$ (4.15)
$\displaystyle R_{-(2j)}^{(r_{2})}(y)$
$\displaystyle=\frac{1}{(r_{2}-1)!}\lim_{s\rightarrow-2j}\frac{{\rm
d}^{r_{2}-1}}{{\rm
d}s^{r_{2}-1}}\left((s+2j)^{r_{2}}\Lambda_{\mathbb{F},k}(s)y^{-s}\right).$
(4.16)

As we know the poles of $\Lambda_{\mathbb{F},k}(s)y^{-s}$ at $s=-(2j-1)$ and
$s=-2j$ are of higher order, so it is a difficult task to simplify the above
residual terms. However, one can say that these residual terms will be
polynomials in $\mathbb{C}[y,\log(y)]$ with degree being $2j+r_{1}+r_{2}-2$
and $2j+r_{2}-1$, respectively. This is because for the residual term (4.15)
the highest degree of $y$ will be $2j-1$ and the highest power of $\log(y)$
will be $r=r_{1}+r_{2}-1$. Similarly, for the residual term (4.16) the highest
degree of $y$ will be $2j$ and the highest degree of $\log(y)$ will be
$r_{2}-1$. In a similar fashion, one can verify that the residue at $s=0$,
i.e., the residual term (4.9) is a polynomial in $\mathbb{C}[\log(y)]$ of
degree $r_{2}$. This completes the calculations of all residues for $k>0$. Now
we shall compute residues for $k<0$.
Case 2: When $k<0$, from (4.4), we know that the residual term $\mathcal{R}$
is given by

$\displaystyle\mathcal{R}=R_{0}^{(1)}(y)+R_{1}^{(1)}(y)+R_{-2k}^{(1)}(y).$

Recall that, in this case, poles of $\Lambda_{\mathbb{F},k}(s)$ are at
$0,1,-2k$ and all are simple. Therefore, we have

$\displaystyle R_{0}^{(1)}(y)$ $\displaystyle=\lim_{s\rightarrow
0}~{}s\Lambda_{\mathbb{F},k}(s)y^{-s}=\lim_{s\rightarrow
0}s\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}$
$\displaystyle=\lim_{s\rightarrow
0}(s\Gamma(s))^{d}\frac{\zeta_{\mathbb{F}}(s)}{s^{r_{1}+r_{2}-1}}\frac{\zeta_{\mathbb{F}}(s+2k+1)}{s^{r_{2}}}\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}$
$\displaystyle=\lim_{s\rightarrow
0}\Gamma(s+1)^{d}\frac{\zeta_{\mathbb{F}}(s)}{s^{r_{1}+r_{2}-1}}\frac{\zeta_{\mathbb{F}}(s+2k+1)}{s^{r_{2}}}\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}$
$\displaystyle=\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!}.$
(4.17)

To obtain the final step we used (1.12) and $s=0$ is a zero of order $r_{2}$
for $\zeta_{\mathbb{F}}(s+2k+1)$ as $k<0$ and negative odd integers are zeros
of $\zeta_{\mathbb{F}}(s)$ of order $r_{2}$. Now we calculate the residual
term, at $s=-2k$, which is given by

$\displaystyle R_{-2k}^{(1)}(y)$
$\displaystyle=\lim_{s\rightarrow-2k}(s+2k)\Lambda_{\mathbb{F},k}(s)y^{-s}.$

To obtain a simplified form, we use Lemma 3.2. Thus, we get

$\displaystyle R_{-2k}^{(1)}(y)$
$\displaystyle=(-1)^{kr_{1}+r_{2}}\lim_{s\rightarrow-2k}(s+2k)\Lambda_{\mathbb{F},k}(-s-2k)y^{-s}$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}\lim_{s\rightarrow
0}~{}s\Lambda_{\mathbb{F},k}(s)y^{s}$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}R_{0}^{(1)}\left(\frac{1}{y}\right)$
$\displaystyle=(-1)^{kr_{1}+r_{2}+1}y^{2k}\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!},$
(4.18)

here in the penultimate step and final step, we have used the definition of
$R_{0}^{1}(y)$ and its final expression ( Proof of Theorem 2.1.). This
indicates that the residues at $s=0$ and $s=-2k$ are associated with each
other. Thus, from ( Proof of Theorem 2.1.) and ( Proof of Theorem 2.1.), we
have

$\displaystyle
R_{0}^{(1)}(y)+R_{-2k}^{(1)}(y)=\left\\{1+(-1)^{kr_{1}+r_{2}+1}y^{2k}\right\\}\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!}.$
(4.19)

Finally, the residue at $s=1$, a simple pole of
$\Lambda_{\mathbb{F},k}(s)y^{-s}$, can be evaluated as follows:

$\displaystyle R_{1}^{(1)}(y)=\lim_{s\rightarrow
1}(s-1)\Lambda_{\mathbb{F},k}(s)y^{-s}$ $\displaystyle=\lim_{s\rightarrow
1}(s-1)\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{D}{(2\pi)^{d}y},$
$\displaystyle=\begin{cases}-\frac{1}{4\pi
y},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,1,0),\\\
\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)D}{(2\pi)^{2}y},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,0,1),\\\
0,&\quad\text{otherwise.}\end{cases}$ (4.20)

Till now, we assumed that $y>0$. However, by analytic continuation, one can
show that the identity can be extended for $\operatorname{Re}(y)>0$. Thus, we
substitute $y=-iz$ with $z\in\mathbb{H}$ in (4.8) to get the following form,
for $z\in\mathbb{H}$,

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{-(2\pi)^{d}niz}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=(-1)^{k(r_{1}+1)+r_{2}}z^{2k}\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(\frac{(2\pi)^{d}ni}{Dz}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle+\mathcal{R},$

where, for $k>0$, the term $\mathcal{R}$ is given by

$\displaystyle\mathcal{R}$
$\displaystyle=\Big{\\{}R_{0}^{(r_{2}+1)}(-iz)+(-1)^{k(r_{1}+1)+r_{2}+1}z^{2k}R_{0}^{(r_{2}+1)}\left(\frac{i}{z}\right)\Big{\\}}$
$\displaystyle+\Big{\\{}R_{1}^{(1)}(-iz)+(-1)^{k(r_{1}+1)+r_{2}+1}z^{2k}R_{1}^{(1)}\left(\frac{i}{z}\right)\Big{\\}}$
$\displaystyle+{\displaystyle\sum_{j=1}^{k}R_{-(2j-1)}^{(r_{1}+r_{2})}(-iz)+\sum_{j=1}^{k-1}R_{-2j}^{(r_{2})}}(-iz).$

Here we note that to get the above form we have used the relations (4.12) and
(4.13) among residues. To simplicity further, we define

$\displaystyle\mathfrak{R}_{0}(z):=R_{0}^{(r_{2}+1)}(-iz),\mathfrak{R}_{1}(z):=R_{1}^{(1)}(-iz),$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z):=R_{-(2j-1)}^{(r_{1}+r_{2})}(-iz),\mathfrak{R}_{-(2j)}(z):=R_{-2j}^{(r_{2})}(-iz).$

Then the above the term $\mathcal{R}$ becomes, for $k>0$,

$\displaystyle\mathcal{R}=$
$\displaystyle\Big{\\{}\mathfrak{R}_{0}(z)+(-1)^{k(r_{1}+1)+r_{2}+1}z^{2k}\mathfrak{R}_{0}\left(-\frac{1}{z}\right)\Big{\\}}$
$\displaystyle+$
$\displaystyle\Big{\\{}\mathfrak{R}_{1}(z)+(-1)^{k(r_{1}+1)+r_{2}+1}z^{2k}\mathfrak{R}_{1}\left(-\frac{1}{z}\right)\Big{\\}}$
$\displaystyle{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)+\sum_{j=1}^{k-1}\mathfrak{R}_{-(2j)}(z).}$

In a similar way, after substituting $y=-iz$, for $k<0$, from ( Proof of
Theorem 2.1.)-(4.19), the residual term $\mathcal{R}$ can be written as

$\displaystyle\mathcal{R}=$
$\displaystyle\Big{\\{}1+(-1)^{k(r_{1}+1)+r_{2}+1}z^{2k}\Big{\\}}\frac{C_{\mathbb{F}}\zeta_{\mathbb{F}}^{(r_{2})}(2k+1)}{(r_{2})!}$
$\displaystyle+\mathfrak{R}(z),$

where the term $\mathfrak{R}(z)$ is given by

$\displaystyle\mathfrak{R}(z)=\begin{cases}-\frac{i}{4\pi
z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,1,0),\\\
\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)Di}{(2\pi)^{2}z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,0,1),\\\
0,&\quad\text{otherwise.}\end{cases}$

Finally, combining all these residual terms and using the definition (1.20) of
$\mathfrak{F}_{\mathbb{F},k}(z)$, we obtain the final identity. ∎

  * Proof of Corollary 2.2.

If we consider $\mathbb{F}=\mathbb{Q}$, then one can easily check that

$(r_{1},r_{2},d,D,H_{\mathbb{F}},C_{\mathbb{F}})=(1,0,1,1,1,-1/2).$

Moreover, the number field analogous generalized divisor function
$\sigma_{\mathbb{F},-k}(n)$ reduces to the usual generalized divisor function
$\sigma_{-k}(n)$. Therefore, for $k>0$, we have

$\displaystyle\mathfrak{F}_{\mathbb{F},2k+1}(z)$
$\displaystyle=\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{2\pi inz},$
$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{2\pi
inz}-\mathfrak{R}_{0}(z)-\mathfrak{R}_{1}(z),$

where $\mathfrak{R}_{0}(z)=-\frac{\zeta(2k+1)}{2}$ and
$\mathfrak{R}_{1}(z)=\frac{i\zeta(2k+2)}{2\pi z}=\frac{(2\pi
i)^{2k+1}}{2z}\frac{B_{2k+2}}{(2k+2)!}$. Again, one can check that

$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\lim_{s\rightarrow-(2j-1)}(s+2j-1)\Gamma(s)\zeta(s)\zeta(s+2k+1)(-2\pi
iz)^{-s}$ $\displaystyle=\frac{(2\pi
i)^{2k+1}}{2z}\frac{B_{2j}}{(2j)!}\frac{B_{2k-2j+2}}{(2k-2j+2)!}z^{2j}.$

Here we have used the fact that $\zeta(1-2j)=-\frac{B_{2j}}{(2j)!}$ and
Euler’s formula (1.1). Substituting these values in Theorem 2.1, we get

$\displaystyle\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{2\pi
inz}+\frac{\zeta(2k+1)}{2}-\frac{(2\pi i)^{2k+1}}{2z}\frac{B_{2k+2}}{(2k+2)!}$
$\displaystyle=z^{2k}\left\\{\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{-2\pi
in/z}+\frac{\zeta(2k+1)}{2}+\frac{(2\pi
i)^{2k+1}}{2z}\frac{z^{2}B_{2k+2}}{(2k+2)!}\right\\}$
$\displaystyle+\frac{(2\pi
i)^{2k+1}}{2z}\sum_{j=1}^{k}\frac{B_{2j}}{(2j)!}\frac{B_{2k-2j+2}}{(2k-2j+2)!}z^{2j}.$

This implies that

$\displaystyle\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{2\pi
inz}+\frac{\zeta(2k+1)}{2}$
$\displaystyle=z^{2k}\left\\{\sum_{n=1}^{\infty}\sigma_{-(2k+1)}(n)e^{-2\pi
in/z}+\frac{\zeta(2k+1)}{2}\right\\}$ $\displaystyle+\frac{(2\pi
i)^{2k+1}}{2z}\sum_{j=0}^{k+1}\frac{B_{2j}}{(2j)!}\frac{B_{2k-2j+2}}{(2k-2j+2)!}z^{2j},$

which is exactly same as the identity (1.6). This completes the proof. ∎

  * Proof of Theorem 2.3.

Given that $\mathbb{F}$ is a totally real number field of degree $r_{1}$, so
we have $r_{2}=0$ and $d=r_{1}$. Thus, in Theorem 2.1, we have

$\displaystyle\mathfrak{R}_{0}(z)$ $\displaystyle=\lim_{s\rightarrow
0}s\Gamma^{r_{1}}(s)\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+2k+1)\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)^{-s}$
$\displaystyle=\lim_{s\rightarrow
0}\Gamma^{r_{1}}(s+1)\frac{\zeta_{\mathbb{F}}(s)}{s^{r_{1}-1}}\zeta_{\mathbb{F}}(s+2k+1)\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)^{-s}$
$\displaystyle=C_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+1),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{iD}{(2\pi)^{r_{1}}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\frac{1}{(r_{1}-1)!}\lim_{s\rightarrow(1-2j)}\frac{{\rm
d}^{r_{1}-1}}{{\rm
d}s^{r_{1}-1}}\left((s+2j-1)^{r_{1}}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

Note that we have used (1.12) to evaluate $\mathfrak{R}_{0}(z)$. Moreover, we
point out that the terms $\mathfrak{R}_{-2j}(z)$ would not appear in this case
since we are dealing with totally real fields. Substituting the above values
in Theorem 2.1, we finish the proof of Theorem 2.3. ∎

  * Proof of Corollary 2.4.

Given that $m$ is a square-free positive integer and
$\mathbb{F}=\mathbb{Q}(\sqrt{m})$. In this case, we have $r_{1}=2,r_{2}=0,d=2$
and $D=4m$. In this case, we use (1.18) to see that

$\displaystyle
V\left(-\frac{(2\pi)^{2}niz}{4m}\bigg{|}\bar{0}_{2}\right)=2K_{0}\left(2\pi\sqrt{\frac{nz}{m}}e^{-\frac{i\pi}{4}}\right).$
(4.21)

Moreover, the residual terms become

$\displaystyle\mathfrak{R}_{0}(z)$
$\displaystyle=\zeta_{\mathbb{F}}^{{}^{\prime}}(0)\zeta_{\mathbb{F}}(2k+1),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{im}{\pi^{2}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\lim_{s\rightarrow(1-2j)}\frac{{\rm d}}{{\rm
d}s}\left((s+2j-1)^{2}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right).$

To calculate $\mathfrak{R}_{0}(z)$ we have used the fact
$C_{\mathbb{F}}=\zeta_{\mathbb{F}}^{{}^{\prime}}(0)$ as we are working on real
quadratic field. Putting these terms in Theorem 2.3, one can complete the
proof of this result. ∎

  * Proof of Theorem 2.5.

This is result is an implication of our main Theorem 2.1 for negative integers
$k$ and totally real fields. Mainly, from (2.3) with $r_{2}=0$, we have

$\displaystyle\mathfrak{F}_{\mathbb{F},2k+1}(z)-C_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+1)$
$\displaystyle=(-1)^{k(r_{1}+1)}z^{2k}\left\\{\mathfrak{F}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)-C_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+1)\right\\}$
$\displaystyle+\begin{cases}-\frac{i}{4\pi
z},&\quad\text{if}\,\,\,(k,r_{1},r_{2})=(-1,1,0),\\\
0,&\quad\text{otherwise.}\end{cases}$

Now replacing $k$ by $-k$ and simplifying, we complete the proof. ∎

  * Proof of Corollary 2.6.

Given that $k\geq 3$ and $r_{1}$ both are positive odd integers. Substituting
$z=i$ in Theorem 2.5, we obtain

$\displaystyle\mathfrak{F}_{\mathbb{F},-2k+1}(i)=C_{\mathbb{F}}\zeta_{\mathbb{F}}(1-2k).$

Now the proof of this corollary follows by using the value (1.12) of
$C_{\mathbb{F}}$ and the definition (1.20) of
$\mathfrak{F}_{\mathbb{F},-2k+1}(i)$. ∎

  * Proof of Corollary 2.7.

Substituting $\mathbb{F}=\mathbb{Q}(\sqrt{m})$, i.e., $r_{1}=2,r_{2}=0,D=4m$
in Theorem 2.5, we can see that

$\displaystyle(iz)^{2k}\left\\{\mathfrak{F}_{\mathbb{F},-2k+1}(z)-\zeta_{\mathbb{F}}^{{}^{\prime}}(0)\zeta_{\mathbb{F}}(1-2k)\right\\}$
$\displaystyle=\left\\{\mathfrak{F}_{\mathbb{F},-2k+1}\left(-\frac{1}{z}\right)-\zeta_{\mathbb{F}}^{{}^{\prime}}(0)\zeta_{\mathbb{F}}(1-2k)\right\\},$
(4.22)

where

$\mathfrak{F}_{\mathbb{F},-2k+1}(z)=\sum_{n=1}^{\infty}\sigma_{\mathbb{F},2k-1}(n)V\left(-\frac{(2\pi)^{2}niz}{4m}\bigg{|}\bar{0}_{2}\right).$

Now employing the expression (4.21) for
$V\left(-\frac{(2\pi)^{2}niz}{4m}\bigg{|}\bar{0}_{2}\right)$ in the above
series and then putting it in (4.22), the proof of Corollary 2.7 follows. ∎

  * Proof of Theorem 2.8.

In this case, we have taken $\mathbb{F}$ to be a purely imaginary number field
with degree $d=2r_{2}$. Thus, the proof of this theorem immediately follows by
substituting $r_{1}=0$ in Theorem 2.1. ∎

  * Proof of Corollary 2.9.

Given that $\mathbb{F}=\mathbb{Q}(\sqrt{-m})$ is a quadratic imaginary field
and $k>0$. Therefore, substituting $(r_{1},r_{2},d,D)=(0,1,2,4m)$ in Theorem
2.8, it yields that

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=(-1)^{k+1}z^{2k}\mathfrak{S}_{\mathbb{F},2k+1}\left(-\frac{1}{z}\right)+{\displaystyle\sum_{j=1}^{k}\mathfrak{R}_{-(2j-1)}(z)+\sum_{j=1}^{k-1}\mathfrak{R}_{-2j}(z)},$

where

$\displaystyle\mathfrak{S}_{\mathbb{F},2k+1}(z)$
$\displaystyle=\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-2k-1}(n)V\left(-\frac{\pi^{2}niz}{m}\bigg{|}\bar{0}_{2}\right)-\mathfrak{R}_{0}(z)-\mathfrak{R}_{1}(z),$

and the residual terms are defined as

$\displaystyle\mathfrak{R}_{0}(z)$ $\displaystyle=\lim_{s\rightarrow
0}\frac{{\rm d}^{2}}{{\rm
d}s^{2}}\left(s^{2}\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=H_{\mathbb{F}}\zeta_{\mathbb{F}}(2k+2)\frac{im}{\pi^{2}z},$
$\displaystyle\mathfrak{R}_{-(2j-1)}(z)$
$\displaystyle=\lim_{s\rightarrow-(2j-1)}\left((s+2j-1)\Lambda_{\mathbb{F},k}(s)(-iz)^{-s}\right),$
$\displaystyle=\lim_{s\rightarrow-(2j-1)}\left((s+2j-1)^{2}\Gamma(s)^{2}\frac{\zeta_{\mathbb{F}}(s)}{s+2j-1}\zeta_{\mathbb{F}}(s+2k+1)\left(\frac{-i\pi^{2}z}{m}\right)^{-s}\right)$
$\displaystyle=-\frac{\zeta_{\mathbb{F}}^{{}^{\prime}}(1-2j)}{((2j-1)!)^{2}}\zeta_{\mathbb{F}}(2k-2j+2)\left(\frac{\pi^{2}iz}{m}\right)^{2j-1}.$

In a similar way, one can show that

$\displaystyle\mathfrak{R}_{-2j}(z)$
$\displaystyle=\frac{\zeta_{\mathbb{F}}^{{}^{\prime}}(-2j)}{((2j)!)^{2}}\zeta_{\mathbb{F}}(2k-2j+1)\left(\frac{\pi^{2}iz}{m}\right)^{2j}.$

Now combining all these residual terms and using the expression (4.21) for
Steen function, we complete the proof of (2.12). ∎

  * Proof of Corollary 2.10.

Substitute $k=-1$ and $\mathbb{F}=\mathbb{Q}(\sqrt{-m})$, that is,
$r_{1}=0,r_{2}=1$ in Theorem 2.8. Thus, from (2.9)-(2.11), we arrive at

$\displaystyle\mathfrak{U}_{\mathbb{F},-1}(z)$
$\displaystyle=z^{-2}\mathfrak{U}_{\mathbb{F},-1}\left(-\frac{1}{z}\right)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)im}{\pi^{2}z},$
(4.23)

where

$\displaystyle\mathfrak{U}_{\mathbb{F},-1}(z)=\mathfrak{F}_{\mathbb{F},-1}(z)-\zeta_{\mathbb{F}}(0)\zeta_{\mathbb{F}}^{{}^{\prime}}(-1).$

Here we used (3.6) with the fact that $C_{\mathbb{F}}=\zeta_{\mathbb{F}}(0)$
as $\mathbb{F}$ is quadratic imaginary field. Now multiplying by $z^{2}$ on
both sides of (4.23), we finish the proof of (2.13). Further, to obtain
(2.14), substituting $z=i$ in (4.23) it yields that

$\displaystyle\mathfrak{F}_{\mathbb{F},-1}(i)=\zeta_{\mathbb{F}}(0)\zeta_{\mathbb{F}}^{{}^{\prime}}(-1)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)m}{2\pi^{2}}.$

Now using the definition (1.20) of $\mathfrak{F}_{\mathbb{F},-1}(i)$ we derive
that

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},1}(n)V\left(\frac{n\pi^{2}}{m}|\bar{0}_{2}\right)=\zeta_{\mathbb{F}}(0)\zeta_{\mathbb{F}}^{{}^{\prime}}(-1)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(0)m}{2\pi^{2}}.$

Finally, using the fact that
$V\left(\frac{n\pi^{2}}{m}|\bar{0}_{2}\right)=2K_{0}\left(2\pi\sqrt{\frac{n}{m}}\right)$,
we complete the proof of (2.14). ∎

  * Proof of Corollary 2.11.

The proof of this corollary immediately follows from (2.9)-(2.11) by replacing
$k$ by $-k$. As it is given that $(k,r_{2})\neq(1,1)$, so we have to use the
fact that $\mathfrak{R}(z)=0$ and multiply by $z^{2k}$ on both sides of (2.9)
after replacing $k$ by $-k$. ∎

  * Proof of Theorem 2.12.

The proof of this theorem goes along the same line as in Theorem 2.1, however,
we give brief outline. This identity is due to the case $k=0$. In the way as
we proceeded in ( Proof of Theorem 2.1.) and (4.2), one can show that

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)$
$\displaystyle=\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+1)\left(\frac{(2\pi)^{d}y}{D}\right)^{-s}{\rm
d}s$ $\displaystyle=\frac{1}{2\pi
i}\int_{(c)}\Lambda_{\mathbb{F},0}(s)y^{-s}{\rm d}s,$ (4.24)

where
$\Lambda_{\mathbb{F},0}(s)=\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+1)(2\pi)^{-ds}D^{s}$,
which is exactly same as we defined in (2.1), and
$1<\operatorname{Re}(s)=c<1+\epsilon$ with $0<\epsilon<1$. Further proceeding
in a similar way as in Theorem 2.1, here also we set up a rectangular contour
$\mathcal{C}$ made up of the vertices $c-iT,c+iT,\alpha+iT$ and $\alpha-iT$,
with $-2<\alpha<-1$ and $T$ is some large positive quantity.

$c-iT$$c+iT$$\alpha+iT$$\alpha-iT$01-12-2$\mathrm{Re}(s)$$\mathrm{Im}(s)$

We now examine the poles of our integrand function
$\Lambda_{\mathbb{F},0}(s)$. At $s=0$, $\Gamma(s)^{d}$ has a pole order $d$,
$\zeta_{\mathbb{F}}(s)$ has a zero of order $r_{1}+r_{2}-1$, and
$\zeta_{\mathbb{F}}(s+1)$ has a simple pole. Therefore, the order of the pole
of the integrand function $\Lambda_{\mathbb{F},0}(s)$ at $s=0$ is
$d-(r_{1}+r_{2}-1)+1=r_{2}+2$ since $d=r_{1}+2r_{2}$. It is easy to see that
the integrand function has a simple pole at $s=1$. From Lemma 3.2, we can see
that $\Lambda_{\mathbb{F},0}(s)=(-1)^{r_{2}}\Lambda_{\mathbb{F},0}(-s)$. This
indicates that $s=-1$ is also a simple pole of the integrand function. Now
utilizing Cauchy residue theorem, we arrive

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)=(-1)^{r_{2}}\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}n}{Dy}\bigg{|}\bar{0}_{d}\right)+\mathcal{R},$
(4.25)

where the residual term $\mathcal{R}$ is given by

$\displaystyle\mathcal{R}=R_{0}^{(r_{2}+2)}(y)+R_{1}^{(1)}(y)+R_{-1}^{(1)}(y).$
(4.26)

The above term $\mathcal{R}$ makes the main difference with Theorem 2.1, see
(4.4). The terms of $\mathcal{R}$ can be calculated as follows:

$\displaystyle R_{0}^{(r_{2}+2)}(y)$
$\displaystyle=\frac{1}{(r_{2}+1)!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}+1}}{{\rm
d}s^{r_{2}+1}}\left(s^{r_{2}+2}\Lambda_{\mathbb{F},0}(s)y^{-s}\right),$ (4.27)
$\displaystyle R_{1}^{(1)}(y)$ $\displaystyle=\lim_{s\rightarrow
1}(s-1)\Lambda_{\mathbb{F},0}(s)y^{-s}$ $\displaystyle=\lim_{s\rightarrow
1}(s-1)\Gamma(s)^{d}\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+1)(2\pi)^{-ds}D^{s}y^{-s}$
$\displaystyle=\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)D}{(2\pi)^{d}y},$
(4.28) $\displaystyle R_{-1}^{(1)}(y)$
$\displaystyle=\lim_{s\rightarrow-1}(s+1)\Lambda_{\mathbb{F},0}(s)y^{-s}$
$\displaystyle=\lim_{s\rightarrow-1}(s+1)(-1)^{r_{2}}\Lambda_{\mathbb{F},0}(-s)y^{-s}$
$\displaystyle=(-1)^{r_{2}+1}\lim_{s\rightarrow
1}(s-1)\Lambda_{\mathbb{F},0}(s)y^{s}$
$\displaystyle=(-1)^{r_{2}+1}\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)Dy}{(2\pi)^{d}}.$
(4.29)

Here we used class number formula (1.11) and Lemma 3.2 to simplify the above
residual terms $R_{1}^{(1)}(y)$ and $R_{-1}^{(1)}(y)$. Substituting the above
residual terms in (4.26) and together with (4.25), we get

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}ny}{D}\bigg{|}\bar{0}_{d}\right)-\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)D}{(2\pi)^{d}y}$
$\displaystyle=(-1)^{r_{2}}\left\\{\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}n}{Dy}\bigg{|}\bar{0}_{d}\right)-\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)Dy}{(2\pi)^{d}}\right\\}$
$\displaystyle+\frac{1}{(r_{2}+1)!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}+1}}{{\rm
d}s^{r_{2}+1}}\left(s^{r_{2}+2}\Lambda_{\mathbb{F},0}(s)y^{-s}\right).$

The above identity holds for $y>0$, however, by analytic continuation, one can
show that this identity is also true for $\operatorname{Re}(y)>0$. We
substitute $y=-iz$ to get the following form, for $z\in\mathbb{H}$,

$\displaystyle\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(-\frac{(2\pi)^{d}niz}{D}\bigg{|}\bar{0}_{d}\right)-\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)iD}{(2\pi)^{d}z}$
$\displaystyle=(-1)^{r_{2}}\left\\{\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-1}(n)V\left(\frac{(2\pi)^{d}ni}{Dz}\bigg{|}\bar{0}_{d}\right)+\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)Diz}{(2\pi)^{d}}\right\\}$
$\displaystyle+\frac{1}{(r_{2}+1)!}\lim_{s\rightarrow 0}\frac{{\rm
d}^{r_{2}+1}}{{\rm
d}s^{r_{2}+1}}\left(s^{r_{2}+2}\Lambda_{\mathbb{F},0}(s)(-iz)^{-s}\right).$

Now if we use the definition (1.20) of $\mathfrak{F}_{\mathbb{F},1}(z)$, then
the above expression can be rewritten as

$\displaystyle\mathfrak{F}_{\mathbb{F},1}(z)-\mathfrak{R}_{1}(z)=(-1)^{r_{2}}\left\\{\mathfrak{F}_{\mathbb{F},1}\left(-\frac{1}{z}\right)-\mathfrak{R}_{1}\left(-\frac{1}{z}\right)\right\\}+\mathfrak{R}_{0}(z),$

where

$\displaystyle\mathfrak{R}_{1}(z)=\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)iD}{(2\pi)^{d}z},\quad{\rm
and}\quad\mathfrak{R}_{0}(z)=\frac{1}{(r_{2}+1)!}\lim_{s\rightarrow
0}\frac{{\rm d}^{r_{2}+1}}{{\rm
d}s^{r_{2}+1}}\left(s^{r_{2}+2}\Lambda_{\mathbb{F},0}(s)(-iz)^{-s}\right).$

Letting
$\mathfrak{T}_{\mathbb{F}}(z):=\mathfrak{F}_{\mathbb{F},1}(z)-\mathfrak{R}_{1}(z)$,
one can complete the proof of Theorem 2.12. ∎

  * Proof of Corollary 2.13.

Letting $\mathbb{F}$ to be a totally real number field of degree $d=r_{1}$ in
Theorem 2.12, we get

$\displaystyle\mathfrak{F}_{\mathbb{F},1}(z)-\mathfrak{F}_{\mathbb{F},1}\left(-\frac{1}{z}\right)$
$\displaystyle=\mathfrak{R}_{1}(z)-\mathfrak{R}_{1}\left(-\frac{1}{z}\right)+\mathfrak{R}_{0}(z),$
(4.30)

where

$\mathfrak{R}_{1}(z)=\frac{H_{\mathbb{F}}\zeta_{\mathbb{F}}(2)iD}{(2\pi)^{r_{1}}z},\quad\mathfrak{R}_{0}(z)=\lim_{s\rightarrow
0}\frac{{\rm d}}{{\rm
d}s}\left(s^{2}\Gamma^{r_{1}}(s)\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+1)\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)^{-s}\right).$

Now we shall try to simplify the term $\mathfrak{R}_{0}(z)$. Utilizing the
fact that

$\displaystyle\Gamma(s)=\frac{1}{s}-\gamma+O(s),$

one can see that

$\displaystyle(s\Gamma(s))^{r_{1}}=1-r_{1}\gamma s+O(s^{2}).$ (4.31)

Since $\mathbb{F}$ is a totally real number field of degree $r_{1}$, so from
(3.6), the Laurent series expansion of $\zeta_{\mathbb{F}}(s)$ at $s=0$, one
has

$\displaystyle\frac{\zeta_{\mathbb{F}}(s)}{s^{r_{1}-1}}=a_{0}+a_{1}s+O(s^{2}),$
(4.32)

where

$a_{0}=\frac{\zeta_{\mathbb{F}}^{(r_{1}-1)}(0)}{(r_{1}-1)!}=C_{\mathbb{F}},\quad
a_{1}=\frac{\zeta_{\mathbb{F}}^{(r_{1})}(0)}{(r_{1})!}.$

Further, utilizing (3.5), the Laurent series expansion of
$\zeta_{\mathbb{F}}(s)$, it yields that

$\displaystyle
s\zeta_{\mathbb{F}}(s+1)=H_{\mathbb{F}}+\gamma_{\mathbb{F}}s+O(s^{2}).$ (4.33)

Moreover, we have

$\displaystyle\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)^{-s}=1-s\log\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)+O(s^{2}).$
(4.34)

Now combining all the above Laurent series expansions (4.31)-(4.34), one can
check that the coefficient of $s$ in the Laurent series expansion of
$s^{2}\Gamma^{r_{1}}(s)\zeta_{\mathbb{F}}(s)\zeta_{\mathbb{F}}(s+1)(-2\pi
iz)^{-s}$ is
$a_{0}\gamma_{\mathbb{F}}-a_{0}H_{\mathbb{F}}\left\\{r_{1}\gamma+\log\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)\right\\}+a_{1}H_{\mathbb{F}}$,
which shows that

$\displaystyle\mathfrak{R}_{0}(z)=a_{0}\gamma_{\mathbb{F}}-a_{0}H_{\mathbb{F}}\left\\{r_{1}\gamma+\log\left(-\frac{(2\pi)^{r_{1}}iz}{D}\right)\right\\}+a_{1}H_{\mathbb{F}}.$

Finally, substituting the above value of $\mathfrak{R}_{0}(z)$ in (4.30), we
complete the proof of (2.18).

∎

  * Proof of Corollary 2.14.

We know that $\zeta_{\mathbb{Q}}(s)=\zeta(s)$. Therefore, considering
$\mathbb{F}=\mathbb{Q}$, Corollary 2.13 reduces to

$\displaystyle\mathfrak{F}_{\mathbb{Q},1}(z)-\mathfrak{F}_{\mathbb{Q},1}\left(-\frac{1}{z}\right)=\mathfrak{R}_{1}(z)-\mathfrak{R}_{1}\left(-\frac{1}{z}\right)+\mathfrak{R}_{0}(z),$
(4.35)

where

$\displaystyle\mathfrak{F}_{\mathbb{Q},1}(z)$
$\displaystyle=\sum_{n=1}^{\infty}\sigma_{\mathbb{Q},-1}(n)V\left(-2\pi
niz\bigg{|}\bar{0}\right)=\sum_{n=1}^{\infty}\sigma_{-1}(n)\exp(2\pi inz),$
$\displaystyle\mathfrak{R}_{1}(z)$
$\displaystyle=\frac{H_{\mathbb{Q}}\zeta_{\mathbb{Q}}(2)iD}{2\pi
z}=\frac{i\pi}{12z},$ $\displaystyle\mathfrak{R}_{0}(z)$
$\displaystyle=a_{0}\gamma_{\mathbb{Q}}-a_{0}H_{\mathbb{Q}}\left\\{r_{1}\gamma+\log(-2\pi
iz)\right\\}+a_{1}H_{\mathbb{Q}}=\frac{1}{2}\log(-iz).$

The above simplified forms have been obtained by using the following well-
known values:

$H_{\mathbb{Q}}=1,\gamma_{\mathbb{Q}}=\gamma,\zeta_{\mathbb{Q}}(2)=\frac{\pi^{2}}{6},\\\
a_{0}=\zeta(0)=-\frac{1}{2},a_{1}=\zeta^{\prime}(0)=-\frac{1}{2}\log(2\pi).$

Eventually, putting the above values of
$\mathfrak{F}_{\mathbb{Q},1}(z),\mathfrak{R}_{1}(z),\mathfrak{R}_{0}(z)$ in
(4.35) and using the fact that $n\sigma_{-1}(n)=\sigma(n)$, one can finish the
proof of (2.20). ∎

##  5\. Concluding Remarks

Ramanujan’s formula (1.2) involves the following Lambert series

$\displaystyle\sum_{n=1}^{\infty}\frac{n^{-k}}{e^{ny}-1}=\sum_{n=1}^{\infty}\sigma_{-k}(n)e^{-ny},\quad\operatorname{Re}(y)>0,$
(5.1)

associated with the generalized divisor function
$\sigma_{k}(n)=\sum_{d|n}d^{k},k\in\mathbb{C}$. Upon suitable change of
variable, one can easily observe that the above series is exactly same as the
following series:

$\displaystyle F_{k}(z)=\sum_{n=1}^{\infty}\sigma_{-k}(n)e^{2\pi inz},\quad
z\in\mathbb{H},$ (5.2)

which is also present in Grosswald’s identity (1.6). Recently, Banerjee, Gupta
and Kumar [4, Equation (1.7)] have generalized the above Lambert series (5.1),
while obtaining a number field analogue of Ramanujan’s identity, in the
following way:

$\sum_{\mathfrak{a}\subset\mathcal{O}_{\mathbb{K}}}\mathfrak{N}(\mathfrak{a})^{k}\Omega_{\mathbb{F}}\left(\frac{\mathfrak{N}(\mathfrak{a})y}{D}\right)=\sum_{n=1}^{\infty}\mathtt{a}_{\mathbb{F}}(n)n^{k}\Omega_{\mathbb{F}}\left(\frac{ny}{D}\right),$

where the function $\Omega_{\mathbb{F}}(x)$, which involves the Meijer
$G$-function, is defined as

$\Omega_{\mathbb{F}}(x):=\frac{2^{1-r_{1}-r_{2}}}{\pi^{1-\frac{r_{1}}{2}}}\sum_{j=1}^{\infty}\mathtt{a}_{\mathbb{F}}(j)G_{0,2d}^{d+1,0}\left(\begin{array}[]{c}-\\\
(0)_{r_{1}+r_{2}},\left(\frac{1}{2}\right)_{r_{2}+1};\left(\frac{1}{2}\right)_{r_{1}+r_{2}-1},(0)_{r_{2}}\end{array}\bigg{|}\frac{x^{2}j^{2}}{4^{d}}\right),$

with $d=r_{1}+2r_{2}$ (the degree of the field $\mathbb{F}$ over
$\mathbb{Q}$), and $D$ is the absolute value of the discriminant of
$\mathbb{F}$. In this paper, we obtained a new number field analogue of
Ramanujan’s identity by generalizing the Lambert series $F_{k}(z)$, defined in
(5.2), in a different way than Banerjee et. al. Mainly, we studied modular
transformation formula for the following infinite series:

$\displaystyle\mathfrak{F}_{\mathbb{F},k}(z)=\sum_{n=1}^{\infty}\sigma_{\mathbb{F},-k}(n)V\left(-\frac{(2\pi)^{d}niz}{D}\bigg{|}\bar{0}_{d}\right),$

where $\sigma_{\mathbb{F},k}(n)$ is the number field analogue (1.13) of the
generalized divisor function $\sigma_{k}(n)$ and $V(z|\bar{0}_{d})$ is the
Steen function (1.16). By obtaining a number field analogue of Ramanujan-
Grosswald identity for odd zeta values, we are able to find a formula for
Dedekind zeta function at odd arguments. As an application our main identity
i.e., Theorem 2.1, we obtain many interesting modular transformation formulae
that are true for totally real number fields and purely imaginary fields, see
Theorem 2.3, Theorem 2.8. In particular, for real quadratic fields and
imaginary quadratic fields, we obtained modular transformation formulae, i.e.,
Corollary 2.7, Corollary 2.11 that are perfect analogue of the transformation
formula (1.5) for the Eisenstien series. Further, we found an exact evaluation
of $\mathfrak{F}_{\mathbb{F},2k-1}(i)$ in terms of class number and the values
of the Dedekind zeta function at negative odd integers, see (2.7), (2.16) and
(2.17). We also obtain a modular transformation formula which generalizes
transformation formula for the Dedekind eta function $\eta(z)$, see Theorem
2.12. As one of the interesting applications of Theorem 2.12, we derived a
formula (2.19) for the class number of a totally real field which also
indicates a relation with the Kronceker’s limit formula for the Dedekind zeta
function. This observation might be of an independent interest to the readers
of this article.

Acknowledgement The authors would like to thank Prof. Bruce Berndt for going
through the manuscript and giving valuable suggestions. The first author’s
research is supported by the Prime Minister Research Fellowship (PMRF), Govt.
of India, Grant No. 2102227. The last author wants to thank Science and
Engineering Research Board (SERB), India, for giving MATRICS grant (File No.
MTR/2022/000545) and SERB CRG grant (File No. CRG/CRG/2023/002122). Both
authors sincerely thank IIT Indore for providing conductive research
environment.

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