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Spectral decomposition formula and moments of symmetric square O. G. Balkanova Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
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     § 1. IntroductionThe aim of this paper is to prove a spectral decomposition formula for the average
$$
\begin{equation}
\sum_{l=1}^{\infty}\omega(l)\mathscr{L}_{n^2-4l^2}(s),
\end{equation}
\tag{1.1}
$$
where $\omega$ is a suitable test function and the $L$-series is defined as
$$
\begin{equation}
\mathscr{L}_{n}(s)=\frac{\zeta(2s)}{\zeta(s)}\sum_{q=1}^{\infty} \frac{b_q(n)}{q^{s}}
\end{equation}
\tag{1.2}
$$
for $\operatorname{Re}{s}>1$ and can be meromorphically continued to the whole complex plane (see [1], Proposition 3). Here $\zeta(s)$ denotes the Riemann zeta function and
$$
\begin{equation*}
b_q(n):=\#\{x\ (\mathrm{mod}\ 2q)\colon x^2\equiv n\ (\mathrm{mod}\ 4q)\}.
\end{equation*}
\notag
$$
The dual problem of investigating the average over $n$
$$
\begin{equation}
\sum_{n=1}^{\infty}\omega(n)\mathscr{L}_{n^2-4l^2}(s),
\end{equation}
\tag{1.3}
$$
was studied in [2] in connection with the prime geodesic theorem. For related results, see also [3]–[7]. Furthermore, sums of the form (1.3) appear in the explicit formulas for the first moments of symmetric square $L$-functions associated to holomorphic (see [1], [8]) or Maass (see [9]) cusp forms. These explicit formulas can serve as a starting point for analyzing second moments of symmetric square $L$-functions. For example, if we take the first moment of Maass form symmetric square $L$-functions for $\operatorname{SL}_{2}(\mathbf{Z})$, twisted by the Fourier coefficient $\rho_j(l^2)$, namely, multiply it by $\zeta(2s)l^{-s}$, and sum with respect to $l$ from $1$ to $\infty$, we obtain the second moment Applying these manipulations to the explicit formula proved in [9], we discover expressions of the following form
$$
\begin{equation}
\sum_{l=1}^{\infty}\sum_{n=1}^{\infty}\mathscr{L}_{n^2-4l^2}(s)f(n,l;s)
\end{equation}
\tag{1.6}
$$
on the right-hand side of the explicit formula for the second moment. A possible approach to evaluating (1.6) calls forspectral methods. However, spectral decomposition for the inner sum over $n$ results in a loop bringing us back to the moment we started from. To avoid this issue, we can change the order of summation and investigate the average over $l$ first. This is the main reason behind our interest in a spectral decomposition formula for (1.1). Even though the averages (1.1) and (1.3) look similar, there are some important differences. Spectral decomposition of (1.3) is a relatively straightforward application of the Kuznetsov trace formula to the generalized Kloosterman sums, while in case of (1.1) our method relies heavily on various properties of Gauss sums. Evaluating these Gauss sums, we obtain sums of Kloosterman sums for $ \Gamma_0(N)$ with $N=4,16, 64$ at various cusps and twisted by $\chi_4$ (non-trivial Dirichlet character modulo $4$). Finally, applying the Kuznetsov trace formula, we derive a spectral decomposition formula for (1.1), which contains moments of symmetric square $L$-functions for $ \Gamma_0(N)$ with $N=4,16, 64$ twisted by Fourier coefficients at the cusps $0$ and $\infty$. To state the main results rigorously, we introduce the function
$$
\begin{equation}
\psi(x)=\psi(x;n;s)=\frac{2}{\sqrt{\pi}}\biggl(\frac{x}{n}\biggr)^s \int_{0}^{\infty}\omega(y)\cos\biggl( \frac{2xy}{n}\biggr)\, dy
\end{equation}
\tag{1.7}
$$
and denote by $\psi_{H}(x)$ and $\psi_{D}(x)$ the Bessel integral transforms of $\psi(x)$ appearing in the Kuznetsov trace formula (see (2.41) and (2.42)). These transforms can be expressed in terms of the Gauss hypergeometric function as shown in Lemmas 4.1 and 4.2. Let $\omega \in C^{\infty}$ be a function of compact support on $[a_1,a_2]$ for some $0<a_1<a_2<\infty$ and let $\widehat{\omega}$ stand for its Mellin transform. It is also required to define the generalized divisor function For a cusp $\mathfrak{a}$ of $\Gamma_0(N)$, let us introduce the following notation
$$
\begin{equation}
\mathfrak{M}_{\mathfrak{a}}(n,N,s)= \mathfrak{M}^{\mathrm{hol}}_{\mathfrak{a}}(n,N,s)+ \mathfrak{M}^{\mathrm{disc}}_{\mathfrak{a}}(n,N,s),
\end{equation}
\tag{1.9}
$$
where
$$
\begin{equation}
\mathfrak{M}^{\mathrm{hol}}_{\mathfrak{a}}(n,N,s) := \sum_{\substack{k>1\\k \text{ odd}}}\psi_{H}(k)\Gamma(k) \sum_{f\in H_k(N,\chi_4)} \rho_{f_{\mathfrak{a}}}(n) \overline{L(s,\operatorname{sym}^2 f_{\infty})},
\end{equation}
\tag{1.10}
$$
$$
\begin{equation}
\mathfrak{M}^{\mathrm{disc}}_{\mathfrak{a}}(n,N,s) := \sum_{f\in H(N,\chi_4)}\frac{\psi_{D}(t_f)}{\cosh(\pi t_f)} \rho_{f_{\mathfrak{a}}}(n)\overline{L(s,\operatorname{sym}^2 f_{\infty})}
\end{equation}
\tag{1.11}
$$
are the moments of symmetric square $L$-functions associated with holomorphic and Maass cusp forms of level $N$ with nebentypus $\chi_4$ twisted by the Fourier coefficient $\rho_{f_{\mathfrak{a}}}(n)$ of $f$ at a cusp $\mathfrak{a}$. The bar over $L$-functions means complex conjugation, and $f_{\infty}$ means that the $L$-functions are formed using the Fourier coefficients of $f$ around the parabolic cusp $\infty$. Theorem 1.1. Assume that $n$ is even. For $0<\operatorname{Re}{s}<1$ the following explicit formula holds Theorem 1.2. Assume that $n$ is odd. For $0<\operatorname{Re}{s}<1$ the following explicit formula holds Remark 1.1. Note that the main terms The paper is organized as follows. In § 2 we collect all required tools and preliminary results. In § 3, assuming that $\operatorname{Re}{s}$ is sufficiently large, we isolate the diagonal and non-diagonal terms for (1.1), compute the diagonal term explicitly, and prove an expression for the non-diagonal term which is suitable for application of the Kuznetsov trace formula. Section 4 is devoted to the analysis of the Bessel integral transforms $\psi_{H}(x)$ and $\psi_{D}(x)$ appearing after the Kuznetsov trace formula is applied. More precisely, we show how to express $\psi_{H}(x)$ and $\psi_{D}(x)$ in terms of the Gauss hypergeometric functions. In §§ 5 and 6, we are concerned with evaluation of the continuous spectrum, while the holomorphic and discrete spectra are studied in § 7. Finally, in § 8 we complete the proof of Theorems 1.1 and 1.2 and compute the main terms at the central point. § 2. Preliminaries2.1. Generalized divisor functionIn this subsection, we collect various results related to the function $\sigma_s(\chi;n)$ defined by (1.8). Note that the derivative of $\sigma_{s}(\chi;n)$ with respect to $s$ is equal to
$$
\begin{equation}
\sigma'_{s}(\chi;n)=\sum_{d\,|\,n}\chi(d)d^s\log{d}.
\end{equation}
\tag{2.1}
$$
Let $\chi_4$ be a non-trivial Dirichlet character modulo $4$, so that and
$$
\begin{equation}
\sigma_s\biggl(\chi_4; \biggl(\frac{n}{2} \biggr)^2\biggr)= \sigma_s(\chi_4;n^2).
\end{equation}
\tag{2.3}
$$
Proof. Let $n^2=bd$. Since $n$ is odd, Hence $\chi_4(b)=\chi_4(d)$. Consequently, proving the lemma. Consider the Dirichlet series
$$
\begin{equation}
Z(z,s):=\sum_{n=1}^{\infty}\frac{\sigma_s(\chi_4;n^2)}{n^z}.
\end{equation}
\tag{2.7}
$$
Proof. First, assume that $\operatorname{Re}{z}>1+2\operatorname{Re}{s}$. The Euler product for $Z(z,s)$ is
$$
\begin{equation}
\begin{aligned} \, Z(z,s) &=\prod_{p}\biggl( 1+\frac{\sigma_s(\chi_4;p^2)}{p^z} + \frac{\sigma_s(\chi_4;p^4)}{p^{2z}} +\cdots\biggr) \nonumber \\ &=\biggl( 1+\sum_{k=1}^{\infty}\frac{\sigma_s(\chi_4;2^{2k})}{2^{kz}}\biggr) \prod_{p>2}\biggl(1+\sum_{k=1}^{\infty} \frac{\sigma_s(\chi_4;p^{2k})}{p^{kz}} \biggr). \end{aligned}
\end{equation}
\tag{2.9}
$$
Note that $\chi_4(d)=0$ for even $d$, and therefore,
$$
\begin{equation*}
\sigma_{s}(\chi_4;2^{2k})=\sum_{d\,|\,2^{2k}}d^s\chi_4(d)=1.
\end{equation*}
\notag
$$
Consequently,
$$
\begin{equation}
1+\sum_{k=1}^{\infty}\frac{\sigma_s(\chi_4;2^{2k})}{2^{kz}}= \frac{1}{1-2^{-z}}.
\end{equation}
\tag{2.10}
$$
Next, we evaluate the second multiple on the right hand side of (2.9). Since $p^2\equiv 1$ $ (\operatorname{mod} 4)$ we have
$$
\begin{equation}
\chi_4(p^{2m})=1, \quad \chi_4(p^{2m+1})=\chi_4(p) \quad \forall\, m \in \mathbf{N}.
\end{equation}
\tag{2.11}
$$
Therefore,
$$
\begin{equation}
\sigma_s(\chi_4;p^{2k})=\frac{(p^{2s})^{k+1}-1}{p^{2s}-1} + \frac{(p^{2s})^k-1}{p^{2s}-1}\chi_4(p)p^{s}.
\end{equation}
\tag{2.12}
$$
This implies that
$$
\begin{equation}
\sum_{k=1}^{\infty}\frac{\sigma_s(\chi_4;p^{2k})}{p^{kz}} = \frac{p^{2s}+\chi_4(p)p^{s}}{(p^{2s}-1)(p^{z-2s}-1)}- \frac{1+\chi_4(p)p^s}{(p^{2s}-1)(p^z-1)}.
\end{equation}
\tag{2.13}
$$
Since $\chi_{4}^2(p)=1$, we have
$$
\begin{equation}
\begin{aligned} \, 1+\sum_{k=1}^{\infty}\frac{\sigma_s(\chi_4;p^{2k})}{p^{kz}} &= \frac{1+\chi_4(p)/p^{z-s}}{(1-1/p^{z-2s})(1-1/p^z)} \nonumber \\ &=\frac{1-1/p^{2(z-s)}}{(1-1/p^{z-2s})(1-1/p^z)(1-\chi_4(p)/p^{z-s})}. \end{aligned}
\end{equation}
\tag{2.14}
$$
Finally, substituting (2.10) and (2.14) in (2.9), we prove the lemma. 2.2. Cusps and Kloosterman sumsFor a positive integer $N$, let $\Gamma=\Gamma_0(N)$ denote the Hecke congruence subgroup of level $N$. The stabilizer of the cusp $\mathfrak{a}$ in $\Gamma$ is defined by
$$
\begin{equation}
\Gamma_{\mathfrak{a}}:=\{\gamma \in \Gamma \colon \gamma \mathfrak{a}= \mathfrak{a}\}.
\end{equation}
\tag{2.15}
$$
A scaling matrix for the cusp $\mathfrak{a}$ is a matrix $\sigma_{\mathfrak{a}}\in \mathbf{SL}_2(\mathbf{R})$ such that
$$
\begin{equation}
\sigma_{ \mathfrak{a}}\infty= \mathfrak{a}, \qquad \sigma_{\mathfrak{a}}^{-1}\Gamma_{\mathfrak{a}}\sigma_{ \mathfrak{a}}= \biggl\{\pm \begin{pmatrix} 1&n\\ 0&1 \end{pmatrix}\colon n \in \mathbf{Z}\biggr\}:=B.
\end{equation}
\tag{2.16}
$$
Note that the choice of scaling matrix is not unique. Let $\chi$ be a Dirichlet character modulo $N$. This can be extended to $\Gamma$ as follows:
$$
\begin{equation}
\chi(\gamma)=\chi(d), \qquad \gamma=\begin{pmatrix} a&b\\ cN&d \end{pmatrix} \in \Gamma.
\end{equation}
\tag{2.17}
$$
Let $\lambda_{\mathfrak{a}}$ be defined by
$$
\begin{equation*}
\sigma_{\mathfrak{a}}^{-1}\lambda_{\mathfrak{a}}\sigma_{\mathfrak{a}} = \begin{pmatrix}1&1\\0&1\end{pmatrix}.
\end{equation*}
\notag
$$
The cusp $\mathfrak{a}$ is called singular for $\chi$ if $\chi(\lambda_{\mathfrak{a}})=1$. Suppose that $N=rs$, $(r,s)=1$. Then a cusp of the form $\mathfrak{a}=1/r$ is called an Atkin–Lehner cusp. Note that Atkin–Lehner cusps are singular with respect to any Dirichlet character modulo $N$, see [10], p. 395. Let $\kappa$ be defined by $\chi(-1)=(-1)^{\kappa}$ and let $\mathfrak{a}$, $\mathfrak{b}$ be two singular cusps for $\chi$ with corresponding scaling matrices $\sigma_{\mathfrak{a}}$, $\sigma_{\mathfrak{b}}$. Similarly to [10], (2.3), we define the Kloosterman sum associated to $\mathfrak{a}$, $\mathfrak{b}$ as
$$
\begin{equation*}
S_{\mathfrak{a}\mathfrak{b}}(m,n;c;\chi):=\sum_{\gamma= \left(\begin{smallmatrix} a & b\\ c & d \end{smallmatrix}\right) \in \Gamma_{\infty}\setminus \sigma^{-1}_{\mathfrak{a}} \Gamma\sigma_{\mathfrak{b}}/\Gamma_{\infty}} \chi(\operatorname{sgn}(c)) \overline{\chi(\sigma_{\mathfrak{a}}\gamma \sigma_{\mathfrak{b}}^{-1})}e \biggl( \frac{am+dn}{c}\biggr).
\end{equation*}
\notag
$$
The set of allowed moduli is given by
$$
\begin{equation}
C_{\mathfrak{a},\mathfrak{b}}(N)=\biggl\{\gamma>0\colon \begin{pmatrix} * &* \\ \gamma &* \end{pmatrix}\in \sigma^{-1}_{\mathfrak{a}}\Gamma\sigma_{\mathfrak{b}}\biggr\}.
\end{equation}
\tag{2.18}
$$
2.3. Holomorphic and Maass cusp formsLet $H_{k}(N,\chi)$ be an orthonormal basis of holomorphic cusp forms of weight $k>0$, $k\equiv \kappa\ (\operatorname{mod}2)$, level $N$, and nebentypus $\chi$. The Fourier expansion of $f \in H_{k}(N,\chi)$ around a singular cusp $\mathfrak{a}$ with a scaling matrix $\sigma_{\mathfrak{a}}$ is given by
$$
\begin{equation}
f(\sigma_{\mathfrak{a}}z)i(\sigma_{\mathfrak{a}},z)^{-k}=\sum_{m\geqslant 1} \frac{\rho_{f_{\mathfrak{a}}}(m)}{\sqrt{m}}(4\pi m)^{k/2}e(mz),
\end{equation}
\tag{2.19}
$$
where $i(\sigma_{\mathfrak{a}},z):=cz+d$ for $\sigma_{\mathfrak{a}}= \left(\begin{smallmatrix} * &*\\ c &d \end{smallmatrix}\right)$. Let $H(N,\chi)$ be an orthonormal basis of the space of Maass cusp forms of weight $\kappa\in\{0,1\}$. For the function $f \in H(N,\chi)$ (which is an eigenfunction of the Laplace–Beltrami operator with eigenvalue $1/4+t_{f}^{2}$), the following Fourier–Whittaker expansion holds around the cusp $\mathfrak{a}$ with scaling matrix $\sigma_{\mathfrak{a}}$
$$
\begin{equation}
f(\sigma_{\mathfrak{a}}z)e^{-i\kappa\arg i(\sigma_{\mathfrak{a}},z)}= \sum_{m\neq 0} \frac{\rho_{f_{\mathfrak{a}}}(m)}{\sqrt{m}} W_{(|m|/m)(\kappa/2),\, it_f}(4\pi|m|y)e(mx),
\end{equation}
\tag{2.20}
$$
where $z=x+iy$ and the Whittaker function $W_{\lambda,\mu}(z)$ is defined in [11], § 9.22. For $f \in H_{k}(N,\chi)$ or $f \in H(N,\chi)$, we define
$$
\begin{equation}
L(s,\operatorname{sym}^2 f_{\infty}) =\zeta^{(N)}(2s)\sum_{l=1}^{\infty} \frac{\rho_{f_{\infty}}(l^2)}{l^s},\qquad \operatorname{Re}{s}>1,
\end{equation}
\tag{2.21}
$$
where the superscript in $\zeta^{(N)}(2s)$ means that Euler factors at primes dividing $N$ have been removed. Shimura [12] proved an analytic continuation and a functional equation for (2.21). 2.4. Eisenstein seriesWe fix $\kappa=1$. For $\Gamma=\Gamma_0(N)$, the Eisenstein series associated to a singular cusp $\mathfrak{c}$ for the nebentypus $\chi$ is defined as
$$
\begin{equation}
E_{\mathfrak{c}}(z,s):=\sum_{\gamma \in \Gamma_{\mathfrak{c}}\setminus \Gamma}\overline{\chi}(\gamma)j_{\sigma_{\mathfrak{c}}^{-1}\gamma}(z)^{-1} \bigl(\operatorname{Im}{(\sigma_{\mathfrak{c}}^{-1}\gamma z )}\bigr)^s,
\end{equation}
\tag{2.22}
$$
where $\sigma_{\mathfrak{c}}$ is a scaling matrix for $\mathfrak{c}$ and
$$
\begin{equation}
j_{\gamma}(z):=\frac{cz+d}{|cz+d|}=e^{i\operatorname{arg}(cz+d)}, \qquad \gamma=\begin{pmatrix} a & b\\ c & d \end{pmatrix}.
\end{equation}
\tag{2.23}
$$
Theorem 2.1. Let $\mathfrak{c}$ be a singular cusp for the nebentypus $\chi$ and $\mathfrak{a}$ be the Aktin–Lehner cusp. Them the following Fourier–Whittaker expansion holds Proof. Consider the Eisenstein series
$$
\begin{equation}
E_{\mathfrak{c}}(\sigma_{\mathfrak{a}}z,s)=\sum_{\gamma \in \Gamma_{\mathfrak{c}}\setminus \Gamma}\overline{\chi}(\gamma) j_{\sigma_{\mathfrak{c}}^{-1}\gamma}(\sigma_{\mathfrak{a}}z)^{-1} \bigl(\operatorname{Im}{(\sigma_{\mathfrak{c}}^{-1} \gamma\sigma_{\mathfrak{a}}z)} \bigr)^s.
\end{equation}
\tag{2.28}
$$
Making the change of variables $\tau:=\sigma_{\mathfrak{c}}^{-1}\gamma\sigma_{\mathfrak{a}}$ (so that $\tau \in B\setminus \sigma_{\mathfrak{c}}^{-1}\Gamma\sigma_{\mathfrak{a}}$), we infer
$$
\begin{equation}
E_{\mathfrak{c}}(\sigma_{\mathfrak{a}}z,s)=\sum_{\tau \in B\setminus \sigma_{\mathfrak{c}}^{-1}\Gamma\sigma_{\mathfrak{a}}} \overline{\chi}(\sigma_{\mathfrak{c}} \tau\sigma_{\mathfrak{a}}^{-1})j_{\tau\sigma_{\mathfrak{a}}^{-1}} (\sigma_{\mathfrak{a}}z)^{-1} \bigl(\operatorname{Im}{(\tau z)} \bigr)^s.
\end{equation}
\tag{2.29}
$$
Using the property
$$
\begin{equation}
j_{\tau\sigma_{\mathfrak{a}}^{-1}}(\sigma_{\mathfrak{a}}z)^{-1} j_{\sigma_{\mathfrak{a}}}(z)^{-1} =j_{\tau}(z)^{-1},
\end{equation}
\tag{2.30}
$$
we find that
$$
\begin{equation}
\begin{aligned} \, &E_{\mathfrak{c}}(\sigma_{\mathfrak{a}}z,s)j_{\sigma_{\mathfrak{a}}}(z)^{-1} = \sum_{\tau \in B\setminus \sigma_{\mathfrak{c}}^{-1} \Gamma\sigma_{\mathfrak{a}}} \overline{\chi}(\sigma_{\mathfrak{c}} \tau\sigma_{\mathfrak{a}}^{-1})j_{\tau}(z)^{-1} \bigl(\operatorname{Im}{(\tau z)} \bigr)^s \nonumber \\ &\qquad=\delta_{\mathfrak{a}\mathfrak{c}} y^s+\sum_{\gamma \in B\setminus \sigma_{\mathfrak{c}}^{-1} \Gamma\sigma_{\mathfrak{a}}/B} \, \sum_{\tau \in B} \overline{\chi}(\sigma_{\mathfrak{c}}\gamma \tau\sigma_{\mathfrak{a}}^{-1}) j_{\gamma\tau}(z)^{-1} \bigl(\operatorname{Im}{(\gamma \tau z)} \bigr)^s. \end{aligned}
\end{equation}
\tag{2.31}
$$
Note that $\overline{\chi}(\sigma_{\mathfrak{c}}\gamma \tau \sigma_{\mathfrak{a}}^{-1})=\overline{\chi}(\sigma_{\mathfrak{c}} \gamma \sigma_{\mathfrak{a}}^{-1})$ since $\mathfrak{a}$ is singular. Furthermore, taking $\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ and $\tau=\left(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right)$, we obtain
$$
\begin{equation}
\gamma\tau z=\frac{a}{c}-\frac{1}{c(c(z+n)+d)}, \qquad j_{\gamma \tau}=\frac{cz+cn+d}{|cz+cn+d|}.
\end{equation}
\tag{2.32}
$$
Consequently, for $z=x+iy$, we have
$$
\begin{equation}
\operatorname{Im}{(\gamma \tau z)}=\frac{y}{c^2}\, \frac{1}{(x+n+d/c)^2+y^2}
\end{equation}
\tag{2.33}
$$
and
$$
\begin{equation}
\begin{aligned} \, &E_{\mathfrak{c}}(\sigma_{\mathfrak{a}}z,s)j_{\sigma_{\mathfrak{a}}}(z)^{-1} = \delta_{\mathfrak{a}\mathfrak{c}} y^s+\sum_{\gamma \in B\setminus \sigma_{\mathfrak{c}}^{-1}\Gamma\sigma_{\mathfrak{a}}/B}\overline{\chi} (\sigma_{\mathfrak{c}} \gamma \sigma_{\mathfrak{a}}^{-1}) \nonumber \\ &\qquad \times \sum_{n\in \mathbf{Z}} \biggl(\frac{cz+cn+d}{|cz+cn+d|} \biggr)^{-1}\biggl(\frac{y}{c^2} \frac{1}{(x+n+d/c)^2+y^2} \biggr)^s. \end{aligned}
\end{equation}
\tag{2.34}
$$
In order to evaluate the sum over $n$, we apply the Poisson summation formula, showing that
$$
\begin{equation}
\begin{aligned} \, &\sum_{n\in \mathbf{Z}}\biggl(\frac{cz+cn+d}{|cz+cn+d|} \biggr)^{-1} \biggl(\frac{y}{c^2}\frac{1}{(x+n+d/c)^2+y^2} \biggr)^s \nonumber \\ &\qquad= \sum_{m\in \mathbf{Z}}\int_{-\infty}^{\infty} \biggl( \frac{c(z+v)+d}{|c(z+v)+d|}\biggr)^{-1} \frac{(yc^{-2})^se(-mv)}{((x+d/c+v)+y^2)^s}\, dv. \end{aligned}
\end{equation}
\tag{2.35}
$$
Introducing the new variable $t:=x+d/c+v$, we obtain
$$
\begin{equation}
\sum_{m\in \mathbf{Z}} e\biggl(mx+\frac{md}{c}\biggr) \int_{-\infty}^{\infty} \biggl(\frac{t+iy}{|t+iy|} \biggr)^{-1} \biggl( \frac{yc^{-2}}{t^2+y^2}\biggr)^s e(-mt)\, dt.
\end{equation}
\tag{2.36}
$$
We first assume that $m=0$. Then
$$
\begin{equation}
\begin{aligned} \, &\int_{-\infty}^{\infty}\frac{|t+iy|}{t+iy}\, \frac{(yc^{-2})^s}{(t^2+y^2)^s}\, dt =\frac{1}{i}\int_{-\infty}^{\infty} \frac{(yc^{-2})^s\, dt}{(y+it)^{s-1/2}(y-it)^{s+1/2}} \nonumber \\ &\qquad=\frac{(yc^{-2})^s}{i}\, \frac{2\pi (2y)^{1-2s}\Gamma(2s)}{(2s-1) \Gamma(s-1/2)\Gamma(s+1/2)} =-\frac{\sqrt{\pi}\, i \Gamma(s)}{\Gamma(s+1/2)}\, \frac{y^{1-s}}{c^{2s}}, \end{aligned}
\end{equation}
\tag{2.37}
$$
where we used (8.381.1) in [11] to evaluate the integral. If $m \neq 0$, the integral in (2.36) can be evaluated using (3.384.9) in [11] as follows:
$$
\begin{equation}
\begin{aligned} \, &\frac{(yc^{-2})^s}{i}\int_{-\infty}^{\infty}(y+it)^{1/2-s} (y-it)^{-1/2-s}e(-mt)\, dt \nonumber \\ &\qquad=\frac{(2\pi)^s2^{-s}|m|^{s-1}}{ic^{2s}\Gamma(1/2+s)} W_{|m|/(2m),\, 1/2-s}(4\pi y|m|). \end{aligned}
\end{equation}
\tag{2.38}
$$
Consequently, (2.36) is equal to
$$
\begin{equation}
\begin{aligned} \, &-\frac{\pi^{1/2}i\Gamma(s)}{\Gamma(s+1/2)}y^{1-s}\frac{1}{c^{2s}} \nonumber \\ &\qquad+\frac{\pi^s}{i\Gamma(1/2+s)}\frac{1}{c^{2s}}\sum_{m\neq 0} |m|^{s-1} e\biggl(mx+\frac{md}{c}\biggr)W_{|m|/(2m),\,1/2-s}(4\pi y|m|). \end{aligned}
\end{equation}
\tag{2.39}
$$
Now to complete the proof is suffices replace the sum over $n$ in (2.34) by (2.39). Theorem 2.1 is proved. 2.5. Kuznetsov trace formulaIn this section, we state the Kuznetsov trace formula for Dirichlet multiplier system and general cusps. To this end, we follow [13], [14], § 3.3, and [15], § 4.1.3, assuming that $\kappa=1$ (that is, $\chi(1)=-1$). Consider a function $\psi \in C^{\infty}$ such that
$$
\begin{equation}
\psi(0)=\psi'(0)=0, \qquad \psi^{(j)}(x)\ll (1+x)^{-2-\eta}, \quad j=0,1,2,3,
\end{equation}
\tag{2.40}
$$
for some $\eta>0$. We also introduce the integral transforms
$$
\begin{equation}
\psi_H(k) :=4i^k\int_{0}^{\infty}J_{k-1}(x)\psi(x)\, \frac{dx}{x},
\end{equation}
\tag{2.41}
$$
$$
\begin{equation}
\psi_D(t) :=\frac{2\pi i t}{\sinh(\pi t)}\int_{0}^{\infty}(J_{2it}(x)+ J_{-2it}(x))\psi(x)\, \frac{dx}{x},
\end{equation}
\tag{2.42}
$$
where $J_{\alpha}$ denotes the $J$-Bessel function of order $\alpha$. For $m,n \geqslant 1$,
$$
\begin{equation}
\sum_{c\in C_{\mathfrak{a},\mathfrak{b}}(N)} \frac{S_{\mathfrak{a}\mathfrak{b}}(m,n;c;\chi)}{c} \psi\biggl(\frac{4\pi\sqrt{mn}}{c} \biggr)= H+D+C,
\end{equation}
\tag{2.43}
$$
where
$$
\begin{equation}
H :=\sum_{\substack{k>1\\ k\equiv 1 \ (\operatorname{mod}2)}} \sum_{f \in H_k(N,\chi)} \psi_H(k)\Gamma(k) \overline{\rho_{f_{\mathfrak{a}}}(m)}\rho_{f_{\mathfrak{b}}}(n),
\end{equation}
\tag{2.44}
$$
$$
\begin{equation}
D :=\sum_{f\in H(N,\chi)}\frac{\psi_D(t_f)}{\cosh(\pi t_f)} \overline{\rho_{f_{\mathfrak{a}}}(m)} \rho_{f_{\mathfrak{b}}}(n),
\end{equation}
\tag{2.45}
$$
$$
\begin{equation}
C :=\sum_{\mathfrak{c} \text{ sing.}} \frac{\sqrt{mn}}{4\pi} \int_{-\infty}^{\infty}\frac{\psi_D(t)}{\cosh(\pi t)} \overline{\rho_{\mathfrak{a,c}}\biggl(m,\,\frac12+it\biggr)} \rho_{\mathfrak{b,c}}\biggl(n,\, \frac12+it\biggr)\, dt.
\end{equation}
\tag{2.46}
$$
According to (2.26), the continuous part can be written as
$$
\begin{equation}
\begin{aligned} \, C &=\sum_{\mathfrak{c} \text{ sing.}} \frac{1}{4\pi}\int_{-\infty}^{\infty} \frac{\psi_D(t)}{\cosh(\pi t)}\, \frac{\pi}{|\Gamma(1+it)|^2}m^{-it}n^{it} \nonumber \\ &\qquad \times \overline{\phi_{\mathfrak{a,c}} \biggl(m,\, \frac12+it,\, \chi\biggr)} \phi_{\mathfrak{b,c}} \biggl(n,\, \frac12+it,\, \chi\biggr)\, dt. \end{aligned}
\end{equation}
\tag{2.47}
$$
Using the identity (see (5.4.3) in [16])
$$
\begin{equation}
\Gamma(1+it)\Gamma(1-it)=\frac{\pi t}{\sinh(\pi t)},
\end{equation}
\tag{2.48}
$$
we find that
$$
\begin{equation}
C = \sum_{\mathfrak{c} \text{ sing.}} \frac{1}{4\pi}\int_{-\infty}^{\infty} \frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)}m^{-it}n^{it} \overline{\phi_{\mathfrak{a,c}}\biggl(m,\, \frac12+it,\, \chi\biggr)} \phi_{\mathfrak{b,c}}\biggl(n,\, \frac12+it,\, \chi\biggr)\, dt.
\end{equation}
\tag{2.49}
$$
2.6. Gauss sumsFor a Dirichlet character $\chi$ modulo $q$, the Gauss sum of $\chi$ is defined by
$$
\begin{equation}
g(\chi;q;m):=\sum_{\substack{u\ (\operatorname{mod}q)\\(u,q)=1}}\chi(u) e\biggl(\frac{mu}{q} \biggr), \qquad \tau(\chi):=g(\chi;q;1).
\end{equation}
\tag{2.50}
$$
Let $\chi$ be a character modulo $q$ induced from a primitive character $\chi^*$ modulo $q^*$. Then according to [17], Lemma 3.1.3, (2), we have
$$
\begin{equation}
g(\chi;q;m)=\tau(\chi^*)\sum_{d\,|\,(m,q/q^*)}d\chi^* \biggl( \frac{q}{q^*d}\biggr) \overline{\chi^*} \biggl( \frac{m}{d}\biggr)\mu\biggl(\frac{q}{q^{*}d}\biggr).
\end{equation}
\tag{2.51}
$$
The generalized quadratic Gauss sums is given by
$$
\begin{equation}
G(a,n;q):=\sum_{x\ (\operatorname{mod}q)}e\biggl(\frac{ax^2+nx}{q} \biggr), \qquad (a,q)=1.
\end{equation}
\tag{2.52}
$$
Let $G(a;q):=G(a,0;q)$. The notation $\overline{a}_q$ means that $\overline{a}_qa\equiv 1\ (\operatorname{mod}q)$. Proof. If $q$ is odd, we have by (23) in [18]
$$
\begin{equation}
G^2(a;q)=\biggl(\biggl( \frac{a}{q}\biggr)i^{((q-1)/2)^2}\sqrt{q} \biggr)^2 = q(-1)^{((a-1)/2)^2}=q\chi_4(q).
\end{equation}
\tag{2.54}
$$
If $q\equiv 2\ (\operatorname{mod}4)$, then using (25) in [18] we find that $G(a;q)=0$. Finally, if $q\equiv 0$ $ (\operatorname{mod}4)$, then (25) in [18] implies that For a proof, see (26) in [18]. Lemma 2.5. Suppose that $n$ and $q$ are even. If $q\equiv2\ (\operatorname{mod}4)$, then $G(a,n;q)=0$. If $q\equiv0\ (\operatorname{mod}4)$, then Proof. Relation (2.57) follows from formula (26) in [18] for any even $n$ and $q$. However, $G(a;q)=0$ for $q\equiv2\ (\operatorname{mod}4)$ by (25) in [18]. Proof. In this case, according to (27) in [18] we have From the relation we have This completes the proof of the lemma. Lemma 2.7. Let $n$ be odd and $q$ be even. If $q\equiv 0\ (\operatorname{mod}4)$, then $G(a,n;q)=0$. Otherwise, we can write $q=2r$ with $r$ odd. In this case, Proof. If $n$ is odd and $q\equiv 0\ (\operatorname{mod}4)$, then $G(a,n;q)=0$ by (28) in [18]. Assume that $q=2r$ with $r$ odd. Using the twisted multiplicativity of Gauss sums, we find
$$
\begin{equation}
G(a,n;2r)=G(a\overline{2}_r,n\overline{2}_r;r) G(a\overline{r}_2,n\overline{r}_2;2).
\end{equation}
\tag{2.63}
$$
By direct computations $G(a\overline{r}_2,n\overline{r}_2;2)=2$. Finally,
$$
\begin{equation}
G(a\overline{2}_r,n\overline{2}_r;r)=G(2a,n;r)= e\biggl(-\frac{\overline{(8a)}_rn^2}{r} \biggr)G(2a;r),
\end{equation}
\tag{2.64}
$$
where we used (2.58) to evaluate $G(2a,n;r)$. § 3. Diagonal and non-diagonal termsAssuming that $s$ is sufficiently large, we prove in this section an explicit formula for the sum
$$
\begin{equation*}
\sum_{l=1}^{\infty}\omega(l)\mathscr{L}_{n^2-4l^2}(s),
\end{equation*}
\notag
$$
with diagonal and non-diagonal terms. Applying the results of § 2.6, we compute the diagonal term explicitly and prove an expression for the non-diagonal term in terms of sums of Kloosterman sums suitable for application of the Kuznetsov trace formula. Lemma 3.1. For $\operatorname{Re}{s}>3/2$, Proof. Using the Mellin transform for the function $\omega$, we have
$$
\begin{equation}
\sum_{l=1}^{\infty}\omega(l)\mathscr{L}_{n^2-4l^2}(s)=\frac{1}{2\pi i} \int_{(a)}\widehat{\omega}(\alpha) \sum_{l=1}^{\infty} \frac{\mathscr{L}_{n^2-4l^2}(s)}{l^{\alpha}}\, d\alpha,
\end{equation}
\tag{3.3}
$$
where $a>1$. According to (4.9) in [8], the function $\mathscr{L}_{n^2-4l^2}(s)$ can be written in terms of sums of Kloosterman sums
Substituting this expression into (3.3) we can change the order of summation, summing with respect to $q$ first and over $l$ second, as long as $\operatorname{Re}{s}>3/2$. Furthermore, dividing the range of summation for $l$ into arithmetic progressions, we obtain
$$
\begin{equation}
\begin{aligned} \, \sum_{l=1}^{\infty}\omega(l)\mathscr{L}_{n^2-4l^2}(s) &= \frac{\zeta(2s)}{2\pi i} \int_{(a)}\widehat{\omega}(\alpha) \sum_{q=1}^{\infty}\frac{1}{q^{1+s}} \nonumber \\ &\qquad \times \sum_{c,d\ (\operatorname{mod}q)}S(d^2,c^2;q) e\biggl( \frac{nc}{q}\biggr)\sum_{\substack{l\geqslant 1\\ l\equiv d\ (\operatorname{mod}q)}}\frac{1}{l^{\alpha}}\, d\alpha. \end{aligned}
\end{equation}
\tag{3.5}
$$
The inner sum on the right-hand side of (3.5) can be written in terms of the Lerch zeta function $\zeta(a,b;\alpha)$ with the following functional equation
$$
\begin{equation}
\begin{aligned} \, &\sum_{\substack{l\geqslant 1\\ l\equiv d\ (\operatorname{mod}q)}} \frac{1}{l^{\alpha}} =\frac{1}{q^{\alpha}}\sum_{l=1}^{\infty} \frac{1}{(l+d/q)^{\alpha}} =\frac{\zeta(d/q,0;\alpha)}{q^{\alpha}} \nonumber \\ &\qquad=\frac{\Gamma(1-\alpha)}{q^{\alpha}(2\pi)^{1-\alpha}} \biggl[-ie\biggl(\frac{\alpha}4\biggr)\sum_{l=1}^{\infty} \frac{e(ld/q)}{l^{1-\alpha}} +ie\biggl(-\frac{\alpha}4\biggr) \sum_{l=1}^{\infty}\frac{e(-ld/q)}{l^{1-\alpha}}\biggr]. \end{aligned}
\end{equation}
\tag{3.6}
$$
In order to apply this functional equation, we move the contour of integration in (3.5) to $\operatorname{Re}{\alpha}<0$, crossing a simple pole of $\zeta(d/q,0;\alpha)$ at $\alpha=1$. The contribution of this pole is equal to
$$
\begin{equation}
\widehat{\omega}(1)\zeta(2s)\sum_{q=1}^{\infty}\frac{1}{q^{2+s}} \sum_{c,d\ (\operatorname{mod}q)}S(d^2,c^2;q)e\biggl(\frac{nc}{q} \biggr).
\end{equation}
\tag{3.7}
$$
Next, applying the functional equation (3.6) and using the fact that
$$
\begin{equation}
ie\biggl(-\frac{\alpha}4\biggr)-ie\biggl(\frac{\alpha}4\biggr)= 2\sin\biggl(\frac{\pi \alpha}{2}\biggr),
\end{equation}
\tag{3.8}
$$
we find that (3.5) is equal to (3.7) plus
$$
\begin{equation}
\begin{aligned} \, &\frac{\zeta(2s)}{2\pi i} \int_{(a)}\widehat{\omega}(\alpha) \sum_{q=1}^{\infty}\frac{1}{q^{1+s+\alpha}}\sum_{l=1}^{\infty} \frac{1}{l^{1-\alpha}} \nonumber \\ &\qquad \times \sum_{c,d\ (\operatorname{mod}q)}S(d^2,c^2;q) e\biggl( \frac{nc+ld}{q}\biggr)\frac{2\Gamma(1-\alpha) \sin(\pi \alpha/2)}{(2\pi)^{1-\alpha}}\, d\alpha, \qquad a<0. \end{aligned}
\end{equation}
\tag{3.9}
$$
It follows from (5.5.5), (5.5.3) in [16]
$$
\begin{equation}
\Gamma(1-\alpha)\sin\biggl(\frac{\pi \alpha}2\biggr) = \pi^{1/2}2^{-\alpha}\frac{\Gamma(1/2-\alpha/2)}{\Gamma(\alpha/2)}.
\end{equation}
\tag{3.10}
$$
Substituting this into (3.9), we prove the lemma. 3.1. The inner sumIn order to evaluate the inner sum in (3.1), we express it as a sum of a product of two Gauss sums:
$$
\begin{equation}
\begin{aligned} \, K(n,l;q) &:=\sum_{c,d\ (\operatorname{mod}q)}S(c^2,d^2,q) e\biggl( \frac{nc+ld}{q}\biggr) \nonumber \\ &\,=\sum_{\substack{a,b\ (\operatorname{mod}q)\\ab\equiv 1\ (\operatorname{mod}q)}}\sum_{c,d\ (\operatorname{mod}q)} e\biggl(\frac{ac^2+db^2}{q} \biggr) e\biggl(\frac{nc+ld}{q} \biggr) \nonumber \\ &\,=\sum_{\substack{a,b\ (\operatorname{mod}q)\\ab\equiv 1\ (\operatorname{mod}q)}}G(a,n;q)G(b,l;q). \end{aligned}
\end{equation}
\tag{3.11}
$$
It is required to examine different cases depending on the parity of the parameters $n$, $l$, and $q$. To this end, we use the results of § 2.6. Proof. There are four different cases to consider depending on the parity of $n$ and $l$. Assume first that $n$ and $l$ are even. Using (3.11) and (2.56), we have
$$
\begin{equation}
K(n,l;q)=\sum_{\substack{a,b\ (\operatorname{mod}q)\\ab\equiv1\ (\operatorname{mod}q)}}e\biggl(-\frac{\overline{4}_q bn^2+ \overline{4}_q al^2}{q} \biggr)G(a;q)G(b;q).
\end{equation}
\tag{3.13}
$$
Note that $G(a;q)=G(b;q)$ since $ab\equiv 1\ (\operatorname{mod}q)$. In this case, the required result is secured by Lemma 2.3 with even $n$ and $l$. The other three cases can be treated similarly by evaluating $G(a,n;q)$ via (2.56) if $n$ is even and via (2.58) if $n$ is odd. Proof. According to (3.11),
$$
\begin{equation}
K(n,l;q)=\sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv 1\ (\operatorname{mod}q)}}G(a,n;q)G(b,l;q),
\end{equation}
\tag{3.15}
$$
where one of the parameters $n$ and $l$ is even and another one is odd. Without loss of generality, we can assume that $n$ is even and $l$ is odd. Now Lemma 2.5 implies that $G(a,n;q)$ is non-zero only if $q\equiv0\ (\operatorname{mod}4)$, but in that case $G(b,l;q)=0$ by Lemma 2.7. Lemma 3.4. Suppose that $q$, $n$ and $l$ are even. Then $K(n,l;q)=0$ if $q\equiv 2$ $ (\operatorname{mod}4)$, and otherwise Proof. It follows directly from (3.11) and Lemma 2.5 that $K(n,l;q)\neq 0$ only if $q\equiv 0\ (\operatorname{mod}4)$. In that case, (3.16) is a consequence of Lemmas 2.5 and 2.3. Lemma 3.5. Suppose that $q$ is even and $n$, $l$ are odd. Then $K(n,l;q)=0$ if $q\equiv 0$ $ (\operatorname{mod}4)$. If $q\equiv 2\ (\operatorname{mod}4)$, then $r:=q/2$ is odd and Proof. If $q\equiv 0\ (\operatorname{mod}4)$, then (3.11) and Lemma 2.7 imply that $K(n,l;q)=0$. In the opposite case, it follows from (2.62) and Lemma 2.3 that
$$
\begin{equation}
K(n,l;q)=2q\chi_4(r)\sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv1\ (\operatorname{mod}q)}}e\biggl(-\frac{\overline{(8a)}_r n^2+ \overline{(8b)}_r l^2}{r} \biggr),
\end{equation}
\tag{3.18}
$$
where $r:=q/2$ is odd. This is equal to
$$
\begin{equation}
K(n,l;q)=2q\chi_4(r) S\bigl(2\overline{(8)}_r n^2,2\overline{(8)}_r l^2;q\bigr).
\end{equation}
\tag{3.19}
$$
Writing $q=2r$ and applying the multiplicity property of Kloosterman sums, we infer
$$
\begin{equation*}
K(n,l;q)=2q\chi_4(r)S\bigl(\overline{(8)}_r n^2,\overline{(8)}_r l^2;r\bigr) S\bigl(2\overline{(8)}_r \overline{(r)}_2 n^2,2\overline{(8)}_r \overline{(r)}_2 l^2;2\bigr).
\end{equation*}
\notag
$$
Now the required assertion follows by using the fact that
$$
\begin{equation*}
S\bigl(2\overline{(8)}_r \overline{(r)}_2 n^2,2\overline{(8)}_r \overline{(r)}_2 l^2;2\bigr)=1,
\end{equation*}
\notag
$$
proving Lemma 3.5. 3.2. The diagonal termNow we are ready to evaluate the diagonal main term in (3.1), namely
$$
\begin{equation}
M^D(n,s):=\widehat{\omega}(1)\zeta(2s)\sum_{q=1}^{\infty} \frac{1}{q^{2+s}}K(n,0;q).
\end{equation}
\tag{3.20}
$$
We use the subscripts “even” and “odd” in $M^{\mathrm{D}}_{\mathrm{even}}(n,s)$ and $M^{\mathrm{D}}_{\mathrm{odd}}(n,s)$ to mark the parity of $n$. Proof. Assume first that $n$ is even. The the sum over $q$ in (3.20) is split as follows:
$$
\begin{equation}
\sum_{q=1}^{\infty}=\sum_{q\equiv 0\ (\operatorname{mod}2)}+ \sum_{q\equiv 1\ (\operatorname{mod}2)}.
\end{equation}
\tag{3.23}
$$
For $n$ and $q$ even, we have $K(n,0;q)=0$ unless $q\equiv 0\ (\operatorname{mod}4)$. If $q\equiv 0\ (\operatorname{mod}4)$ the following identity holds $\chi_4(a)=\chi_4(b)$ for $ab\equiv 1\ (\operatorname{mod}q)$. Then by Lemma 3.4
$$
\begin{equation}
\frac{K(n,0;q)}{2iq}=\sum_{b\ (\operatorname{mod}q)}^{*}\chi_4(b) e\biggl(-\frac{b(n/2)^2}{q} \biggr)= \chi_4(-1)g\biggl(\chi_4;q;\biggl(\frac{n}2\biggr)^2\biggr),
\end{equation}
\tag{3.24}
$$
where the star over the sum above means that we are summing over $b\ (\operatorname{mod}q)$ such that $(b,q)=1$. Consequently, using (2.51) we find that
$$
\begin{equation}
\begin{aligned} \, &\sum_{q \equiv 0\ (\operatorname{mod}2)}\frac{K(n,0;q)}{q^{2+s}}= 2i\chi_4(-1)\sum_{q\equiv 0\ (\operatorname{mod}4)} \frac{g(\chi_4;q;(n/2)^2)}{q^{1+s}} \nonumber \\ &\qquad=\frac{2i\chi_4(-1)}{4^{1+s}}\tau(\chi_4)\sum_{q=1}^{\infty} \frac{1}{q^{1+s}} \sum_{d\,|\,((n/2)^2,q)}d\chi_4 \biggl(\frac{q}{d}\biggr) \chi_4\biggl(\frac{(n/2)^2}{d}\biggr) \mu\biggl(\frac{q}{d}\biggr). \end{aligned}
\end{equation}
\tag{3.25}
$$
Computing the sum over $q$, we obtain
$$
\begin{equation}
\begin{aligned} \, &\sum_{q \equiv 0\ (\operatorname{mod}2)}\frac{K(n,0;q)}{q^{2+s}} = \frac{2i\chi_4(-1)\tau(\chi_4)}{4^{1+s}L(\chi_4,1+s)} \sum_{d\,|\,(n/2)^2}d^{-s}\chi_4 \biggl(\frac{(n/2)^2}{d}\biggr) \nonumber \\ &\qquad=\frac{2i\chi_4(-1)\tau(\chi_4)}{4^{1+s}L(\chi_4,1+s)} \biggl(\frac{n}{2} \biggr)^{-2s} \sigma_s \biggl(\chi_4;\biggl(\frac{n}2\biggr)^2\biggr) = \frac{\sigma_s(\chi_4;(n/2)^2)}{L(\chi_4,1+s)}n^{-2s}. \end{aligned}
\end{equation}
\tag{3.26}
$$
Now consider the sum over odd $q$ in (3.23). Applying Lemma 3.2 to compute $K(n,0;q)$ and making the change of variables $-\overline{4}_qb\to b$, we evaluate the second sum as follows:
$$
\begin{equation}
\begin{aligned} \, \sum_{q \equiv 1\ (\operatorname{mod}2)}\frac{K(n,0;q)}{q^{2+s}} &= \sum_{q=1}^{\infty}\frac{\chi_4(q)}{q^{1+s}} \sum_{b\ (\operatorname{mod}q)} e\biggl(\frac{bn^2}{q} \biggr) = \sum_{q=1}^{\infty}\frac{\chi_4(q)}{q^{1+s}} \sum_{d\,|\,(q,n^2)}d\mu\biggl(\frac{q}{d} \biggr) \nonumber \\ &=\sum_{d\,|\,n^2}d\sum_{q\equiv 0\ (\operatorname{mod}d)} \frac{\chi_4(q)\mu(q/d)}{q^{1+s}} = \frac{\sigma_{-s}(\chi_4;n^2)}{L(\chi_4,1+s)}. \end{aligned}
\end{equation}
\tag{3.27}
$$
Combining (3.27) and (3.26) and using the fact that $\sigma_s(\chi_4;(n/2)^2)=\sigma_s(\chi_4;n^2)$, we arrive at (3.21). Similarly, for odd $n$, the identity (3.22) follows from 3.3. The non-diagonal termIn this subsection, we study the non-diagonal term in (3.1), namely
$$
\begin{equation}
M^{\mathrm{ND}}:=\frac{\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty} \frac{1}{l^s}\sum_{q=1}^{\infty} \frac{f(\omega,s;4\pi n l/q)}{q^{2}} K(n,l;q).
\end{equation}
\tag{3.29}
$$
For simplicity, let $\psi(x):=f(\omega,s;4x)$. Consider the following two Kloosterman sums (see, for example, (2.20), (2.23) in [10]):
$$
\begin{equation}
S_{\infty,0}\bigl(m,n;c\sqrt{N};\chi\bigr) = \overline{\chi}(c)S(\overline{N}m,n;c), \qquad (c,N)=1,
\end{equation}
\tag{3.30}
$$
$$
\begin{equation}
S_{\infty,\infty}(m,n;c;\chi) = \sum_{ab\equiv 1\ (\operatorname{mod}c)} e\biggl(\frac{am+bn}{c} \biggr)\overline{\chi}(b).
\end{equation}
\tag{3.31}
$$
Using (2.18) and (2.15) in [10] we find that
$$
\begin{equation}
C_{\infty,\infty}(4) =\{\gamma=q> 0, \ q\equiv 0\ (\operatorname{mod}4)\},
\end{equation}
\tag{3.32}
$$
Lemma 3.7. If $n$ is even (let $n_1:=n/2$), then Proof. First, assume that $n$ is even. By applying Lemmas 3.2–3.4, we have
$$
\begin{equation}
\begin{aligned} \, &M^{\mathrm{ND}}= \frac{\zeta(2s)}{\pi^{s-1/2}} \sum_{l\equiv 0\ (\operatorname{mod}2)}\frac{1}{l^s} \sum_{q\equiv 0\ (\operatorname{mod}4)} \frac{2i}{q}f\biggl(\frac{4\pi nl}{q} \biggr) \nonumber \\ &\qquad\qquad \times \sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv 1\ (\operatorname{mod}q)}}\chi_4(a) e\biggl(-\frac{a(l/2)^2+b(n/2)^2}{q} \biggr) \nonumber \\ &\qquad+\frac{\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{q=1}^{\infty}\frac{\chi_4(q)}{q}f\biggl(\frac{4\pi nl }{q} \biggr) \sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv 1\ (\operatorname{mod}q)}}e\biggl( \frac{\overline{4}_q bn^2+ \overline{4}_q al^2}{q}\biggr). \end{aligned}
\end{equation}
\tag{3.38}
$$
Since $\chi_4(a)=\chi_4(b)$ for $q\equiv 0\ (\operatorname{mod}4)$, the first summand in (3.38) is equal to
$$
\begin{equation*}
-\frac{2i\zeta(2s)}{\pi^{s-1/2}}\sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{1}{l^s}\sum_{\gamma \in C_{\infty, \infty}(4)}\frac{1}{\gamma} \psi\biggl(\frac{\pi ln}{\gamma} \biggr) S_{\infty\infty} \biggl(\frac{l^2}{4},\frac{n^2}{4};\gamma;\chi_4\biggr).
\end{equation*}
\notag
$$
Using (3.30), we obtain
$$
\begin{equation}
\chi_4(q)\sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv 1\ (\operatorname{mod}q)}}e\biggl( \frac{\overline{4}_q bn^2+ \overline{4}_q al^2}{q}\biggr)= S_{\infty 0}\bigl(l^2,n_{1}^{2};q\sqrt{4};\chi_4\bigr).
\end{equation}
\tag{3.39}
$$
Therefore, the second summand in (3.38) is equal to
$$
\begin{equation*}
\frac{2\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{\gamma\in C_{\infty, 0}(4)}\frac{1}{\gamma}\psi \biggl(\frac{4\pi ln_1}{\gamma} \biggr)S_{\infty 0} (l^2,n_{1}^{2};\gamma;\chi_{4}).
\end{equation*}
\notag
$$
This completes the proof of (3.36). Now we assume that $n$ is odd. Using Lemmas 3.2, 3.3, 3.5 we obtain
$$
\begin{equation}
\begin{aligned} \, &M^{\mathrm{ND}}=\frac{\zeta(2s)}{\pi^{s-1/2}}\sum_{(l,2)=1} \frac{1}{l^s}\sum_{q\equiv 2\ (\operatorname{mod}4)}\frac{2\chi_4(q/2)}{q} f\biggl(\frac{4\pi nl}{q} \biggr) \nonumber \\ &\qquad\qquad\qquad \times \sum_{\substack{a,b\ (\operatorname{mod}r)\\ ab\equiv 1\ (\operatorname{mod}r)}}e\biggl(\frac{a\overline{8}_r n^2+ b\overline{8}_r l^2}{r} \biggr) \nonumber \\ &\qquad+\frac{\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{q=1}^{\infty}\frac{\chi_4(q)}{q}f\biggl(\frac{4\pi nl}{q} \biggr) \sum_{\substack{a,b\ (\operatorname{mod}q)\\ ab\equiv 1\ (\operatorname{mod}q)}}e\biggl( \frac{\overline{4}_q bn^2+ \overline{4}_q al^2}{q}\biggr), \end{aligned}
\end{equation}
\tag{3.40}
$$
where $r=q/2$. Note that $\chi_4$ can be extended to $\chi_{16}$ as follows
$$
\begin{equation}
\chi_{16}(q)=\begin{cases} \chi_4(q) &\text{if }(q,16)=1, \\ 0 &\text{otherwise}. \end{cases}
\end{equation}
\tag{3.41}
$$
Consequently, $\chi_4(q)=\chi_{16}(q)$ for all $(q,2)=1$. Therefore, the second summand in (3.40) is equal to
$$
\begin{equation}
\frac{4\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{\gamma\in C_{\infty, 0}(16)}\frac{1}{\gamma}\psi \biggl(\frac{4\pi ln}{\gamma} \biggr)S_{\infty 0}(l^2,n^2;\gamma,\chi_{4}).
\end{equation}
\tag{3.42}
$$
In order to evaluate the first summand in (3.40), we split the sum over $l$ as
$$
\begin{equation*}
\sum_{(l,2)=1}=\sum_{l=1}^{\infty}-\sum_{l\equiv 0\ (\operatorname{mod}2)},
\end{equation*}
\notag
$$
which yields
$$
\begin{equation}
\begin{aligned} \, &\frac{8\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{\gamma\in C_{\infty,0}(64)}\frac{1}{\gamma}\psi \biggl(\frac{4\pi ln}{\gamma} \biggr)S_{\infty 0}(l^2,n^2;\gamma,\chi_{4}) \nonumber \\ &\qquad- \frac{4\zeta(2s)}{2^{s}\pi^{s-1/2}}\sum_{l=1}^{\infty}\frac{1}{l^s} \sum_{\gamma\in C_{\infty, 0}(16)}\frac{1}{\gamma}\psi \biggl(\frac{4\pi ln}{\gamma} \biggr)S_{\infty 0}(l^2,n^2;\gamma,\chi_{4}). \end{aligned}
\end{equation}
\tag{3.43}
$$
This completes the proof. § 4. Special functionsIn this section, we study the Bessel transforms $\psi_H(k)$ and $\psi_D(t)$ defined by (2.41) and (2.42) respectively. Our goal is to prove integral representations for these functions in terms of the Gauss hypergeometric function. First, recall that in our case $\kappa=1$ and $k\equiv \kappa \ (\operatorname{mod}2)$. Therefore, we can write $k=2m+1$ for $m \in \mathbf{N}$ and
$$
\begin{equation}
\psi_H(k) =4i^{2m+1}\int_{0}^{\infty}J_{2m}(x)\psi(x)\, \frac{dx}{x},
\end{equation}
\tag{4.1}
$$
$$
\begin{equation}
\psi_D(t) =\frac{2\pi i t}{\sinh(\pi t)}\int_{0}^{\infty}(J_{2it}(x)+ J_{-2it}(x))\psi(x)\, \frac{dx}{x},
\end{equation}
\tag{4.2}
$$
where for $a<0$ (see (3.2))
$$
\begin{equation}
\psi(x)=f(\omega,s;4x)=\frac{1}{2\pi i}\int_{(a)}\frac{\Gamma(1/2-\alpha/2)} {\Gamma(\alpha/2)}\widehat{\omega}(\alpha) \biggl( \frac{x}{n}\biggr)^{\alpha+s-1}\, d\alpha.
\end{equation}
\tag{4.3}
$$
The function $\psi(x)$ has another integral representation:
$$
\begin{equation}
\psi(x)=\frac{2}{\sqrt{\pi}}\biggl(\frac{x}{n}\biggr)^s\int_{0}^{\infty} \omega(y)\cos\biggl( \frac{2xy}{n}\biggr)\, dy
\end{equation}
\tag{4.4}
$$
(for a proof. see (1.7) in [2] and [2], p. 1984). Lemma 4.1. The following identity holds Proof. Substituting (4.4) into (4.1) we have
$$
\begin{equation}
\psi_H(k)=\frac{8i^{2m+1}}{\sqrt{\pi}\, n^s} \int_{0}^{\infty} \omega(y)\int_{0}^{\infty}J_{2m}(x)x^{s-1} \cos \biggl(x\, \frac{2y}{n} \biggr)\, dx\, dy.
\end{equation}
\tag{4.6}
$$
The outer integral over $y$ can be split in two parts: $\int_{0}^{\infty}=\int_{0}^{n/2}+\int_{n/2}^{\infty}$. For each of these parts we evaluate the inner integral over $x$. For $2y/n < 1$, we apply (6.699), (2) in [11] and Euler’s reflection formula, so that
$$
\begin{equation}
\begin{aligned} \, &\int_{0}^{\infty}J_{2m}(x)x^{s-1}\cos \biggl(x\, \frac{2y}{n} \biggr)\, dx = \frac{2^{s-1}(-1)^m\sin(\pi s/2)}{\pi} \nonumber \\ &\qquad \times \Gamma\biggl(\frac{s}2+m\biggr)\Gamma\biggl(\frac{s}2-m\biggr) F\biggl(\frac{s}{2}+m,\, \frac{s}{2}-m;\, \frac{1}{2};\, \biggl(\frac{2y}{n}\biggr)^2\biggr). \end{aligned}
\end{equation}
\tag{4.7}
$$
This gives the first summand in (4.5). The second summand is obtained similarly using (6.699), (2) in [11] and the fact that $2y/n\geqslant 1$. Lemma 4.2. The following identity holds Proof. Substituting (4.4) into (4.2) we show that
$$
\begin{equation}
\psi_D(t)=\frac{2\pi i t}{\sinh(\pi t)}\bigl(h_1(t)+h_1(-t)+h_2(t)\bigr),
\end{equation}
\tag{4.11}
$$
where
$$
\begin{equation*}
\begin{aligned} \, h_1(t) &=\frac{2}{\sqrt{\pi}} \, \frac{1}{n^s} \int_{n/2}^{\infty}\omega(y) \int_{0}^{\infty}J_{2it}(x)x^{s-1}\cos\biggl(\frac{2xy}{n} \biggr)\, dx\, dy, \\ h_2(t) &=\frac{2}{\sqrt{\pi}} \, \frac{1}{n^s} \int_{0}^{n/2}\omega(y) \int_{0}^{\infty}\bigl(J_{2it}(x)+J_{-2it}(x)\bigr)x^{s-1} \cos\biggl(\frac{2xy}{n} \biggr)\, dx\, dy. \end{aligned}
\end{equation*}
\notag
$$
Evaluating the inner integrals with respect to $x$ in $h_1(t)$ and $h_2(t)$ using (6.699), (2) in [11], we complete the proof. Lemma 4.3. We have Proof. The identity (4.12) follows directly from (4.10). In order to prove (4.13) and (4.14), we apply (15.4.6) in [16]. Accordingly, proving the lemma. Corollary 4.1. The following identity holds § 5. Fourier coefficients of Eisenstein seriesAs a preliminary step towards understanding the continuous spectrum arising from the Kuznetsov trace formula, we compute explicitly some Fourier coefficients of Eisenstein series for the groups $\Gamma_0(4)$, $\Gamma_0(16)$ and $\Gamma_0(64)$. To this end, it is required to determine a list of singular cusps for the considered groups and to compute various characters appearing as a part of Fourier coefficients. 5.1. Singular cusps for $\Gamma_0(4)$, $\Gamma_0(16)$ and $\Gamma_0(64)$Let $\sigma_{\mathfrak{a}}$ be a scaling matrix for a cusp $\mathfrak{a}$, and let $\lambda_{\mathfrak{a}}$ be defined by $\sigma_{\mathfrak{a}}^{-1}\lambda_{\mathfrak{a}}\sigma_{\mathfrak{a}} = \left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$. Recall that $\mathfrak{a}$ is called singular for $\chi$ if $\chi(\lambda_{\mathfrak{a}})=1$. It follows from Proposition 3.3 in [10] that if $\mathfrak{c}=1/\omega$ is a cusp of $\Gamma=\Gamma_0(N)$ and
$$
\begin{equation}
N=(N,\omega)N', \qquad \omega=(N,\omega)\omega', \qquad N'=(N',\omega)N'',
\end{equation}
\tag{5.1}
$$
then
$$
\begin{equation}
\lambda_{1/\omega}=\begin{pmatrix} 1-\omega N''& N'' \\ -\omega^2N''&1+\omega N'' \end{pmatrix}.
\end{equation}
\tag{5.2}
$$
Lemma 5.1. The following cusps are singular for $\Gamma_0(4)$: The following cusps are singular for $\Gamma_0(16)$:
$$
\begin{equation*}
0,\quad \frac12,\quad \frac14,\quad \frac18,\quad \frac1{12},\quad \infty.
\end{equation*}
\notag
$$
The following cusps are singular for $\Gamma_0(64)$: Proof. According to [10], Corollary 3.2, a complete set of representatives for the set of inequivalent cusps of $\Gamma=\Gamma_0(N)$ is given by $1/w=1/(uf)$, where $f$ runs over divisors of $N$ and $u$ runs modulo $(f,N/f)$ such that $u$ is coprime to $(f,N/f)$, where $u$ is chosen so that $(u,N)=1$ after adding a suitable multiple of $(f,N/f)$.
Consequently, the group $\Gamma_0(4)$ has the following inequivalent cusps:
$$
\begin{equation*}
\frac{1}{1}\sim 0,\quad \frac{1}{2},\quad \frac{1}{4}\sim \infty.
\end{equation*}
\notag
$$
For the group $\Gamma_0(16)$, there are six inequivalent cusps:
$$
\begin{equation*}
\frac{1}{1}\sim 0,\quad \frac{1}{2},\quad \frac{1}{4},\quad \frac{1}{8},\quad \frac{1}{12},\quad \frac{1}{16} \sim \infty;
\end{equation*}
\notag
$$
and for the group $\Gamma_0(64)$, there are twelve cusps:
$$
\begin{equation*}
\frac{1}{1}\sim 0,\quad \frac{1}{2},\quad \frac{1}{4},\quad \frac{1}{8},\quad \frac{1}{12},\quad \frac{1}{16},\quad \frac{1}{24},\quad \frac{1}{32},\quad \frac{1}{40},\quad \frac{1}{48},\quad \frac{1}{56},\quad \frac{1}{64}\sim \infty.
\end{equation*}
\notag
$$
In order to check which of these cusps are singular, we will employ (5.2). Accordingly, the cusp is singular if Verifying this condition, we find that, for $\Gamma_0(16)$ and $\Gamma_0(64)$, all the cusps listed above are singular. For $\Gamma_0(4)$ the cusps $0$, $\infty$ are singular, but $1/2$ is not because in this case $\chi_4(1+wN'')=\chi_4(3)=-1$. This proves the lemma.5.2. Computations with characters Lemma 5.2. Let $\mathfrak{c}=1/w$ be any cusp and $\mathfrak{a}=1/r$ be an Atkin–Lehner cusp of $\Gamma_0(N)$. Let us choose the scaling matrices as follows: Proof. For a proof of (5.6), see Lemma 3.5 in [10].
A direct computation shows that
$$
\begin{equation}
\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1}=\begin{pmatrix} *&* \\ *&(wA+C)\dfrac{1-\overline{s}s}{r}+WB+D \end{pmatrix}.
\end{equation}
\tag{5.10}
$$
This implies (5.7). If $\mathfrak{a}=\infty$, this formula can be simplified further by noting that in this case $r=N$, $s=1$, and therefore, Similarly, for $\mathfrak{a}=0$, we have $r=1$, $s=N$ and Now using the relation $wB+D\equiv 0\ (\operatorname{mod}N)$ in (5.7) we arrive at (5.9), thereby proving the lemma. Proof. If $\mathfrak{a}=\mathfrak{c}=\infty$, then $w=N=4$ and by (5.8) we have $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})=\chi_4(D)$.
If $\mathfrak{a}=\infty$, $\mathfrak{c}=0$, then $w=1$, and by (5.8) we have $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})=\chi_4(B+D)$. According to (5.6),
$$
\begin{equation}
\begin{cases} AD-BC=1, \\ C\equiv-A\ (\operatorname{mod}N). \end{cases}
\end{equation}
\tag{5.13}
$$
Therefore, $A(B+D)\equiv 1\ (\operatorname{mod}N)$ and $B+D\equiv \overline{A}\ (\operatorname{mod}N)\equiv - \overline{C}\ (\operatorname{mod}N)$. Consequently, $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})= \chi_4(- \overline{C})=\chi_4(- C).$
If $\mathfrak{a}=0$, $\mathfrak{c}=\infty$, then $w=N=4$, and by (5.9) we have $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})=\chi_4(C)$. If $\mathfrak{a}=\mathfrak{c}=0$, then $w=1$ and by (5.9) we have $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})=\chi_4(A+C)$. Using (5.6) we find that
$$
\begin{equation}
\begin{cases} AD-BC=1, \\ D\equiv-B\ (\operatorname{mod}N). \end{cases}
\end{equation}
\tag{5.14}
$$
Therefore, $D(A+C)\equiv 1\ (\operatorname{mod}N)$ and $\chi_4(\sigma_{\mathfrak{c}}\rho\sigma_{\mathfrak{a}}^{-1})= \chi_4(\overline{D})=\chi_4(D)$. This proves Corollary 5.1. Corollary 5.2. For the groups $\Gamma_0(16)$ and $\Gamma_0(64)$, Proof. Note that if $w\equiv 0\ (\operatorname{mod}4)$, then This is the case for all singular cusps of $\Gamma_0(16)$ and $\Gamma_0(64)$ except $\mathfrak{c}=1/1\sim 0$ and $\mathfrak{c}=1/2$.
For $\mathfrak{c}=1/1\sim 0$ all computations are exactly the same as in Corollary 5.1. Consider $\mathfrak{c}=1/2$. It follows from (5.8) that
$$
\begin{equation}
\chi_4(\sigma_{1/2}\rho\sigma_{\infty}^{-1})=\chi_4(2B+D).
\end{equation}
\tag{5.19}
$$
According to (5.6),
$$
\begin{equation}
\begin{cases} AD-BC=1, \\ C\equiv -2A\ (\operatorname{mod}N), \end{cases}
\end{equation}
\tag{5.20}
$$
which implies that $A(D+2B)\equiv 1\ (\operatorname{mod}N)$ and $\chi_4(\sigma_{1/2}\rho\sigma_{\infty}^{-1})=\chi_4(\overline{A})$. Since $A\overline{A}\equiv 1\ (\operatorname{mod}4)$ and $A\equiv -C/2\ (\operatorname{mod}N/2)$ we have
$$
\begin{equation}
\chi_4(\sigma_{1/2}\rho\sigma_{\infty}^{-1})=\chi_4(\overline{A})= \chi_4(A)=\chi_4\biggl(-\frac{C}2 \biggr).
\end{equation}
\tag{5.21}
$$
As a consequence of (5.9),
$$
\begin{equation}
\chi_4(\sigma_{1/2}\rho\sigma_{0}^{-1})=\chi_4(2A+C).
\end{equation}
\tag{5.22}
$$
According to (5.6),
$$
\begin{equation}
\begin{cases} AD-BC=1, \\ D\equiv -2B\ (\operatorname{mod}N). \end{cases}
\end{equation}
\tag{5.23}
$$
Therefore, $-B(2A+C)\equiv 1\ (\operatorname{mod}N)$ and
$$
\begin{equation}
\chi_4(\sigma_{1/2}\rho\sigma_{0}^{-1})=\chi_4(-\overline{B})= \chi_4(-B)=\chi_4\biggl(\frac{D}2 \biggr).
\end{equation}
\tag{5.24}
$$
This proves Corollary 5.2. 5.3. Fourier coefficientsIn this section, we assume that $m$ is positive. For brevity, we write
$$
\begin{equation}
\delta_{n}(m):=\begin{cases} 1 &\text{if }n\,|\,m, \\ 0 &\text{otherwise}, \end{cases}
\end{equation}
\tag{5.25}
$$
$$
\begin{equation}
s(m):=\frac{\sigma_{1-2s}(\chi_4;m)}{L(\chi_4,2s)},
\end{equation}
\tag{5.26}
$$
$$
\begin{equation}
t(m):=\frac{\tau(\chi_4)\sigma_{2s-1}(\chi_4;m)m^{1-2s}}{L(\chi_4,2s)},
\end{equation}
\tag{5.27}
$$
$$
\begin{equation}
\sum_{a\ (\operatorname{mod}m)}^{*}:= \sum_{\substack{a\ (\operatorname{mod}m)\\(a,m)=1}}.
\end{equation}
\tag{5.28}
$$
In this section, we will frequently use the following lemma 5.3, which follows directly from the Chinese remainder theorem. Lemma 5.3. For $(m,n)=1$, where $n\overline{n}_m\equiv 1\ (\operatorname{mod}m)$ and $m\overline{m}_n\equiv 1\ (\operatorname{mod}n)$. Now we are ready to compute the Fourier coefficients (2.27) for $N=4$, $16$, $64$ and $\mathfrak{a}=0, \infty$. We provide a complete proof for the case $N=64$ and $\mathfrak{a}=\infty$. The other cases are dealt with similarly. Lemma 5.4. Let $N=64$ and $\mathfrak{a}=\infty$. Then Proof. We need to evaluate (2.27) for $\mathfrak{a}=\infty$. Consequently, (3.20) in [10] can be simplified as follows for $r=N=64$:
$$
\begin{equation}
\begin{aligned} \, &\Gamma_{\infty}\setminus\sigma_{\mathfrak{c}}^{-1} \Gamma\sigma_{\mathfrak{\infty}}/\Gamma_{\infty} \nonumber \\ &=\left\{ \begin{pmatrix} *& * \\ C\sqrt{N''}&D\sqrt{N''} \end{pmatrix}\colon \begin{aligned} \, D\ (\operatorname{mod}C), \ (D,C)=1, \ (C,N)=f, \\ D\equiv -\overline{\dfrac{C}{f}}\, u\ \biggl(\operatorname{mod}\biggl(f,\dfrac{N}{f}\biggr)\biggr) \end{aligned} \right\}. \end{aligned}
\end{equation}
\tag{5.36}
$$
If $\mathfrak{c}=0$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(-C)$ by Corollary 5.2, and also
$$
\begin{equation*}
f=1, \qquad \biggl(f,\frac{N}{f}\biggr)=1, \qquad N''=64, \qquad D\equiv -\overline{C}\quad (\operatorname{mod}1).
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation}
\phi_{\infty,0}(m,s,\chi_4)=\sum_{(C,2)=1}\frac{\chi_4(-C)}{(8C)^{2s}} \sum_{D\ (\operatorname{mod}C)}^{*}e\biggl( \frac{mD}{C}\biggr).
\end{equation}
\tag{5.37}
$$
The condition $(C,2)=1$ can be omitted since $\chi_4(-C)=0$ if this does not hold. Furthermore, we use the following identity for the inner sum
$$
\begin{equation}
\sum_{D\ (\operatorname{mod}C)}^{*}e\biggl( \frac{mD}{C}\biggr)= \sum_{d\,|\,(m,C)}d\mu\biggl(\frac{C}{d}\biggr)
\end{equation}
\tag{5.38}
$$
and interchange the order of summations, getting
$$
\begin{equation}
\begin{aligned} \, \phi_{\infty,0}(m,s,\chi_4) &=\frac{\chi_4(-1)}{8^{2s}} \sum_{d\,|\,m}d\sum_{C\equiv0\ (\operatorname{mod}d)} \frac{\chi_4(C)}{C^{2s}}\mu\biggl(\frac{C}{d}\biggr) \nonumber \\ &=\frac{\chi_4(-1)}{8^{2s}}\sum_{d\,|\,m}\frac{\chi_4(d)}{d^{2s-1}} \sum_{C}\frac{\chi_4(C)}{C^{2s}}\mu(C)=\chi_4(-1)\frac{s(m)}{8^{2s}}. \end{aligned}
\end{equation}
\tag{5.39}
$$
If $\mathfrak{c}=1/2$, then $\chi_4(\sigma_{1/2}\rho\sigma_{\infty}^{-1}) =\chi_4(-\overline{C/2})$ by Corollary 5.2, and
$$
\begin{equation*}
f=2, \qquad \biggl(f,\frac{N}{f}\biggr)=2, \qquad N''=16, \qquad D\equiv -\overline{\frac{C}2}\quad (\operatorname{mod}2).
\end{equation*}
\notag
$$
This yields that
$$
\begin{equation}
\phi_{\infty,1/2}(m,s,\chi_4)=\sum_{(C,2)=1}\frac{\chi_4(-C)}{(8C)^{2s}} \sum_{D\ (\operatorname{mod}2C)}^{*}e\biggl( \frac{mD}{2C}\biggr).
\end{equation}
\tag{5.40}
$$
By Lemma 5.3,
$$
\begin{equation}
\begin{aligned} \, \sum_{D\ (\operatorname{mod}2C)}^{*}e\biggl( \frac{mD}{2C}\biggr) &= \sum_{d_1\ (\operatorname{mod}2)}^{*}\sum_{d_2(C)}^{*} e\biggl(\frac{md_1\overline{C}_2}{2} \biggr) e\biggl(\frac{md_2\overline{2}_C}{C} \biggr) \nonumber \\ &=e\biggl( \frac{m}{2}\biggr)\sum_{d_2(C)}^{*} e\biggl(\frac{md_2\overline{2}_C}{C} \biggr) = e\biggl( \frac{m}{2}\biggr)\sum_{d\,|\,(m,C)}d\mu\biggl(\frac{C}{d}\biggr). \end{aligned}
\end{equation}
\tag{5.41}
$$
Consequently,
$$
\begin{equation}
\phi_{\infty,1/2}(m,s,\chi_4)= \chi_4(-1)e\biggl(\frac{m}{2}\biggr)\frac{s(m)}{8^{2s}}.
\end{equation}
\tag{5.42}
$$
If $\mathfrak{c}=1/(4u)$ with $ u=1,3$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(D)$ by Corollary 5.2, and
$$
\begin{equation*}
f=4, \qquad \biggl(f,\frac{N}{f}\biggr)=4, \qquad N''=4, \qquad D\equiv -u\overline{\frac{C}4}\quad (\operatorname{mod}4).
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation}
\phi_{\infty,1/(4u)}(m,s,\chi_4)=\sum_{(C,2)=1} \frac{\chi_4(-u\overline{C})}{(8C)^{2s}} \sum_{\substack{D\ (\operatorname{mod}4C)\\D\equiv -u\overline{C}\ (\operatorname{mod}4)}}^{*}e\biggl( \frac{mD}{4C}\biggr).
\end{equation}
\tag{5.43}
$$
Applying Lemma 5.3, we infer
$$
\begin{equation}
\begin{aligned} \, \sum_{\substack{D\ (\operatorname{mod}4C)\\D\equiv -u\overline{C}\ (\operatorname{mod}4)}}^{*}e\biggl( \frac{mD}{4C}\biggr) &= \sum_{\substack{d_1\ (\operatorname{mod}4)\\ d_1\overline{C}_4\equiv -u\ (\operatorname{mod}4) }}e\biggl(\frac{md_1\overline{C}_4}{4} \biggr) \sum_{d_2\ (\operatorname{mod}C)}e\biggl(\frac{md_2}{C} \biggr) \nonumber \\ &=e\biggl( -\frac{mu}{4}\biggr)\sum_{d\,|\,(m,C)}d\mu \biggl(\frac{C}{d}\biggr). \end{aligned}
\end{equation}
\tag{5.44}
$$
This implies that, for $ u=1,3$,
$$
\begin{equation}
\phi_{\infty,1/(4u)}(m,s,\chi_4)= \chi_4(-u)e\biggl(-\frac{mu}{4}\biggr)\frac{s(m)}{8^{2s}}.
\end{equation}
\tag{5.45}
$$
If $\mathfrak{c}=1/(8u)$ with $ u=1,3,5,7$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(D)$ by Corollary 5.2, and
$$
\begin{equation*}
f=8, \qquad \biggl(f,\frac{N}{f}\biggr)=8, \qquad N''=1, \qquad D\equiv -u\,\overline{\frac{C}8}\quad (\operatorname{mod}8).
\end{equation*}
\notag
$$
Similarly to the previous cases
$$
\begin{equation}
\begin{aligned} \, &\phi_{\infty,1/(8u)}(m,s,\chi_4) =\sum_{(C,8)=1} \frac{\chi_4(-u\overline{C})}{(8C)^{2s}} \sum_{\substack{D\ (\operatorname{mod}8C)\\ D\equiv -u\overline{C}\ (\operatorname{mod}8)}}^{*} e\biggl( \frac{mD}{8C}\biggr) \nonumber \\ &\qquad= \chi_4(-u)e\biggl( -\frac{mu}{8}\biggr) \sum_{C=1}^{\infty}\frac{\chi_4(C)}{(8C)^{2s}} \sum_{d\,|\,(m,C)}d\mu\biggl(\frac{C}{d} \biggr) = \chi_4(-u)e\biggl(-\frac{mu}{8}\biggr)\frac{s(m)}{8^{2s}}. \end{aligned}
\end{equation}
\tag{5.46}
$$
If $\mathfrak{c}=1/(16u)$ with $u=1,3$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(D)$ by Corollary 5.2, and
$$
\begin{equation*}
f=16, \qquad \biggl(f,\frac{N}{f}\biggr)=4, \qquad N''=1, \qquad D\equiv -u\,\overline{\frac{C}{16}}\quad (\operatorname{mod}4).
\end{equation*}
\notag
$$
From this we derive that
$$
\begin{equation}
\phi_{\infty,1/(16u)}(m,s,\chi_4)=\sum_{(C,4)=1} \frac{\chi_4(-u\overline{C})}{(16C)^{2s}} \sum_{\substack{D\ (\operatorname{mod}16C)\\ D\equiv -u\overline{C}\ (\operatorname{mod}4)}}^{*} e\biggl( \frac{mD}{16C}\biggr).
\end{equation}
\tag{5.47}
$$
The inner sum can be evaluated by applying Lemma 5.3 as follows
$$
\begin{equation}
\begin{aligned} \, \sum_{\substack{D\ (\operatorname{mod}16C)\\ D\equiv -u\overline{C}\ (\operatorname{mod}4)}}^{*} e\biggl( \frac{mD}{16C}\biggr)&= \sum_{\substack{d_1\ (\operatorname{mod}16)\\ d_1\equiv-u\overline{C}_4\ (\operatorname{mod}4)}}^{*} \sum_{d_2\ (\operatorname{mod}C)}^{*} e\biggl(\frac{md_1\overline{C}_{16}}{16} \biggr) e\biggl(\frac{md_2\overline{16}_{C}}{C} \biggr) \nonumber \\ &= \sum_{\substack{d_1\ (\operatorname{mod}16)\\ Cd_1\equiv-u\overline{C}_4\ (\operatorname{mod}4)}}^{*} e\biggl(\frac{md_1}{16} \biggr) \sum_{d_2\ (\operatorname{mod}C)}^{*}e\biggl(\frac{md_2}{C} \biggr). \end{aligned}
\end{equation}
\tag{5.48}
$$
Note that $\overline{C}_4\equiv C\ (\operatorname{mod}4)$, and therefore, the condition $Cd_1\equiv-u\overline{C}_4\ (\operatorname{mod}4)$ can be replaced by $d_1\equiv -u\ (\operatorname{mod}4)$.
Making the change of variables $d_3:=-u+4d_3$, $d_3\ (\operatorname{mod}4)$ we infer
$$
\begin{equation}
\sum_{\substack{d_1\ (\operatorname{mod}16)\\ d_1\equiv -u\ (\operatorname{mod}4)}}e\biggl( \frac{md_1}{16}\biggr)= \sum_{d_3\ (\operatorname{mod}4)}e\biggl(-\frac{mu}{16}\biggr) e\biggl( \frac{md_3}{4}\biggr)=4\delta_{4}(m)e\biggl(-\frac{mu}{16}\biggr).
\end{equation}
\tag{5.49}
$$
Consequently,
$$
\begin{equation}
\phi_{\infty,1/(16u)}(m,s,\chi_4) = \chi_4(-u)4\delta_{4}(m)e\biggl(-\frac{mu}{16}\biggr)\frac{s(m)}{16^{2s}}.
\end{equation}
\tag{5.50}
$$
If $\mathfrak{c}=1/32$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(D)$ by Corollary 5.2, and
$$
\begin{equation*}
f=32, \qquad \biggl(f,\frac{N}{f}\biggr)=2, \qquad N''=1, \qquad D\equiv -u\,\overline{\frac{C}{32}}\quad (\operatorname{mod}2).
\end{equation*}
\notag
$$
In these settings,
$$
\begin{equation}
\phi_{\infty,1/32}(m,s,\chi_4)=\sum_{(C,2)=1}\frac{1}{(32C)^{2s}} \sum_{D\ (\operatorname{mod}32C)}^{*}\chi_4(D)e\biggl( \frac{mD}{32C}\biggr),
\end{equation}
\tag{5.51}
$$
where the inner sum is the Gauss sum $g(\chi_4;32C;m)$ defined by (2.50). Now an application of (2.51) shows that
$$
\begin{equation*}
\begin{aligned} \, &\phi_{\infty,1/32}(m,s,\chi_4) =\frac{\tau(\chi_4)}{(32)^{2s}} \sum_{d\,|\,m}d\chi_4\biggl(\frac{m}{d}\biggr) \sum_{\substack{(C,2)=1\\C\equiv0\ (\operatorname{mod}d/(8,d))}} \frac{\chi_4(8C/d)\mu(8C/d)}{C^{2s}} \\ &\qquad=\frac{\tau(\chi_4)}{(32)^{2s}}\sum_{\substack{d\,|\,m\\ (d/(8,d),2)=1}}d\chi_4 \biggl(\frac{m}{d}\biggr) \biggl(\frac{(8,d)}{d} \biggr)^{2s}\sum_{(C,2)=1} \frac{\chi_4(8C/(8,d))\mu(8C/(8,d))}{C^{2s}}. \end{aligned}
\end{equation*}
\notag
$$
Since $\chi_4(8C/(8,d))=0$ unless $(8,d)=8$, the expression above simplifies to
$$
\begin{equation*}
\begin{aligned} \, \phi_{\infty,1/32}(m,s,\chi_4) &=\frac{\tau(\chi_4)}{(32)^{2s}} \sum_{\substack{d\,|\,m\\d\equiv 0\ (\operatorname{mod}8)\\ (d/8,2)=1}}d\chi_4\biggl(\frac{m}{d} \biggr)\biggl(\frac{8}{d} \biggr)^{2s} \sum_{(C,2)=1}\frac{\chi_4(C)\mu(C)}{C^{2s}} \\ &=\frac{\tau(\chi_4)}{(32)^{2s}L(\chi_4,2s)}\, 8\delta_8(m) \sum_{\substack{d\,|\,m/8\\(d,2)=1}}d^{1-2s}\chi_4\biggl(\frac{m/8}{d}\biggr). \end{aligned}
\end{equation*}
\notag
$$
If $m/8$ is odd, then the condition $(d,2)=1$ can be removed, and if $m/8$ is even, we have $\chi_4((m/8)/d)=0$ for $d$ odd. Consequently,
$$
\begin{equation}
\phi_{\infty,1/32}(m,s,\chi_4)=\frac{8}{(32)^{2s}}\, \delta_{8}(m) (1-\delta_{16}(m))t \biggl(\frac{m}{8}\biggr).
\end{equation}
\tag{5.52}
$$
For $m \equiv 0 \ (\operatorname{mod}16)$, the following identity holds
$$
\begin{equation}
\biggl( \frac{m}{8}\biggr)^{1-2s}\sigma_{2s-1}\biggl(\chi_4;\frac{m}8\biggr)= 2^{1-2s}\biggl( \frac{m}{16}\biggr)^{1-2s}\sigma_{2s-1} \biggl(\chi_4;\frac{m}{16}\biggr),
\end{equation}
\tag{5.53}
$$
and, therefore,
$$
\begin{equation}
\phi_{\infty,1/32}(m,s,\chi_4)=\frac{8}{(32)^{2s}}\, \delta_{8}(m)t \biggl(\frac{m}{8}\biggr) -\frac{16}{(64)^{2s}}\, \delta_{16}(m) t\biggl(\frac{m}{16}\biggr).
\end{equation}
\tag{5.54}
$$
If $\mathfrak{c}=1/64$, then $\chi_4(\sigma_{0}\rho\sigma_{\infty}^{-1})=\chi_4(D)$ by Corollary 5.2, and
$$
\begin{equation*}
f=64, \qquad \biggl(f,\frac{N}{f}\biggr)=1, \qquad N''=1, \qquad D\equiv -u\, \overline{\frac{C}{64}}\quad (\operatorname{mod}1).
\end{equation*}
\notag
$$
This implies that
$$
\begin{equation}
\phi_{\infty,\infty}(m,s,\chi_4)=\sum_{C}\frac{1}{(64C)^{2s}} \sum_{D\ (\operatorname{mod}64C)}^{*}\chi_4(D)e\biggl( \frac{mD}{64C}\biggr).
\end{equation}
\tag{5.55}
$$
The inner sum is the Gauss sum $g(\chi_4;64C;m)$, as defined by (2.50). Applying the representation (2.51), for this sum, we have
$$
\begin{equation*}
\begin{aligned} \, &\phi_{\infty,\infty}(m,s,\chi_4)=\frac{\tau(\chi_4)}{(64)^{2s}} \sum_{C=1}^{\infty}\frac{1}{C^{2s}}\sum_{d\,|\,(m,16C)}d\chi_4 \biggl(\frac{16C}{d}\biggr) \chi_4\biggl(\frac{n}{d} \biggr) \mu\biggl( \frac{16C}{q}\biggr) \\ &\qquad=\frac{\tau(\chi_4)}{(64)^{2s}}\sum_{d\,|\,m}d\chi_4 \biggl( \frac{m}{d}\biggr) \biggl(\frac{(16,d)}{d} \biggr)^{2s} \sum_{C=1}^{\infty}\frac{\chi_4(16C/(16,d)) \mu(16C/(16,d))}{C^{2s}}. \end{aligned}
\end{equation*}
\notag
$$
We remark that $\chi_4(16C/(16,d))=0$ unless $(16,d)/d =1$. Therefore, we can assume that $d\equiv 0\ (\operatorname{mod}16)$ and
$$
\begin{equation*}
\begin{aligned} \, &\phi_{\infty,\infty}(m,s,\chi_4)=\frac{\tau(\chi_4)}{(64)^{2s}} \sum_{\substack{d\,|\,m\\d\equiv 0\ (\operatorname{mod}16)}}d\chi_4 \biggl( \frac{m}{d}\biggr)\biggl(\frac{16}{d} \biggr)^{2s} \sum_{C=1}^{\infty}\frac{\chi_4(C)\mu(C)}{C^{2s}} \\ &\qquad=\frac{\tau(\chi_4)}{(64)^{2s}}\delta_{16}(m)\sum_{d\,|\,m/16} \chi_4 \biggl(\frac{m/16}{d} \biggr)\frac{16d^{-2s+1}}{L(\chi_4,2s)} =\frac{16}{(64)^{2s}}\delta_{16}(m)t\biggl(\frac{m}{16}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Lemma 5.5. Let $N=64$ and $\mathfrak{a}=0$. Then Lemma 5.6. Let $N=16$ and $\mathfrak{a}=0$. Then Lemma 5.7. Let $N=16$ and $\mathfrak{a}=\infty$. Then Lemma 5.8. Let $N=4$. Then For $N=16$, Proof. The identity (5.76) is a consequence of (5.52), (5.34) and (5.57). Similarly, (5.77) follows from (5.71) and (5.64).
§ 6. Contribution of the continuous spectrumApplying the Kuznetsov trace formula (2.43) to the sums of Kloosterman sums in (3.36), we find that, for even $n$, the continuous spectrum (2.49) can be written as a sum of
$$
\begin{equation}
\begin{aligned} \, C^{1}_{\mathrm{even}} &:=-\frac{2i\zeta(2s)}{2^s\pi^{s-1/2}} \sum_{l=1}^{\infty}\frac{1}{l^s}\sum_{\mathfrak{c} \text{ sing. } \Gamma_0(4)} \frac{1}{4\pi}\int_{-\infty}^{\infty} \frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \nonumber \\ &\qquad \times l^{-2it} \overline{\phi_{\infty,\mathfrak{c}} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n^{2it}_{1} \phi_{\infty,\mathfrak{c}}\biggl(n^{2}_{1},\, \frac12+it,\, \chi_4\biggr)\, dt \end{aligned}
\end{equation}
\tag{6.1}
$$
and
$$
\begin{equation}
\begin{aligned} \, C^{2}_{\mathrm{even}} &:=\frac{2\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty} \frac{1}{l^s}\sum_{\mathfrak{c} \text{ sing. } \Gamma_0(4)} \frac{1}{4\pi}\int_{-\infty}^{\infty}\frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \nonumber \\ &\qquad \times l^{-2it} \overline{\phi_{\infty,\mathfrak{c}} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n_{1}^{2it}\phi_{0,\mathfrak{c}} \biggl(n_{1}^{2},\, \frac12+it,\, \chi_4\biggr)\, dt. \end{aligned}
\end{equation}
\tag{6.2}
$$
Similarly, applying (3.37) for odd $n$ it is required to investigate
$$
\begin{equation}
\begin{aligned} \, C^{1}_{\mathrm{odd}} &:=\frac{8\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty} \frac{1}{l^s}\sum_{\mathfrak{c} \text{ sing. }\Gamma_0(64)} \frac{1}{4\pi} \int_{-\infty}^{\infty}\frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \nonumber \\ &\qquad \times l^{-2it} \overline{\phi_{\infty,\mathfrak{c}}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n^{2it}\phi_{0,\mathfrak{c}} \biggl(n^2,\, \frac12+it,\, \chi_4\biggr)\, dt \end{aligned}
\end{equation}
\tag{6.3}
$$
and
$$
\begin{equation}
\begin{aligned} \, C^{2}_{\mathrm{odd}} &:=\frac{4\zeta(2s)}{\pi^{s-1/2}}\sum_{l=1}^{\infty} \frac{1}{l^s}\sum_{\mathfrak{c} \text{ sing. } \Gamma_0(16)} \frac{1}{4\pi} \int_{-\infty}^{\infty}\frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \nonumber \\ &\qquad \times (1-2^{-s}) l^{-2it} \overline{\phi_{\infty,\mathfrak{c}} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n^{2it}\phi_{0,\mathfrak{c}} \biggl(n^2,\, \frac12+it,\, \chi_4\biggr)\, dt. \end{aligned}
\end{equation}
\tag{6.4}
$$
In order to compute the sums over $l$ and $\mathfrak{c}$ in the expressions above we use the results of the previous section. Even and odd cases require separate treatment. 6.1. Even case Lemma 6.1. For $\operatorname{Re}{s}>1$, Proof. There are two nonequivalent singular cusps for $\Gamma_0(4)$: $0$ and $\infty$.
Consider first $\mathfrak{c}=\infty$. Applying (5.75), (5.27), (2.8) with $z:=s-2it$ and $s:=-2it$, we compute the sum over $l$ in (6.1):
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{1}{l^s} l^{-2it} \overline{\phi_{\infty,\mathfrak{\infty}}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n_{1}^{2it}\phi_{\infty,\mathfrak{\infty}} \biggl(n_{1}^{2},\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &=\frac{|\tau(\chi_4)|^2}{16}\, \frac{1-2^{-2it-s}}{1-2^{-2s}} n_{1}^{-2it}\sigma_{2it}(\chi_4;n_{1}^{2}) \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1+2it)L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.6}
$$
Now let us consider the case $\mathfrak{c}=0$. Using (5.74), (5.26), (2.8) with $z:=s+2it$ $s:=2it$, we infer
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{1}{l^s} l^{-2it} \overline{\phi_{\infty,0} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n_{1}^{2it}\phi_{\infty,0} \biggl(n_{1}^{2},\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad= \frac{1}{4} \frac{1-2^{2it-s}}{1-2^{-2s}}n_{1}^{2it} \sigma_{-2it} (\chi_4;n_{1}^{2}) \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1+2it)L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.7}
$$
To complete the proof it suffices to sum the last two expressions and note that $|\tau(\chi_4)|^2=4$. Lemma 6.2. For $\operatorname{Re}{s}>1$, the following identity holds Proof. Similarly to the previous lemma, the sum over $\mathfrak{c}$ in (6.2) contains only two summands: $0$ and $\infty$.
Let us start with $\mathfrak{c}=\infty$. Applying (5.75), (5.73), (5.27), (5.26), (2.8) with $z:=s-2it$ and $s:=-2it$, we obtain
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{1}{l^s} l^{-2it} \overline{\phi_{\infty,\mathfrak{\infty}}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n_{1}^{2it} \phi_{0,\mathfrak{\infty}} \biggl(n_{1}^{2},\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad=\frac{\overline{\tau(\chi_4)}}{2^{3-2it}}\, \frac{1-2^{-2it-s}}{1-2^{-2s}}n_{1}^{2it}\sigma_{-2it}(\chi_4;n_{1}^{2}) \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)}{\zeta(2s) L(\chi_4,1+2it)L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.9}
$$
Next let us consider $\mathfrak{c}=0$. Applying (5.74), (5.73), (5.27), (5.26), (2.8) with $z:=s+2it$ and $s:=2it$, we find that
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{1}{l^s} l^{-2it} \overline{\phi_{\infty,0} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n_{1}^{2it}\phi_{0,0} \biggl(n_{1}^{2},\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad=\frac{\tau(\chi_4)\chi_{4}(-1)}{2^{3+2it}}\, \frac{1-2^{2it-s}}{1-2^{-2s}}n_{1}^{-2it}\sigma_{2it}(\chi_4;n_{1}^{2}) \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)}{\zeta(2s)L(\chi_4,1+2it) L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.10}
$$
To complete the proof it suffices to sum the last two expressions and note that $\tau(\chi_4)=2i$ and $\chi_4(-1)=-1$. This proves Lemma 6.2. Using (2.3), as a direct consequence of Lemmas 6.1 and 6.2, we obtain the following corollary. Corollary 6.1. For $\operatorname{Re} s>1$, Finally, we extend this result to the critical strip using the following lemma. Lemma 6.3. Suppose that the function $F(s)$ is defined for $\operatorname{Re}{s}>1$ by where $f(s,z)$ has two simple poles at the points $z_{1}=1-s$ and $z_2=s-1$. Then, for $\operatorname{Re}{s}<1$, Proof. Assume that $1<\operatorname{Re}{s}<1+\epsilon/2$ and $\operatorname{Im}{s}>0$. Consider the new integration contour where $C_{\epsilon}^{-}$ is a semicircle in the left half-plane of radius $\epsilon$, and $C_{\epsilon}^{+}$ is a semicircle in the right half-plane of radius $\epsilon$. By changing the contour of integration to $\gamma_1$ we cross poles at the points $z_1$, $z_2$. Therefore, Now if $\operatorname{Re}{s}<1$, we can change the contour back to $\operatorname{Re}{z}=0$. This completes the proof. Lemma 6.4. For $0<\operatorname{Re}{s}<1$, $s\neq 1/2$, Proof. Making the change of variables $z:=2it$ in (6.11), we have
$$
\begin{equation}
\begin{aligned} \, C^{1}_{\mathrm{even}}\,{+}\,C^{2}_{\mathrm{even}} &= \frac{L(\chi_4,s)}{4\pi^{s-1/2}}\, \frac{1}{2\pi i} \! \int_{(0)}\!\! \frac{\psi_D(z/(2i))\sinh(\pi z/(2i))}{z\cosh(\pi z/(2i))} \, \frac{\zeta(s+z)\zeta(s-z)}{L(\chi_4,1+z)L(\chi_4,1-z)} \nonumber \\ &\qquad\times \bigl( n^{z}\sigma_{-z}(\chi_4;n^2)+ n^{-z}\sigma_{z}(\chi_4;n^2) \bigr)\, dt. \end{aligned}
\end{equation}
\tag{6.16}
$$
Now (6.15) follows from Lemma 6.3 and (4.17). 6.2. Odd case Lemma 6.5. For $\operatorname{Re}{s}>1$, Proof. Let us decompose the sum over $l$ in (6.3) as follows so that $C^{1}_{\mathrm{odd}}=C^{1,1}_{\mathrm{odd}}+C^{1,2}_{\mathrm{odd}}$.
Consider first $C^{1,1}_{\mathrm{odd}}$. By Lemma 5.5, for odd $n$ we have
$$
\begin{equation}
\phi_{0,\mathfrak{c}}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr)=0, \qquad \mathfrak{c}=0, \, \frac{1}{2},\, \frac{1}{4},\, \frac{1}{12}.
\end{equation}
\tag{6.18}
$$
If $l$ is odd, then, by Lemma 5.4
$$
\begin{equation}
\phi_{\infty,\mathfrak{c}}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)=0, \qquad \mathfrak{c}=\infty,\, \frac{1}{32},\, \frac{1}{16},\, \frac{1}{48}.
\end{equation}
\tag{6.19}
$$
It remains to consider the case $\mathfrak{c}=1/(8u)$, $u=1,3,5,7$. Note that $e(n^2u/8)=e(u/8)$ for odd $n$. Consequently, applying (5.26), (5.32) and (5.59) we obtain
$$
\begin{equation}
\begin{aligned} \, &\sum_{(l,2)=1}\frac{1}{l^{s+2it}} \overline{\phi_{\infty,1/(8u)} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n^{2it}\phi_{0,1/(8u)} \biggl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad= \frac{n^{2it}\sigma_{-2it}(\chi_4;n^2)\chi_4(-u)e(u/4)} {64L(\chi_4,1+2it)L(\chi_4,1-2it)} \sum_{(l,2)=1} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.20}
$$
Using (2.8) with $z:=s+2it$, $s:=2it$ we show that
$$
\begin{equation}
\begin{aligned} \, \sum_{(l,2)=1}\frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} &=(1-2^{-s-2it}) \sum_{l=1}^{\infty}\frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} \nonumber \\ &=(1-2^{-s-2it})\frac{1-2^{2it-s}}{1-2^{-2s}}\, \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)}{\zeta(2s)}. \end{aligned}
\end{equation}
\tag{6.21}
$$
Substituting (6.21) into (6.20), summing over $u=1,3,5,7$ and using the identity
$$
\begin{equation}
\sum_{u=1,3,5,7}\chi_4(-u)e\biggl(\frac{u}4\biggr)=-4i,
\end{equation}
\tag{6.22}
$$
we infer
$$
\begin{equation}
\begin{aligned} \, C^{1,1}_{\mathrm{odd}} &=\frac{L(\chi_4,s)}{4\pi^{s-1/2}(1-2^{-2s})} \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \sigma_{-2it}(\chi_4;n^2) \nonumber \\ &\qquad\times n^{2it} \frac{\zeta(s+2it)\zeta(s-2it)}{L(\chi_4,1+2it)L(\chi_4,1-2it)} (1-2^{2it-s})(1-2^{-2it-s})\, dt. \end{aligned}
\end{equation}
\tag{6.23}
$$
Now let us evaluate $C^{1,2}_{\mathrm{odd}}$. In view of (6.18) we have to consider the following cases: $\mathfrak{c}=1/(8u)$, $u=1,3,5,7$ and $\mathfrak{c}=1/16, 1/48$, $\mathfrak{c}=1/32$, $\mathfrak{c}=\infty$. Assume that $\mathfrak{c}=1/(8u)$. Using the fact that $e(n^2u/8)=e(u/8)$ for odd $n$, and applying (5.26), (5.32) and (5.59) we obtain
$$
\begin{equation}
\begin{aligned} \, &\sum_{l\equiv 0\ (\operatorname{mod}2)}\frac{1}{l^{s+2it}} \overline{\phi_{\infty,1/(8u)}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} n^{2it}\phi_{0,1/(8u)}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad= \frac{n^{2it}\sigma_{-2it}(\chi_4;n^2)\chi_4(-u)e(u/8)} {64L(\chi_4,1+2it)L(\chi_4,1-2it)} \sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{e(l^2u/8)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.24}
$$
Note that $e(m^2u/2)=1$ if $m$ is even and $e(m^2u/2)=-1$ if $m$ is odd. Therefore,
$$
\begin{equation}
\begin{aligned} \, &\sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{e(l^2u/8)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} = \frac{1}{2^{s+2it}}\sum_{l=1}^{\infty} \frac{e(l^2u/2)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} \nonumber \\ &\qquad=\frac{1}{2^{s+2it}}\sum_{l\equiv 0 \ (\operatorname{mod}2)} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}-\frac{1}{2^{s+2it}} \sum_{l\equiv 1 \ (\operatorname{mod}2)} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.25}
$$
Since
$$
\begin{equation}
\sum_{u=1,3,5,7}\chi_4(-u)e\biggl(\frac{u}8\biggr)=0,
\end{equation}
\tag{6.26}
$$
the contribution of the part with $\mathfrak{c}=1/(8u)$, $u=1,3,5,7$ to $C^{1,2}_{\mathrm{odd}}$ is zero.
Now assume that $\mathfrak{c}=1/(16u)$, $u=1,3$. By (5.26), (5.33) and (5.60), we have
$$
\begin{equation}
\begin{aligned} \, &\sum_{(l,2)=1}\frac{n^{2it}}{l^{s+2it}} \overline{\phi_{\infty,1/(16u)} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} \phi_{0,1/(16u)} \bigl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad =\frac{n^{2it}\sigma_{-2it}(\chi_4;n^2)\chi_4(-u)e(u/4)} {2^{5-2it}L(\chi_4,1+2it)L(\chi_4,1-2it)} \sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{e(l^2u/16)\sigma_{2it} (\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.27}
$$
Using (2.3), this gives
$$
\begin{equation}
\begin{aligned} \, &\sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{e(l^2u/16)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} = \frac{1}{2^{s+2it}}\sum_{l=1}^{\infty} \frac{e(l^2u/4)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} \nonumber \\ &\qquad=\frac{1}{4^{s+2it}}\sum_{l=1}^{\infty} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} +\frac{e(u/4)}{2^{s+2it}} \sum_{(l,2)=1}\frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.28}
$$
Rewriting the sum over $l$ as
$$
\begin{equation*}
\sum_{(l,2)=1}=\sum_{l=1}^{\infty}-\sum_{l \equiv 0\ (\operatorname{mod}2)}
\end{equation*}
\notag
$$
and applying (2.8) with $z:=s+2it$, $s:=2it$, we show that
$$
\begin{equation}
\begin{aligned} \, \sum_{l\equiv 0\ (\operatorname{mod}2)} \frac{e(l^2u/16)\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} &=\frac{1-2^{2it-s}}{2^{s+2it}(1-2^{-2s})} \, \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)}{\zeta(2s)} \nonumber \\ &\qquad\times \biggl(2^{-s-2it}+e\biggl(\frac{u}4\biggr) \bigl(1-2^{-s-2it}\bigr) \biggr). \end{aligned}
\end{equation}
\tag{6.29}
$$
Substituting (6.29) into (6.27) and summing over $u=1,2$, we conclude that the contribution of the part with $\mathfrak{c}=1/(16u)$, $u=1,2$ to $C^{1,2}_{\mathrm{odd}}$ is equal to
$$
\begin{equation}
\frac{n^{2it}\sigma_{-2it}(\chi_4;n^2)}{16i}\, \frac{1-2^{2it-s}}{2^{2s+2it}(1-2^{-2s})} \, \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1+2it)L(\chi_4,1-2it)}.
\end{equation}
\tag{6.30}
$$
Next, assume that $\mathfrak{c}=1/32$. By Corollary 5.3,
$$
\begin{equation*}
\phi_{\infty,1/32}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)=0.
\end{equation*}
\notag
$$
Finally, assume that $\mathfrak{c}=\infty$. An appeal to of (5.35), (5.62), (5.26), (5.27), (2.8) shows that
$$
\begin{equation}
\begin{aligned} \, &\sum_{l \equiv 0\ (\operatorname{mod}2)}\frac{n^{2it}}{l^{s+2it}} \overline{\phi_{\infty,\infty}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)} \phi_{0,\infty}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\quad=\frac{n^{2it}\sigma_{-2it}(\chi_4;n^2)}{16i}\, \frac{1-2^{-2it-s}}{2^{2s-2it}(1-2^{-2s})}\, \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1+2it)L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.31}
$$
Combining (6.3), (6.30), (6.31), we find that
$$
\begin{equation}
\begin{aligned} \, C^{1,2}_{\mathrm{odd}} &=\frac{L(\chi_4,s)}{4\pi^{s-1/2}(1-2^{-2s})}\, \frac{1}{2\pi i} \int_{-\infty}^{\infty} \frac{\psi_D(t)\sinh(\pi t)}{t\cosh(\pi t)} \sigma_{-2it}(\chi_4;n^2) \nonumber \\ &\qquad \times n^{2it} \frac{\zeta(s+2it)\zeta(s-2it)}{L(\chi_4,1+2it)L(\chi_4,1-2it)} \biggl( \frac{2^{-2it}+2^{2it}-2^{1-s}}{2^{2s}} \biggr)\, dt. \end{aligned}
\end{equation}
\tag{6.32}
$$
The required result now follows by summing (6.23) and (6.32). Lemma 6.6. For $\operatorname{Re}{s}>1$, the following identity holds Proof. The list of singular cusps for $\Gamma_0(16)$ is given in Lemma 5.1.
Consider $\mathfrak{c}=0$ and $\mathfrak{c}=1/2$. As a consequence of (5.63) and Corollary 5.3, we have, for odd $n$,
$$
\begin{equation}
\phi_{0,0}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr)= \phi_{0,1/2}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr)=0.
\end{equation}
\tag{6.34}
$$
Consider $\mathfrak{c}=1/8$. By Corollary 5.3,
$$
\begin{equation}
\phi_{\infty,1/8}\biggl(l^2,\, \frac12+it,\, \chi_4\biggr)=0.
\end{equation}
\tag{6.35}
$$
Consider $\mathfrak{c}=1/(4u)$, $u=1,3$. Using (5.70), (5.26), we compute the sum over $l$ in (6.4)
$$
\begin{equation}
\sum_{l=1}^{\infty} \frac{\overline{\phi_{\infty,1/(4u)}(l^2,1/2+it,\chi_4)}}{l^{s+2it}} = \frac{\chi_4(-u)}{4^{1-2it}L(\chi_4,1-2it)}\sum_{l=1}^{\infty} \frac{\sigma_{2it}(\chi_4;l^2)e(l^2u/4)}{l^{s+2it}}.
\end{equation}
\tag{6.36}
$$
The last sum can be split into two parts
$$
\begin{equation}
\begin{aligned} \, \sum_{l=1}^{\infty}\frac{\sigma_{2it}(\chi_4;l^2)e(l^2u/4)}{l^{s+2it}} &= \frac{1}{2^{s+2it}}\sum_{l=1}^{\infty} \frac{\sigma_{2it}(\chi_4;4l^2)}{l^{s+2it}} \nonumber \\ &\qquad+e\biggl(\frac{u}4\biggr)\sum_{l\equiv 1\ (\operatorname{mod}2)} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}}. \end{aligned}
\end{equation}
\tag{6.37}
$$
Furthermore,
$$
\begin{equation}
\sum_{l\equiv 1\ (\operatorname{mod}2)} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} =\sum_{l=1}^{\infty} \frac{\sigma_{2it}(\chi_4;l^2)}{l^{s+2it}} - \frac{1}{2^{s+2it}}\sum_{l=1}^{\infty} \frac{\sigma_{2it}(\chi_4;4l^2)}{l^{s+2it}}.
\end{equation}
\tag{6.38}
$$
Note that $\sigma_{2it}(\chi_4;4l^2)=\sigma_{2it}(\chi_4;l^2)$, and therefore,
$$
\begin{equation}
\sum_{l=1}^{\infty}\frac{\sigma_{2it}(\chi_4;l^2)e(l^2u/4)}{l^{s+2it}} = \biggl( \frac{1}{2^{s+2it}}+ e\biggl(\frac{u}4\biggr)(1-2^{-s-2it})\biggr) \sum_{l=1}^{\infty}\frac{\sigma_{2it}(\chi_4;4l^2)}{l^{s+2it}}.
\end{equation}
\tag{6.39}
$$
Substituting (6.39) into (6.36) and applying (2.8) with $z:=s+2it$ and $s:=2it$ yields
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty} \frac{\overline{\phi_{\infty,1/(4u)}(l^2,1/2+it,\chi_4)}}{l^{s+2it}} = \biggl( \frac{1}{2^{s+2it}}+ e\biggl(\frac{u}4\biggr) \bigl(1-2^{-s-2it}\bigr)\biggr)\ \nonumber \\ &\qquad \times \frac{\chi_4(-u)}{4^{1-2it}L(\chi_4,1-2it)} \, \frac{1-2^{2it-s}}{1-2^{-2s}}\, \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)}{\zeta(2s)}. \end{aligned}
\end{equation}
\tag{6.40}
$$
Using (5.65), (5.26), and the fact that $n$ is odd, we infer
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{\overline{\phi_{\infty,1/(4u)}(l^2,1/2+it,\chi_4)}} {l^{s+2it}} n^{2it}\phi_{0,1/(4u)}\biggl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\qquad= \biggl( \frac{1}{2^{s+2it}}+ e\biggl(\frac{u}4\biggr)(1-2^{-s-2it})\biggr)n^{2it}\sigma_{-2it}(\chi_4;n^2) \frac{\chi_4(-u)e(u/4)}{16} \nonumber \\ &\qquad\qquad \times \frac{1-2^{2it-s}}{1-2^{-2s}} \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1-2it)L(\chi_4,1+2it)}. \end{aligned}
\end{equation}
\tag{6.41}
$$
Consider $\mathfrak{c}=\infty$. In order to evaluate the sum over $l$ in this case we apply (5.72), (5.27), and (2.8) with $z:=s-2it$, $s:=-2it$. Furthermore, using (5.67), (5.26) and the fact that $\tau(\chi_4)=2i$, we obtain
$$
\begin{equation}
\begin{aligned} \, &\sum_{l=1}^{\infty}\frac{1}{l^s} l^{-2it} \overline{\phi_{\infty,\infty} \biggl(l^2,\, \frac12+it,\, \chi_4\biggr)}n^{2it}\phi_{0,\infty} \biggl(n^2,\, \frac12+it,\, \chi_4\biggr) \nonumber \\ &\quad =\frac{-i}{2^{3+s-2it}}\, \frac{1-2^{-2it-s}}{1-2^{-2s}}n^{2it} \sigma_{-2it}(\chi_4;n^2) \frac{L(\chi_4,s)\zeta(s+2it)\zeta(s-2it)} {\zeta(2s)L(\chi_4,1+2it)L(\chi_4,1-2it)}. \end{aligned}
\end{equation}
\tag{6.42}
$$
we prove the lemma. Combining the previous two lemmas, we prove the following result. Corollary 6.2. For $\operatorname{Re}{s}>1$, Lemma 6.7. For $0<\operatorname{Re}{s}<1$, $s\neq 1/2$, § 7. Contribution of the discrete and homomorphic spectraApplying the Kuznetsov trace formula (2.43) to the sums of Kloosterman sums in (3.36) for even $n$, we find that the discrete spectrum is a sum of two twisted moments of symmetric square $L$-functions associated to Maass cusp forms of level $4$ with nebentypus $\chi_4$, namely
$$
\begin{equation}
D_{1}^{\mathrm{even}}=-\frac{2^{1-s}\pi^{1/2-s}i}{1-2^{-2s}} \sum_{f\in H(4,\chi_4)} \frac{\psi_{D}(t_f)}{\cosh(\pi t_f)} \rho_{f_{\infty}} \biggl(\frac{n^{2}}4\biggr) \overline{L(s,\operatorname{sym}^2 f_{\infty})}
\end{equation}
\tag{7.1}
$$
and
$$
\begin{equation}
D_{2}^{\mathrm{even}}=\frac{2\pi^{1/2-s}}{1-2^{-2s}}\sum_{f\in H(4,\chi_4)} \frac{\psi_{D}(t_f)}{\cosh(\pi t_f)}\rho_{f_{0}}\biggl(\frac{n^{2}}4\biggr) \overline{L(s,\operatorname{sym}^2 f_{\infty})}.
\end{equation}
\tag{7.2}
$$
Similarly, the holomorphic spectrum is a sum of two twisted moments of symmetric square $L$-functions associated to holomorphic cusp forms of level $4$ with nebentypus $\chi_4$ given by
$$
\begin{equation}
H_{1}^{\mathrm{even}}=-\frac{2^{1-s}\pi^{1/2-s} i}{1-2^{-2s}} \sum_{\substack{k>1\\k \text{ odd}}}\psi_{H}(k)\Gamma(k) \sum_{f\in H_k(4,\chi_4)}\rho_{f_{\infty}} \biggl(\frac{n^{2}}{4}\biggr) \overline{L(s,\operatorname{sym}^2 f_{\infty})}
\end{equation}
\tag{7.3}
$$
and
$$
\begin{equation}
H_{2}^{\mathrm{even}}=\frac{2\pi^{1/2-s}}{1-2^{-2s}} \sum_{\substack{k>1\\k \text{ odd}}}\psi_{H}(k)\Gamma(k) \sum_{f\in H_k(4,\chi_4)}\rho_{f_{0}} \biggl(\frac{n^{2}}{4}\biggr) \overline{L(s,\operatorname{sym}^2 f_{\infty})}.
\end{equation}
\tag{7.4}
$$
Next let us consider the case of odd $n$. The main difference with the previous case is that now we obtain moments of $L$-functions associated to forms of levels $64$ and $16$. Indeed, applying the Kuznetsov trace formula (2.43) to the sums of Kloosterman sums in (3.37), we infer that the discrete spectrum consists of
$$
\begin{equation}
D_{1}^{\mathrm{odd}}=\frac{8\pi^{1/2-s}}{1-2^{-2s}}\sum_{f\in H(64,\chi_4)} \frac{\psi_{D}(t_f)}{\cosh(\pi t_f)}\rho_{f_{0}}(n^{2}) \overline{L(s,\operatorname{sym}^2 f_{\infty})}
\end{equation}
\tag{7.5}
$$
and
$$
\begin{equation}
D_{2}^{\mathrm{odd}}=\frac{4\pi^{1/2-s}}{1+2^{-s}}\sum_{f\in H(16,\chi_4)} \frac{\psi_{D}(t_f)}{\cosh(\pi t_f)}\rho_{f_{0}}(n^{2}) \overline{L(s,\operatorname{sym}^2 f_{\infty})}.
\end{equation}
\tag{7.6}
$$
Similarly, the holomorphic spectrum is a sum of these two parts:
$$
\begin{equation}
H_{1}^{\mathrm{odd}}=\frac{8\pi^{1/2-s}}{1-2^{-2s}} \sum_{\substack{k>1\\k \text{ odd}}}\psi_{H}(k)\Gamma(k) \sum_{f\in H_k(64,\chi_4)} \rho_{f_{0}}(n^{2}) \overline{L(s,\operatorname{sym}^2 f_{\infty})}
\end{equation}
\tag{7.7}
$$
and
$$
\begin{equation}
H_{2}^{\mathrm{odd}}=\frac{4\pi^{1/2-s}}{1+2^{-s}} \sum_{\substack{k>1\\k \text{ odd}}} \psi_{H}(k)\Gamma(k) \sum_{f\in H_k(16,\chi_4)}\rho_{f_{0}}(n^{2}) \overline{L(s,\operatorname{sym}^2 f_{\infty})}.
\end{equation}
\tag{7.8}
$$
§ 8. Proof of the main theorems8.1. Proof of Theorems 1.1 and 1.2The diagonal main terms in Theorems 1.1 and 1.2 are computed in Lemma 3.6. In order to evaluate the non-diagonal part, we apply the Kuznetsov trace formula (2.43) to the sums of Kloosterman sums in (3.36) and (3.37). It follows from Lemmas 6.4 and 6.7 that the contribution of the continuous spectrum is equal to $M^{\mathrm{C}}(n,s)+\mathfrak{C}(n,s)$ in case of even $n$ and to $(1/2)M^{\mathrm{C}}(n,s)+(1/2)\mathfrak{C}(n,s)$ in case of odd $n$, where
$$
\begin{equation}
\mathfrak{C}(n,s)=C_{\mathrm{even}}^{1}+ C_{\mathrm{even}}^{2}-M^{\mathrm{C}}(n,s).
\end{equation}
\tag{8.1}
$$
Finally, the contribution of the discrete and holomorphic spectra is given in § 7. For the sake of brevity, we express the final result in terms of sums of moments (1.9) using the following identities:
$$
\begin{equation}
D_{\mathrm{even}}^{1}+H_{\mathrm{even}}^{1} = -\frac{2^{1-s}\pi^{1/2-s}i}{1-2^{-2s}}\mathfrak{M}_{\infty} \biggl(\frac{n^2}4,4,s\biggr),
\end{equation}
\tag{8.2}
$$
$$
\begin{equation}
D_{\mathrm{even}}^{2}+H_{\mathrm{even}}^{2} = \frac{2\pi^{1/2-s}}{1-2^{-2s}}\mathfrak{M}_{0}\biggl(\frac{n^2}4,4,s\biggr),
\end{equation}
\tag{8.3}
$$
$$
\begin{equation}
D_{\mathrm{odd}}^{1}+H_{\mathrm{odd}}^{1} = \frac{8\pi^{1/2-s}}{1-2^{-2s}}\mathfrak{M}_{0}(n^2,64,s),
\end{equation}
\tag{8.4}
$$
$$
\begin{equation}
D_{\mathrm{odd}}^{2}+H_{\mathrm{odd}}^{2} = \frac{4\pi^{1/2-s}}{1+2^{-s}}\mathfrak{M}_{0}(n^2,16,s).
\end{equation}
\tag{8.5}
$$
8.2. Main terms at the central pointAs the final step, we show that the main terms in Theorems 1.1 and 1.2 are holomorphic at the central point. Lemma 8.1. For even $n$, the following identity holds Proof. As a consequence of the functional equations for the Riemann zeta function and for the Gamma function, we obtain
$$
\begin{equation}
\zeta(2u)\Gamma(u)=(2\pi)^{2u}\frac{\Gamma(1-2u)}{\Gamma(1-u)}\zeta(1-2u).
\end{equation}
\tag{8.7}
$$
Combining (1.13), (1.14) and (8.7), we conclude that
$$
\begin{equation}
\begin{aligned} \, &M^{\mathrm{C}}\biggl(n,\, \frac12+u\biggr)+M^{\mathrm{D}}_{\mathrm{even}} \biggl(n,\, \frac12+u\biggr) \nonumber \\ &\qquad=\zeta(1+2u)\frac{\sigma_{-1/2-u}(\chi_4;n^2)+n^{-1-2u}\sigma_{1/2+u} (\chi_4;n^2)} {L(\chi_4,3/2+u)}\int_{0}^{\infty}\omega(y)\, dy \nonumber \\ &\qquad\qquad+\zeta(1-2u)(2\pi)^{u}\frac{\Gamma(1-2u)}{\Gamma(1-u)}\, \frac{\sigma_{-1/2+u}(\chi_4;n^2)+n^{-1+2u}\sigma_{1/2-u} (\chi_4;n^2)}{L(\chi_4,3/2-u)} \nonumber \\ &\qquad\qquad\qquad\times \sqrt{2} \biggl( \sin \biggl(\frac{\pi}4+ \frac{\pi u}2\biggr) \int_{0}^{n/2}\omega(y) \biggl( \frac{n^2}{4}-y^2\biggr)^{-u}\, dy \nonumber \\ &\qquad\qquad\qquad\qquad\qquad+\cos\biggl(\frac{\pi}4+ \frac{\pi u}2\biggr) \int_{n/2}^{\infty}\omega(y)\biggl(y^2- \frac{n^2}{4}\biggr)^{-u}\, dy\biggr). \end{aligned}
\end{equation}
\tag{8.8}
$$
The expression above is holomorphic at $u=0$. Consequently, letting $u$ tend to zero and applying the L’Hôspital rule, we prove the lemma. Lemma 8.2. For odd $n$, the following identity holds Proof. Using (1.14), (1.17), (2.4) and (8.7), we obtain
$$
\begin{equation}
\begin{aligned} \, &\frac{1}{2}M^{\mathrm{C}}\biggl(n,\, \frac12+u\biggr) + M^{\mathrm{D}}_{\mathrm{odd}}\biggl(n,\, \frac12+u\biggr) =\zeta(1+2u)\frac{\sigma_{-1/2-u}(\chi_4;n^2)}{L(\chi_4,3/2+u)} \int_{0}^{\infty}\omega(y)\, dy \nonumber \\ &\qquad+\zeta(1-2u)(2\pi)^{u}\frac{\Gamma(1-2u)}{\Gamma(1-u)}\, \frac{\sigma_{-1/2+u}(\chi_4;n^2)}{L(\chi_4,3/2-u)} \nonumber \\ &\qquad\qquad\times \sqrt{2} \biggl( \sin \biggl(\frac{\pi}4+ \frac{\pi u}2\biggr) \int_{0}^{n/2}\omega(y) \biggl( \frac{n^2}{4}-y^2\biggr)^{-u}\, dy \nonumber \\ &\qquad\qquad\qquad\qquad+\cos\biggl(\frac{\pi}4+\frac{\pi u}2\biggr) \int_{n/2}^{\infty}\omega(y) \biggl(y^2- \frac{n^2}{4}\biggr)^{-u}\, dy\biggr). \end{aligned}
\end{equation}
\tag{8.10}
$$
The expression above is holomorphic at $u=0$. Now the required assertion follows by letting $u$ tend to zero and applying the L’Hôspital rule. 
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