# The Second Moment of the Product of the Riemann Zeta and Dirichlet
$L$-Functions

Ikuya Kaneko  Department of Mathematics, California Institute of Technology,
1200 E California Blvd, Pasadena, CA 91125, USA<EMAIL_ADDRESS>https://sites.google.com/view/ikuyakaneko/

###### Abstract.

We establish Motohashi’s formula for the second moment of the product of the
Riemann zeta function and a Dirichlet $L$-function associated to an arbitrary
primitive Dirichlet character modulo $q\in\mathbb{N}$. If the conductor $q$ is
an odd prime, our reciprocity formula is consistent with Motohashi’s result in
the early 1990s. The cubic moment side features two versions of central
$L$-values of automorphic forms on $\Gamma_{0}(q)\backslash\mathbb{H}$
weighted by an integral transform of a test function on the second moment
side. Our techniques involve a combination of analytic number theory and
automorphic forms.

###### Key words and phrases:

Motohashi’s formula; second moment; Riemann zeta function; Dirichlet
$L$-functions

###### 2010 Mathematics Subject Classification:

Primary: 11M06; Secondary: 11F03, 11F72

The author is supported in part by the Masason Foundation and the Spirit of
Ramanujan STEM Talent Initiative.

##  1\. Introduction

In a series of work culminating in [6], Motohashi constructed an explicit
formula relating the fourth moment of the Riemann zeta function to the cubic
moment of $L$-functions relevant to modular forms (Maaß and holomorphic forms
and Eisenstein series) for the full modular group
$\operatorname{SL}_{2}(\mathbb{Z})$. The aim of this article is to deduce a
version of Motohashi’s formula for the second moment of the product of the
Riemann zeta and Dirichlet $L$-functions

$\mathcal{Z}_{2}(g;\psi)\coloneqq\int_{\mathbb{R}}|\zeta(1/2+it)L(1/2+it,\psi)|^{2}g(t)dt$

for an arbitrary primitive Dirichlet character $\psi$ modulo $q$. This problem
was proposed by Motohashi [7] in 1992 at the conference of the Research
Institute for Mathematical Sciences, Kyoto University. It behoves us to
mention that he handled the case where $q$ is an odd prime in loc. cit. and
extended this to the case of squarefree conductors. Our formula is valid for
all $q\in\mathbb{N}$ and describes all spectral terms on the cubic moment side
in terms of newforms. A heuristic deduction that we sketch in §1.2 manifests
that we should expect Motohashi’s formula looking like

$\mathcal{Z}_{2}(g;\psi)\leftrightsquigarrow\sum_{f\in\mathcal{B}^{\ast}(q)}L(1/2,f)^{2}L(1/2,f\otimes\overline{\psi})\check{g}(t_{f}),$
(1.1)

where $\mathcal{B}^{\ast}(q)$ stands for an orthonormal Hecke basis of Maaß
newforms of level $q$ and trivial central character,
$t_{f}\in\mathbb{R}\cup[-i\vartheta,i\vartheta]$ is the spectral parameter
with $\vartheta=7/64$ being chosen at the current state of knowledge, and
$\check{g}$ is an elaborate integral transform of a fixed test function $g$,
which involves the Gauß hypergeometric function.

###  1.1. Statement of Main Results

Let $\psi$ be an arbitrary primitive Dirichlet character modulo a positive
integer $q$ and let $\mathcal{R}_{4}^{+}$ be the subdomain of $\mathbb{C}^{4}$
where all four parameters have real parts greater than one. In this article,
we generalise the classical formula of Motohashi shown in [9]. For a test
function $g$, we consider the second moment

$\mathcal{M}_{2}(s_{1},s_{2},s_{3},s_{4};g;\psi)\coloneqq\int_{\mathbb{R}}\zeta(s_{1}+it)L(s_{2}+it,\psi)\zeta(s_{3}-it)L(s_{4}-it,\overline{\psi})g(t)dt,$
(1.2)

where the parameters (or shifts) satisfy
$(s_{1},s_{2},s_{3},s_{4})\in\mathcal{R}_{4}^{+}$. To deduce a reciprocity
formula for (1.2), we initially work with (1.2) by exploiting Dirichlet series
expansions for the integrand in the region of absolute convergence, and then
we can take the limit $(s_{1},s_{2},s_{3},s_{4})\to(1/2,1/2,1/2,1/2)$ after
the meromorphic continuation.

Pertaining to the cubic moment side, we define the twisted spectral mean
values

$\mathcal{J}_{\pm}^{\text{Maa{\ss}}}\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}^{\ast}(q_{0})}\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2},f\right)L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2},f\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2},f\otimes\overline{\psi}\right)}{L(1,\mathrm{Ad}^{2}f)}\Theta_{q}^{\text{Maa{\ss}}}(s_{1},s_{2},s_{3},s_{4},f),$
$\mathcal{J}_{\pm}^{\text{Eis}}\coloneqq\sum_{c_{\chi}^{2}\mid M\mid
q}\int_{\mathbb{R}}\frac{L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}+it,\chi\right)L^{q/M}\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+it,\chi\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}+it,\chi\overline{\psi}\right)}{L^{q}(1+2it,\chi^{2})}\\\
\times\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-it,\overline{\chi}\right)L^{q/M}\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}-it,\overline{\chi}\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}-it,\overline{\chi\psi}\right)}{L^{q}(1-2it,\overline{\chi}^{2})}\Theta_{\chi,q}^{\text{Eis}}(s_{1},s_{2},s_{3},s_{4},t)\frac{dt}{2\pi},$
$\mathcal{J}_{+}^{\text{hol}}\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}_{\text{hol}}^{\ast}(q_{0})}\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2},f\right)L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2},f\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2},f\otimes\overline{\psi}\right)}{L(1,\mathrm{Ad}^{2}f)}\Theta_{q}^{\text{hol}}(s_{1},s_{2},s_{3},s_{4},f),$

where $\Theta_{q}^{\Diamond}(s_{1},s_{2},s_{3},s_{4},f)$ with
$\Diamond\in\\{\text{Maa{\ss}},\text{hol}\\}$ and
$\Theta_{\chi,q}^{\text{Eis}}(s_{1},s_{2},s_{3},s_{4},t)$ are complicated, but
completely explicit expressions defined in (4.23) and the dependence on $\psi$
and $M$ is suppressed from $\Theta$. Here the notation $L^{q}$ denotes the
usual $L$-functions with the Euler factors dividing $q$ elided. The Dirichlet
series expansion for the adjoint square $L$-function $L(s,\mathrm{Ad}^{2}f)$
in $\Re(s)>1$ (which serves as s definition) is given in (3.3). For short, we
write

$\mathcal{J}_{\pm}(s_{1},s_{2},s_{3},s_{4};g;\psi)=\mathcal{J}_{\pm}^{\text{Maa{\ss}}}(s_{1},s_{2},s_{3},s_{4};g;\psi)+\mathcal{J}_{\pm}^{\text{Eis}}(s_{1},s_{2},s_{3},s_{4};g;\psi)+\delta_{\pm=+}\mathcal{J}_{+}^{\text{hol}}(s_{1},s_{2},s_{3},s_{4};g;\psi).$

We are now ready to state our new reciprocity formula to which we have already
alluded.

######  Theorem 1.1.

Let $(s_{1},s_{2},s_{3},s_{4})\in\mathbb{C}^{4}$. Let $\psi$ be a primitive
Dirichlet character modulo a positive integer $q$. If a test function $g$
satisfies the basic assumption in Convention 1.4, then we have that

$\mathcal{M}_{2}(s_{1},s_{2},s_{3},s_{4};g;\psi)=\mathcal{N}(s_{1},s_{2},s_{3},s_{4};g;\psi)+\sum_{\pm}\left(\mathcal{J}_{\pm}(s_{1},s_{2}s_{3},s_{4};g;\psi)+\overline{\mathcal{J}_{\pm}(\overline{s_{3}},\overline{s_{4}},\overline{s_{2}},\overline{s_{1}};g;\psi)}\right),$
(1.3)

where $\mathcal{N}$ is an explicitly computable main term.

In the special case in which $q$ is an odd prime, a beautiful formula for the
second moment was envisioned by Motohashi [7]. With regard to this
restriction, he mentioned that ‘this is to avoid unnecessary complexity, and
in fact its elimination is by no means difficult’. The difficulties stem from
the formulation of the Kuznetsov formula, which is simple to elucidate when
the conductor is squarefree (see [8]). Indeed, the lack of precise description
of the Fourier expansion of Eisenstein series for arbitrary composite
conductors caused a problem at that time. The Kuznetsov formula for
nonsquarefree conductors has recently been devised in the work of Blomer–Khan
[2] with the aid of explicit formulæ for the Fourier coefficients of a
complete orthonormal basis.

######  Remark 1.2.

Let $\zeta_{K}(s)$ be the Dedekind zeta function of a quadratic number field
$K$ with discriminant $D$. We then have the factorisation

$\zeta_{K}(s)=\zeta(s)L(s,\psi_{D})$

with a real primitive Dirichlet character $\psi_{D}$ modulo $|D|$. The
reciprocity formula shown in Theorem 1.1 expresses the second moment of
Dedekind zeta functions in terms of the cubic moment of automorphic
$L$-functions on $\operatorname{GL}_{2}$.

######  Corollary 1.3.

Let $T\geqslant 1$. Let $\psi$ be a quadratic Dirichlet character modulo
$q\leqslant T^{1-\epsilon}$. Then we have that

$\int_{T}^{T+q^{1/3}T^{2/3}}|\zeta(1/2+it)L(1/2+it,\psi)|^{2}dt\ll_{\epsilon}q^{1/3+\epsilon}T^{2/3+\epsilon}.$

Corollary 1.3 is seen as a variant of Iwaniec’s [5] short interval fourth
moment bound

$\int_{T}^{T+T^{2/3}}|\zeta(1/2+it)|^{4}dt\ll_{\epsilon}T^{2/3+\epsilon}.$

###  1.2. Sketch

We here produce a heuristic overview on the genesis of the reciprocity shown
in Theorem 1.1. This is a high-level sketch geared towards experts. For the
sake of argument, we may rely on approximate tools, although our proof will be
based upon more precise inspection of $L$-functions in the region of absolute
convergence. One should also ignore all polar terms, correction factors and
the $t$-average whose purpose is to achieve convergence.

We start with (1.1). An astute reader understands our strategies from the
shape of (1.1). The overall method is inspired by Motohashi’s seminal work [9,
§4.3–4.7]. In a single phrase, we open the four zeta values as Dirichlet
series and apply Atkinson’s dissection, and hence we deal with the shifted
convolution problem with the Voronoĭ summation and the Kloosterman summation
formula (the latter being the Kuznetsov formula in the formulation that
expresses sums of Kloosterman sums in terms of Fourier coefficients of
automorphic forms). An initial form of the off-diagonal term after Atkinson’s
dissection looks like

$\sum_{n,m}\tau(n,\overline{\psi})\tau(n+m,\psi),$

where the twisted divisor function is defined as111This notation is not to be
confused with the Gauß sum.

$\tau(m,\psi)=\sum_{d\mid m}\psi(d).$

We then make use of the approximate functional equation (Lemma 2.3) for the
twisted divisor function

$\tau(m,\psi)=\frac{1+\psi(m)}{\tau(\psi)}\sum_{\ell=1}^{\infty}\frac{1}{\ell}f\left(\frac{\ell}{\sqrt{m}}\right)\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell q}}}\psi(h)e_{\ell
q}(hm),$ (1.4)

where $f(x)$ and its derivatives are certain functions of rapid decay. The
formula (1.4) is seen as a simple alternative to the $\delta$-symbol method of
Duke–Friedlander–Iwaniec [4]. One then utilises the $\operatorname{GL}_{2}$
Voronoĭ summation formula to the $n$-sum. In other words, the functional
equation for the twisted Estermann zeta function is invoked. We are eventually
left with sums of Kloosterman sums of the form

$\sum_{q\mid c}\frac{S(m,\pm n;c)}{c}h\left(\frac{4\pi\sqrt{mn}}{c}\right).$
(1.5)

This is basically the same as the formula (1.6) of Motohashi [8]. We now use
the Kloosterman summation formula (Theorem 3.1) of level $q$ and trivial
central character obtained by Blomer et al. [1]. Two classical automorphic
$L$-functions and a twisted automorphic $L$-function then emerge when we
calculate the resulting sums over $m$ and $n$, thereby deriving our
reciprocity formula (1.1). Note that Motohashi [7, 8] has not dealt with
general test functions.

###  1.3. Notation and Conventions

The Vinogradov symbol $A\ll B$ or the big $\mathrm{O}$ notation
$A=\mathrm{O}(B)$ signifies that $|A|\leqslant C|B|$ for some constant $C$. We
use the usual exponential notation $e(z)\coloneqq\exp(2\pi iz)$ and
$e_{p}(z)\coloneqq e(z/p)$ with $z\in\mathbb{C}$ and $p\in\mathbb{R}$. The
Kronecker symbol $\delta_{S}$ takes the value $1$ if the statement $S$ holds
true and $0$ otherwise. We denote the Gauß sum associated to a Dirichlet
character $\psi$ modulo $q$ by

$\tau(\psi,h)\coloneqq\sum_{a{\@displayfalse\pmod{q}}}\psi(a)e_{q}(ah),$

where we set $\tau(\psi)=\tau(\psi,1)$ as usual. Other frequently occurring
exponential sums are the Ramanujan sums and Kloosterman sums, for which we
shall use the notation

$r_{q}(h)=\sum_{d\mid(h,q)}\mu\left(\frac{q}{d}\right)d\quad\text{and}\quad
S(m,n;q)=\sideset{}{{}^{\ast}}{\sum}_{a{\@displayfalse\pmod{q}}}e_{q}(ma+n\overline{a}),$

where $\overline{a}$ indicates the multiplicative inverse modulo $q$, namely a
solution to the congruence $a\overline{a}\equiv 1{\@displayfalse\pmod{q}}$. At
last, we assume that a test function $g$ satisfies the basic assumption as
follows:

######  Convention 1.4.

The function $g$ is real valued on $\mathbb{R}$ and there exists a large
positive constant $A$ such that $g(t)$ is regular and $\ll(1+|t|)^{-A}$ on a
sufficiently wide horizontal strip $|\Im(t)|\leqslant A$. All implicit
constants in Vinogradov symbols and big $\mathrm{O}$ notation may possibly
depend on $A$ (where applicable).

##  2\. Preliminaries

We introduce the twisted Estermann zeta function and mention its properties
including the functional equation. The Ramanujan expansion and an approximate
functional equation for the twisted divisor function are also proven.

###  2.1. Estermann Zeta Function

For any positive integers $h,\ell,q$ with $(h,\ell q)=1$ and $\Re(s)>1$, we
define

$D_{\psi}(s,\xi;h/\ell q)=\sum_{n=1}^{\infty}\sigma_{\xi}(n,\psi)e(hn/\ell
q)n^{-s},$ (2.1)

where $\psi$ is an arbitrary primitive Dirichlet character modulo $q$ and

$\sigma_{\xi}(n,\psi)=\sum_{d\mid n}\psi(d)d^{\xi}.$ (2.2)

Most of the analytic properties of $D_{\psi}(s,\xi;h/\ell)$ follow from the
identity

$D_{\psi}(s,\xi;h/\ell q)=(\ell q)^{\xi-2s}\sum_{a,b{\@displayfalse\pmod{\ell
q}}}\psi(b)e(abh/\ell q)\zeta(s,a/\ell q)\zeta(s-\xi,b/\ell q),$ (2.3)

where for $\alpha\in\mathbb{R}$ and $\Re(s)>1$,

$\zeta(s,\alpha)\coloneqq\sum_{n+\alpha>0}(n+\alpha)^{-s}$

is the Hurwitz zeta function. It has a meromorphic continuation to the whole
complex plane $\mathbb{C}$ with a simple pole at $s=1$ of residue 1 and
satisfies the functional equation

$\zeta(s,\alpha)=\sum_{\pm}G^{\mp}(1-s)\zeta^{(\pm\alpha)}(1-s),$ (2.4)

where $G^{\pm}(s)=(2\pi)^{-s}\Gamma(s)\exp(\pm i\pi s/2)$ and
$\zeta^{(\alpha)}(s)$ is the meromorphic continuation of $\sum_{n\geqslant
1}e(\alpha n)n^{-s}$. For $\alpha\in\mathbb{Q}$, the formula (2.4) is seen as
a restatement of the Poisson summation in residue classes. Hence, as a
function of the single variable $s$, the twisted Estermann zeta function (2.1)
also has a meromorphic continuation to all $s\in\mathbb{C}$ with a simple pole
at $s=1+\xi$ with the residue

$\frac{\overline{\psi}(h)\tau(\psi)L(1+\xi,\overline{\psi})}{\ell^{1+\xi}q}$

provided $\xi\neq 0$. This can be deduced from our routine calculations. As an
initial manoeuvre, we find that

$\displaystyle\mathop{\mathrm{res}}_{s=1+\xi}D_{\psi}(s,\xi;h/\ell q)$
$\displaystyle=(\ell q)^{-2-\xi}\sum_{a,b{\@displayfalse\pmod{\ell
q}}}\psi(b)e(abh/\ell q)\zeta(1+\xi,a/\ell q)$ $\displaystyle=(\ell
q)^{-2-\xi}\sum_{a{\@displayfalse\pmod{\ell
q}}}\sum_{b_{1}=1}^{q}\psi(b_{1})\sum_{b_{2}=0}^{\ell-1}e(ab_{1}h/\ell
q)e(ab_{2}h/\ell)\zeta(1+\xi,a/\ell q).$

By the orthogonality relation and the definition of the Gauß sum, the right-
hand side equals

$\ell^{-1-\xi}q^{-2-\xi}\sum_{a,b{\@displayfalse\pmod{q}}}\psi(b)e(abh/q)\zeta(1+\xi,a/q)=\frac{\overline{\psi}(h)\tau(\psi)L(1+\xi,\overline{\psi})}{\ell^{1+\xi}q}.$

This finishes our computation of the residue. The functional equation for
$D_{\psi}(s,\xi;h/\ell)$ reads as follows:

######  Theorem 2.1.

The twisted Estermann zeta function satisfies the functional equation

$D_{\psi}(s,\xi;h/\ell q)=\psi(-\overline{h})(2\pi)^{2s-\xi-2}(\ell
q)^{\xi-2s+1}\Gamma(1-s)\Gamma(1+\xi-s)\\\
\times\bigg{[}\left\\{e(\xi/4)+\psi(-1)e(-\xi/4)\right\\}D_{\psi}(1+\xi-s,\xi;\overline{h}/\ell
q)\\\
-\left\\{e((2s-\xi)/4)+\psi(-1)e(-(2s-\xi)/4)\right\\}D_{\psi}(1+\xi-s;\xi;-\overline{h}/\ell
q)\bigg{]},$ (2.5)

where $\overline{h}$ is the multiplicative inverse of $h$ modulo $\ell q$,
i.e. $h\overline{h}\equiv 1{\@displayfalse\pmod{\ell q}}$.

This is essentially tantamount to the twisted $\operatorname{GL}_{2}$ Voronoĭ
summation formula.

###### Proof.

With the formula (2.3) in mind, the twisted Estermann zeta function
$D_{\psi}(s,\xi;h/q\ell)$ equals

$\displaystyle(2\pi)^{2s-\xi-2}(\ell q)^{\xi-2s}\Gamma(1-s)\Gamma(1+\xi-s)$
$\displaystyle\times\bigg{[}\left\\{\psi(-1)e(\xi/4)+e(-\xi/4)\right\\}\sum_{a,b{\@displayfalse\pmod{\ell
q}}}\psi(b)e(-abh/\ell q)\zeta^{(a/\ell q)}(1-s)\zeta^{(b/\ell q)}(1+\xi-s)$
$\displaystyle-\\{e((2s-\xi)/4)+\psi(-1)e(-(2s-\xi)/4)\\}\sum_{a,b{\@displayfalse\pmod{\ell
q}}}\psi(b)e(abh/\ell q)\zeta^{(a/\ell q)}(1-s)\zeta^{(b/\ell
q)}(1+\xi-s)\bigg{]}.$

We then utilise

$\zeta^{(a/\ell q)}(1-s)\zeta^{(b/\ell q)}(1+\xi-s)=(\ell
q)^{2s-\xi-2}\sum_{c,d{\@displayfalse\pmod{\ell q}}}e\bigg{(}\frac{ac+bd}{\ell
q}\bigg{)}\zeta(1-s,c/\ell q)\zeta(1+\xi-s,d/\ell q).$ (2.6)

This follows from decomposing $\zeta^{(a/\ell q)}(1-s)$ and $\zeta^{(b/\ell
q)}(1+\xi-s)$ into sums over residue classes modulo $\ell q$. We have four
sums over $a,b,c,d\pmod{\ell q}$ and proceed to compute the $a,b$-sums to
simplify the expression. It then holds that

$\sum_{a,b{\@displayfalse\pmod{\ell q}}}\psi(b)e\left(\frac{ac+bd-abh}{\ell
q}\right)=\sum_{b{\@displayfalse\pmod{\ell q}}}\psi(b)e\left(\frac{bd}{\ell
q}\right)\delta_{c\equiv bh{\@displayfalse\pmod{\ell
q}}}=\psi(c\overline{h})e\left(\frac{cd\overline{h}}{\ell q}\right).$

This concludes the proof of Theorem 2.1. ∎

###  2.2. Ramanujan Expansion

In this subsection, the Ramanujan expansion is shown.

######  Lemma 2.2 (Twisted Ramanujan expansion).

Let $\psi$ be a Dirichlet character modulo $q$. For any positive integer $n$
and $\Re(\xi)>0$, we have

$\frac{\sigma_{\xi}(n,\psi)}{n^{\xi}}=\frac{L(1+\xi,\overline{\psi})}{\tau(\overline{\psi})}\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1+\xi}}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell
q}}}\overline{\psi}(h)e_{\ell q}(hn).$ (2.7)

###### Proof.

One expresses the condition $(h,q\ell)=1$ in terms of the Möbius function and
verifies that the coefficients of $\ell^{-\xi}$ are equal on both sides of the
identity

$\frac{\sigma_{\xi}(n,\psi)\tau(\overline{\psi})}{L(1+\xi,\overline{\psi})n^{\xi}}=\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1+\xi}}\sum_{d\mid\ell}\mu(d)\overline{\psi}(d)\sum_{h{\@displayfalse\pmod{\ell
q/d}}}\overline{\psi}(h)e_{\ell q/d}(hn).$ (2.8)

There is a direct way to prove (2.7). Assuming that $\psi$ is primitive, we
see that the right-hand side of (2.8) equals

$\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1+\xi}}\sum_{d=1}^{\infty}\mu(d)\frac{\overline{\psi}(d)}{d^{1+\xi}}\sum_{h{\@displayfalse\pmod{\ell
q}}}\overline{\psi}(h)e_{\ell
q}(hn)=\frac{1}{L(1+\xi,\overline{\psi})}\sum_{\ell=1}^{\infty}\ell^{-1-\xi}\sum_{h{\@displayfalse\pmod{\ell
q}}}\overline{\psi}(h)e_{\ell q}(hn).$

With the formula $\tau(\psi,m)=\overline{\psi}(m)\tau(\psi)$ in mind, this
equals

$\frac{1}{L(1+\xi,\overline{\psi})}\sum_{\ell=1}^{\infty}\ell^{-1-\xi}\sum_{h_{1}=1}^{q}\overline{\psi}(h_{1})e_{\ell
q}(h_{1}n)\sum_{h_{2}=0}^{\ell-1}e_{\ell}(h_{2}n)=\frac{\psi(n)\sigma_{-\xi}(n,\overline{\psi})\tau(\overline{\psi})}{L(1+\xi,\overline{\psi})},$

Moreover, a simple consideration renders

$\sigma_{\xi}(n,\psi)n^{-\xi}=\psi(n)\sigma_{-\xi}(n,\overline{\psi}).$ (2.9)

This finishes the proof of Lemma 2.2. ∎

We now state the following approximate functional equation since it is of
independent interest:

######  Lemma 2.3.

Let $G(s)$ be an even entire holomorphic function with rapid decay in vertical
strips, satisfying $G(0)=1$ (e.g. $G(s)=\exp(s^{2})$). For $\xi\in\mathbb{C}$,
we then have

$\sigma_{\xi}(n,\psi)=\frac{\psi(n)}{\tau(\psi)}\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1-\xi}}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell q}}}\psi(h)e_{\ell
q}(hn)f_{\xi}\left(\frac{\ell}{\sqrt{n}}\right)+\frac{n^{\xi}}{\tau(\psi)}\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1+\xi}}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell q}}}\psi(h)e_{\ell
q}(hn)f_{-\xi}\left(\frac{\ell}{\sqrt{n}}\right),$

where

$f_{\xi}(x)=\int_{(a)}x^{-w}L(1-\xi+w,\psi)\frac{G(w)}{w}\frac{dw}{2\pi i}$

with $a>|\Re(\xi)|$.

###### Proof.

We start with shifting the contour, getting

$\int_{(a)}\sigma_{\xi-w}(n,\psi)n^{w/2}\frac{G(w)}{w}\frac{dw}{2\pi
i}=\sigma_{\xi}(n,\psi)+\int_{(-a)}\sigma_{\xi-w}(n,\psi)n^{w/2}\frac{G(w)}{w}\frac{dw}{2\pi
i}.$

The application of (2.9) and the change of variables $w\mapsto-w$ give the
integral representation

$\sigma_{\xi}(n,\psi)=\int_{(a)}\sigma_{\xi-w}(n,\psi)n^{w/2}\frac{G(w)}{w}\frac{dw}{2\pi
i}+\psi(n)n^{\xi}\int_{(a)}\sigma_{-\xi-w}(n,\overline{\psi})n^{w/2}\frac{G(w)}{w}\frac{dw}{2\pi
i}.$

Inserting the Ramanujan expansion

$\sigma_{\xi-w}(n,\psi)=\frac{\psi(n)}{\tau(\psi)}L(1-\xi+w,\psi)\sum_{\ell=1}^{\infty}\frac{1}{\ell^{1-\xi+w}}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell q}}}\psi(h)e_{\ell
q}(hn),$

one sees that the above integral representation matches the desired
expression. ∎

###  2.3. Multiplicativity

The aim of this subsection is to establish the Hecke relation for
$\sigma_{\xi}(n,\psi)$.

######  Lemma 2.4.

The twisted divisor function satisfies the multiplicativity relation

$\sigma_{\xi}(mn,\psi)=\sum_{c\mid(m,n)}\mu(c)\psi(c)c^{\xi}\sigma_{\xi}(m/c,\psi)\sigma_{\xi}(n/c,\psi).$
(2.10)

###### Proof.

If $m=p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{r}^{k_{r}}$,
$n=p_{1}^{\ell_{1}}p_{2}^{\ell_{2}}\cdots p_{r}^{\ell_{r}}$ and such a formula
is shown for powers of primes, then

$\displaystyle\begin{split}\sigma_{\xi}(mn,\psi)&=\sigma_{\xi}(p_{1}^{k_{1}+\ell_{1}},\psi)\sigma_{\xi}(p_{2}^{k_{2}+\ell_{2}},\psi)\cdots\sigma_{\xi}(p_{r}^{k_{r}+\ell_{r}},\psi)\\\
&=\prod_{i=1}^{r}\sum_{j=0}^{\min(k_{i},\ell_{i})}\mu(p_{i}^{j})\psi(p_{i}^{j})p_{i}^{j\xi}\sigma_{\xi}(p_{i}^{k_{i}-j})\sigma_{\xi}(p_{i}^{\ell_{i}-j})\\\
&=\sum_{c\mid(m,n)}\mu(c)\psi(c)c^{w}\sigma_{\xi}(m/c)\sigma_{\xi}(n/c).\end{split}$
(2.11)

By the multiplicativity, it suffices to prove the result for $m=p^{k}$,
$n=p^{\ell}$. The formula is obvious if $\min(k,\ell)=0$, so we assume both
the variables are at least $1$. Setting $X=\psi(p)p^{\xi}$, we find that the
right-hand side of (2.11) equals

$\displaystyle\sigma_{\xi}(p^{k},\psi)\sigma_{\xi}(p^{\ell},\psi)-\psi(p)p^{\xi}\sigma_{\xi}(p^{k-1},\psi)\sigma_{\xi}(p^{\ell-1},\psi)$
$\displaystyle\quad=(1+X+\cdots+X^{k})(1+X+\cdots+X^{\ell})-X(1+X+\cdots+X^{k-1})(1+X+\cdots+X^{\ell-1})$
$\displaystyle\quad=\frac{X^{k+\ell+1}-1}{X-1}=(1+\psi(p)p^{\xi}+\psi(p)^{2}p^{2\xi}+\cdots+\psi(p)^{k+\ell}p^{(k+\ell)\xi})=\sigma_{\xi}(p^{k+\ell},\psi).$

This concludes the proof of Lemma 2.4. ∎

##  3\. The Kuznetsov Formula

Our spectral summation formula is an asymmetric trace formula relating sums of
Kloosterman sums to Fourier coefficients of automorphic forms in the sense
that the spectral side and the geometric side have a quite different shape.
This technology can show that there is considerable cancellation in sums of
Kloosterman sums. Moreover, it provides a separation of variables of $m$ and
$n$ in the sum of the shape $\sum_{c}S(m,n;c)/c$.

Before stating the Kuznetsov formula, we formulate Fourier expansions of Maaß
and holomorphic forms and Eisenstein series in the same manner as in [2, §3]
and refer to this source for more details. Let $N\in\mathbb{N}$ and write

$\nu(N)=\prod_{p\mid N}\left(1+\frac{1}{p}\right).$

We equip $\Gamma_{0}(N)\backslash\mathbb{H}$ with the Petersson inner product

$\langle
f,g\rangle\coloneqq\int_{\Gamma_{0}(N)\backslash\mathbb{H}}f(z)\overline{g(z)}\frac{dxdy}{y^{2}}.$

###  3.1. Fourier Expansions

The cuspidal spectrum is parametrised by pairs of $\Gamma_{0}(N)$-normalised
newforms $f$ of level $N_{0}\mid N$ and integers $M\mid N/N_{0}$. This comes
from orthonormalising the set $\\{z\mapsto f(Mz):M\mid N/N_{0}\\}$ by
Gram–Schmidt. If $f$ is a Maaß form, then we write the Fourier expansion of
the pair $(f,M)$ as

$\sqrt{2\cosh(\pi t)y}\sum_{n\neq 0}\rho_{f,M,N}(n)K_{it}(2\pi|n|y)e(nx),$

where

$\rho_{f,M,N}(n)=\frac{1}{L(1,\mathrm{Ad}^{2}f)^{1/2}(MN\nu(N))^{1/2}}\prod_{p\mid
N_{0}}\left(1-\frac{1}{p^{2}}\right)^{1/2}\sum_{d\mid(M,n)}d\xi_{f}(M,d)\lambda_{f}(n/d)$
(3.1)

for $n\in\mathbb{N}$ with the convention $\lambda_{f}(n)=0$ for
$n\not\in\mathbb{Z}$. Here $\xi_{f}(M,d)$ is a certain multiplicative function
defined in the article of Blomer–Khan [2, §3.2] and all we need to know is the
bound

$\xi_{f}(M,d)\ll_{\epsilon}M^{\epsilon}(M/d)^{\theta}.$ (3.2)

Moreover, we exploit the notation

$L(s,\mathrm{Ad}^{2}f)=\zeta^{(N)}(2s)\sum_{n=1}^{\infty}\frac{\lambda_{f}(n^{2})}{n^{s}}$
(3.3)

(which might differ from the corresponding Langlands $L$-function by finitely
many Euler factors). For $-n\in\mathbb{N}$, we have
$\rho_{f,M,N}(n)=\epsilon_{f}\rho_{f,M,N}(-n)$ if $f$ is a Maaß form of parity
$\epsilon_{f}\in\\{\pm 1\\}$.

We now proceed to the description of the case of holomorphic newforms. If $f$
is a holomorphic newform of weight $k$ and level $N_{0}\mid N$, we write the
Fourier expansion of the pair $(f,M)$ as

$\left(\frac{2\pi^{2}}{\Gamma(k)}\right)^{1/2}\sum_{n\geqslant
1}\rho_{f,M,N}(n)(4\pi n)^{(k-1)/2}e(nz).$

Then (3.1) remains true for $n\in\mathbb{N}$ and we define $\rho_{f,M,N}(n)=0$
for negative $n$.

Unitary Eisenstein series for $\Gamma_{0}(N)$ are parametrised by a continuous
parameter $s=1/2+it$, $t\in\mathbb{R}$ and pairs $(\chi,M)$ where $\chi$ is a
primitive Dirichlet character of conductor $c_{\chi}$ and $M\in\mathbb{N}$
satisfies $c_{\chi}^{2}\mid M\mid N$. We define

$\tilde{\mathfrak{n}}_{N}(M)=\left(\prod_{\begin{subarray}{c}p\mid N\\\
p\nmid(M,N/M)\end{subarray}}\frac{p}{p+1}\prod_{p\mid(M,N/M)}\frac{p-1}{p+1}\right)^{1/2}$

and write

$M=c_{\chi}M_{1}M_{2},\quad\text{where}\quad(M_{2},c_{\chi})=1,\quad M_{1}\mid
c_{\chi}^{\infty}.$

The Fourier coefficients of the Eisenstein series associated to the data
$(N,M,t,\chi)$ are

$\rho_{\chi,M,N}(n,t)=\frac{C(\chi,M,t)|n|^{it}}{(N\nu(N))^{1/2}\tilde{\mathfrak{n}}_{N}(M)L^{N}(1+2it,\chi^{2})}\left(\frac{M_{1}}{M_{2}}\right)^{1/2}\sum_{\delta\mid
M_{2}}\delta\mu\left(\frac{M_{2}}{\delta}\right)\overline{\chi}(\delta)\sum_{\begin{subarray}{c}cM_{1}\delta
f=n\\\ (c,N/M)=1\end{subarray}}\frac{\chi(c)}{c^{2it}}\overline{\chi}(f)$

for a constant $C(\chi,M,t)$ with $|C(\chi,M,t)|=1$.

###  3.2. Kloosterman Summation Formula

We devote this subsection to expounding on a variant of the Kuznetsov formula
due to Blomer et al. [1] (that only involves Fourier expansions at infinity).
This is seen as a Kloosterman summation formula since it expresses sums of
Kloosterman sums weighted by a certain test function in terms of sums of
automorphic forms weighted by transformed functions. For $x>0$, we introduce
the integral kernels

$\displaystyle\mathcal{J}^{+}(x,t)$ $\displaystyle\coloneqq\frac{\pi
i}{\sinh(\pi t)}(J_{2it}(x)-J_{-2it}(x)),$ $\displaystyle\mathcal{J}^{-}(x,t)$
$\displaystyle\coloneqq\frac{\pi i}{\sinh(\pi
t)}(I_{2it}(x)-I_{-2it}(x))=4\cosh(\pi t)K_{2it}(x),$
$\displaystyle\mathcal{J}^{\text{hol}}(x,k)$ $\displaystyle\coloneqq 2\pi
i^{k}J_{k-1}(x)=\mathcal{J}^{+}(x,(k-1)/2i),\qquad k\in 2\mathbb{N}.$

If $H\in C^{3}((0,\infty))$ satisfies $x^{j}H^{(j)}(x)\ll\min(x,x^{-3/2})$ for
$0\leqslant j\leqslant 3$, we define

$\mathscr{L}^{\Diamond}H=\int_{0}^{\infty}\mathcal{J}^{\Diamond}(x,\cdot)H(x)\frac{dx}{x}$

for $\Diamond\in\\{+,-,\text{hol}\\}$. For notational convenience, we define

$\mathcal{O}_{q}(m,\pm n;h)=\sum_{q\mid c}\frac{S(m,\pm
n;c)}{c}h\left(\frac{4\pi\sqrt{mn}}{c}\right).$ (3.4)

The Kuznetsov formula reads in our case as follows:

######  Theorem 3.1 (Blomer et al. [1]).

Let $x^{j}H^{(j)}(x)\ll_{s}\min(x,x^{-3/2})$ for $0\leqslant j\leqslant 3$.
For $m,n,q\in\mathbb{N}$, we then have

$\mathcal{O}_{q}(m,\pm n;H)=\mathcal{A}_{q}^{\text{Maa{\ss}}}(m,\pm
n;\mathscr{L}^{\pm}H)+\mathcal{A}_{q}^{\text{Eis}}(m,\pm
n;\mathscr{L}^{\pm}H)+\delta_{\pm=+}\mathcal{A}_{q}^{\text{hol}}(m,n;\mathscr{L}^{\text{hol}}H),$
(3.5)

where

$\displaystyle\mathcal{A}_{q}^{\text{Maa{\ss}}}(m,n;h)$
$\displaystyle\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}^{\ast}(q_{0})}\rho_{f,M,q}(m)\overline{\rho_{f,M,q}(n)}h(t_{f}),$
$\displaystyle\mathcal{A}_{q}^{\text{Eis}}(m,n;h)$
$\displaystyle\coloneqq\sum_{c_{\chi}^{2}\mid M\mid
q}\int_{\mathbb{R}}\rho_{\chi,M,q}(m,t)\overline{\rho_{\chi,M,q}(n,t)}h(t)\frac{dt}{2\pi},$
$\displaystyle\mathcal{A}_{q}^{\text{hol}}(m,n;h)$
$\displaystyle\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}_{\text{hol}}^{\ast}(q_{0})}\rho_{f,M,q}(m)\overline{\rho_{f,M,q}(n)}h(k_{f}).$

The above formulation mimics the Kuznetsov formula that Motohashi [9] used to
prove a spectral reciprocity for the smoothed fourth moment of the Riemann
zeta function. One must stress that there is no delta term in the Kloosterman
summation formula.

##  4\. Proof of Theorem 1.1

In this section, we establish Theorem 1.1 via the shifted convolution problem.
To achieve this aim, we invoke Convention 1.4 and the definition (1.2) of the
second moment.

###  4.1. Initial Manipulations

Let $\mathcal{R}_{4}^{+}$ (resp. $\mathcal{R}_{4}^{-}$) be the subdomain of
$\mathbb{C}^{4}$ where all four parameters have real parts greater than (resp.
less than) one, namely

$\displaystyle\mathcal{R}_{4}^{+}$
$\displaystyle\coloneqq\\{(s_{1},s_{2},s_{3},s_{4})\in\mathbb{C}^{4}:\Re(s_{i})>1,\,1\leqslant
i\leqslant 4\\},$ $\displaystyle\mathcal{R}_{4}^{-}$
$\displaystyle\coloneqq\\{(s_{1},s_{2},s_{3},s_{4})\in\mathbb{C}^{4}:\Re(s_{i})<1,\,1\leqslant
i\leqslant 4\\}.$

For notational convenience, we write
$\mathcal{M}_{2}=\mathcal{M}_{2}(\texttt{s};g;\psi)=\mathcal{M}_{2}(s_{1},s_{2},s_{3},s_{4};g;\psi)$
for a vector $\texttt{s}=(s_{1},s_{2},s_{3},s_{4})$. Moving the contour
appropriately, one sees that the second moment $\mathcal{M}_{2}$ is
meromorphic over the whole complex plane $\mathbb{C}$. Taking
$(s_{1},s_{2},s_{3},s_{4})\in\mathcal{R}_{4}^{-}\cap\mathcal{B}_{4}$ and
shifting the contour back to the original one, we deduce the following
meromorphic continuation to the domain
$\mathcal{R}_{4}^{-}\cap\mathcal{B}_{4}$ as follows:

$\displaystyle\begin{split}\mathcal{M}_{2}(s_{1},s_{2},s_{3},s_{4};g;\psi)&=\int_{\mathbb{R}}\zeta(s_{1}+it)L(s_{2}+it,\psi)\zeta(s_{3}-it)L(s_{4}-it,\overline{\psi})g(t)dt\\\
&\quad+2\pi\zeta(s_{1}+s_{3}-1)L(1-s_{1}+s_{2},\psi)L(s_{1}+s_{4}-1,\overline{\psi})g((s_{1}-1)i)\\\
&\quad+2\pi\zeta(s_{1}+s_{3}-1)L(s_{2}+s_{3}-1,\psi)L(1-s_{3}+s_{4},\overline{\psi})g((1-s_{3})i).\end{split}$
(4.1)

To continue $\mathcal{M}_{2}$ from $\mathcal{R}_{4}^{+}$ to the vicinity of
the central point $\texttt{s}=(1/2,1/2,1/2,1/2)$, we open up the Dirichlet
series in the region of absolute convergence to recast

$\mathcal{M}_{2}=\sum_{n_{1},n_{2},n_{3},n_{4}=1}^{\infty}\frac{\psi(n_{2})\overline{\psi(n_{4})}}{n_{1}^{s_{1}}n_{2}^{s_{2}}n_{3}^{s_{3}}n_{4}^{s_{4}}}\int_{\mathbb{R}}g(t)\left(\frac{n_{1}n_{2}}{n_{3}n_{4}}\right)^{-it}dt=\sum_{m,n=1}^{\infty}\frac{\sigma_{s_{1}-s_{2}}(m,\psi)\sigma_{s_{3}-s_{4}}(n,\overline{\psi})}{m^{s_{1}}n^{s_{3}}}\hat{g}\bigg{(}\frac{1}{2\pi}\log\frac{m}{n}\bigg{)},$
(4.2)

where $\hat{g}$ is the Fourier transform

$\hat{g}(\xi)=\int_{\mathbb{R}}g(t)e(-\xi t)dt.$

This is where Atkinson’s dissection comes into play and we decompose (4.2)
into three parts. We write

$\mathcal{M}_{2}=\mathcal{D}_{2}+\mathcal{OD}_{2}^{\dagger}+\mathcal{OD}_{2}^{\ddagger}$
(4.3)

with $\mathcal{D}_{2}$ being the diagonal contribution associated to $m=n$,
whereas $\mathcal{OD}_{2}^{\dagger}$ (resp. $\mathcal{OD}_{2}^{\ddagger}$)
being the off-diagonal contribution which involves terms with $m>n$ (resp.
$m<n$). Note that the case where $m<n$ is symmetrical to the case where $m>n$
and it follows that

$\mathcal{OD}_{2}^{\ddagger}(s_{1},s_{2},s_{3},s_{4};g;\psi)=\overline{\mathcal{OD}_{2}^{\dagger}(\overline{s_{3}},\overline{s_{4}},\overline{s_{1}},\overline{s_{2}};g;\psi)}.$
(4.4)

We initially calculate the diagonal term, mimicking the method of Motohashi
[10].

######  Lemma 4.1.

With the notation as above, we have that

$\mathcal{D}_{2}=\hat{g}(0)\frac{\zeta(s_{1}+s_{3})\zeta_{q}(s_{2}+s_{4})L(s_{2}+s_{3},\psi)L(s_{1}+s_{4},\overline{\psi})}{\zeta_{q}(s_{1}+s_{2}+s_{3}+s_{4})},$
(4.5)

where, for any $L$-function, the notation $L^{q}$ signifies the removal of the
Euler factor at primes dividing $q$.

###### Proof.

We need to calculate

$\mathcal{D}_{2}=\hat{g}(0)\sum_{m=1}^{\infty}\frac{\sigma_{s_{1}-s_{2}}(m,\psi)\sigma_{s_{3}-s_{4}}(m,\overline{\psi})}{m^{s_{1}+s_{3}}},$

where the Dirichlet coefficient is a multiplicative function of $m$. We then
write

$\sigma_{s_{1}-s_{2}}(m,\psi)\sigma_{s_{3}-s_{4}}(m,\overline{\psi})=\sum_{[u,v]\mid
m}\psi(u)\overline{\psi}(v)u^{s_{1}-s_{2}}v^{s_{3}-s_{4}},$

where $[u,v]$ stands for the least common multiple of $u$ and $v$. Hence the
Dirichlet series equals

$\zeta(s_{1}+s_{3})\sum_{u,v=1}^{\infty}\psi(u)\overline{\psi}(v)\frac{u^{s_{1}-s_{2}}v^{s_{3}-s_{4}}}{[u,v]^{s_{1}+s_{3}}}=\zeta(s_{1}+s_{3})\sum_{u,v=1}^{\infty}\psi(u)\overline{\psi}(v)\frac{(u,v)^{s_{1}+s_{3}}}{u^{s_{2}+s_{3}}v^{s_{1}+s_{4}}},$
(4.6)

where $(u,v)$ stands for the greatest common divisor of $u$ and $v$. On the
other hand, the Möbius inversion gives

$\sum_{f\mid g}\eta_{s}(f)=g^{s}\quad\text{with}\quad\eta_{s}(m)=\sum_{d\mid
m}\mu(d)(m/d)^{s}$ (4.7)

and the right-hand side of (4.6) equals

$\zeta(s_{1}+s_{3})\sum_{u,v=1}^{\infty}\frac{\psi(u)\overline{\psi}(v)}{u^{s_{2}+s_{3}}v^{s_{1}+s_{4}}}\sum_{d\mid(u,v)}\eta_{s_{1}+s_{3}}(d)=\zeta(s_{1}+s_{3})L(s_{2}+s_{3},\psi)L(s_{1}+s_{4},\overline{\psi})\sum_{(d,q)=1}\frac{\eta_{s_{1}+s_{3}}(d)}{d^{s_{1}+s_{2}+s_{3}+s_{4}}}.$

The last sum is obviously equal to
$\zeta_{q}(s_{2}+s_{4})/\zeta_{q}(s_{1}+s_{2}+s_{3}+s_{4})$. This concludes
the proof. ∎

Treatments of (4.5) published thus far in articles and books, except for
Motohashi [10], are essentially the same as Ramanujan’s original proof
depending on Euler products. In contrast, the above proof is independent of
Euler products and the application of the Möbius inversion after (4.6) plays
an alternative rôle. The extension of our method to general situations would
be an interesting arithmetical issue, in particular in the automorphic
context.

###  4.2. Off-Diagonal Terms

The continuation of the off-diagonal terms requires the full machinery of the
spectral theory of automorphic forms associated to congruence subgroups. We
compute the second term in (4.3) and the contribution of the third term will
be combined at the end. For notational convenience, we call

$G(y,s)=(1+y)^{-s}\hat{g}\bigg{(}\frac{1}{2\pi}\log(1+y)\bigg{)}=\int_{(\alpha)}\mathring{g}(\tau,s)y^{-\tau}\frac{d\tau}{2\pi
i}$ (4.8)

for suitable $\alpha>1$, where $\mathring{g}$ is the Mellin transform

$\mathring{g}(\tau,s)=\int_{0}^{\infty}y^{\tau-1}(1+y)^{-s}\hat{g}\left(\frac{1}{2\pi}\log(1+y)\right)dy=\Gamma(\tau)\int_{\mathbb{R}}\frac{\Gamma(s-\tau+it)}{\Gamma(s+it)}g(t)dt,$
(4.9)

provided $\Re(s)>\Re(\tau)>0$. We have the following lemma:

######  Lemma 4.2.

As a function of two complex variables, $\mathring{g}(\tau,s)/\Gamma(\tau)$ is
entire in $\tau$ and $s$. Besides, $\mathring{g}(\tau,s)$ is of rapid decay in
$\tau$ as far as $s$ and $\Re(\tau)$ are bounded, say

$\mathring{g}(\tau,s)\ll(1+|\tau|)^{-A}.$

###### Proof.

Shifting the contour $\Im(t)=0$ in (4.9) downwards appropriately, we prove the
first assertion. The second claim is a consequence of an upward shift. See [9,
Lemma 4.1] for an analogous statement. ∎

We embark on the computation of $\mathcal{OD}_{2}^{\dagger}$ to resolve the
shifted convolution problem. One initially observes that

$\mathcal{OD}_{2}^{\dagger}=\sum_{n,m=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\sigma_{s_{1}-s_{2}}(n+m,\psi)}{n^{s_{2}+s_{3}}(n+m)^{s_{1}-s_{2}}}G(m/n,s_{2}).$
(4.10)

The substitution of (2.7) to the right-hand side of (4.10) yields

$\mathcal{OD}_{2}^{\dagger}=\frac{L(1+s_{1}-s_{2},\overline{\psi})}{\tau(\overline{\psi})}\sum_{\ell=1}^{\infty}\ell^{s_{2}-s_{1}-1}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell
q}}}\overline{\psi}(h)\sum_{n,m=1}^{\infty}e_{\ell
q}(h(n+m))\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})}{n^{s_{2}+s_{3}}}G(m/n,s_{2}).$

Moreover, we introduce

$\mathcal{R}_{4,\alpha}=\\{(s_{1},s_{2},s_{3},s_{4})\in\mathcal{R}_{4}^{+}:\Re(s_{1})>\Re(s_{2})+1,\,\Re(s_{2})>\alpha,\,\Re(s_{3})>1,\,\Re(s_{4})>1\\},$

where $\alpha>1$. We deduce in this domain that

$\mathcal{OD}_{2}^{\dagger}=\frac{L(1+s_{1}-s_{2},\overline{\psi})}{\tau(\overline{\psi})}\sum_{\ell=1}^{\infty}\ell^{s_{2}-s_{1}-1}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell
q}}}\overline{\psi}(h)\int_{(\alpha)}D_{\overline{\psi}}(s_{2}+s_{3}-\tau,s_{3}-s_{4};h/\ell
q)\zeta^{(h/\ell q)}(\tau)\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.$
(4.11)

This is now in shape to make use of the Poisson summation formula twice, that
is the Voronoĭ summation formula once. In anticipation of shifting the contour
in (4.11) to the right, we define the new domain

$\mathcal{E}_{4,\beta}=\\{(s_{1},s_{2},s_{3},s_{4}):\Re(s_{1}+s_{3})<\beta,\
\Re(s_{1}+s_{4})<\beta,\ \Re(s_{1}+s_{2}+s_{3}+s_{4})>3\beta\\},$ (4.12)

where $\beta$ is assumed to be sufficiently large. For the moment, we restrict
ourselves to
$(s_{1},s_{2},s_{3},s_{4})\in\mathcal{R}_{4,\alpha}\cap\mathcal{E}_{4,\beta}$,
which is not empty if $\beta>1+\alpha$. Then the contour $(\alpha)$ in (4.11)
is shifted to $(\beta)$ and the residue is calculated as

$\frac{\mathring{g}(s_{2}+s_{4}-1,s_{2})}{q}L(1+s_{1}-s_{2},\overline{\psi})L(1+s_{3}-s_{4},\psi)\sum_{(\ell,q)=1}\ell^{-2-s_{1}+s_{2}-s_{3}+s_{4}}\sum_{n=1}^{\infty}r_{\ell
q}(n)n^{1-s_{2}-s_{4}}\\\
=\frac{\mathring{g}(s_{2}+s_{4}-1,s_{2})}{q}\frac{L(1+s_{1}-s_{2},\overline{\psi})L(1+s_{3}-s_{4},\psi)}{\zeta_{q}(2+s_{1}-s_{2}+s_{3}-s_{4})}\sum_{n=1}^{\infty}\frac{\sigma_{1+s_{1}-s_{2}+s_{3}-s_{4}}(n)r_{q}(n)}{n^{s_{1}+s_{3}}},$

where we exploited the coprimality condition between $\ell$ and $q$ (with the
aid of the version (2.8) of the twisted Ramanujan expansion) and the classical
Ramanujan expansion. One sees that the $n$-sum becomes

$\sum_{n=1}^{\infty}\frac{\sigma_{1+s_{1}-s_{2}+s_{3}-s_{4}}(n)r_{q}(n)}{n^{s_{1}+s_{3}}}=\zeta(s_{1}+s_{3})\zeta(s_{2}+s_{4}-1)\sum_{c\mid
q}\mu(c)c^{2-s_{1}-s_{2}-s_{3}-s_{4}}\sum_{d\mid
q/c}\mu\left(\frac{q}{cd}\right)d^{1-s_{1}-s_{3}}\sigma_{1+s_{1}-s_{2}+s_{3}-s_{4}}(d).$

This procedure renders the polar term222It seems that Motohashi [8] made a
mistake in his formula (1.5).

$\mathcal{P}\coloneqq
A_{q}(s_{1},s_{2},s_{3},s_{4})\mathring{g}(s_{2}+s_{4}-1,s_{2})\frac{\zeta(s_{1}+s_{3})\zeta(s_{2}+s_{4}-1)L(1+s_{1}-s_{2},\overline{\psi})L(1+s_{3}-s_{4},\psi)}{\zeta_{q}(2+s_{1}-s_{2}+s_{3}-s_{4})},$
(4.13)

where $A_{q}$ is the multiplicative function defined by

$A_{q}(s_{1},s_{2},s_{3},s_{4})=\frac{1}{q}\sum_{c\mid
q}\mu(c)c^{2-s_{1}-s_{2}-s_{3}-s_{4}}\sum_{d\mid
q/c}\mu\left(\frac{q}{cd}\right)d^{1-s_{1}-s_{3}}\sigma_{1+s_{1}-s_{2}+s_{3}-s_{4}}(d).$

######  Remark 4.3.

Using the multiplicativity, we derive that

$A_{q}(1/2,1/2,1/2,1/2)=\prod_{p\mid q}\left(1-\frac{1}{p}\right).$

###  4.3. Application of the Voronoĭ Summation Formula

This subsection is devoted to applying the functional equation of the twisted
Estermann zeta function and simplifying our expression to a certain sum of
Kloosterman sums of the form $S(m,\pm n;cq)$. By Theorem 2.5, the off-diagonal
contribution $\mathcal{OD}_{2}^{\dagger}$ equals the polar term $\mathcal{P}$
plus

$\displaystyle\frac{\psi(-1)}{\tau(\overline{\psi})q}L(1+s_{1}-s_{2},\overline{\psi})\sum_{\ell=1}^{\infty}\ell^{s_{2}-s_{1}-2}\
\sideset{}{{}^{\ast}}{\sum}_{h{\@displayfalse\pmod{\ell
q}}}\int_{(\beta)}\mathring{g}(\tau,s_{2})\zeta^{(h/\ell
q)}(\tau)\left(\frac{2\pi}{\ell
q}\right)^{2s_{2}+s_{3}+s_{4}-2\tau-2}\Gamma(1+\tau-s_{2}-s_{3})$
$\displaystyle\quad\times\Gamma(1+\tau-
s_{2}-s_{4})\bigg{[}\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}D_{\overline{\psi}}(1+\tau-
s_{2}-s_{4},s_{3}-s_{4};\overline{h}/\ell q)$
$\displaystyle\quad-\left\\{e\left(\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)+\psi(-1)e\left(-\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)\right\\}D_{\overline{\psi}}(1+\tau-
s_{2}-s_{4},s_{3}-s_{4};-\overline{h}/\ell q)\bigg{]}\frac{d\tau}{2\pi i}.$

Performing the sum over $h$, we eventually arrive at the following
proposition:

######  Proposition 4.4.

For a primitive character $\psi$ modulo $q$, the off-diagonal contribution
$\mathcal{OD}_{2}^{\dagger}$ is meromorphically continued to the domain
$\mathcal{E}_{4,\beta}$ and there we have

$\mathcal{OD}_{2}^{\dagger}=\mathcal{P}+\sum_{\pm}\mathcal{J}_{\pm},$

where the polar term $\mathcal{P}$ is defined in (4.13) and

$\mathcal{J}_{\pm}\coloneqq\frac{\tau(\psi)}{q}\left(\frac{q}{2\pi}\right)^{1+s_{1}-s_{2}}L(1+s_{1}-s_{2},\overline{\psi})\sum_{m,n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\mathcal{O}_{q}(m,\pm
n;\Psi_{\texttt{s}}^{\pm})}{m^{(s_{1}+s_{2}+s_{3}+s_{4}-1)/2}n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2}}.$

Here $\mathcal{O}_{q}(m,n;h)$ is introduced in (3.4) and we set

$\displaystyle\begin{split}\Psi_{\texttt{s}}^{+}(x)&\coloneqq\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\\\
&\quad\times\int_{(\beta)}\left(\frac{x}{2}\right)^{s_{1}+s_{2}+s_{3}+s_{4}-1-2\tau}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i},\end{split}$ (4.14)
$\displaystyle\begin{split}\Psi_{\texttt{s}}^{-}(x)&\coloneqq-\int_{(\beta)}\left\\{e\left(\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)+\psi(-1)e\left(-\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)\right\\}\\\
&\quad\times\left(\frac{x}{2}\right)^{s_{1}+s_{2}+s_{3}+s_{4}-1-2\tau}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.\end{split}$ (4.15)

Our problem has now been reduced to a sum of Kloosterman sums whose moduli are
divisible by $q$. Hence one can appeal to the spectral theory of automorphic
forms over the Hecke congruence subgroup $\Gamma_{0}(q)$.

###  4.4. Application of the Kuznetsov Formula

We are in a position to apply the Kloosterman summation formula (Theorem 3.1)
to our sums of Kloosterman sums with the careful analysis of integrals
involving Bessel functions. Following Motohashi’s book [9], we first observe
how the case of the plus sign contributes. In order to reduce the integral
transform $\mathscr{L}^{+}$ in the light of the definition (4.14), we consider

$\int_{0}^{\infty}J_{2it}(x)\int_{(\beta)}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\Big{(}\frac{x}{2}\Big{)}^{s_{1}+s_{2}+s_{3}+s_{4}-2-2\tau}\mathring{g}(\tau,s_{2})d\tau
dx.$ (4.16)

In a suitable domain such as [9, (4.4.8)], this double integral is equal to

$\int_{(\beta)}\frac{\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau+it\right)}{\Gamma\left(\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau+it\right)}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-s_{2}-s_{4})\mathring{g}(\tau,s_{2})d\tau,$ (4.17)

where we have utilised the formula

$\int_{0}^{\infty}J_{\nu}(x)\left(\frac{x}{2}\right)^{-\mu}dx=\Gamma\left(\frac{1+\nu-\mu}{2}\right)\Gamma\left(\frac{1+\nu+\mu}{2}\right)^{-1}$

available for $1/2<\Re(\mu)<1+\Re(\nu)$. Since the integral (4.17) is regular
in that domain, the analytic continuation implies that the double integral
(4.16) equals (4.17) throughout $\mathcal{E}_{4,\beta}$. Then we use the
formula

$\frac{\Gamma(a-\tau+it)}{\Gamma(1-a+\tau+it)}-\frac{\Gamma(a-\tau-
it)}{\Gamma(1-a+\tau-it)}=\frac{2}{\pi i}\sinh(\pi
t)\cos(\pi(a-\tau))\Gamma(a-\tau+it)\Gamma(a-\tau-it),$ (4.18)

which follows from the functional equation $\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi
s)$. Substituting $a=(s_{1}+s_{2}+s_{3}+s_{4}-1)/2$, we have after some
rearrangements that

$\displaystyle\mathscr{L}^{+}\Psi_{\texttt{s}}^{+}(t)$
$\displaystyle=\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\int_{(\beta)}\sin\left(\frac{\pi(s_{1}+s_{2}+s_{3}+s_{4}-2\tau)}{2}\right)$
$\displaystyle\qquad\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+it-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-it-\tau\right)$
$\displaystyle\qquad\times\Gamma(1+\tau-s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.$

In a similar fashion, the holomorphic contribution turns into

$\mathscr{L}^{\text{hol}}\Psi_{\texttt{s}}^{+}(k)=i^{k}\pi\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\\\
\times\int_{(\beta)}\frac{\Gamma\left(\frac{k-2+s_{1}+s_{2}+s_{3}+s_{4}}{2}-\tau\right)}{\Gamma\left(\frac{k+2-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.$

To proceed, we concentrate on the context of the minus sign. In this case, the
bound shown in Lemma 4.2 plays an important rôle. By definition, the integral
transform is written as

$\mathscr{L}^{-}\Psi_{\texttt{s}}^{-}(t)=4\cosh(\pi
t)\int_{0}^{\infty}\Psi_{\texttt{s}}^{-}(x)K_{2it}(x)\frac{dx}{x}.$

Inserting the definition (4.15), we obtain an absolutely convergent integral,
since we have

$K_{2it}(x)\ll\begin{cases}|\log x|&\text{as $x\to 0$},\\\ \exp(-x)&\text{as
$x\to\infty$}.\end{cases}$

Substituting Heaviside’s integral formula

$\int_{0}^{\infty}K_{2v}(x)\Big{(}\frac{x}{2}\Big{)}^{2s-1}dx=\frac{\Gamma(s+v)\Gamma(s-v)}{2}\quad\text{for}\quad\Re(s)>|\Re(v)|,$
(4.19)

we are left with the expression

$\displaystyle\mathscr{L}^{-}\Psi_{\texttt{s}}^{-}(t)$
$\displaystyle=-\cosh(\pi
t)\int_{(\beta)}\left\\{e\left(\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)+\psi(-1)e\left(-\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)\right\\}$
$\displaystyle\quad\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+it-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-it-\tau\right)$
$\displaystyle\quad\times\Gamma(1+\tau-s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.$

In order to give precise expressions of
$\mathscr{L}^{\pm}\Psi_{\texttt{s}}^{\pm}(t)$, we define the following three
functions:

$\displaystyle\Phi_{\texttt{s}}^{+}(\xi)$
$\displaystyle=-i(2\pi)^{s_{1}-s_{2}-2}\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}$
$\displaystyle\quad\times\int_{-i\infty}^{i\infty}\sin\left(\frac{\pi(s_{1}+s_{2}+s_{3}+s_{4}-2\tau)}{2}\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)$
$\displaystyle\quad\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right)\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-s_{2}-s_{4})\mathring{g}(\tau,s_{2})d\tau,$
$\displaystyle\Phi_{\texttt{s}}^{-}(\xi)$
$\displaystyle=i(2\pi)^{s_{1}-s_{2}-2}\cos(\pi\xi)\int_{-i\infty}^{i\infty}\left\\{e\left(\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)+\psi(-1)e\left(-\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)\right\\}$
$\displaystyle\quad\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right)$
$\displaystyle\quad\times\Gamma(1+\tau-s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})d\tau,$
$\displaystyle\Xi_{\texttt{s}}(\xi)$
$\displaystyle=\int_{-i\infty}^{i\infty}\frac{\Gamma\left(\xi+\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau\right)}{\Gamma\left(\xi+\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-
s_{2}-s_{4})\mathring{g}(\tau,s_{2})\frac{d\tau}{2\pi i}.$

Here the contour in the formula for $\Phi_{\texttt{s}}^{+}(\xi)$ is curved to
ensure that the poles of the first two gamma factors in the integrand lie to
the right of the path and those of other factors are on the left of the path;
moreover the variables $\xi,s_{1},s_{2},s_{3},s_{4}$ are assumed to be such
that the path can be drawn. The path in the definition of
$\Phi_{\texttt{s}}^{-}(\xi)$ is chosen in just the same way. On the other
hand, the contour in $\Xi$ separates the poles of
$\Gamma(\xi+(s_{1}+s_{2}+s_{3}+s_{4}-1)/2-\tau)$ and those of $\Gamma(1+\tau-
s_{2}-s_{3})\Gamma(1+\tau-s_{2}-s_{4})\mathring{g}(\tau,s_{2})$ to the left
and the right of the path, respectively.

######  Proposition 4.5.

With the notation above, we have that

$\Phi_{\texttt{s}}^{+}(\xi)=-\frac{(2\pi)^{s_{1}-s_{2}}}{4\sin(\pi\xi)}\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\Big{[}\Xi_{\texttt{s}}(\xi)-\Xi_{\texttt{s}}(-\xi)\Big{]},$
(4.20)

and

$\Phi_{\texttt{s}}^{-}(\xi)=\frac{(2\pi)^{s_{1}-s_{2}}}{4i\sin(\pi\xi)}\bigg{[}\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}+2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}+2\xi}{4}\right)\right\\}\Xi_{\texttt{s}}(\xi)\\\
-\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}-2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}-2\xi}{4}\right)\right\\}\Xi_{\texttt{s}}(-\xi)\bigg{]},$
(4.21)

provided the left-hand sides are well-defined. We also have for real $t$ and
$\texttt{s}=(s_{1},s_{2},s_{3},s_{4})\in\mathcal{E}_{4,\beta}$ that

$\displaystyle\mathscr{L}^{+}\Psi_{\texttt{s}}^{+}(t)=(2\pi)^{1-s_{1}+s_{2}}\Phi_{\texttt{s}}^{+}(it),$
$\displaystyle\mathscr{L}^{-}\Psi_{\texttt{s}}^{-}(t)=(2\pi)^{1-s_{1}+s_{2}}\Phi_{\texttt{s}}^{-}(it).$

For $k\equiv 0{\@displayfalse\pmod{2}}$ and
$(s_{1},s_{2},s_{3},s_{4})\in\mathcal{E}_{4,\beta}$, we have that

$\mathscr{L}^{\mathrm{hol}}\Psi_{\texttt{s}}^{+}(k)=i^{k}\pi\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\Xi_{\texttt{s}}\left(\frac{k-1}{2}\right).$

###### Proof.

First of all, we aim to establish (4.20). One derives from (4.18) that

$\frac{\Gamma\left(\xi+\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau\right)}{\Gamma\left(\xi+\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}-\frac{\Gamma\left(-\xi+\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau\right)}{\Gamma\left(-\xi+\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}=-\frac{2}{\pi}\sin(\pi\xi)\sin\left(\frac{\pi(s_{1}+s_{2}+s_{3}+s_{4}-2\tau)}{2}\right)\\\
\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right).$

In order to verify (4.21), the use of the relation
$\Gamma(s)\Gamma(1-s)=\pi/\sin(\pi s)$ transforms the expression as follows:

$\displaystyle\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}+2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}+2\xi}{4}\right)\right\\}\frac{\Gamma\left(\xi+\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau\right)}{\Gamma\left(\xi+\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}$
$\displaystyle\quad-\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}-2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}-2\xi}{4}\right)\right\\}\frac{\Gamma\left(-\xi+\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\tau\right)}{\Gamma\left(-\xi+\frac{3-s_{1}-s_{2}-s_{3}-s_{4}}{2}+\tau\right)}$
$\displaystyle=\frac{1}{\pi}\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right)$
$\displaystyle\quad\times\bigg{[}\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}+2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}+2\xi}{4}\right)\right\\}\sin\left(\pi\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right)\right)$
$\displaystyle\quad-\left\\{\psi(-1)e\left(\frac{s_{1}-s_{2}-2\xi}{4}\right)-e\left(-\frac{s_{1}-s_{2}-2\xi}{4}\right)\right\\}\sin\left(\pi\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)\right)\bigg{]}.$

Via a simple manipulation of trigonometrics, the right-hand side equals

$\frac{\sin(2\pi\xi)}{\pi
i}\left\\{e\left(\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)+\psi(-1)e\left(-\frac{2s_{2}+s_{3}+s_{4}-2\tau}{4}\right)\right\\}\\\
\times\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+\xi-\tau\right)\Gamma\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-\xi-\tau\right)$

Since $\sin(2\pi\xi)/\sin(\pi\xi)=2\cos(\pi\xi)$, this concludes the proof of
Lemma 4.5. ∎

###  4.5. Cubic Moment of Central $L$-Values

This subsection is devoted to the derivation of the cubic moment side. The
absolute convergence we would like to check is apparent as long as the double
summation over the variables $m,n$ is concerned, since we have a trivial bound
on the Hecke eigenvalues and
$\texttt{s}=(s_{1},s_{2},s_{3},s_{4})\in\mathcal{E}_{4,\beta}$. Hence, the
main problem is reduced to bounding $\mathscr{L}^{\pm}\Psi_{\texttt{s}}^{\pm}$
and then Proposition 4.5 reduces our task to the analysis of the function
$\Xi$. If real $t$ and $k\in 2\mathbb{N}$ tend to infinity, we have

$\Xi_{\texttt{s}}(it)\ll|t|^{-A},\qquad\Xi_{\texttt{s}}\left(\frac{k-1}{2}\right)\ll
k^{-A}$

uniformly for bounded $(s_{1},s_{2},s_{3},s_{4})$. Hence we deduce the
meromorphic continuation of $\mathcal{J}_{\pm}$ to the whole of
$\mathbb{C}^{4}$. This proves that the decomposition (4.3) holds throughout
$\mathbb{C}^{4}$.

Finally, we are able to execute the resulting summations over $m$ and $n$. In
order to calculate the contribution from Maaß newforms, one initially derives

$\sum_{n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\overline{\rho_{f,M,q}(\pm
n)}}{n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2}}=\frac{\epsilon_{f}^{(1\mp
1)/2}}{L(1,\mathrm{Ad}^{2}f)^{1/2}(Mq\nu(q))^{1/2}}\prod_{p\mid
q_{0}}\left(1-\frac{1}{p^{2}}\right)^{1/2}\\\ \times\sum_{d\mid
M}\xi_{f}(M,d)d^{(1-s_{1}+s_{2}-s_{3}+s_{4})/2}\sum_{n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\lambda_{f}(n)}{n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2}},$

where $f\in\mathcal{B}^{\ast}(q_{0})$ with $q_{0}M\mid q$. We here used the
property that
$\sigma_{s_{3}-s_{4}}(nd,\overline{\psi})=\sigma_{s_{3}-s_{4}}(n,\overline{\psi})$
as $d$ is not relatively prime to $q$. It then follows from the
multiplicativity of Hecke eigenvalues that

$\sum_{n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\lambda_{f}(n)}{n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2}}=\frac{L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2},f\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2},f\otimes\overline{\psi}\right)}{L(1+s_{1}-s_{2},\overline{\psi})}.$

There is structural beauty in the above quotient, because we have
$L(1+s_{1}-s_{2},\overline{\psi})^{-1}$, which completely cancels out with the
term $L(1+s_{1}-s_{2},\overline{\psi})$ appearing in the expression for
$\mathcal{J}_{\pm}$ in Proposition 4.4. One can calculate the Dirichlet series
for the $m$-sum in a similar manner, getting

$\sum_{m=1}^{\infty}\frac{\rho_{f,M,q}(m)}{m^{(s_{1}+s_{2}+s_{3}+s_{4}-1)/2}}=\frac{1}{L(1,\mathrm{Ad}^{2}f)^{1/2}(Mq\nu(q))^{1/2}}\prod_{p\mid
q_{0}}\left(1-\frac{1}{p^{2}}\right)^{1/2}\\\ \times\sum_{d\mid
M}\xi_{f}(M,d)d^{(3-s_{1}-s_{2}-s_{3}-s_{4})/2}L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2},f\right).$

Note that we obtain the central $L$-values
$L(1/2,f)^{2}L(1/2,f\otimes\overline{\psi})$ if we take
$(s_{1},s_{2},s_{3},s_{4})\mapsto(1/2,1/2,1/2,1/2)$. A similar procedure
applies to the Eisenstein term and one has

$\sum_{n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})\overline{\rho_{\chi,M,q}(n,t)}}{n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2}}=\frac{\overline{C(\chi,M,t)}}{(q\nu(q))^{1/2}\tilde{\mathfrak{n}}_{q}(M)L^{q}(1-2it,\overline{\chi}^{2})}\left(\frac{M_{1}}{M_{2}}\right)^{1/2}\sum_{\delta\mid
M_{2}}\delta\mu\left(\frac{M_{2}}{\delta}\right)\chi(\delta)\\\
\times\sum_{n=1}^{\infty}\frac{\sigma_{s_{3}-s_{4}}(n,\overline{\psi})}{n^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2+it}}\sum_{\begin{subarray}{c}cM_{1}\delta
f=n\\\ (c,q/M)=1\end{subarray}}\overline{\chi}(c)\chi(f)c^{2it}.$

The second line of the above formula is equal to

$(M_{1}\delta)^{-(1+s_{1}-s_{2}+s_{3}-s_{4})/2-it}\sum_{(c,q/M)=1}\sum_{f=1}^{\infty}\frac{\overline{\chi}(c)\chi(f)\sigma_{s_{3}-s_{4}}(cM_{1}\delta
f,\overline{\psi})}{c^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2-it}f^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2+it}}$
(4.22)

and it follows from Lemma 2.10 that

$\sigma_{s_{3}-s_{4}}(cM_{1}\delta
f,\overline{\psi})=\mathop{\sum\sum\sum}_{e_{1}\mid c,\,e_{1}e_{2}\mid
f,\,e_{2}\mid\delta}\mu(e_{1})\mu(e_{2})\overline{\psi}(e_{1}e_{2})(e_{1}e_{2})^{s_{3}-s_{4}}\sigma_{s_{3}-s_{4}}(c/e_{1},\overline{\psi})\sigma_{s_{3}-s_{4}}(f/e_{1}e_{2},\overline{\psi})\sigma_{s_{3}-s_{4}}(\delta/e_{2},\overline{\psi}),$

where we exploited the fact that
$\sigma_{s_{3}-s_{4}}(M_{1},\overline{\psi})=1$. In anticipation of future
simplifications, we make the changes of variables $c\mapsto ce_{1}$ and
$f\mapsto fe_{1}e_{2}$. Then the double sum in (4.22) becomes

$\sum_{(c,q/M)=1}\sum_{f=1}^{\infty}\frac{\overline{\chi}(c)\chi(f)\sigma_{s_{3}-s_{4}}(c,\overline{\psi})\sigma_{s_{3}-s_{4}}(f,\overline{\psi})}{c^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2-it}f^{(1+s_{1}-s_{2}+s_{3}-s_{4})/2+it}}\sum_{e_{1}=1}^{\infty}\sum_{e_{2}\mid\delta}\frac{\mu(e_{1})\mu(e_{2})\chi(e_{2})\overline{\psi}(e_{1}e_{2})\sigma_{s_{3}-s_{4}}(\delta/e_{2},\overline{\psi})}{e_{1}^{1+s_{1}-s_{2}}e_{2}^{(1+s_{1}-s_{2}-s_{3}+s_{4})/2+it}}.$

Consequently, the sums over $c$, $e_{1}$, $f$ are equal to

$\frac{L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}+it,\chi\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}+it,\chi\overline{\psi}\right)L^{q/M}\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}-it,\overline{\chi}\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}-it,\overline{\chi\psi}\right)}{L(1+s_{1}-s_{2},\overline{\psi})}.$

The $m$-sum is calculated as

$\sum_{m=1}^{\infty}\frac{\rho_{\chi,M,q}(m)}{m^{(s_{1}+s_{2}+s_{3}+s_{4}-1)/2}}=\frac{C(\chi,M,t)}{(q\nu(q))^{1/2}\tilde{\mathfrak{n}}_{q}(M)L^{q}(1+2it,\chi^{2})}\left(\frac{M_{1}}{M_{2}}\right)^{1/2}\sum_{\delta\mid
M_{2}}\delta\mu\left(\frac{M_{2}}{\delta}\right)\overline{\chi}(\delta)\\\
\times(M_{1}\delta)^{(1-s_{1}-s_{2}-s_{3}-s_{4})/2+it}\sum_{(c,q/M)=1}\sum_{f=1}^{\infty}\frac{\chi(c)\overline{\chi}(f)}{c^{(s_{1}+s_{2}+s_{3}+s_{4}-1)/2+it}f^{(s_{1}+s_{2}+s_{3}+s_{4}-1)/2-it}}.$

The second line of the above formula turns into

$(M_{1}\delta)^{(1-s_{1}-s_{2}-s_{3}-s_{4})/2+it}L^{q/M}\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+it,\chi\right)L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-it,\overline{\chi}\right).$

The contribution from holomorphic newforms can be treated in the same manner.

Hence, the eventual form of the cubic moment becomes

$\mathcal{J}_{\pm}=\mathcal{J}_{\pm}^{\text{Maa{\ss}}}+\mathcal{J}_{\pm}^{\text{Eis}}+\delta_{\pm=+}\mathcal{J}_{+}^{\text{hol}}$

with

$\mathcal{J}_{\pm}^{\text{Maa{\ss}}}\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}^{\ast}(q_{0})}\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2},f\right)L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2},f\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2},f\otimes\overline{\psi}\right)}{L(1,\mathrm{Ad}^{2}f)}\Theta_{q}^{\text{Maa{\ss}}}(s_{1},s_{2},s_{3},s_{4},f),$
$\mathcal{J}_{\pm}^{\text{Eis}}\coloneqq\sum_{c_{\chi}^{2}\mid M\mid
q}\int_{\mathbb{R}}\frac{L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}+it,\chi\right)L^{q/M}\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}+it,\chi\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}+it,\chi\overline{\psi}\right)}{L^{q}(1+2it,\chi^{2})}\\\
\times\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2}-it,\overline{\chi}\right)L^{q/M}\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2}-it,\overline{\chi}\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2}-it,\overline{\chi\psi}\right)}{L^{q}(1-2it,\overline{\chi}^{2})}\Theta_{\chi,q}^{\text{Eis}}(s_{1},s_{2},s_{3},s_{4},t)\frac{dt}{2\pi},$
$\mathcal{J}_{+}^{\text{hol}}\coloneqq\sum_{q_{0}M\mid
q}\sum_{f\in\mathcal{B}_{\text{hol}}^{\ast}(q_{0})}\frac{L\left(\frac{s_{1}+s_{2}+s_{3}+s_{4}-1}{2},f\right)L\left(\frac{1+s_{1}-s_{2}+s_{3}-s_{4}}{2},f\right)L\left(\frac{1+s_{1}-s_{2}-s_{3}+s_{4}}{2},f\otimes\overline{\psi}\right)}{L(1,\mathrm{Ad}^{2}f)}\Theta_{q}^{\text{hol}}(s_{1},s_{2},s_{3},s_{4},f).$

Here we have defined the local factors as

$\displaystyle\begin{split}\Theta_{q}^{\text{Maa{\ss}}}(s_{1},s_{2},s_{3},s_{4},f)&=\epsilon_{f}^{(1\mp
1)/2}\frac{q^{s_{1}-s_{2}-1}\tau(\psi)}{M\nu(q)}\prod_{p\mid
q_{0}}\left(1-\frac{1}{p^{2}}\right)\sum_{d_{1}\mid M}\sum_{d_{2}\mid
M}\frac{\xi_{f}(M,d_{1})\xi_{f}(M,d_{2})\Phi_{\texttt{s}}^{\pm}(it_{f})}{d_{1}^{(s_{1}+s_{2}+s_{3}+s_{4}-3)/2}d_{2}^{(s_{1}-s_{2}+s_{3}-s_{4}-1)/2}},\\\
\Theta_{\chi,q}^{\text{Eis}}(s_{1},s_{2},s_{3},s_{4},t)&=\frac{q^{s_{1}-s_{2}-1}\tau(\psi)}{\nu(q)\tilde{\mathfrak{n}}_{q}(M)^{2}}M_{1}^{1-s_{1}-s_{3}}M_{2}^{-1}\sum_{\delta_{1}\mid
M_{2}}\sum_{\delta_{2}\mid
M_{2}}\frac{\mu(M_{2}/\delta_{1})\mu(M_{2}/\delta_{2})\overline{\chi}(\delta_{1})\chi(\delta_{2})}{\delta_{1}^{(s_{1}+s_{2}+s_{3}+s_{4}-3)/2-it}\delta_{2}^{(s_{1}-s_{2}+s_{3}-s_{4}-1)/2+it}}\\\
&\quad\times\sum_{e\mid\delta_{2}}\mu(e)\chi(e)\overline{\psi}(e)e^{-(1+s_{1}-s_{2}-s_{3}+s_{4})/2-it}\sigma_{s_{3}-s_{4}}(\delta/e,\overline{\psi})\Phi_{\texttt{s}}^{\pm}(it),\\\
\Theta_{q}^{\text{hol}}(s_{1},s_{2},s_{3},s_{4},f)&=\frac{i^{k}\tau(\psi)}{2Mq\nu(q)}\left(\frac{q}{2\pi}\right)^{s_{1}-s_{2}}\left\\{e\left(\frac{s_{3}-s_{4}}{4}\right)+\psi(-1)e\left(-\frac{s_{3}-s_{4}}{4}\right)\right\\}\\\
&\quad\times\prod_{p\mid q_{0}}\left(1-\frac{1}{p^{2}}\right)\sum_{d_{1}\mid
M}\sum_{d_{2}\mid
M}\frac{\xi_{f}(M,d_{1})\xi_{f}(M,d_{2})}{d_{1}^{(s_{1}+s_{2}+s_{3}+s_{4}-3)/2}d_{2}^{(s_{1}-s_{2}+s_{3}-s_{4}-1)/2}}\Xi_{\texttt{s}}\left(\frac{k-1}{2}\right).\end{split}$
(4.23)

This establishes Theorem 1.1.∎

##  5\. Proof of Corollary 1.3

In this section, we establish Corollary 1.3. One chooses the test function in
Theorem 1.1 as

$g(t)=\frac{1}{\sqrt{\pi}H}\exp\left(-\left(\frac{t-T}{H}\right)^{2}\right)$
(5.1)

with the parameter $H$ at one’s disposal. We assume that

$T^{1/2}\leqslant H\leqslant T(\log T)^{-1}.$

Motohashi [9, (5.1.40)–(5.1.42)] evaluated the corresponding integral
transform $\Xi$ and his expression can be used in our context. We then derive
an asymptotic formula that is almost identical to [9, Theorem 5.1]. Estimating
the spectral sum by absolute values, one sees that an oscillatory component in
the summand evanishes, obtaining the bound

$\int_{T}^{T+H}|\zeta(1/2+it)L(1/2+it,\psi)|^{2}dt\ll_{\epsilon}H^{1+\epsilon}q^{\epsilon}+\frac{H^{3/2}}{T}q^{-1/2+\epsilon}\sum_{t_{f}\ll
T/H}|L(1/2,f)^{2}L(1/2,f\otimes\psi)|,$ (5.2)

where the level of Maaß newforms is $q$ in the spectral sum on the right-hand
side. Here we have also evaluated the finite sums in the local factors using
(3.2). The truncation of the spectral sum is justified since the Taylor series
expansion of $\exp(-(Ht_{f}/T)^{2}/4)$ in [9, (5.1.44)] implies that the rest
of the spectral sum is smaller than the main term
$H^{1+\epsilon}q^{\epsilon}$. As $\psi$ is quadratic, the central $L$-values
in (5.2) are nonnegative and one can use Hölder’s inequality with the exponent
$(3/2,3)$. It therefore follows from the method of Conrey–Iwaniec [3] or
Petrow–Young [11] that

$H^{1+\epsilon}q^{\epsilon}+\frac{H^{3/2}}{T}q^{-1/2+\epsilon}\left(\frac{qT^{2}}{H^{2}}\right)^{2/3+\epsilon}\left(\frac{qT^{2}}{H^{2}}\right)^{1/3+\epsilon}\ll
H^{1+\epsilon}q^{\epsilon}+\left(\frac{qT^{2}}{H}\right)^{1/2+\epsilon}.$

Optimising $H=q^{1/3}T^{2/3}$ with the restriction $q\leqslant
T^{1-\epsilon}$, we conclude the proof of Corollary 1.3. ∎

## Acknowledgements

The author would like to thank Eren Mehmet Kıral and Peter Humphries for
illuminative discussions.

## References

  * [1] V. Blomer, P. Humphries, R. Khan, and M. B. Milinovich.  Motohashi’s fourth moment identity for non-archimedean test functions and applications.  Compositio Math., 156(5):1004–1038, 2020.
  * [2] V. Blomer and R. Khan.  Twisted moments of $L$-functions and spectral reciprocity.  Duke Math. J., 168(6):1109–1177, 2019.
  * [3] J. B. Conrey and H. Iwaniec.  The cubic moment of central values of automorphic $L$-functions.  Ann. of Math. (2), 151(3):1175–1216, 2000.
  * [4] W. Duke, J. Friedlander, and H. Iwaniec.  Bounds for automorphic $L$-functions.  Invent. Math., 112(1):1–8, 1993.
  * [5] H. Iwaniec.  Fourier coefficients of cusp forms and the Riemann zeta-function.  Séminaire de Théorie des Nombres (Année 1979–1980), pages 1–36, 1980.  Exp. No. 18, Univ. Bordeaux I, Talence.
  * [6] Y. Motohashi.  An explicit formula for the fourth power mean of the Riemann zeta-function.  Acta Math., 170(2):181–220, 1993.
  * [7] Y. Motohashi.  On the mean square of the product of the zeta- and $L$-functions.  In Interdisciplinary series on number theory (Kyoto, 1992), volume 837 of Sūrikaisekikenkyūsho Kōkyūroku, pages 57–62, 1993.
  * [8] Y. Motohashi.  The mean square of Dedekind zeta-functions of quadratic number fields.  In G. R. H. Greaves, G. Harman, and M. N. Huxley, editors, Sieve Methods, Exponential Sums, and their Applications in Number Theory, Proceedings of a symposium held in Cardiff, July 1995, volume 237 of London Math. Soc. Lecture Note Ser., pages 309–324. Cambridge Univ. Press, Cambridge, 1997.
  * [9] Y. Motohashi.  Spectral theory of the Riemann zeta-function, volume 127 of Cambridge Tracts in Math. Cambridge Univ. Press, Cambridge, 1997.
  * [10] Y. Motohashi.  On an identity of Ramanujan.  Hardy-Ramanujan J., 41:48–49, 2018.
  * [11] I. Petrow and M. P. Young.  The fourth moment of Dirichlet $L$-functions along a coset and the Weyl bound.  arXiv e-prints, 54 pages, 2019.  https://arxiv.org/abs/1908.10346.