# Extended Higher Herglotz function II

Rajat Gupta and Rahul Kumar  Discipline of Mathematics, Indian Institute of
Technology, Gandhinagar, Palaj, Gandhinagar 382355, Gujarat, India
<EMAIL_ADDRESS>Center for Geometry and Physics, Institute for Basic
Science (IBS), Pohang 37673, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Very recently, Radchenko and Zagier revived the theory of Herglotz functions.
The main goal of the article is to show that one of the formulas on page 220
of Ramanujan’s Lost Notebook actually lives in the realms of this theory. As a
consequence of our general theorem, we derive an interesting identity
analogous to Ramanujan’s formula for $\zeta(2m+1)$. We also introduce a
character analogue of the Herglotz function and initiate its theory by
obtaining an elegant functional equation governed by it.

2020 Mathematics Subject Classification. Primary 11M06; Secondary 33E20.
Keywords and phrases. Herglotz function, functional equations, Ramanujan’s
lost notebook, Dirichlet character

##  1\. Introduction

In his seminal work on the Kronecker limit formula for a real quadratic field,
Zagier [19] found an interesting function which is now known as the Herglotz
function. It is defined by

$\displaystyle F(x):=\sum_{n=1}^{\infty}\frac{\psi(nx)-\log(nx)}{n},\qquad
x\in\mathbb{C}\backslash(-\infty,0],$ (1.1)

where $\psi(x):=\Gamma^{\prime}(x)/\Gamma(x)$ is the digamma function. Prior
to Zagier, a function quite similar to $F(x)$ is also appeared in the work of
Herglotz [10]. This is why Radchenko and Zagier call it the _Herglotz
function_ in [14]. Zagier establishes beautiful functional equations for
$F(x)$, namely, [19, Equations (7.4), (7.8)], for
$x\in\mathbb{C}\backslash(-\infty,0]$,

$\displaystyle F(x)-F(x+1)-F\left(\frac{x}{x+1}\right)$
$\displaystyle=-F(1)+\textup{Li}_{2}\left(\frac{1}{1+x}\right),$
$\displaystyle F(x)+F\left(\frac{1}{x}\right)$
$\displaystyle=2F(1)+\frac{1}{2}\log^{2}(x)-\frac{\pi^{2}}{6x}(x-1)^{2},$
(1.2)

where $\mathrm{Li}_{2}(x):=\sum_{n=1}^{\infty}t^{n}/n^{2}$ is dilogarithm
function and [19, Equation (7.12)]

$F(1)=-\frac{1}{2}\gamma^{2}-\frac{\pi^{2}}{12}-\gamma_{1}.$

Here $\gamma$ and $\gamma_{1}$ denote the Euler and first Stieltjes constants
respectively. In the same paper [19, Section 8], Zagier used these functional
equations to prove Meyer’s theorem [13].

Recently, Radchenko and Zagier [14] studied $F(x)$ extensively and unearthed a
myriad of connections that $F(x)$ has with other areas of number theory. They
provided many exciting properties of $F(x)$, including functional equations,
special values at rational or quadratic irrational arguments, connections with
Stark’s conjecture and with 1-cocycles for the modular group
PSL$(2,\mathbb{Z})$.

Many generalizations of $F(x)$ or its variants have been studied in the
literature. Vlasenko and Zagier [18], in their work on higher Kronecker limit
formula, considered the study of the higher Herglotz function:

$\displaystyle F_{k}(x):=\sum_{n=1}^{\infty}\frac{\psi(nx)}{n^{k}},\qquad
k\in\mathbb{N},\ k>1,\ x\in\mathbb{C}\backslash(-\infty,0].$ (1.3)

They found the following functional equations for $F_{k}(x)$ [18, Equations
(11), (12)]:

$\displaystyle F_{k}(x)+(-x)^{k-1}F_{k}\left(\frac{1}{x}\right)$
$\displaystyle=-\gamma\zeta(k)\left(1+(-x)^{k-1}\right)-\sum_{r=2}^{k-1}\zeta(r)\zeta(k+1-r)(-x)^{r-1}$
$\displaystyle\quad+\zeta(k+1)\left((-x)^{k}-\frac{1}{x}\right),$ (1.4)

and111The interpretation of $\zeta(1,k-1)$ given in [18, p. 28] has a minus
sign missing in front of the whole term, that is, the correct form is
$-\left(\zeta(k-1,1)+\zeta(k)-\gamma\zeta(k-1)\right)$.

$\displaystyle F_{k}(x)-F_{k}(x+1)+(-x)^{k-1}F_{k}\left(\frac{x+1}{x}\right)$
$\displaystyle=(-x)^{k-1}\left(\zeta(k,1)+\zeta(k+1)-\gamma\zeta(k)\right)-\sum_{r=1}^{k-1}\zeta(k+1-r,r)(-x)^{r-1}$
$\displaystyle\hskip
99.58464pt+\zeta(k+1)\left(\frac{(-x)^{k}}{x+1}-\frac{1}{x}\right),$

where, $\zeta(k)$ is the Riemann zeta function and

$\zeta(m,n):=\sum_{p>q>0}\frac{1}{p^{m}q^{n}}\hskip 14.22636pt(m\geq 2,n\geq
1).$

Vlasenko and Zagier [18, p. 54, Section 3] also dealt with another twisted
extension of (1.3) to treat the zeta functions of ray classes. Ishibashi
studied $j^{th}$ order Herglotz function to find an explicit representation of
the Laurent series coefficients of the zeta function associated to indefinite
quadratic forms [11, Theorem 3].

Let $\alpha:\mathbb{Z}\to\mathbb{C}$ be a periodic function with period $M$.
Masri [12] defined the following $L$-series generalization of $F(x)$:

$\displaystyle
F(\alpha,s,x):=\sum_{n=1}^{\infty}\frac{\alpha(n)\left(\psi(nx)-\log(nx)\right)}{n^{s}},\qquad\mathrm{Re}(s)>0,\
x>0,$

and provided its analytic continuation and evaluated it at integer points.

Very recently, the authors along with Dixit [7] studied the _extended higher
Herglotz function_

$\displaystyle\mathscr{F}_{k,N}(x):=\sum_{n=1}^{\infty}\frac{\psi(n^{N}x)-\log(n^{N}x)}{n^{k}},\qquad
x\in\mathbb{C}\backslash(-\infty,0],$

where $k$ and $N$ are positive real numbers such that $k+N>1$. Note that the
function $\mathscr{F}_{k,N}(x)$ subsumes Herglotz function $F(x)$ as well as
Vlasenko and Zagier’s higher Herglotz function $F_{k}(x)$. One of the nice
functional equations which $\mathscr{F}_{k,N}(x)$ satisfies is

$\displaystyle
x^{\frac{1-k}{N}}\mathscr{F}_{k,N}(x)-\frac{(-1)^{k}}{N}\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}e^{i\pi
j(k-1)/N}\mathscr{F}_{\frac{N+k-1}{N},\frac{1}{N}}\left(\frac{e^{-\frac{i\pi
j}{N}}}{x^{1/N}}\right)$
$\displaystyle=\frac{\pi}{N}\frac{\zeta\left(1+\frac{k-1}{N}\right)}{\sin\left(\frac{\pi}{N}(k-1)\right)}+x^{\frac{1-k}{N}}\left(-\left(\gamma+\log
x\right)\zeta(k)+N\zeta^{\prime}(k)\right)-\frac{1}{x^{\frac{N+k-1}{N}}}\zeta(k+N)+\mathscr{B}(k,N,x),$
(1.5)

where $k,N\in\mathbb{N}$ such that $1<k\leq N$ and
$x\in\mathbb{C}\backslash(-\infty,0]$ and

$\mathscr{B}(k,N,x):=\left\\{\begin{array}[]{ll}(-1)^{k+N+1}x^{1/N}\zeta\left(1+\frac{k}{N}\right)&\mbox{if
}k=N\\\ 0&\mbox{if }k\neq N\end{array}\right..$

Here and throughout this paper, the notation
$\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}$ denotes a sum over
over $j=-(N-1),-(N-3),\cdots,N-3,N-1$. For many other nice properties of
$\mathscr{F}_{k,N}(x)$ and non-trivial applications of its functional
equations, we refer the reader to [7].

In the last section of their paper, the present authors with Dixit [7] posed
several questions. Here in this article, we answer some of those questions
affirmatively. Before we do this though, we record an elegant formula from
Ramanujan’s Lost Notebook on page 220. If

$\displaystyle\varphi(x):=\psi(x)+\frac{1}{2x}-\log x,$

then for $\alpha,\beta$ positive such that $\alpha\beta=1$, we have [16, p.
220]

$\displaystyle\sqrt{\alpha}$
$\displaystyle\left\\{\frac{\gamma-\log(2\pi\alpha)}{2\alpha}+\sum_{n=1}^{\infty}\varphi(n\alpha)\right\\}=\sqrt{\beta}\left\\{\frac{\gamma-\log(2\pi\beta)}{2\beta}+\sum_{n=1}^{\infty}\varphi(n\beta)\right\\}.$
(1.6)

He also provided the following surprising integral representation which is
equal to either sides of (1.6):

$\displaystyle-\frac{1}{\pi^{3/2}}\int_{0}^{\infty}\left|\Xi\left(\frac{t}{2}\right)\Gamma\left(\frac{-1+it}{4}\right)\right|^{2}\frac{\cos\left(\frac{t}{2}\log\alpha\right)}{1+t^{2}}dt,$
(1.7)

where $\Xi(t)$ denotes Riemann’s function defined by [17, p. 16, Equations
(2.1.14), (2.1.12)]

$\displaystyle\Xi(t)$ $\displaystyle:=\xi(\tfrac{1}{2}+it),$

and

$\displaystyle\xi(s)$
$\displaystyle:=\frac{s}{2}(s-1)\pi^{-\frac{1}{2}s}\Gamma(\tfrac{1}{2}s)\zeta(s).$

The proof of the above formula can be found in [3]. The above transformation
and various other transformations of its type have generated interesting
mathematics. We refer the reader to recent papers [6], [9], and references
therein for the extensive literature related to this formula.

It was commented in [7, p. 27] that “ _it seems that ( 1.6) lives in the
realms of the theory of Herglotz function_” by observing the similarity
between the definition of $F(x)$ in (1.1) and series appearing on both sides
of (1.6). One of the main objectives of this paper is to show that this is in-
fact the case, that is, Ramanujan’s formula (1.6) is actually a special case
of two of our more general results on the Herglotz functions (Theorems 2.1,
and 2.4).

Note that the functional equation (1) is valid only when $1<k\leq N$. It was
also asked in [7, Section 5, p. 28] if there exists a functional equation for
$\mathscr{F}_{k,N}(x)$ where $k$ is greater than $N$? We answer this question
affirmatively in Theorem 2.5.

Although $\mathscr{F}_{k,N}(x)$ is a generalization of Valsenko-Zagier’s
higher Herglotz function $\mathscr{F}_{k}(x)$, the functional equation of
$\mathscr{F}_{k}(x)$ in (1) could not be obtained from (1) as a special case
due to the restriction $1<k\leq N$. This problem is now circumvented through
our Theorem 2.5.

To achieve our goals, we work with the following version of the extended
higher Herglotz function:

$\displaystyle\mathfrak{F}_{k,N}(x):=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\left(\psi\left(n^{N}x\right)-\log\left(n^{N}x\right)+\frac{1}{2n^{N}x}\right),$
(1.8)

for $x\in\mathbb{C}\backslash(-\infty,0],\ \mathrm{and}\ k+2N>1$. The series
is absolutely and uniformly convergent for $k+2N>1$ since [1, p. 259, formula
6.3.18]

$\psi(x)\sim\log(x)-\frac{1}{2x}-\frac{1}{12x^{2}}+\cdots,\qquad\mathrm{as}\
x\to\infty,\ |\mathrm{arg}(x)|<\pi.$

Note that for $k=N=1$, $\mathfrak{F}_{k,N}(x)$ reduces to the function
$F^{*}(x)$:

$\displaystyle
F^{*}(x):=\sum_{n=1}^{\infty}\frac{1}{n}\left(\psi\left(nx\right)-\log\left(nx\right)+\frac{1}{2nx}\right).$

The function $F^{*}(x)$ was used by Radchenko and Zagier [14, p. 17, Section
7.3] to reveal the cocycle nature of the Herglotz function $F(x)$ by finding
an Eichler-type integral for it.

The Herglotz function $F(x)$ is a special case of $\mathfrak{F}_{k,N}(x)$:

$\displaystyle\mathfrak{F}_{1,1}(x)=F(x)+\frac{\pi^{2}}{12x}.$

Also,

$\displaystyle\mathfrak{F}_{k,N}(x)=\mathscr{F}_{k,N}(x)+\frac{1}{2x}\zeta(k+N).$

######  Remark 1.1.

Ramanujan’s formula (1.6) can be rephrased in terms of $\mathfrak{F}_{k,1}(x)$
by

$\displaystyle\mathfrak{F}_{k,1}(x)-\frac{1}{x}\mathfrak{F}_{k,1}\left(\frac{1}{x}\right)=\frac{1}{2}\left(\gamma-\log\left(\frac{2\pi}{x}\right)\right)-\frac{1}{2x}\left(\gamma-\log(2\pi
x)\right).$ (1.9)

Thus Ramanujan understood the importance of such a function. Therefore, it
would not be unfair to say that Ramanujan studied Hergltoz-type functions even
before Herglotz and Zagier. Further, he went beyond the functional equation to
find a surprising intergal representation for this Herglotz-type function
which involves the Riemann $\Xi$-function.

The next objective of this paper is to initiate the theory of a character
analogue of the Herglotz function. We define it by

$\displaystyle
F_{k}(x,\chi):=\sum_{n=1}^{\infty}\frac{\chi(n)\psi(nx,\chi)}{n^{k}},\quad
x\in\mathbb{C}\backslash(-\infty,0]$ (1.10)

here and throughout the article $\chi(n)$ is a primitive, nonprincipal
character modulo $d$, $k$ is a non-negative integer, and

$\displaystyle\psi(x,\chi):=-\sum_{k=1}^{\infty}\frac{\chi(k)}{k+x}.$

For a real character $\chi$, the function $\psi(n,\chi)$ is equivalent to the
character analogue of the digamma function obtained by the logarithmic
differentiation of the Weierstrass product form of the character analogue of
the gamma function for real characters introduced by Berndt [2]:

$\displaystyle\Gamma(s,\chi):=e^{-sL(1,\chi)}\prod_{n=1}^{\infty}\left(1+\frac{s}{n}\right)^{-\chi(n)}e^{s\chi(n)/n},\hskip
14.22636pts\in\mathbb{C}\backslash\mathbb{Z}_{<0}.$

Note that the series in (1.10) is convergent for any $k\geq 0$ since [5, p.
334, Corollary 4.4]

$\displaystyle\psi(x,\chi)\sim-\frac{L(0,\chi)}{x}-\chi(-1)\sum_{j=2}^{\infty}\frac{B_{j}(\overline{\chi})}{jx^{j}},\quad|\mathrm{arg}(x)|<\pi,$

as $x\to\infty$. Here $B_{j}(\chi)$ is the generalized Bernoulli numbers [2,
p. 426].

##  2\. Main Results

We divide this section into two parts. The first part is devoted to extending
the modular relation of Ramanujan (1.6), and the other is dedicated to
initiating the theory of $F_{k}(x,\chi)$ defined in (1.10).

###  2.1. Extension of the formula on page 220 of Ramanujan’s Lost Notebook

In this subsection, we state three of our main theorems. The first is Theorem
2.1, which is valid for $k\in\mathbb{N}\cup\\{0\\}$ and $N=1$. The second is
Theorem 2.4 which holds for $k=1$ and $N\in\mathbb{N}$. We conclude this
subsection by providing Theorem 2.5 which holds for any natural numbers $k$
and $N$ greater than 1.

######  Theorem 2.1.

Let $\mathfrak{F}_{k,N}(x)$ be defined in (1.8). For any
$k\in\mathbb{N}\cup\\{0\\}$ and $\mathrm{Re}(x)>0$, the following identity
holds:

$\displaystyle\mathfrak{F}_{k,1}(x)=(-1)^{k}x^{k-1}\left\\{\mathfrak{F}_{k,1}\left(\frac{1}{x}\right)-\mathcal{B}_{k}(x)\right\\},$

where, for $k=0$,

$\displaystyle\mathcal{B}_{k}(x):=\frac{x}{2}\left(\log\left(\frac{2\pi}{x}\right)-\gamma\right)+\frac{1}{2}\left(\gamma-\log(2\pi
x)\right),$

for $k=1$,

$\displaystyle\mathcal{B}_{k}(x):=\frac{1}{6}\left(3\log^{2}(x)+\pi^{2}-6\gamma^{2}-12\gamma_{1}\right)-\frac{\pi^{2}}{12}\left(x+\frac{1}{x}\right),$

and for $k\geq 2$,

$\displaystyle\mathcal{B}_{k}(x):$
$\displaystyle=\zeta(k)\left(\log(x)-\gamma\right)+\zeta^{\prime}(k)+(-1)^{k}x^{1-k}\left((\log(x)+\gamma)\zeta(k)-\zeta^{\prime}(k)\right)+\frac{(-1)^{k}}{2x^{k}}\zeta(k+1)$
$\displaystyle\qquad-\frac{x}{2}\zeta(1+k)+\sum_{j=1}^{k-2}(-1)^{j-1}\zeta(j+1)\zeta(k-j)x^{-j}.$

######  Remark 2.2.

It is easy to see that Ramanujan’s formula (1.6)(or (1.9)), Zagier’s
functional equation (1.2), and Vlasenko-Zagier’s functional equation (1) are
special cases of the above theorem for $k=0,~{}k=-1,$ and $k\geq 2$,
respectively.

An immediate consequence of the above theorem is the following beautiful
symmetric formula.

######  Theorem 2.3.

Let $k$ be any natural number greater than 1. Let $\mathrm{Re}(\alpha)>0$ and
$\mathrm{Re}(\beta)>0$ such that $\alpha\beta=1$, then the following relation
holds:

$\displaystyle\alpha^{\frac{1-k}{2}}\left\\{(\gamma+\log(\alpha))\zeta(k)-\zeta^{\prime}(k)+\frac{1}{2\alpha}\zeta(k+1)+\mathfrak{F}_{k,1}(\alpha)\right\\}$
$\displaystyle=(-1)^{k}\beta^{\frac{1-k}{2}}\left\\{(\gamma+\log(\beta))\zeta(k)-\zeta^{\prime}(k)+\frac{1}{2\beta}\zeta(k+1)+\mathfrak{F}_{k,1}(\beta)\right\\}$
$\displaystyle\qquad+\sum_{j=1}^{k-2}(-1)^{j}\zeta(j+1)\zeta(k-j)\alpha^{\frac{1-k+j}{2}}\beta^{-\frac{j}{2}}.$
(2.1)

This formula is analogous to the Ramanujan’s formula for $\zeta(2m+1)$ [15, p.
173, Ch. 14, Entry 21(i)], namely, for
$\textup{Re}(\alpha)>0,\textup{Re}(\beta)>0,\alpha\beta=\pi^{2}$ and
$m\in\mathbb{Z}\backslash\\{0\\}$, by

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

where $B_{n}$ is the $n^{\textup{th}}$ Bernoulli number.

The next theorem is another extension of (1.6) in the second parameter $N$.

######  Theorem 2.4.

Let $N$ be any natural number and $\mathrm{Re}(x)>0$. Then

$\displaystyle\mathfrak{F}_{0,N}(x)=\frac{x^{-1/N}}{N}\sideset{}{{}^{\prime\prime}}{\sum}_{j=-(N-1)}^{(N-1)}e^{-\frac{i\pi
j}{N}}\mathfrak{F}_{\frac{N-1}{N},\frac{1}{N}}\left(\frac{e^{-\frac{i\pi
j}{N}}}{x^{1/N}}\right)-\frac{1}{2}\left(\log\left(\frac{(2\pi)^{N}}{x}\right)-\gamma\right)+R(N,x),$
(2.2)

where

$\displaystyle R(N,x):=\begin{cases}\frac{1}{2x}\left(\log(2\pi
x)-\gamma\right),&\mathrm{if}\ N=1\\\ -\frac{\pi
x^{-1/N}\zeta\left(1-1/N\right)}{N\sin(\pi/N)}-\frac{1}{2x}\zeta(N),&\mathrm{if}\
N>1.\end{cases}$ (2.3)

Now we state a general theorem which is valid for any
$k,N\in\mathbb{N}\backslash{\\{1\\}}.$

######  Theorem 2.5.

Let $k,\ N$ be any natural numbers greater $1$. Then for $\mathrm{Re}(x)>0$,
we have

$\displaystyle\mathfrak{F}_{k,N}(x)=\frac{(-1)^{k}}{N}x^{\frac{k-1}{N}}\sideset{}{{}^{\prime\prime}}{\sum}_{j=-(N-1)}^{N-1}e^{\frac{i\pi
j(k-1)}{N}}\mathfrak{F}_{\frac{N+k-1}{N},\frac{1}{N}}\left(\frac{e^{-\frac{i\pi
j}{N}}}{x^{1/N}}\right)-\frac{x^{\frac{k-1}{N}}}{N}\mathscr{R}_{N,k}(x),$

where, for $N\nmid(k-1)$ and $N\nmid k$,

$\displaystyle\mathscr{R}_{N,k}(x)$
$\displaystyle=\frac{N}{2}x^{\frac{1-k-N}{N}}\zeta(k+N)-\frac{\pi\zeta\left(\frac{N+k-1}{N}\right)}{\sin\left(\frac{\pi}{N}(k-1)\right)}-Nx^{\frac{1-k}{N}}\left(N\zeta^{\prime}(k)-(\gamma+\log
x)\zeta(k)\right)$
$\displaystyle\qquad+N\sum_{j=1}^{\lfloor\frac{k-1}{N}\rfloor}(-1)^{j}\zeta(1+j)\zeta(k-N)x^{\frac{k-1}{N}+j},$

and, for $N|k$,

$\displaystyle\mathscr{R}_{N,k}(x)$
$\displaystyle=\frac{N}{2}x^{\frac{1-k-N}{N}}\zeta(k+N)-\frac{\pi\zeta\left(\frac{N+k-1}{N}\right)}{\sin\left(\frac{\pi}{N}(k-1)\right)}-Nx^{\frac{1-k}{N}}\left(N\zeta^{\prime}(k)-(\gamma+\log
x)\zeta(k)\right)$
$\displaystyle\quad-\frac{N}{2}(-1)^{\frac{k}{N}}x^{\frac{1}{N}}\zeta\left(\frac{k+N}{N}\right)+N\sum_{j=1}^{\lfloor\frac{k-1}{N}\rfloor}(-1)^{j}\zeta(1+j)\zeta(k-N)x^{\frac{k-1}{N}+j},$

and, for $N|(k-1)$,

$\displaystyle\mathscr{R}_{N,k}(x)$
$\displaystyle=\frac{N}{2}x^{\frac{1-k-N}{N}}\zeta(k+N)+(-1)^{\frac{k-1}{N}}\left((N\gamma-\log
x)\zeta\left(\frac{N+k-1}{N}\right)-\zeta^{\prime}\left(\frac{N+k-1}{N}\right)\right)$
$\displaystyle\quad-Nx^{\frac{1-k}{N}}\left(N\zeta^{\prime}(k)-(\gamma+\log
x)\zeta(k)\right)+N\sum_{j=1}^{\lfloor\frac{k-1}{N}\rfloor-1}(-1)^{j}\zeta(1+j)\zeta(k-N)x^{\frac{k-1}{N}+j}.$

When we restrict $1<k\leq N$ in the above theorem, it gives (1) as a special
case:

######  Corollary 2.6.

The functional equation (1) holds true.

###  2.2. Character analogue of the Herglotz function

For a non-principal Dirichlet character $\chi(n)$, let $L(s,\chi)$ denote the
Dirichlet $L$-function defined by

$\displaystyle L(s,\chi)=\sum_{n=1}^{\infty}\frac{\chi(n)}{n^{s}}$

for Re$(s)>1$. This series converges conditionally for $0<\textup{Re}(s)<1$.
Also, it can be analytically continued to an entire function of $s$.

The two-term functional equation governed by $F_{k}(x,\chi)$ is given in the
following theorem.

######  Theorem 2.7.

Let $F_{k}(x,\chi)$ be defined in (1.10). Let $\chi(n)$ be a primitive,
nonprincipal character modulo $d$. Then $k\in\mathbb{N}\cup\\{0\\}$ and
Re$(x)>0$, we have

$\displaystyle
F_{k}(x,\chi)-(-1)^{k}x^{k-1}F_{k}\left(x,\frac{1}{\chi}\right)=\sum_{j=0}^{k-1}(-x)^{k-j-1}L(1+j,\chi)L(-j+k,\chi).$

Now to state our next theorem, we define

$\displaystyle b:=\left\\{\begin{array}[]{ll}0&\mbox{if }\chi(-1)=1\\\
1&\mbox{if }\chi(-1)=-1.\end{array}\right.$

Then the $\Xi(s,\chi)$ can be defined as

$\displaystyle\Xi(t,\chi):=\xi\left(\frac{1}{2}+it,\chi\right),$ (2.4)

where

$\displaystyle\xi(s,\chi):=\left(\frac{\pi}{d}\right)^{-(s+b)/2}\Gamma\left(\frac{s+b}{2}\right)L(s,\chi).$
(2.5)

Our next theorem gives an elegant modular relation along with the integral
comprising $\Xi(t,\chi)$-function.

######  Theorem 2.8.

Let $\chi(n)$ be a primitive, nonprincipal character modulo $d$. Further
assume $\mathrm{Re}(\alpha)>0,\ \mathrm{Re}(\beta)>0$ such that
$\alpha\beta=1$, then

$\displaystyle\sqrt{\alpha}F_{0}(\alpha,\chi)$
$\displaystyle=\sqrt{\beta}F_{0}(\beta,\chi)$
$\displaystyle=-\frac{1}{2\pi}\left(\frac{\pi}{d}\right)^{b+\frac{1}{2}}\int_{0}^{\infty}\left|\frac{\Gamma\left(\frac{1}{2}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4}+\frac{it}{4}+\frac{b}{2}\right)}\right|^{2}\Xi\left(-\frac{t}{2},\chi\right)\Xi\left(\frac{t}{2},\chi\right)\cos\left(\frac{t}{2}\log\alpha\right)dt.$

The above theorem reduces to [5, Corollaries 5.1 and 5.2] depending on the
parity of the character.

##  3\. Proofs

We first present the proof of our general Theorem 2.5. Throughout the proofs,
the notation $\int_{(c)}ds$ will denote the line integral
$\int_{c-i\infty}^{c+i\infty}ds$ with $c=\mathrm{Re}(s)$.

  * Proof of Theorem 2.5.

Kloosterman’s formula is given by [17, p. 25, Equation (2.9.2)]

$\displaystyle\psi(x+1)-\log(x)=\frac{1}{2\pi
i}\int_{(c)}\frac{-\pi\zeta(1-z)}{\sin(\pi z)}x^{-z}dz,$ (3.1)

which is valid for $0<c=\mathrm{Re}(z)<1$. An application of the functional
equation

$\displaystyle\psi(x+1)=\psi(x)+1/x$

in (3.1) implies that

$\displaystyle\psi(x)-\log(x)+\frac{1}{x}=\frac{1}{2\pi
i}\int_{(c)}\frac{-\pi\zeta(1-z)}{\sin(\pi z)}x^{-z}dz.$ (3.2)

For our purpose we need the integral representation for the function
$\psi(x)-\log(x)+\frac{1}{2x}$. For that, we shift the line of the integration
to $1<d=\mathrm{Re}(z)<2$. Let $\mathfrak{C}$ be a positively oriented
rectangular contour formed by the points $c-iT,d-iT,d+iT,c+iT$. Observe that
the integrand has only one simple pole at $s=1$ due to $\sin(\pi z)$ and its
residue is $-1/(2x)$ as $\zeta(0)=-1/2$. Therefore, residue theorem
immediately yields

$\displaystyle\frac{1}{2\pi
i}\left(\int_{d-iT}^{d+iT}+\int_{d+iT}^{c+iT}-\int_{c-iT}^{c+iT}+\int_{c-iT}^{d-iT}\right)\frac{-\pi\zeta(1-z)}{\sin(\pi
z)}x^{-z}dz=-\frac{1}{2x}.$ (3.3)

Note that the integrals along the horizontal lines goes to zero by using the
Stirling formula in the vertical strip $p\leq\sigma\leq q$ [4, p. 224]:

$|\Gamma(s)|=\sqrt{2\pi}|t|^{\sigma-\frac{1}{2}}e^{-\frac{1}{2}\pi|t|}\left(1+O\left(\frac{1}{|t|}\right)\right)$
(3.4)

as $|t|\to\infty$. Thus, from (3.3), we have

$\displaystyle\frac{1}{2\pi i}\int_{(d)}\frac{-\pi\zeta(1-z)}{\sin(\pi
z)}x^{-z}dz=\frac{1}{2\pi i}\int_{(c)}\frac{-\pi\zeta(1-z)}{\sin(\pi
z)}x^{-z}dz-\frac{1}{2x}.$ (3.5)

Combining (3.2) and (3.5) together, we obtain

$\displaystyle\psi(x)-\log(x)+\frac{1}{2x}=\frac{1}{2\pi
i}\int_{(d)}\frac{-\pi\zeta(1-z)}{\sin(\pi z)}x^{-z}dz.$ (3.6)

Replacing $x$ by $n^{N}x$ in the above equation and invoking (1.8), we obtain

$\displaystyle\mathfrak{F}_{k,N}(x)$
$\displaystyle=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\frac{1}{2\pi
i}\int_{(d)}\frac{-\pi\zeta(1-z)}{\sin(\pi z)}(n^{N}x)^{-z}dz$
$\displaystyle=\frac{1}{2\pi
i}\int_{(d)}\frac{-\pi\zeta(1-z)\zeta(k+Nz)}{\sin(\pi z)}x^{-z}dz,$ (3.7)

where in the last step we interchanged the order of the summation and
integration and used the series definition of $\zeta(s)$. We now make change
of variable $s=1-k-Nz$ in ( Proof of Theorem 2.5.) to deduce, for
$1-k-2N<c<1-k-N$,

$\displaystyle\mathfrak{F}_{k,N}(x)$
$\displaystyle=\frac{x^{\frac{k-1}{N}}}{N}\frac{1}{2\pi
i}\int_{(c)}\frac{-\pi\zeta\left(1+\frac{s+k-1}{N}\right)\zeta(1-s)}{\sin\left(\frac{\pi}{N}(1-s-k)\right)}x^{\frac{s}{N}}ds$
(3.8)

We next shift the line of integration to $1<\lambda<2$. Consider the contour
formed by the line segments $[\lambda-iT,\lambda+iT],\ [\lambda+iT,c+iT],\
[c+iT,c-iT]$ and $[c-iT,\lambda-iT]$. In order to do so, we encounter several
poles of different orders depending upon $k$ and $N$. We denote the sum of the
residues by $\mathscr{R}_{k,N}(x)$. By invoking Stirling’s formula (3.4), we
see that the integrals along the horizontal lines vanish as $T\to\infty$.
Thus, by residue theorem and (3.8), we have

$\displaystyle\mathfrak{F}_{k,N}(x)$
$\displaystyle=\frac{x^{\frac{k-1}{N}}}{N}\left(\frac{1}{2\pi
i}\int_{(\lambda)}\frac{\pi\zeta\left(1+\frac{s+k-1}{N}\right)\zeta(1-s)}{\sin\left(\frac{\pi}{N}(s+k-1)\right)}x^{\frac{s}{N}}ds-\mathscr{R}_{k,N}(x)\right).$
(3.9)

Now, the main task is to evaluate the line integral present on the right-hand
side of the above equation. Let us denote it by $\mathcal{I}_{k,N}(x)$. To
that end, we invoke following result [8, Lemma 4.1]

$\displaystyle\frac{1}{\sin(z)}=\frac{1}{\sin(Nz)}\sideset{}{{}^{\prime\prime}}{\sum}_{j=-(N-1)}^{N-1}e^{ijz}$

with $z=\frac{\pi}{N}(s+k-1)$ in $\mathcal{I}_{k,N}(x)$, so that

$\displaystyle\mathcal{I}_{k,N}(x)=\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}e^{\frac{i\pi
j(k-1)}{N}}\frac{1}{2\pi
i}\int_{(\lambda)}\frac{\pi\zeta(1-s)\zeta\left(1+\frac{s+k-1}{N}\right)}{\sin\left(\pi(s+k-1)\right)}\left(\frac{e^{-\frac{i\pi
j}{N}}}{x^{1/N}}\right)^{-s}.$

We can use the series definition of $\zeta\left(1+\frac{s+k-1}{N}\right)$
since $1<\lambda=\mathrm{Re}(s)<2$. Thus, we arrive at

$\displaystyle\mathcal{I}_{k,N}(x)=(-1)^{k}\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}e^{\frac{i\pi
j(k-1)}{N}}\sum_{n=1}^{\infty}\frac{1}{n^{1+\frac{k-1}{N}}}\frac{1}{2\pi
i}\int_{(\lambda)}\frac{-\pi\zeta(1-s)}{\sin\left(\pi
s\right)}\left(\left(\frac{n}{x}\right)^{\frac{1}{N}}e^{-\frac{i\pi
j}{N}}\right)^{-s}ds.$

Invoking (3.6) in the above equation so as to obtain

$\displaystyle\mathcal{I}_{k,N}(x)$
$\displaystyle=(-1)^{k}\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}e^{\frac{i\pi
j(k-1)}{N}}\sum_{n=1}^{\infty}\frac{1}{n^{1+\frac{k-1}{N}}}\left\\{\psi\left(\left(\frac{n}{x}\right)^{\frac{1}{N}}e^{-\frac{i\pi
j}{N}}\right)-\log\left(\left(\frac{n}{x}\right)^{\frac{1}{N}}e^{-\frac{i\pi
j}{N}}\right)\right.$
$\displaystyle\left.\qquad\qquad\qquad+\frac{1}{2}\left(\frac{n}{x}\right)^{-\frac{1}{N}}e^{\frac{i\pi
j}{N}}\right\\}$
$\displaystyle=(-1)^{k}\sum_{j=-(N-1)}^{(N-1)}{\vphantom{\sum}}^{\prime\prime}e^{\frac{i\pi
j(k-1)}{N}}\mathfrak{F}_{\frac{N+k-1}{N},\frac{1}{N}}\left(\frac{e^{-\frac{i\pi
j}{N}}}{x^{1/N}}\right),$ (3.10)

where in the last step we employed the definition of $\mathfrak{F}_{k,N}(x)$
from (1.8).

We next calculate the residues. We break this calculation into different cases
depending on whether $N$ divides $k$ and $k-1$ or not. It boils down to only
three cases: $N\nmid k$ and $N\nmid(k-1)$, $N|k$ and $N|(k-1)$ as if $N$
divides $k$ then it cannot divide $k-1$ and vice-versa.

Case I: Let $N\nmid k$ and $N\nmid(k-1)$. Observe that the integrand has
simple poles at $s=0$ due to $\zeta(1-s),$ and at $s=1-k-N,1-k+Nj$, where
$1\leq j\leq\lfloor\frac{k-1}{N}\rfloor$, due to
$\sin\left(\frac{\pi}{N}(1-s-k)\right)$. It also has double pole at $s=1-k$
due to $\sin\left(\frac{\pi}{N}(1-s-k)\right)$ and
$\zeta\left(1+\frac{s+k-1}{N}\right)$. These can be evaluated to as:

$\displaystyle R_{0}$
$\displaystyle=-\frac{\pi\zeta\left(\frac{N+k-1}{N}\right)}{\sin\left(\frac{\pi}{N}(k-1)\right)}$
$\displaystyle R_{1-k-N}$
$\displaystyle=\frac{N}{2}\zeta(k+N)x^{\frac{1-k-N}{N}}$ $\displaystyle
R_{1-k}$
$\displaystyle=N\left\\{N\zeta^{\prime}(k)-(\gamma+\log(x))\zeta(k)\right\\}x^{\frac{1-k}{N}}$
$\displaystyle R_{1-k+Nj}$
$\displaystyle=N\sum_{j=1}^{\lfloor\frac{k-1}{N}\rfloor}(-1)^{j}\zeta(1+j)\zeta(k-Nj)x^{\frac{k-1}{N}+j}.$
(3.11)

Case II: Let $N|k$. Therefore, $k=mN$ for some $m\in\mathbb{N}$. This implies
$\sin\left(\frac{\pi}{N}(1-s-k)\right)=(-1)^{m+1}\sin\left(\frac{\pi}{N}(1-s)\right)$.
Hence, the integrand has an extra simple pole at $s=1$ besides all other poles
in the previous case. The residue at the pole $s=1$ is:

$\displaystyle
R_{1}=-\frac{N}{2}(-1)^{k/N}x^{1/N}\zeta\left(\frac{k+N}{N}\right).$ (3.12)

Case III: Let $N|(k-1)$. In this case, $\sin\left(\frac{\pi}{N}(1-s-k)\right)$
also has pole at $s=0$ which was not the case before (this was
$j=\lfloor\frac{k-1}{N}\rfloor$ case in Case I). This implies that the
integrand has now double pole at $s=0$ along with other poles in the Case I,
but now $j$ runs between $1\leq j\leq\lfloor\frac{k-1}{N}\rfloor-1$.
Therefore, we only need to calculate the residue at $s=0$, that is:

$\displaystyle
R_{0}=(-1)^{\frac{k-1}{N}}\left\\{\big{(}N\gamma-\log(x)\big{)}\zeta\left(\frac{N+k-1}{N}\right)-\zeta^{\prime}\left(\frac{N+k-1}{N}\right)\right\\}.$
(3.13)

We now combine all of the residues from ( Proof of Theorem 2.5.), (3.12), and
(3.13) together, then it is nothing but the term $\mathscr{R}_{k,N}(x)$
defined in the statement of the theorem according to the different cases.

Theorem now follows upon using the above facts and ( Proof of Theorem 2.5.) in
(3.9). ∎

  * Proof of Theorem 2.1.

Note that up to (3.8), the calculation holds for any $N\in\mathbb{N}$ and
$k\in\mathbb{N}\cup\\{0\\}$. Thus, letting $N=1$ in (3.8) gives for
$-1-k<c<-k$,

$\displaystyle\mathfrak{F}_{k,1}(x)=(-1)^{k}x^{k-1}\frac{1}{2\pi
i}\int_{(c)}\frac{-\pi\zeta(s+k)\zeta(1-s)}{\sin(\pi s)}x^{s}ds.$

We shift the line of integration to $1<\lambda<2$ in the similar manner as we
did in after (3.8). Thus, we have

$\displaystyle\mathfrak{F}_{k,1}(x)=(-1)^{k}x^{k-1}\left(\frac{1}{2\pi
i}\int_{(\lambda)}\frac{-\pi\zeta(s+k)\zeta(1-s)}{\sin(\pi
s)}x^{s}ds-\mathscr{R}_{k,1}(x)\right).$ (3.14)

The calculation of the residues depends on the values of $k$ is $0$, $1$ and
$k\geq 2$.
Case I: When $k=0$, the integrand has double pole at $s=0$ due to $\zeta(1-s)$
and $\sin(\pi s)$ and $s=1$ due to $\zeta(s)$ and $\sin(\pi s)$. The residues
at these points are:

$\displaystyle R_{0}$ $\displaystyle=\frac{1}{2}\big{(}\gamma-\log(2\pi
x)\big{)}$ $\displaystyle R_{1}$
$\displaystyle=\frac{x}{2}\left(\log\left(\frac{2\pi}{x}\right)-\gamma\right).$
(3.15)

Case II: Let $k=1$. Now integrand has simple pole at $s=1,-1$ due to $\sin(\pi
s)$ and pole of order three at $s=0$ due to $\sin(\pi s),\zeta(s+1)$ and
$\zeta(1-s)$ with residues:

$\displaystyle R_{1}$ $\displaystyle=-\frac{\pi^{2}}{12}x$ $\displaystyle
R_{-1}$ $\displaystyle=\frac{\pi^{2}}{12}\frac{1}{x}$ $\displaystyle R_{0}$
$\displaystyle=\frac{1}{6}\left(3\log^{2}(x)+\pi^{2}-6\gamma^{2}-12\gamma_{1}\right).$
(3.16)

Case III: For $k\geq 2$, the integrand has pole of order 2 at $s=0$ and
$s=1-k$ because of $\sin(\pi s),\zeta(1-s)$ and $\sin(\pi s),\zeta(s+k)$,
respectively. It has also simple poles at $s=1,-k$ and $s=-j,1\leq j\leq k-2$
due to $\sin(\pi s)$. The residues at these poles can be evaluated to

$\displaystyle R_{0}$
$\displaystyle=\zeta(k)(\log(x)-\gamma)+\zeta^{\prime}(k)$ $\displaystyle
R_{1-k}$
$\displaystyle=(-1)^{k}x^{1-k}\left\\{\zeta(k)(\log(x)+\gamma)-\zeta^{\prime}(k)\right\\}$
$\displaystyle R_{1}$ $\displaystyle=-\frac{x}{2}\zeta(1+k)$ $\displaystyle
R_{-k}$ $\displaystyle=\frac{(-1)^{k}}{2x^{k}}\zeta(k+1)$ $\displaystyle
R_{j}$ $\displaystyle=\sum_{j=1}^{k-2}(-1)^{j-1}\zeta(j+1)\zeta(k-j)x^{-j}.$
(3.17)

Upon combining the residues from ( Proof of Theorem 2.1.), ( Proof of Theorem
2.1.) and ( Proof of Theorem 2.1.) together in (3.14), we see that
$\mathscr{R}_{k,1}(x)$ is nothing but $\mathcal{B}_{k}(x)$ defined in the
statement of the theorem.

The same argument, we used to obtain ( Proof of Theorem 2.5.), can be adapted
to show that

$\displaystyle\frac{1}{2\pi
i}\int_{(\lambda)}\frac{-\pi\zeta(s+k)\zeta(1-s)}{\sin(\pi
s)}x^{s}ds=\mathfrak{F}_{k,1}\left(\frac{1}{x}\right).$

Substituting the residues and the above integral in (3.14), we complete the
proof. ∎

  * Proof of Theorem 2.4.

The argument is similar to the proof of Theorem 2.5. One can proceed along the
same lines with letting $k=0$ and then only difference is while calculating
the residual terms. When $N>1$, the integrand has simple poles at $s=0$,
$s=1-N$ and a pole of order two at $s=1$. Whereas, for $N=1$, it has poles of
order two at $s=0$ and $s=1$. Therefore, we have

$\displaystyle\mathfrak{F}_{0,N}(x)$
$\displaystyle=\frac{x^{-\frac{1}{N}}}{N}\left(\frac{1}{2\pi
i}\int_{(\lambda)}\frac{\pi\zeta\left(1+\frac{s-1}{N}\right)\zeta(1-s)}{\sin\left(\frac{\pi}{N}(s-1)\right)}x^{\frac{s}{N}}ds-\mathscr{R}_{0,N}(x)\right).$
(3.18)

It can be easily seen that the residual term $\mathscr{R}_{0,N}$ present in
above equation is

$\displaystyle-\frac{N}{2}x^{1/N}\left(\log\left(\frac{(2\pi)^{N}}{x}\right)-\gamma\right)+R(N,x),$

where $R(N,x)$ is defined in (2.3).

The line integral in (3.18) can be evaluated by letting $k=0$ in ( Proof of
Theorem 2.5.). Hence, now (2.2) follows straightforward upon using the above
facts. ∎

  * Proof of Theorem 2.3.

For $k\geq 2$, Theorem 2.1 gives

$\displaystyle x^{\frac{1-k}{2}}\mathfrak{F}_{k,1}(x)$
$\displaystyle=(-1)^{k}x^{\frac{k-1}{2}}\mathfrak{F}_{k,1}\left(\frac{1}{x}\right)-x^{\frac{1-k}{2}}\left((\log
x+\gamma)\zeta(k)-\zeta^{\prime}(k)\right)$
$\displaystyle\quad+(-1)^{k}x^{\frac{k-1}{2}}\left(\left(\log\left(\frac{1}{x}\right)+\gamma\right)\zeta(k)-\zeta^{\prime}(k)\right)-\frac{1}{2}x^{-\frac{1+k}{2}}\zeta(k+1)$
$\displaystyle\quad+\frac{1}{2}(-1)^{k}x^{\frac{k+1}{2}}\zeta(k+1)+\sum_{j=1}^{k-2}(-1)^{k+j-1}\zeta(j+1)\zeta(k-j)x^{\frac{k-1}{2}-j}.$

Let $x=\alpha$ and $\alpha\beta=1$. Then the above equation can be rephrased
as

$\displaystyle\alpha^{\frac{1-k}{2}}\left\\{(\log\alpha+\gamma)\zeta(k)-\zeta^{\prime}(k)+\frac{1}{2\alpha}\zeta(k+1)+\mathfrak{F}_{k,1}(\alpha)\right\\}$
$\displaystyle=(-1)^{k}\beta^{\frac{1-k}{2}}\left\\{(\log\beta+\gamma)\zeta(k)-\zeta^{\prime}(k)+\frac{1}{2\beta}\zeta(k+1)+\mathfrak{F}_{k,1}(\beta)\right\\}$
$\displaystyle\qquad+\sum_{j=1}^{k-2}(-1)^{k+j-1}\zeta(j+1)\zeta(k-j)\alpha^{\frac{k-1}{2}-j}.$
(3.19)

Replacing $j$ by $k-1-j$ in the finite sum of ( Proof of Theorem 2.3.) and
using the fact $\alpha\beta=1$, we arrive at (2.3). ∎

We now present the proofs of our results on the character analogue of the
Herglotz function $F_{k}(x,\chi)$. We first prove the two-term functional
equation.

  * Proof of Theorem 2.7.

We first evaluate the sum

$\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)\psi(nx,\chi)}{n^{k}},$

by employing [5, Corollary 4.2]222It is easy to see that this result is valid
for Re$(x)>0.$, hence we have for $0<c<1$,

$\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)\psi(nx,\chi)}{n^{k}}$
$\displaystyle=\sum_{n=1}^{\infty}\frac{\chi(n)}{n^{k}}\frac{1}{2\pi
i}\int_{(c)}\frac{-\pi L(1-s,\chi)}{\sin(\pi s)}(nx)^{-s}\ ds$
$\displaystyle=\frac{1}{2\pi i}\int_{(c)}\frac{-\pi
L(1-s,\chi)L(k+s,\chi)}{\sin(\pi s)}x^{-s}\ ds.$

Upon performing the change of variable from $s$ to $1-s-k$, we obtain

$\displaystyle\sum_{n=1}^{\infty}\frac{\chi(n)\psi(nx,\chi)}{n^{k}}$
$\displaystyle=\frac{(-1)^{k}}{2\pi i}\int_{(d)}\frac{-\pi
L(s+k,\chi)L(1-s,\chi)}{\sin(\pi s)}x^{s+k-1}\ ds,$ (3.20)

where $-k<d<1-k$. To evaluate it further, we shift the line of the integration
to $0<c=\mathrm{Re}(s)<1$. Let $\mathfrak{C}$ be a positively oriented
rectangular contour formed by the points $c-iT,d-iT,d+iT,c+iT$. Observe that
the integrand has simple poles at $s=0,1,\cdots,1-k$ due to $\sin(\pi s)$.
Therefore, residue theorem immediately yields

$\displaystyle\frac{1}{2\pi
i}\left(\int_{c-iT}^{c+iT}+\int_{c+iT}^{d+iT}-\int_{d-iT}^{d+iT}+\int_{d-iT}^{c-iT}\right)\frac{(-1)^{k+1}x^{k-1}\pi
L(1-s,\chi)L(s+k,\chi)}{\sin(\pi s)}x^{s}\ ds=-\sum_{j=0}^{k-1}R_{-j}.$

Note that, as $T\to\infty$ the contribution from integrals along the vertical
lines is zero. Hence,

$\displaystyle\frac{(-1)^{k+1}x^{k-1}}{2\pi i}$
$\displaystyle\int_{(d)}\frac{\pi L(1-s,\chi)L(1+s,\chi)}{\sin(\pi s)}x^{s}\
ds$ $\displaystyle=\frac{(-1)^{k+1}x^{k-1}}{2\pi i}\int_{(c)}\frac{\pi
L(1-s,\chi)L(1+s,\chi)}{\sin(\pi s)}x^{s}\ ds-\sum_{j=0}^{k-1}R_{-j},$

where,

$\displaystyle R_{-j}=(-1)^{1+j-k}x^{-1-j+k}L(1+j,\chi)L(-j+k,\chi).$

Finally, from (3.20) and invoking the definition (1.10) we complete our proof.
∎

  * Proof of Theorem 2.8.

The first-equality follows by letting $k=0$ in Theorem 2.7. Now, for the
second equality we evaluate

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi).$

Hence, for $0<c<1,$ from (3.20) and by employing the reflection formula for
the gamma function $\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi s)$ in the second step,
we see that

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi)$
$\displaystyle=\sum_{n=1}^{\infty}\chi(n)\frac{1}{2\pi i}\int_{(c)}\frac{-\pi
L(1-s,\chi)}{\sin(\pi s)}(nx)^{-s}\ ds$ $\displaystyle=-\frac{1}{2\pi
i}\int_{(c)}\Gamma(s)\Gamma(1-s)L(1-s,\chi)L(s,\chi)x^{-s}\ ds.$

Applying the definition (2.5) in the above equation to obtain

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi)$
$\displaystyle=-\frac{1}{2\pi
i}\int_{(c)}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma\left(\frac{1-s+b}{2}\right)}\left(\frac{\pi}{d}\right)^{\frac{1-s+b}{2}}\frac{\xi(1-s,\chi)\xi(s,\chi)}{\Gamma\left(\frac{s+b}{2}\right)}\left(\frac{\pi}{q}\right)^{\frac{s+b}{2}}x^{-s}\
ds.$

Letting $c=\frac{1}{2},$ we see that

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi)$
$\displaystyle=-\left(\frac{\pi}{d}\right)^{\frac{1}{2}+b}\frac{1}{2\pi
i}\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}\frac{\Gamma(s)\Gamma(1-s)}{\Gamma\left(\frac{1-s+b}{2}\right)\Gamma\left(\frac{s+b}{2}\right)}\xi(1-s,\chi)\xi(s,\chi)x^{-s}\
ds.$

Now we perform the change of variable $s=\frac{1}{2}+it$ and employing (2.4),

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi)$
$\displaystyle=-\frac{1}{\sqrt{x}}\left(\frac{\pi}{d}\right)^{\frac{1}{2}+b}\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{\Gamma\left(\frac{1}{2}+it\right)\Gamma\left(\frac{1}{2}-it\right)}{\Gamma\left(\frac{1}{4}-\frac{it}{2}+\frac{b}{2}\right)\Gamma\left(\frac{1}{4}+\frac{it}{2}+\frac{b}{2}\right)}\Xi(-t,\chi)\Xi(t,\chi)x^{-it}\
dt.$

We perform the change of variable by $t\to t/2$, and after some considerable
simplification, we arrive at

$\displaystyle\sum_{n=1}^{\infty}\chi(n)\psi(nx,\chi)$
$\displaystyle=-\frac{1}{2\pi\sqrt{x}}\left(\frac{\pi}{d}\right)^{b+\frac{1}{2}}\int_{0}^{\infty}\frac{\Gamma\left(\frac{1}{2}+\frac{it}{2}\right)\Gamma\left(\frac{1}{2}-\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4}-\frac{it}{4}+\frac{b}{2}\right)\Gamma\left(\frac{1}{4}+\frac{it}{4}+\frac{b}{2}\right)}\Xi\left(-\frac{t}{2},\chi\right)\Xi\left(\frac{t}{2},\chi\right)\cos\left(\frac{t\log
x}{2}\right)dt.$

This completes the proof of the theorem. ∎

##  4\. Acknowledgements

The authors would like to show their sincere gratitude to Prof. Atul Dixit for
fruitful discussions and suggestions on the manuscript. The first author’s
research was supported by the SERB-DST CRG grant CRG/2020/002367 of Prof. Atul
Dixit and partly by his institute IIT Gandhinagar. The second author’s
research was supported by the grant IBS-R003-D1 of the IBS-CGP, POSTECH, South
Korea. Both authors sincerely thank their respective funders for the support.

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# An extension of a supercongruence of Long and Ramakrishna

Victor J. W. Guo  School of Mathematics and Statistics, Huaiyin Normal
University, Huai’an 223300, Jiangsu, People’s Republic of China
<EMAIL_ADDRESS>,  Ji-Cai Liu  Department of Mathematics, Wenzhou
University, Wenzhou 325035, People’s Republic of China<EMAIL_ADDRESS>and  Michael J. Schlosser  Fakultät für Mathematik, Universität Wien, Oskar-
Morgenstern-Platz 1, A-1090 Vienna, Austria<EMAIL_ADDRESS>
###### Abstract.

We prove two supercongruences for specific truncated hypergeometric series.
These include an uniparametric extension of a supercongruence that was
recently established by Long and Ramakrishna. Our proofs involve special
instances of various hypergeometric identities including Whipple’s
transformation and the Karlsson–Minton summation.

###### Key words and phrases:

hypergeometric series; supercongruences; $p$-adic Gamma function; Whipple’s
transformation, Karlsson–Minton’s summation

###### 2010 Mathematics Subject Classification:

Primary 33C20; Secondary 11A07, 11B65

The first author was partially supported by the National Natural Science
Foundation of China (grant 11771175).

The second author was partially supported by the National Natural Science
Foundation of China (grant 11801417).

The third author was partially supported by FWF Austrian Science Fund grant P
32305.

##  1\. Introduction

Let $(a)_{n}=a(a+1)\cdots(a+n-1)$ denote the Pochhammer symbol. For complex
numbers $a_{0},a_{1},\ldots,a_{r}$ and $b_{1},\ldots,b_{r}$, the (generalized)
hypergeometric series ${}_{r+1}F_{r}$ is defined as

${}_{r+1}F_{r}\\!\left[\begin{array}[]{cccc}a_{0},&a_{1},&\ldots,&a_{r}\\\\[5.0pt]
&b_{1},&\ldots,&b_{r}\end{array};z\right]=\sum_{k=0}^{\infty}\frac{(a_{0})_{k}(a_{1})_{k}\cdots(a_{r})_{k}}{k!(b_{1})_{k}\cdots(b_{r})_{k}}z^{k}.$

Summation and transformation formulas for generalized hypergeometric series
play an important part in the investigation of supercongruences. See, for
instance, [7, 11, 12, 15, 16, 17, 19]. In particular, Long and Ramakrishna
[15, Theorems 3 and 2] proved the following two supercongruences:

$\sum_{k=0}^{p-1}\frac{\left(\frac{1}{2}\right)_{k}^{3}}{k!^{3}}\equiv\begin{cases}-\displaystyle\Gamma_{p}\\!\left(\frac{1}{4}\right)^{4}\pmod{p^{3}},&\text{if
$p\equiv 1\pmod{4}$,}\\\\[10.0pt]
-\frac{p^{2}}{16}\displaystyle\Gamma_{p}\\!\left(\frac{1}{4}\right)^{4}\pmod{p^{3}},&\text{if
$p\equiv 3\pmod{4}$,}\end{cases}$ (1.1)

and

$\sum_{k=0}^{p-1}(6k+1)\frac{\left(\frac{1}{3}\right)_{k}^{6}}{k!^{6}}\equiv\begin{cases}-p\,\displaystyle\Gamma_{p}\\!\left(\frac{1}{3}\right)^{9}\pmod{p^{6}},&\text{if
$p\equiv 1\pmod{6}$,}\\\\[10.0pt]
-\frac{10}{27}p^{4}\,\displaystyle\Gamma_{p}\\!\left(\frac{1}{3}\right)^{9}\pmod{p^{6}},&\text{if
$p\equiv 5\pmod{6}$,}\end{cases}$ (1.2)

where $\Gamma_{p}(x)$ is the $p$-adic Gamma function. The restriction of the
supercongruence (1.1) modulo $p^{2}$ was earlier established by Van Hamme [18,
Equation (H.2)]. The supercongruence (1.2) is even stronger than a conjecure
made by Van Hamme [18, Equation (D.2)] who asserted the corresponding
supercongruence modulo $p^{4}$ for $p\equiv 1\pmod{6}$. Long and Ramakrishna
also mentioned that (1.2) does not hold modulo $p^{7}$ in general.

The first purpose of this paper is to prove the following supercongruence.
Note that the $r=\pm 1$ cases partially confirm the $d=5$ and $q\to 1$ case of
[8, Conjectures 1 and 2].

######  Theorem 1.

Let $r\leqslant 1$ be an odd integer coprime with $5$. Let $p$ be a prime such
that $p\equiv-\frac{r}{2}\pmod{5}$ and $p\geqslant\frac{5-r}{2}$. Then

$\sum_{k=0}^{p-1}(10k+r)\frac{\left(\frac{r}{5}\right)_{k}^{5}}{k!^{5}}\equiv
0\pmod{p^{4}},$ (1.3)

Recently, the second author [13] established the following supercongruence
related to (1.2):

$\sum_{k=0}^{p-1}(6k-1)\frac{\left(-\frac{1}{3}\right)_{k}^{6}}{k!^{6}}\equiv\begin{cases}\displaystyle
140p^{4}\,\Gamma_{p}\\!\left(\frac{2}{3}\right)^{9}\pmod{p^{5}},\quad&\text{if
$p\equiv 1\pmod{6}$,}\\\\[15.0pt] \displaystyle
378p\,\Gamma_{p}\\!\left(\frac{2}{3}\right)^{9}\pmod{p^{5}},\quad&\text{if
$p\equiv 5\pmod{6}$,}\end{cases}$ (1.4)

where $p$ is a prime.

The second purpose of this paper is to give the following common
generalization of the second supercongruence in (1.2), restricted to modulo
$p^{5}$, and the first supercongruence in (1.4).

######  Theorem 2.

Let $r\leqslant 1$ be an integer coprime with $3$. Let $p$ be a prime such
that $p\equiv-r\pmod{3}$ and $p\geqslant 3-r$. Then

$\displaystyle\sum_{k=0}^{p-1}(6k+r)\frac{\left(\frac{r}{3}\right)_{k}^{6}}{k!^{6}}$
$\displaystyle\equiv\frac{(-1)^{r+1}\,80rp^{4}}{81}\cdot\frac{\Gamma_{p}\\!\left(1+\frac{r}{3}\right)^{2}}{\Gamma_{p}\\!\left(1+\frac{2r}{3}\right)^{3}\Gamma_{p}\\!\left(1-\frac{r}{3}\right)^{4}}$
$\displaystyle\quad\;\times\sum_{k=0}^{1-r}\frac{(r-1)_{k}\left(\frac{r}{3}\right)_{k}^{3}}{(1)_{k}\left(\frac{2r}{3}\right)_{k}^{3}}\pmod{p^{5}}.$
(1.5)

Letting $r=1$ and $r=-1$ in (1.5) and using (1.14) and (1.16), we arrive at
the $p\equiv 5\pmod{6}$ case of (1.2) modulo $p^{5}$ and the $p\equiv
1\pmod{6}$ case of (1.4), respectively.

Our proof of Theorem 1 will require Whipple’s well-poised ${}_{7}F_{6}$
transformation formula (see [2, p. 28]):

$\displaystyle{}_{7}F_{6}\\!\left[\begin{array}[]{ccccccc}a,&1+\frac{1}{2}a,&b,&c,&d,&e,&-n\\\\[5.0pt]
&\frac{1}{2}a,&1+a-b,&1+a-c,&1+a-d,&1+a-e,&1+a+n\end{array};1\right]$ (1.8)
$\displaystyle=\frac{(a+1)_{n}(a-d-e+1)_{n}}{(a-d+1)_{n}(a-e+1)_{n}}\,{}_{4}F_{3}\\!\left[\begin{array}[]{cccc}1+a-b-c,&d,&e,&-n\\\\[5.0pt]
d+e-a-n,&1+a-b,&1+a-c&\end{array};1\right],$ (1.11)

where $n$ is a non-negative integer, and Karlsson–Minton’s summation formula
(see, for example, [4, Equation (1.9.2)]):

${}_{r+1}F_{r}\\!\left[\begin{array}[]{cccc}-n,&b_{1}+m_{1},&\ldots,&b_{r}+m_{r}\\\\[5.0pt]
&b_{1},&\ldots,&b_{r}\end{array};1\right]=0,$ (1.12)

where $n,m_{1},\ldots,m_{r}$ are non-negative integers and
$n>m_{1}+\cdots+m_{r}$. Our proof of Theorem 2 relies on a ${}_{7}F_{6}$
transformation formula slightly different from Whipple’s ${}_{7}F_{6}$
transformation formula (1.11), obtained as a result from combining (1.11) with
a ${}_{4}F_{3}$ transformation formula. The transformation was already
utilized by the second author to prove (1.4).

Furthermore, in order to prove Theorem 2, we require some properties of the
$p$-adic Gamma function, collected in the following two lemmas.

######  Lemma 1.

[3, Section 11.6] Suppose $p$ is an odd prime and $x\in\mathbb{Z}_{p}$. Then

$\displaystyle\Gamma_{p}(0)=1,\quad\Gamma_{p}(1)=-1,$ (1.13)
$\displaystyle\Gamma_{p}(x)\Gamma_{p}(1-x)=(-1)^{a_{p}(x)},$ (1.14)
$\displaystyle\Gamma_{p}(x)\equiv\Gamma_{p}(y)\pmod{p}\quad\text{for $x\equiv
y\pmod{p}$},$ (1.15)
$\displaystyle\frac{\Gamma_{p}(x+1)}{\Gamma_{p}(x)}=\begin{cases}-x\quad&\text{if
$v_{p}(x)=0$,}\\\ -1\quad&\text{if $v_{p}(x)>0$, }\end{cases}$ (1.16)

where $a_{p}(x)\in\\{1,2,\dots,p\\}$ with $x\equiv a_{p}(x)\pmod{p}$ and
$v_{p}(\cdot)$ denotes the $p$-order.

######  Lemma 2.

[15, Lemma 17, (4)] Let $p$ be an odd prime. If
$a\in\mathbb{Z}_{p},n\in\mathbb{N}$ such that none of $a,a+1,\dots,a+n-1$ are
in $p\mathbb{Z}_{p}$, then

$(a)_{n}=(-1)^{n}\frac{\Gamma_{p}(a+n)}{\Gamma_{p}(a)}.$ (1.17)

In the following Sections 2 and 3, we give proofs of Theorems 1 and 2,
respectively. The final Section 4 is devoted to a discussion and includes two
conjectures.

##  2\. Proof of Theorem 1

Motivated by the work of McCarthy and Osburn [16] and Mortenson [17], we take
the following choice of parameters in (1.11). Let $a=\frac{r}{5}$,
$b=\frac{r+5}{10}$, $c=\frac{r+3p}{5}$, $d=\frac{r+3ip}{5}$,
$e=\frac{r-3ip}{5}$, and $n=\frac{3p-r}{5}$, where $i^{2}=-1$. Then we
conclude that

$\displaystyle{}_{6}F_{5}\\!\left[\begin{array}[]{ccccccc}\frac{r}{5},&1+\frac{r}{10},&\frac{r+3p}{5},&\frac{r+3ip}{5},&\frac{r-3ip}{5},&\frac{r-3p}{5}\\\\[5.0pt]
&\frac{r}{10},&1-\frac{3p}{5},&1-\frac{3ip}{5},&1+\frac{3ip}{5},&1+\frac{3p}{5}\end{array};1\right]$
(2.3)
$\displaystyle=\frac{\left(1+\frac{r}{5}\right)_{\frac{3p-r}{5}}\left(1-\frac{r}{5}\right)_{\frac{3p-r}{5}}}{\left(1-\frac{3ip}{5}\right)_{\frac{3p-r}{5}}\left(1+\frac{3ip}{5}\right)_{\frac{3p-r}{5}}}\,{}_{4}F_{3}\\!\left[\begin{array}[]{cccc}\frac{5-r-6p}{10},&\frac{r+3ip}{5},&\frac{r-3ip}{5},&\frac{r-3p}{5}\\\\[5.0pt]
\frac{2r-3p}{5},&\frac{r+5}{10},&\frac{5-3p}{5}\end{array};1\right],$ (2.6)

It is easy to see that, for $k\geqslant 0$ and any $p$-adic integer $b$,

$\left(a+bp\right)_{k}\left(a-bp\right)_{k}\left(a+bip\right)_{k}\left(a-bip\right)_{k}\equiv(a)_{k}^{4}\pmod{p^{4}}.$
(2.7)

Hence, the left-hand side of (2.6) is congruent to

$\displaystyle\sum_{k=0}^{\frac{3p-r}{5}}\frac{\left(1+\frac{r}{10}\right)_{k}\left(\frac{r}{5}\right)_{k}^{5}}{\left(\frac{r}{10}\right)_{k}(1)_{k}^{5}}$
$\displaystyle=\frac{1}{r}\sum_{k=0}^{\frac{3p-r}{5}}(10k+r)\frac{\left(\frac{r}{5}\right)_{k}^{5}}{k!^{5}}$
$\displaystyle\equiv\frac{1}{r}\sum_{k=0}^{p-1}(10k+r)\frac{\left(\frac{r}{5}\right)_{k}^{5}}{k!^{5}}\pmod{p^{4}},$

where we have used the fact that
$\frac{\left(\frac{r}{5}\right)_{k}}{k!}\equiv 0\pmod{p}$ for
$\frac{3p-r}{5}<k\leqslant p-1$ (the condition $p\geqslant\frac{5-r}{2}$ in
the theorem is to guarantee $\frac{3p-r}{5}\leqslant p-1$). Since
$\frac{3p-r}{5}\geqslant\frac{2p+r}{5}$, we have

$\displaystyle\frac{\left(1+\frac{r}{5}\right)_{\frac{3p-r}{5}}\left(1-\frac{r}{5}\right)_{\frac{3p-r}{5}}}{\left(1-\frac{3ip}{5}\right)_{\frac{3p-r}{5}}\left(1+\frac{3ip}{5}\right)_{\frac{3p-r}{5}}}$
$\displaystyle=\frac{\left(1+\frac{r}{5}\right)_{\frac{3p-r}{5}}\left(1-\frac{r}{5}\right)_{\frac{3p-r}{5}}}{\left(1+\frac{9p^{2}}{25}\right)_{\frac{3p-r}{5}}}$
$\displaystyle\equiv 0\pmod{p^{2}}.$

Finally, by the congruences

$\left(a+bip\right)_{k}\left(a-bip\right)_{k}\equiv\left(a+bp\right)_{k}\left(a-bp\right)_{k}\equiv(a)_{k}^{2}\pmod{p^{2}}$
(2.8)

for any $p$-adic integer $b$, we obtain

$\displaystyle{}_{4}F_{3}\\!\left[\begin{array}[]{cccc}\frac{5-r-6p}{10},&\frac{r+3ip}{5},&\frac{r-3ip}{5},&\frac{r-3p}{5}\\\\[5.0pt]
\frac{2r-3p}{5},&\frac{r+5}{10},&\frac{5-3p}{5}\end{array};1\right]$
$\displaystyle\equiv{}_{4}F_{3}\\!\left[\begin{array}[]{cccc}\frac{5-r-6p}{10},&\frac{r}{5},&\frac{r}{5},&\frac{r-3p}{5}\\\\[5.0pt]
\frac{2r-3p}{5},&\frac{r+5}{10},&\frac{5-3p}{5}\end{array};1\right]$
$\displaystyle\equiv{}_{4}F_{3}\\!\left[\begin{array}[]{cccc}\frac{5-r-6p}{10},&\frac{r+p}{5},&\frac{r-p}{5},&\frac{r-3p}{5}\\\\[5.0pt]
\frac{2r-3p}{5},&\frac{r+5}{10},&\frac{5-3p}{5}\end{array};1\right]$
$\displaystyle=0\pmod{p^{2}},$

where we have utilized Karlsson–Minton’s summation (1.12) with
$n=\frac{3p-r}{5}$, $b_{1}=\frac{2r-3p}{5}$, $b_{2}=\frac{r+5}{10}$,
$b_{3}=\frac{5-3p}{5}$, $m_{1}=\frac{1-r}{2}$, $m_{2}=\frac{2p+r-5}{10}$, and
$m_{3}=\frac{2p+r-5}{5}$ in the last step.

##  3\. Proof of Theorem 2

We can verify (1.5) for $r=1$ and $p=2$ by hand. In what follows, we assume
that $p$ is an odd prime. Recall the following transformation formula [13,
Equation (4.2)]:

$\displaystyle{}_{7}F_{6}\\!\left[\begin{matrix}t,&1+\frac{t}{2},&-n,&t-a,&t-b,&t-c,&1-t-m+n+a+b+c\\\\[5.0pt]
&\frac{t}{2},&1+t+n,&1+a,&1+b,&1+c,&2t+m-n-a-b-c\end{matrix};1\right]$
$\displaystyle=\frac{(1+t)_{n}(a+b+2-m-t)_{n}(a+c+2-m-t)_{n}(b+c+2-m-t)_{n}}{(1+a)_{n}(1+b)_{n}(1+c)_{n}(a+b+c+1-m-2t)_{n}}$
$\displaystyle\quad\;\times\frac{(a+b+1-m-t)(a+c+1-m-t)(b+c+1-m-t)}{(a+b+n+1-m-t)(a+c+n+1-m-t)(b+c+n+1-m-t)}$
$\displaystyle\quad\;\times{}_{4}F_{3}\\!\left[\begin{matrix}-m,&\hskip-25.60747pt-n,&\hskip-25.60747pta+b+c+1-m-2t,&a+b+c+1+n-m-t\\\\[5.0pt]
&\hskip-28.45274pta+b+1-m-t,&a+c+1-m-t,&b+c+1-m-t\end{matrix};1\right].$ (3.1)

Let $\zeta$ be a fifth primitive root of unity. Setting $m=1-r$,
$t=\frac{r}{3}$, $n=\frac{2p-r}{3}$, $a=\frac{2p\zeta}{3}$,
$b=\frac{2p\zeta^{2}}{3}$ and $c=\frac{2p\zeta^{3}}{3}$ in (3.1) and using
$1+\zeta+\zeta^{2}+\zeta^{3}+\zeta^{4}=0$, the left-hand side of (3.1) becomes

$\displaystyle{}_{7}F_{6}\\!\left[\begin{matrix}1+\frac{r}{6},&\frac{r}{3},&\frac{r-2p}{3},&\frac{r-2p\zeta}{3},&\frac{r-2p\zeta^{2}}{3},&\frac{r-2p\zeta^{3}}{3},&\frac{r-2p\zeta^{4}}{3}\\\\[5.0pt]
&\frac{r}{6},&1+\frac{2p}{3},&1+\frac{2p\zeta}{3},&1+\frac{2p\zeta^{2}}{3},&1+\frac{2p\zeta^{3}}{3},&1+\frac{2p\zeta^{4}}{3}\end{matrix};1\right]$
$\displaystyle\equiv\frac{1}{r}\sum_{k=0}^{\frac{2p-r}{3}}(6k+r)\frac{\left(\frac{r}{3}\right)_{k}^{6}}{k!^{6}}\pmod{p^{5}},$

where we have used the facts that none of the denominators in ${}_{7}F_{6}$
contain a multiple of $p$ (the condition $p\geqslant 3-r$ in the theorem is to
guarantee $\frac{2p-r}{3}\leqslant p-1$) and

$\displaystyle\left(u+vp\right)_{k}\left(u+vp\zeta\right)_{k}\left(u+vp\zeta^{2}\right)_{k}\left(u+vp\zeta^{3}\right)_{k}\left(u+vp\zeta^{4}\right)_{k}\equiv(u)_{k}^{5}\pmod{p^{5}}.$

Furthermore, for $\frac{2p-r}{3}<k\leqslant p-1$ we have
$\left(\frac{r}{3}\right)_{k}\equiv 0\pmod{p}$. Thus,

$\displaystyle{}_{7}F_{6}\\!\left[\begin{matrix}1+\frac{r}{6},&\frac{r}{3},&\frac{r-2p}{3},&\frac{r-2p\zeta}{3},&\frac{r-2p\zeta^{2}}{3},&\frac{r-2p\zeta^{3}}{3},&\frac{r-2p\zeta^{4}}{3}\\\\[5.0pt]
&\frac{r}{6},&1+\frac{2p}{3},&1+\frac{2p\zeta}{3},&1+\frac{2p\zeta^{2}}{3},&1+\frac{2p\zeta^{3}}{3},&1+\frac{2p\zeta^{4}}{3}\end{matrix};1\right]$
$\displaystyle\equiv\frac{1}{r}\sum_{k=0}^{p-1}(6k+r)\frac{\left(\frac{r}{3}\right)_{k}^{6}}{k!^{6}}\pmod{p^{5}}.$
(3.2)

On the other hand, we determine the terminating hypergeometric series on the
right-hand side of (3.1) modulo $p$:

$\displaystyle\frac{(a+b+1-m-t)(a+c+1-m-t)(b+c+1-m-t)}{(a+b+n+1-m-t)(a+c+n+1-m-t)(b+c+n+1-m-t)}$
$\displaystyle\times{}_{4}F_{3}\\!\left[\begin{matrix}-m,&\hskip-14.22636pt-n,&\hskip-14.22636pta+b+c+1-m-2t,&a+b+c+1+n-m-t\\\\[5.0pt]
&\hskip-28.45274pta+b+1-m-t,&a+c+1-m-t,&b+c+1-m-t\end{matrix};1\right]$
$\displaystyle\equiv
8\sum_{k=0}^{1-r}\frac{(r-1)_{k}\left(\frac{r}{3}\right)_{k}^{3}}{(1)_{k}\left(\frac{2r}{3}\right)_{k}^{3}}\pmod{p}.$
(3.3)

Moreover,

$\displaystyle\frac{(1+t)_{n}(a+b+2-m-t)_{n}(a+c+2-m-t)_{n}(b+c+2-m-t)_{n}}{(1+a)_{n}(1+b)_{n}(1+c)_{n}(a+b+c+1-m-2t)_{n}}$
$\displaystyle=\frac{\left(1+\frac{r}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{2})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta^{2}+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}}{(-1)^{\frac{2p-r}{3}}\left(1+\frac{2p\zeta}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{2}}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{3}}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{4}}{3}\right)_{\frac{2p-r}{3}}}.$
(3.4)

Note that

$\left(1+\frac{r}{3}\right)_{\frac{2p-r}{3}}=\frac{2p}{3}\left(1+\frac{r}{3}\right)_{\frac{2p-r-3}{3}},$
(3.5)

and

$\displaystyle\left(1+\frac{2r+2p(\zeta+\zeta^{2})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta^{2}+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}$
$\displaystyle=\frac{5p^{3}}{27}\left(1+\frac{2r+2p(\zeta+\zeta^{2})}{3}\right)_{\frac{p-2r-3}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{3})}{3}\right)_{\frac{p-2r-3}{3}}$
$\displaystyle\quad\;\times\left(1+\frac{2r+2p(\zeta^{2}+\zeta^{3})}{3}\right)_{\frac{p-2r-3}{3}}\left(\frac{3+p(2\zeta+2\zeta^{2}+1)}{3}\right)_{\frac{p+r}{3}}$
$\displaystyle\quad\;\times\left(\frac{3+p(2\zeta+2\zeta^{3}+1)}{3}\right)_{\frac{p+r}{3}}\left(\frac{3+p(2\zeta^{2}+2\zeta^{3}+1)}{3}\right)_{\frac{p+r}{3}}.$
(3.6)

Combining (3.5) and (3.6), we arrive at

$\displaystyle\frac{\left(1+\frac{r}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{2})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2r+2p(\zeta^{2}+\zeta^{3})}{3}\right)_{\frac{2p-r}{3}}}{(-1)^{\frac{2p-r}{3}}\left(1+\frac{2p\zeta}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{2}}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{3}}{3}\right)_{\frac{2p-r}{3}}\left(1+\frac{2p\zeta^{4}}{3}\right)_{\frac{2p-r}{3}}}$
$\displaystyle\equiv\frac{(-1)^{\frac{2p-r}{3}}10p^{4}}{81}\cdot\frac{\left(1+\frac{r}{3}\right)_{\frac{2p-r-3}{3}}\left(1+\frac{2r}{3}\right)^{3}_{\frac{p-2r-3}{3}}(1)^{3}_{\frac{p+r}{3}}}{(1)^{4}_{\frac{2p-r}{3}}}\pmod{p^{5}}.$
(3.7)

It follows from (3.2)–(3.4) and (3.7) that

$\displaystyle\sum_{k=0}^{p-1}(6k+r)\frac{\left(\frac{r}{3}\right)_{k}^{6}}{k!^{6}}$
$\displaystyle\equiv\frac{(-1)^{\frac{2p-r}{3}}\,80rp^{4}}{81}\cdot\frac{\left(1+\frac{r}{3}\right)_{\frac{2p-r-3}{3}}\left(1+\frac{2r}{3}\right)^{3}_{\frac{p-2r-3}{3}}(1)^{3}_{\frac{p+r}{3}}}{(1)^{4}_{\frac{2p-r}{3}}}$
$\displaystyle\quad\;\times\sum_{k=0}^{1-r}\frac{(r-1)_{k}\left(\frac{r}{3}\right)_{k}^{3}}{(1)_{k}\left(\frac{2r}{3}\right)_{k}^{3}}\pmod{p^{5}}.$
(3.8)

By Lemmas 1 and 2, we have

$\displaystyle\frac{\left(1+\frac{r}{3}\right)_{\frac{2p-r-3}{3}}\left(1+\frac{2r}{3}\right)^{3}_{\frac{p-2r-3}{3}}(1)^{3}_{\frac{p+r}{3}}}{(1)^{4}_{\frac{2p-r}{3}}}$
$\displaystyle\overset{\eqref{bb-6}}{=}\frac{(-1)^{\frac{2p-r}{3}+r}\,\Gamma_{p}\\!\left(\frac{2p}{3}\right)\Gamma_{p}\\!\left(\frac{p}{3}\right)^{3}\Gamma_{p}\\!\left(1+\frac{p+r}{3}\right)^{3}\Gamma_{p}(1)}{\Gamma_{p}\\!\left(1+\frac{r}{3}\right)\Gamma_{p}\\!\left(1+\frac{2r}{3}\right)^{3}\Gamma_{p}\\!\left(1+\frac{2p-r}{3}\right)^{4}}$
$\displaystyle\overset{\eqref{bb-3}}{\equiv}\frac{(-1)^{\frac{2p-r}{3}+r}\,\Gamma_{p}\\!\left(0\right)^{4}\Gamma_{p}\left(1+\frac{r}{3}\right)^{2}\Gamma_{p}(1)}{\Gamma_{p}\\!\left(1+\frac{2r}{3}\right)^{3}\Gamma_{p}\\!\left(1-\frac{r}{3}\right)^{4}}\pmod{p}$
$\displaystyle\overset{\eqref{bb-1}}{=}\frac{(-1)^{\frac{2p-r}{3}+r+1}\,\Gamma_{p}\\!\left(1+\frac{r}{3}\right)^{2}}{\Gamma_{p}\\!\left(1+\frac{2r}{3}\right)^{3}\Gamma_{p}\\!\left(1-\frac{r}{3}\right)^{4}}.$
(3.9)

The proof of (1.5) then follows from (3.8) and (3.9).

##  4\. Discussion

We know that many supercongruences have nice $q$-analogues (see [5, 6, 8, 9,
10, 14, 20]). For example, we have the following conjectural $q$-analogue of
(1.3): for the same $p$ and $r$ as in Theorem 1,

$\sum_{k=0}^{p-1}[10k+r]\frac{\left(q^{r};q^{5}\right)_{k}^{5}}{\left(q^{5};q^{5}\right)_{k}^{5}}q^{\frac{5(3-r)k}{2}}\equiv
0\pmod{[p]^{4}},$ (4.1)

where $[n]=1+q+\cdots+q^{n-1}$ is the $q$-integer and
$(a;q)_{n}=(1-a)(1-aq)\cdots(1-aq^{n-1})$ denotes the $q$-shifted factorial.

Although there are $q$-analogues of Whipple’s well-poised ${}_{7}F_{6}$
transformation and of Karlsson–Minton’s summation (see [4, Appendix (II.27)
and (III.18)]), we are unable to give a proof of (4.1). This is because we
only know a $q$-analogue of (2.8) (see [9, Lemma 1]) but do not know any
$q$-analogues of (2.7). Besides, we do not know how to prove (4.1) by using
the method of ‘creative microscoping’ devised in [10] either.

While in Theorem 2 we were able to provide a common generalization of the
second supercongruence in (1.2) (restricted to modulo $p^{5}$) and the first
supercongruence in (1.4), it appears to be rather difficult to extend Theorem
1 to a higher supercongruence involving the $p$-adic Gamma function in the
spirit of Theorem 2, even in the special cases $r=1$ or $r=-1$.

We end our paper with two further conjectures for future research. Conjecture
1 concerns a stronger version of Theorem 2 and includes the second
supercongruence in (1.2) as a special case. Conjecture 2 concerns a common
generalization of the first supercongruence in (1.2) and the second
supercongruence in (1.4).

######  Conjecture 1.

The supercongruence (1.5) holds modulo $p^{6}$ for any prime $p>3$.

######  Conjecture 2.

Let $r\leqslant 1$ be an integer coprime with $3$. Let $p\geqslant 7$ be a
prime such that $p\equiv r\pmod{3}$ and $p\geqslant 3-2r$. Then

$\displaystyle\sum_{k=0}^{p-1}(6k+r)\frac{\left(\frac{r}{3}\right)_{k}^{6}}{k!^{6}}$
$\displaystyle\equiv\frac{(-1)^{r}\,8rp}{3}\cdot\frac{\Gamma_{p}\\!\left(1+\frac{r}{3}\right)^{2}}{\Gamma_{p}\\!\left(1+\frac{2r}{3}\right)^{3}\Gamma_{p}\\!\left(1-\frac{r}{3}\right)^{4}}$
$\displaystyle\quad\;\times\sum_{k=0}^{1-r}\frac{(r-1)_{k}\left(\frac{r}{3}\right)_{k}^{3}}{(1)_{k}\left(\frac{2r}{3}\right)_{k}^{3}}\pmod{p^{6}}.$
(4.2)

We remark that by using (3.1) and the same method as in the proof of Theorem
2, we can only show that (4.2) holds modulo $p^{2}$. A new technique is needed
to prove Conjecture 2.

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