2.1 Hecke theory

We generally denote Hecke eigenvalues, with or without subscript, by $\unicode[STIX]{x1D706}(n)$. For newforms of level $N$, we will often use the multiplicativity relation

(2.1)$$\begin{eqnarray}\unicode[STIX]{x1D706}(nm)=\mathop{\sum }_{\substack{ d\mid (m,n) \\ (d,N)=1}}\unicode[STIX]{x1D707}(d)\unicode[STIX]{x1D706}(m/d)\unicode[STIX]{x1D706}(n/d).\end{eqnarray}$$

We have the general upper bound

(2.2)$$\begin{eqnarray}\unicode[STIX]{x1D706}(n)\;\ll _{\unicode[STIX]{x1D700}}\;n^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D700}}.\end{eqnarray}$$

For a newform of level $N=N_{1}N_{2}$ and trivial central character with $N_{1}$ square-free, $(N_{1},N_{2})=1$, and some $n\mid N_{1}$, we have

(2.3)$$\begin{eqnarray}|\unicode[STIX]{x1D706}(n)|=n^{-1/2}\end{eqnarray}$$

via [Reference OggOgg69, Theorem 2].

2.2 Functional equation for the Hurwitz zeta function

For $\unicode[STIX]{x1D6FC}\in \mathbb{R}$, $\Re s>1$, let

$$\begin{eqnarray}\unicode[STIX]{x1D701}(s,\unicode[STIX]{x1D6FC}):=\mathop{\sum }_{n+\unicode[STIX]{x1D6FC}>0}(n+\unicode[STIX]{x1D6FC})^{-s}\end{eqnarray}$$

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

(2.4)$$\begin{eqnarray}\unicode[STIX]{x1D701}(s,\unicode[STIX]{x1D6FC})=\mathop{\sum }_{\pm }G^{\mp }(1-s)\unicode[STIX]{x1D701}^{(\pm \unicode[STIX]{x1D6FC})}(1-s),\end{eqnarray}$$

where

$$\begin{eqnarray}G^{\pm }(s)=(2\unicode[STIX]{x1D70B})^{-s}\unicode[STIX]{x1D6E4}(s)\exp (\pm i\unicode[STIX]{x1D70B}s/2)\end{eqnarray}$$

and $\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC})}(s)$ is (the meromorphic continuation of) $\sum _{n}e(\unicode[STIX]{x1D6FC}n)n^{-s}$. For $\unicode[STIX]{x1D6FC}\in \mathbb{Q}$, this is a reformulation of Poisson summation in residue classes.

2.3 Functional equation for twisted automorphic $L$-functions

For $\unicode[STIX]{x1D6FC}\in \mathbb{R}$, $\Re s>1$, let

(2.5)$$\begin{eqnarray}L(s,\unicode[STIX]{x1D6FC},f):=\mathop{\sum }_{n}\unicode[STIX]{x1D706}_{f}(n)e(\unicode[STIX]{x1D6FC}n)n^{-s},\end{eqnarray}$$

where, as before, $f$ is a Hecke eigenform of the group $\text{SL}_{2}(\mathbb{Z})$, either Maaß with spectral parameter $t$ and parity $\unicode[STIX]{x1D716}\in \{\pm 1\}$, or holomorphic of weight $k$, or the standard Eisenstein series with $\unicode[STIX]{x1D706}_{f}(n)=\unicode[STIX]{x1D70F}(n)$. If $\unicode[STIX]{x1D6FC}=a/c\in \mathbb{Q}$ with $(a,c)=1$, this $L$-function has meromorphic continuation to all $s\in \mathbb{C}$ with a double pole at $s=1$ with Laurent expansion

(2.6)$$\begin{eqnarray}\frac{1}{c}\biggl(\frac{1}{(s-1)^{2}}+\frac{2\unicode[STIX]{x1D6FE}-2\log c}{s-1}+O(1)\biggr)\end{eqnarray}$$

if $f$ is Eisenstein; note that this is independent of $a$. The twisted $L$-function satisfies the functional equation (see, for example, [Reference Harcos and MichelHM06, §2.4])

(2.7)$$\begin{eqnarray}L(s,a/c,f)=\mathop{\sum }_{\pm }G_{f}^{\mp }(1-s)c^{1-2s}L(s,\pm \overline{a}/c,f),\end{eqnarray}$$

where

$$\begin{eqnarray}G_{f}^{+}(s)=i^{k}(2\unicode[STIX]{x1D70B})^{1-2s}\frac{\unicode[STIX]{x1D6E4}(s+(k-1)/2)}{\unicode[STIX]{x1D6E4}(1-s+(k-1)/2)},\quad G_{f}^{-}(s)=0\end{eqnarray}$$

if $f$ is holomorphic of weight $k$ and

(2.8)$$\begin{eqnarray}G_{f}^{\pm }(s)=\unicode[STIX]{x1D716}^{(1\mp 1)/2}\frac{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(s+it))\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(s-it))}{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(1-s+it))\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(1-s-it))}\mp \frac{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(1+s+it))\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(1+s-it))}{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(2-s+it))\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}(2-s-it))}\end{eqnarray}$$

if $f$ is Maaß with spectral parameter $t$ and parity $\unicode[STIX]{x1D716}\in \{\pm 1\}$. This also holds for $f$ equal to the standard Eisenstein series with $t=0$ and $\unicode[STIX]{x1D716}=1$.

2.4 Fourier coefficients

We quote from [Reference Blomer and KhanBK19b, §3] and refer to this source for more details and references. The cuspidal spectrum is parametrized by pairs $(\unicode[STIX]{x1D713},M)$ of $\unicode[STIX]{x1D6E4}_{0}(N)$-normalized newforms $\unicode[STIX]{x1D713}$ of level $N_{0}\mid N$ and integers $M\mid N/N_{0}$. The corresponding Fourier coefficients are

(2.9)$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)=\frac{1}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})^{1/2}(N\unicode[STIX]{x1D708}(N))^{1/2}}\mathop{\prod }_{p\mid N_{0}}\biggl(1-\frac{1}{p^{2}}\biggr)^{1/2}\mathop{\sum }_{d\mid (M,n)}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d)\frac{d}{M^{1/2}}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n/d)\end{eqnarray}$$

for $n\in \mathbb{N}$, where $\unicode[STIX]{x1D708}(N)=\prod _{p\mid N}(1+1/p)$ and the multiplicative function $\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}$ is defined in [Reference Blomer and KhanBK19b, (3.10)] and satisfies in particular

(2.10)$$\begin{eqnarray}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(p,p)=\biggl(1-\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)^{2}}{p(1+1/p)^{2}}\biggr)^{-1/2},\quad \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(p,1)=\frac{-\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)}{1+1/p^{2}},\quad \unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(1,1)=1\end{eqnarray}$$

for $p\nmid N_{0}$ and in general

(2.11)$$\begin{eqnarray}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d)\;\ll _{\unicode[STIX]{x1D700}}\;M^{\unicode[STIX]{x1D700}}(M/d)^{\unicode[STIX]{x1D703}}.\end{eqnarray}$$

For $-n\in \mathbb{N}$ we have $\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)=\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(-n)$ if $\unicode[STIX]{x1D713}$ is Maaß of parity $\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}\in \{\pm 1\}$ and $\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)=0$ if $\unicode[STIX]{x1D713}$ is holomorphic.

If $N$ is square-free, the Fourier coefficients of Eisenstein series of level $N$ are easy to describe. They are parametrized by divisors $v\mid N$ and a continuous parameter $s=1/2+it$. The corresponding Fourier coefficients are given by (see, for example, [Reference Conrey and IwaniecCI00, (3.25)])

(2.12)$$\begin{eqnarray}\unicode[STIX]{x1D70C}_{v,N}(n,t)=\frac{C(v,M,t)}{(Nv)^{1/2}\unicode[STIX]{x1D701}^{(N)}(1+2it)}\mathop{\sum }_{b\mid v}\mathop{\sum }_{\unicode[STIX]{x1D6FE}\mid N/v}\unicode[STIX]{x1D707}(b\unicode[STIX]{x1D6FE})b\biggl(\frac{b}{\unicode[STIX]{x1D6FE}}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{|n|}{b\unicode[STIX]{x1D6FE}},t\biggr),\end{eqnarray}$$

where $\unicode[STIX]{x1D702}(n,t)=\sum _{d_{1}d_{2}=n}(d_{1}/d_{2})^{it}$ for $n\in \mathbb{N}$ (and 0 otherwise) and $|C(v,N,t)|=1$. For general $N$, we follow [Reference Blomer and KhanBK19b, §3] and parametrize unitary Eisenstein series of $\unicode[STIX]{x1D6E4}_{0}(N)$ by a continuous parameter $s=1/2+it$ together with pairs $(\unicode[STIX]{x1D712},M)$, where $\unicode[STIX]{x1D712}$ is a primitive Dirichlet character of conductor $c_{\unicode[STIX]{x1D712}}$ and $M\in \mathbb{N}$ satisfies $c_{\unicode[STIX]{x1D712}}^{2}\mid M\mid N$. Note that the role of $M$ is different than in the cuspidal case. We write

$$\begin{eqnarray}\displaystyle \tilde{\mathfrak{n}}_{N}(M) & = & \displaystyle \biggl(\mathop{\prod }_{\substack{ p\mid N \\ p\nmid (M,N/M)}}\frac{p}{p+1}\mathop{\prod }_{p\mid (M,N/M)}\frac{p-1}{p+1}\biggr)^{1/2},\nonumber\\ \displaystyle M & = & \displaystyle c_{\unicode[STIX]{x1D712}}M_{1}M_{2},\quad \text{where }(M_{2},c_{\unicode[STIX]{x1D712}})=1,\quad M_{1}\mid c_{\unicode[STIX]{x1D712}}^{\infty },\nonumber\end{eqnarray}$$

so that $c_{\unicode[STIX]{x1D712}}\mid M_{1}$ and $(M_{1},M_{2})=1$. The Fourier coefficients of the Eisenstein series attached to the data $(N,M,t,\unicode[STIX]{x1D712})$ are

(2.13)$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D712},M,N}(n,t) & = & \displaystyle \frac{\tilde{C}(\unicode[STIX]{x1D712},M,t)|n|^{it}}{(N\unicode[STIX]{x1D708}(N))^{1/2}\tilde{\mathfrak{n}}_{N}(M)L^{(N)}(1+2it,\unicode[STIX]{x1D712}^{2})}\biggl(\frac{M_{1}}{M_{2}}\biggr)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FF}\mid M_{2}}\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D707}\biggl(\frac{M_{2}}{\unicode[STIX]{x1D6FF}}\biggr)\overline{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D6FF})\mathop{\sum }_{\substack{ cM_{1}\unicode[STIX]{x1D6FF}f=n \\ (c,N/M)=1}}\frac{\unicode[STIX]{x1D712}(c)}{c^{2it}}\overline{\unicode[STIX]{x1D712}}(f),\end{eqnarray}$$

where $|\tilde{C}(\unicode[STIX]{x1D712},M,t)|=1$.

2.5 The Kuznetsov formula

For $x>0$, we define the integral kernels

$$\begin{eqnarray}\displaystyle {\mathcal{J}}^{+}(x,t) & := & \displaystyle \frac{\unicode[STIX]{x1D70B}i}{\sinh (\unicode[STIX]{x1D70B}t)}(J_{2it}(4\unicode[STIX]{x1D70B}x)-J_{-2it}(4\unicode[STIX]{x1D70B}x)),\nonumber\\ \displaystyle {\mathcal{J}}^{-}(x,t) & := & \displaystyle \frac{\unicode[STIX]{x1D70B}i}{\sinh (\unicode[STIX]{x1D70B}t)}(I_{2it}(4\unicode[STIX]{x1D70B}x)-I_{-2it}(4\unicode[STIX]{x1D70B}x))=4\cosh (\unicode[STIX]{x1D70B}t)K_{2it}(4\unicode[STIX]{x1D70B}x),\nonumber\\ \displaystyle {\mathcal{J}}^{\text{hol}}(x,k) & := & \displaystyle 2\unicode[STIX]{x1D70B}i^{k}J_{k-1}(4\unicode[STIX]{x1D70B}x)={\mathcal{J}}^{+}(x,(k-1)/(2i)),\quad k\in 2\mathbb{N}.\nonumber\end{eqnarray}$$

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

$$\begin{eqnarray}\mathscr{L}^{\diamondsuit }H=\int _{0}^{\infty }{\mathcal{J}}^{\diamondsuit }(x,.)H(x)\frac{dx}{x}\end{eqnarray}$$

for $\diamondsuit \in \{+,-,\text{hol}\}$, and for $n,m,N\in \mathbb{N}$, we have

(2.14)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{N\mid c}\frac{S(\pm n,m,c)}{c}H\biggl(\frac{\sqrt{nm}}{c}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad ={\mathcal{A}}_{N}^{\text{Maa}\unicode[STIX]{x00DF}}(\pm n,m;\mathscr{L}^{\pm }H)+{\mathcal{A}}_{N}^{\text{Eis}}(\pm n,m;\mathscr{L}^{\pm }H)+{\mathcal{A}}_{N}^{\text{hol}}(\pm n,m;\mathscr{L}^{\text{hol}}H),\end{eqnarray}$$

where

(2.15)$$\begin{eqnarray}\left.\begin{array}{@{}rcl@{}}{\mathcal{A}}_{N}^{\text{Maa}\unicode[STIX]{x00DF}}(n,m;h)\, & :=\, & \displaystyle \mathop{\sum }_{N_{0}M\mid N}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N_{0})}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(m)}h(t_{\unicode[STIX]{x1D713}}),\\ {\mathcal{A}}_{N}^{\text{Eis}}(n,m;h)\, & :=\, & \displaystyle \mathop{\sum }_{v\mid N}\int _{\mathbb{R}}\unicode[STIX]{x1D70C}_{v,N}(n,t)\overline{\unicode[STIX]{x1D70C}_{v,N}(m,t)}h(t)\frac{dt}{2\unicode[STIX]{x1D70B}}\quad (N\text{ square-free}),\\ {\mathcal{A}}_{N}^{\text{Eis}}(n,m;h)\, & :=\, & \displaystyle \mathop{\sum }_{c_{\unicode[STIX]{x1D712}}^{2}\mid M\mid N}\int _{\mathbb{R}}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D712},M,N}(n,t)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D712},M,N}(m,t)}h(t)\frac{dt}{2\unicode[STIX]{x1D70B}}\quad (\text{in general}),\\ {\mathcal{A}}_{N}^{\text{hol}}(n,m;h)\, & :=\, & \displaystyle \mathop{\sum }_{N_{0}M\mid N}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}_{\text{hol}}^{\ast }(N_{0})}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(m)}h(k_{\unicode[STIX]{x1D713}}).\end{array}\right.\end{eqnarray}$$

(Recall that $\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N_{0})$ and $\unicode[STIX]{x1D713}\in {\mathcal{B}}_{\text{hol}}^{\ast }(N_{0})$ are $\unicode[STIX]{x1D6E4}_{0}(N)$-normalized.) Conversely, if $h$ is holomorphic in an $\unicode[STIX]{x1D700}$-neighbourhood of $|\Im t|\leqslant 1/2$ and satisfies $h(t)\ll (1+|t|)^{-2-\unicode[STIX]{x1D6FF}}$ in this region for some $\unicode[STIX]{x1D6FF}>0$, then for $n,m\in \mathbb{N}$, we have [Reference Blomer and KhanBK19b, (3.14)]

(2.16)$$\begin{eqnarray}{\mathcal{A}}_{N}^{\text{Maa}\unicode[STIX]{x00DF}}(n,m;h)+{\mathcal{A}}_{N}^{\text{Eis}}(n,m;h)=\unicode[STIX]{x1D6FF}_{n,m}\int _{-\infty }^{\infty }h(t)\frac{t\tanh (\unicode[STIX]{x1D70B}t)\,dt}{2\unicode[STIX]{x1D70B}^{2}}+\mathop{\sum }_{N\mid c}\frac{S(n,m,c)}{c}\mathscr{K}h\biggl(\frac{\sqrt{nm}}{c}\biggr),\end{eqnarray}$$

where

(2.17)$$\begin{eqnarray}\mathscr{K}h(x)=\int _{-\infty }^{\infty }{\mathcal{J}}^{+}(x,t)h(t)t\tanh (\unicode[STIX]{x1D70B}t)\frac{dt}{2\unicode[STIX]{x1D70B}^{2}}=\frac{i}{\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\frac{J_{2it}(4\unicode[STIX]{x1D70B}x)}{\cosh (\unicode[STIX]{x1D70B}t)}h(t)t\,dt.\end{eqnarray}$$

2.6 Integral transforms

We generalize (2.17) slightly and define for $s\in \mathbb{C}$ the transform $\mathscr{K}_{s}h$ by $\mathscr{K}_{s}h(x):=x^{s}\mathscr{K}h(x)$.

Proof. We record the formula [Reference Gradshteyn, Ryzhik, Jeffrey and ZwillingerGR07, (8.411.10)]

(2.18)$$\begin{eqnarray}\frac{J_{2it}(4\unicode[STIX]{x1D70B}x)}{\cosh (\unicode[STIX]{x1D70B}t)}=\frac{(2\unicode[STIX]{x1D70B}x)^{2it}}{\sqrt{\unicode[STIX]{x1D70B}}\unicode[STIX]{x1D6E4}(1/2+2it)\cosh (\unicode[STIX]{x1D70B}t)}\int _{-1}^{1}(1-y^{2})^{2it-1/2}\cos (4\unicode[STIX]{x1D70B}xy)\,dy\end{eqnarray}$$

for $\Re (2it)>-1/2$, $x>0$. In particular,

(2.19)$$\begin{eqnarray}\frac{d^{j}}{dx^{j}}\frac{J_{2it}(4\unicode[STIX]{x1D70B}x)}{\cosh (\unicode[STIX]{x1D70B}t)}\;\ll _{j,A,\unicode[STIX]{x1D700}}\;\biggl(\frac{x}{1+|t|}\biggr)^{\Re (2it)}(1+|t|/x)^{j},\quad -1/2+\unicode[STIX]{x1D700}\leqslant \Re (2it)<A,\end{eqnarray}$$

for $j,A\in \mathbb{N}_{0}:=\mathbb{N}\cup \{0\}$, $\unicode[STIX]{x1D700}>0$. Let $s\in \mathbb{C}$ with $\Re s<-10$ be fixed. For $x\geqslant 1$ and $0\leqslant j\leqslant 3$, we obtain

$$\begin{eqnarray}x^{j}H^{(j)}(x)\ll x^{\Re s+j}\int _{-\infty }^{\infty }\biggl(1+\frac{|t|}{x}\biggr)^{j}|h(t)t|\,dt\ll x^{\Re s+j}\ll x^{-3/2},\end{eqnarray}$$

and for $x<1$, we shift the contour to $\Re (2it)=-\Re s+10$ (not passing any pole by our assumption on $h$), getting

Lemma 2. Let $s\in \mathbb{C}$ with $\Re s<1$, and suppose that $h$ is even and holomorphic in $|\Im t|<(-\Re s+15)/2$, satisfying $h(t)\ll e^{-|t|}$ and having zeros at $\pm i(2n-1)/2$, $n\in \mathbb{N}$, in this region.

$(\text{a})$ The transform $\mathscr{L}^{+}\mathscr{K}_{s}h$, defined for $\Re s<-10$ by Lemma 1, has analytic continuation to $\Re s<1$, and we have the Sears–Titchmarsh inversion formula $\mathscr{L}^{+}\mathscr{K}_{0}h=h$.

$(\text{b})$ We have the uniform bounds

$$\begin{eqnarray}\mathscr{L}^{+}\mathscr{K}_{s}h(t)\;\ll _{\Re s}\;\int _{-\infty }^{\infty }\biggl|h\biggl(\unicode[STIX]{x1D70F}-\frac{i}{2}(1-\Re s)\biggr)\biggr|(1+|\unicode[STIX]{x1D70F}|)\displaystyle \mathop{\prod }_{\pm }\biggl(1+\biggl|\frac{1}{2}\Im s-\unicode[STIX]{x1D70F}\pm \Re t\biggr|\biggr)^{-1+\Re s}\,d\unicode[STIX]{x1D70F}\end{eqnarray}$$

for $t\in \mathbb{R}\cup [-i\unicode[STIX]{x1D703},i\unicode[STIX]{x1D703}]$ and $\Re s<1$, and

$$\begin{eqnarray}\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h(k)\;\ll _{\Re s}\;\int _{-\infty }^{\infty }\biggl|h\biggl(\unicode[STIX]{x1D70F}-\frac{i}{2}\max (2-k-\Re s,0)\biggr)\biggr|(1+|\unicode[STIX]{x1D70F}|)\displaystyle \mathop{\prod }_{\pm }\biggl(\biggl|\frac{1}{2}\Im s\pm \unicode[STIX]{x1D70F}\biggr|+k\biggr)^{-1+\Re s}\,d\unicode[STIX]{x1D70F}\end{eqnarray}$$

for $k\in 2\mathbb{N}$, where the implied constants depend only on $\Re s$ (but not on $t$, $k$, $h$, $\Im s$).

Proof. For $\Re s<-10$, we have, by definition,

$$\begin{eqnarray}\mathscr{K}_{s}h(x)=x^{s}\frac{i}{\unicode[STIX]{x1D70B}}\int _{\Re (i\unicode[STIX]{x1D70F})=(10-\Re s)/2}\frac{J_{2i\unicode[STIX]{x1D70F}}(4\unicode[STIX]{x1D70B}x)}{\cosh (\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F})}h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}\,d\unicode[STIX]{x1D70F},\end{eqnarray}$$

and we have an absolutely convergent double integral

$$\begin{eqnarray}\mathscr{L}^{+}\mathscr{K}_{s}h(t)=-\int _{0}^{\infty }\frac{1}{\sinh (\unicode[STIX]{x1D70B}t)}(J_{2it}(4\unicode[STIX]{x1D70B}x)-J_{-2it}(4\unicode[STIX]{x1D70B}x))x^{s}\int _{\Re (i\unicode[STIX]{x1D70F})=(10-\Re s)/2}\frac{J_{2i\unicode[STIX]{x1D70F}}(4\unicode[STIX]{x1D70B}x)}{\cosh (\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F})}h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}\,d\unicode[STIX]{x1D70F}\frac{dx}{x}.\end{eqnarray}$$

To see the absolute convergence, we use (2.19) with $j=0$ to bound $J_{\pm 2it}(x)\;\ll _{t}\;\min (x^{2\unicode[STIX]{x1D703}},x^{-2\unicode[STIX]{x1D703}})$ for $t\in \mathbb{R}\cup [-i\unicode[STIX]{x1D703},i\unicode[STIX]{x1D703}]$, and we combine (2.19) with the bound $J_{\unicode[STIX]{x1D708}}(x)\ll x^{-1/2}$ for $x\gg |\unicode[STIX]{x1D708}|^{2}$ (which follows from the asymptotic formula [Reference Gradshteyn, Ryzhik, Jeffrey and ZwillingerGR07, (8.451.1)]) to bound

$$\begin{eqnarray}J_{2i\unicode[STIX]{x1D70F}}(x)\ll \min (x^{-\Re s+10},x^{-1/2}(1+|\unicode[STIX]{x1D70F}|)^{-\Re s+11}).\end{eqnarray}$$

We can compute the $x$-integral explicitly using [Reference Gradshteyn, Ryzhik, Jeffrey and ZwillingerGR07, (6.574.2)], getting

(2.20)$$\begin{eqnarray}\mathscr{L}^{+}\mathscr{K}_{s}h(t)=\int _{\Re (i\unicode[STIX]{x1D70F})=(10-\Re s)/2}\frac{\unicode[STIX]{x1D6E4}(1-s)\cos (i\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F}+\unicode[STIX]{x1D70B}s/2)h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}}{(2\unicode[STIX]{x1D70B})^{s}\unicode[STIX]{x1D70B}i\cosh (\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F})}\mathop{\prod }_{\pm }\frac{\unicode[STIX]{x1D6E4}(\frac{s}{2}+i\unicode[STIX]{x1D70F}\pm it)}{\unicode[STIX]{x1D6E4}(1-\frac{s}{2}+i\unicode[STIX]{x1D70F}\pm it)}\,d\unicode[STIX]{x1D70F}.\end{eqnarray}$$

Here we can put any $s$ with $\Re s<1$ in the integrand (and also shift the contour to, say, $\Re (i\unicode[STIX]{x1D70F})=5$), in particular $s=0$, so that

$$\begin{eqnarray}\mathscr{L}^{+}\mathscr{K}_{0}h(t)=\int _{\Re (i\unicode[STIX]{x1D70F})=5}\frac{h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}}{t^{2}-\unicode[STIX]{x1D70F}^{2}}\frac{d\unicode[STIX]{x1D70F}}{\unicode[STIX]{x1D70B}i}.\end{eqnarray}$$

The integrand is odd, so the integral equals half the sum of the two residues at $\unicode[STIX]{x1D70F}=\pm t$ and part (a) of the lemma follows. To prove part (b) for $\mathscr{L}^{+}\mathscr{K}_{s}h(t)$, we shift the $\unicode[STIX]{x1D70F}$-contour to $\Re (i\unicode[STIX]{x1D70F})=(1-\Re (s))/2$ and estimate trivially in (2.20) using Stirling’s formula. For $\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h(t)$, we have the similar expression

$$\begin{eqnarray}\displaystyle \mathscr{L}^{\text{hol}}\mathscr{K}_{s}h(k)=i^{k+1}\int _{\Re (i\unicode[STIX]{x1D70F})=a}\frac{\unicode[STIX]{x1D6E4}(1-s)(2\unicode[STIX]{x1D70B})^{-s}\unicode[STIX]{x1D6E4}(\frac{k-1}{2}+\frac{s}{2}+i\unicode[STIX]{x1D70F})}{\unicode[STIX]{x1D6E4}(\frac{k+1}{2}-\frac{s}{2}-i\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6E4}(\frac{3-k}{2}-\frac{s}{2}+i\unicode[STIX]{x1D70F})\unicode[STIX]{x1D6E4}(\frac{k+1}{2}-\frac{s}{2}+i\unicode[STIX]{x1D70F})}\frac{h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}}{\cosh (\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F})}\,d\unicode[STIX]{x1D70F} & & \displaystyle \nonumber\end{eqnarray}$$

for any $a\in \mathbb{R}$ satisfying $k-1+\Re s+2a>0$, say $a=\max ((2-k-\Re s)/2,0)$. We can rewrite this as

$$\begin{eqnarray}\int _{\Re (i\unicode[STIX]{x1D70F})=a}\frac{\unicode[STIX]{x1D6E4}(1-s)\cos (i\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F}+\textstyle \frac{1}{2}\unicode[STIX]{x1D70B}s)h(\unicode[STIX]{x1D70F})\unicode[STIX]{x1D70F}}{(2\unicode[STIX]{x1D70B})^{s}\unicode[STIX]{x1D70B}i\cosh (\unicode[STIX]{x1D70B}\unicode[STIX]{x1D70F})}\mathop{\prod }_{\pm }\frac{\unicode[STIX]{x1D6E4}(\frac{k-1}{2}+\frac{s}{2}\pm i\unicode[STIX]{x1D70F})}{\unicode[STIX]{x1D6E4}(\frac{k+1}{2}-\frac{s}{2}\pm i\unicode[STIX]{x1D70F})}\,d\unicode[STIX]{x1D70F}.\end{eqnarray}$$

The desired bound follows now from $\unicode[STIX]{x1D6E4}(z+w)/\unicode[STIX]{x1D6E4}(z)\;\ll _{w}\;(1+|z|)^{w}$ for $w\in \mathbb{R}$ and $|z|$ sufficiently large; see, for example, [Reference Gradshteyn, Ryzhik, Jeffrey and ZwillingerGR07, (8.328.2)].☐

3.1 The set-up

We recall the definitions (1.6) and (1.9) for a prime $q$ and integers $a,b$ satisfying $(ab,q)=$$(a,b)=1$, $a\asymp b$, and an even holomorphic test function $F$ that is rapidly decaying on vertical lines and is divisible by $(1-u)(v-1)^{2}\prod _{j=1}^{50}(j-s)$. Initially we assume

(3.1)$$\begin{eqnarray}2<\Re s,\quad \Re v<3,\quad 10<\Re u<11.\end{eqnarray}$$

In this section all implicit constants may depend on $s,u,v,\unicode[STIX]{x1D700}$ and $F$. Additional dependencies will be mentioned. We proceed to define the correction polynomial $P_{q}(s,u,v,z)$, and to this end we define three auxiliary quantities

$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{q}^{(1)}(s,u,v) & := & \displaystyle \int _{\mathbb{R}}\unicode[STIX]{x1D701}^{(q)}(s+z)\unicode[STIX]{x1D701}^{(q)}(u+z)L^{(q)}(v-z,f)F(z)\frac{dz}{2\unicode[STIX]{x1D70B}i},\nonumber\\ \displaystyle {\mathcal{T}}_{q}^{(2)}(s,u,v) & := & \displaystyle q^{1-s-v}\int _{\mathbb{R}}\biggl(\unicode[STIX]{x1D706}(q)-\frac{1}{q^{v-z}}\biggr)\unicode[STIX]{x1D701}(s+z)\unicode[STIX]{x1D701}^{(q)}(u+z)L(v-z,f)F(z)\frac{dt}{2\unicode[STIX]{x1D70B}i},\nonumber\\ \displaystyle {\mathcal{T}}_{q}^{(3)}(s,u,v) & := & \displaystyle \int _{\mathbb{R}}(q^{2-s-u-2v}-q^{1-u-2v+z})\unicode[STIX]{x1D701}(s+z)\unicode[STIX]{x1D701}(u+z)L(v-z,f)F(z)\frac{dz}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

Here $\unicode[STIX]{x1D701}^{(q)}(s):=\unicode[STIX]{x1D701}(s)\prod _{p\mid q}(1-p^{-s})$ and $L^{(q)}(s,f)=L(s,f)\prod _{p\mid q}(1-\unicode[STIX]{x1D706}_{f}(p)p^{-s}+p^{-2s})$ are the usual $L$-functions with the Euler factors dividing $q$ omitted. We define

$$\begin{eqnarray}{\mathcal{T}}_{q}^{\text{triv}}(s,u,v):=\mathop{\sum }_{j=1}^{3}{\mathcal{T}}_{q}^{(j)}(s,u,v),\end{eqnarray}$$

so that (1.9) holds with

(3.2)$$\begin{eqnarray}\displaystyle P_{q}(s,u,v,z) & = & \displaystyle \biggl(1-\frac{1}{q^{s+z}}\biggr)\biggl(1-\frac{1}{q^{u+z}}\biggr)\biggl(1-\frac{\unicode[STIX]{x1D706}(q)}{q^{v-z}}+\frac{1}{q^{2v-2z}}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,q^{1-s-v}\biggl(\unicode[STIX]{x1D706}(q)-\frac{1}{q^{v-z}}\biggr)\biggl(1-\frac{1}{q^{u+z}}\biggr)+q^{2-s-u-2v}-q^{1-u-2v+z}.\end{eqnarray}$$

It is easy to see that this satisfies (1.8). In the range (3.1), we can open the Dirichlet series and obtain

$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{a,b,q}(s,u,v)+{\mathcal{T}}_{q}^{(1)}(s,u,v) & = & \displaystyle q\int _{(0)}F(z)\mathop{\sum }_{\substack{ (nmr,q)=1 \\ anm\equiv br\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)}}\frac{\unicode[STIX]{x1D706}(r)}{n^{s+z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i},\nonumber\\ \displaystyle {\mathcal{T}}_{q}^{(2)}(s,u,v) & = & \displaystyle q^{1-s-v}\int _{(0)}F(z)\biggl(\unicode[STIX]{x1D706}(q)-\frac{1}{q^{v-z}}\biggr)\mathop{\sum }_{n,r,(m,q)=1}\frac{\unicode[STIX]{x1D706}(r)}{n^{s+z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i},\nonumber\\ \displaystyle & = & \displaystyle q\int _{(0)}F(z)\mathop{\sum }_{n,r,(m,q)=1}\frac{\unicode[STIX]{x1D706}(qr)}{(qn)^{s+z}m^{u+z}(qr)^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i},\nonumber\\ \displaystyle {\mathcal{T}}_{q}^{(3)}(s,u,v) & = & \displaystyle q^{2-s-u-2v}\int _{(0)}F(z)(1-q^{s+z-1})\mathop{\sum }_{n,r,m}\frac{\unicode[STIX]{x1D706}(r)}{n^{s+z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

We conclude that

$$\begin{eqnarray}\displaystyle \tilde{{\mathcal{T}}}_{a,b,q}(s,u,v) & := & \displaystyle {\mathcal{T}}_{a,b,q}(s,u,v)+{\mathcal{T}}_{q}^{(1)}(s,u,v)+{\mathcal{T}}_{q}^{(2)}(s,u,v)\nonumber\\ \displaystyle & = & \displaystyle q\int _{(0)}F(z)\mathop{\sum }_{\substack{ (m,q)=1 \\ n\equiv \overline{am}br\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)}}\frac{\unicode[STIX]{x1D706}(r)}{n^{s+z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

Eventually the term ${\mathcal{T}}_{q}^{(3)}(s,u,v)$ will remove the last coprimality condition $(m,q)=1$, but this has to wait until the end of argument. Until then, we transform $\tilde{{\mathcal{T}}}_{a,b,q}(s,u,v)$ and ${\mathcal{T}}_{q}^{(3)}(s,u,v)$ in a parallel fashion.

3.2 Poisson summation

We write the $n$-sum in terms of the Hurwitz zeta function with $\unicode[STIX]{x1D6FC}=\overline{am}br/q$ and shift the $z$-contour to the left to $\Re z=-4$. Our assumption on $F$ implies that the potential pole at $z=1-s$ is cancelled. The $m,r$-sums are still absolutely convergent, and we apply the functional equation (2.4), getting

(3.3)$$\begin{eqnarray}\tilde{{\mathcal{T}}}_{a,b,q}(s,u,v)=\mathop{\sum }_{\pm }\int _{(-4)}F(z)G^{\pm }(1-s-z)q^{1-s-z}\mathop{\sum }_{n,r,(m,q)=1}\frac{\unicode[STIX]{x1D706}(r)e(\mp \overline{a}brn\overline{m}/q)}{n^{1-s-z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}\end{eqnarray}$$

and

(3.4)$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{q}^{(3)}(s,u,v) & = & \displaystyle q^{2-s-u-2v}\int _{(-4)}F(z)(1-q^{s+z-1})G^{\pm }(1-s-z)\mathop{\sum }_{n,r,m}\frac{\unicode[STIX]{x1D706}(r)}{n^{1-s-z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}\nonumber\\ \displaystyle & = & \displaystyle q^{2-s-u-2v}\int _{(-4)}F(z)G^{\pm }(1-s-z)\mathop{\sum }_{(n,q)=1,r,m}\frac{\unicode[STIX]{x1D706}(r)}{n^{1-s-z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}.\end{eqnarray}$$

3.3 Reciprocity

For $\unicode[STIX]{x1D6FC}\in \mathbb{R}\setminus \{0\}$, we recall the absolutely convergent Mellin integral

$$\begin{eqnarray}e(\unicode[STIX]{x1D6FC})=\int _{{\mathcal{C}}}G^{\text{sgn}(\unicode[STIX]{x1D6FC})}(s)|\unicode[STIX]{x1D6FC}|^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

where ${\mathcal{C}}$ is the contour

$$\begin{eqnarray}{\mathcal{C}}=\textstyle (-\frac{3}{5}-i\infty ,-\frac{3}{5}-i]\cup [-\frac{3}{5}-i,\frac{1}{10}]\cup [\frac{1}{10},-\frac{3}{5}+i]\cup [-\frac{3}{5}+i,-\frac{3}{5}+i\infty ).\end{eqnarray}$$

In (3.3), we insert

(3.5)$$\begin{eqnarray}1=e\biggl(\pm \frac{brn}{amq}\biggr)\int _{{\mathcal{C}}}G^{\pm }(w)\biggl(\frac{brn}{amq}\biggr)^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i}\end{eqnarray}$$

and apply the additive reciprocity formula (1.16). This gives the absolutely convergent expression

$$\begin{eqnarray}\displaystyle \tilde{{\mathcal{T}}}_{a,b,q}(s,u,v) & = & \displaystyle \mathop{\sum }_{\pm }\int _{(-4)}\int _{{\mathcal{C}}}F(z)\biggl(\frac{b}{a}\biggr)^{-w}G^{\pm }(w)G^{\pm }(1-s-z)q^{1-s-z+w}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n,(m,q)=1,r}\frac{\unicode[STIX]{x1D706}(r)e(\pm \overline{q}brn/(am))}{n^{1-s-z+w}m^{u+z-w}r^{v-z+w}}\frac{dw}{2\unicode[STIX]{x1D70B}i}\frac{dz}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

We temporarily straighten the ${\mathcal{C}}$-contour to $\Re w=-3/5$, picking up the polar term

$$\begin{eqnarray}\mathop{\sum }_{\pm }\int _{(-4)}F(z)G^{\pm }(1-s-z)q^{1-s-z}\mathop{\sum }_{n,(m,q)=1,r}\frac{\unicode[STIX]{x1D706}(r)e(\pm \overline{q}brn/(am))}{n^{1-s-z}m^{u+z}r^{v-z}}\frac{dz}{2\unicode[STIX]{x1D70B}i}.\end{eqnarray}$$

In the remaining double integral, we change variables $w\mapsto w+z$ (so that $\Re z=-4$, $\Re w=17/5$), exchange the two integrals, and in the inner $z$-integral we bend the contour to the right to the contour

$$\begin{eqnarray}\displaystyle {\mathcal{C}}(w)={\mathcal{C}}-w & = & \displaystyle \textstyle (-\frac{3}{5}-w-i\infty ,-\frac{3}{5}-w-i]\cup [-\frac{3}{5}-w-i,\frac{1}{10}-w]\nonumber\\ \displaystyle & & \displaystyle \cup \textstyle [\frac{1}{10}-w,-\frac{3}{5}-w+i]\cup [-\frac{3}{5}-w+i,-\frac{3}{5}-w+i\infty )\nonumber\end{eqnarray}$$

consisting of a union of straight lines. This picks up a polar term

$$\begin{eqnarray}-\mathop{\sum }_{\pm }\int _{(17/5)}F(-w)G^{\pm }(1-s+w)q^{1-s+w}\mathop{\sum }_{n,(m,q)=1,r}\frac{\unicode[STIX]{x1D706}(r)e(\pm \overline{q}brn/(am))}{n^{1-s+w}m^{u-w}r^{v+w}}\frac{dw}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

which cancels the previous one. This shows that

(3.6)$$\begin{eqnarray}\tilde{{\mathcal{T}}}_{a,b,q}(s,u,v)=\mathop{\sum }_{\pm }\int _{(17/5)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)q^{1-s+w}\mathop{\sum }_{n,(m,q)=1,r}\frac{\unicode[STIX]{x1D706}(r)e(\pm \overline{q}brn/(am))}{n^{1-s+w}m^{u-w}r^{v+w}}\biggl(\frac{b}{a}\biggr)^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w) & = & \displaystyle \int _{{\mathcal{C}}(w)}F(z)\biggl(\frac{b}{a}\biggr)^{-z}G^{\pm }(w+z)G^{\pm }(1-s-z)\frac{dz}{2\unicode[STIX]{x1D70B}i}\nonumber\\ \displaystyle & = & \displaystyle (2\unicode[STIX]{x1D70B})^{s-w-1}e^{\pm i\unicode[STIX]{x1D70B}(w+1-s)/2}\int _{{\mathcal{C}}(w)}F(z)\biggl(\frac{b}{a}\biggr)^{-z}\unicode[STIX]{x1D6E4}(w+z)\unicode[STIX]{x1D6E4}(1-s-z)\frac{dz}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

Here we can straighten the contour and shift it to the far left to $\Re z=-A$, say. This gives a sum of polar terms of the shape

$$\begin{eqnarray}p_{n}(w):=\frac{(-1)^{n}}{n!}(2\unicode[STIX]{x1D70B})^{s-w-1}e^{\pm i\unicode[STIX]{x1D70B}(w+1-s)/2}F(-w-n)\biggl(\frac{b}{a}\biggr)^{w+n}\unicode[STIX]{x1D6E4}(1-s+w+n),\quad n\in \mathbb{N}_{0},\end{eqnarray}$$

and a remaining integral that is holomorphic in the half plane $\Re w>-A$ and bounded by$\ll _{\Re w,A}\,(b/a)^{A}(1+|w|)^{\Re w-A-1/2}$. Since $F(1-s)=\cdots =F(50-s)=0$, we conclude that $\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }$ is

(3.7)$$\begin{eqnarray}\text{holomorphic in }|\Re w|<48\text{ and satisfies }\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)\ll (1+|w|)^{-100}\end{eqnarray}$$

as long as $a\asymp b$ and $0<\Re s<3$. We also observe that

(3.8)$$\begin{eqnarray}\mathop{\sum }_{\pm }\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(s)=0.\end{eqnarray}$$

Inserting

$$\begin{eqnarray}1=e\biggl(\pm \frac{brn}{am}\biggr)\int _{{\mathcal{C}}}G^{\pm }(w)\biggl(\frac{brn}{am}\biggr)^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i}\end{eqnarray}$$

into (3.4) (similarly to (3.5), but with different variables), we obtain in the same way

$$\begin{eqnarray}{\mathcal{T}}_{q}^{(3)}(s,u,v)=\mathop{\sum }_{\pm }\int _{(17/5)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)q^{2-s-u-2v}\mathop{\sum }_{m,(n,q)=1,r}\frac{\unicode[STIX]{x1D706}(r)e(\pm brn/(am))}{n^{1-s+w}m^{u-w}r^{v+w}}\biggl(\frac{b}{a}\biggr)^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i}.\end{eqnarray}$$

(The expression is still independent of $a,b$, even though the right-hand side seems to depend on $a,b$.)

3.4 Poisson summation again

We return to (3.6), split the $n$-sum into residue classes modulo $am$, express the $n$-sum in terms of the Hurwitz zeta function, shift the $w$-contour to $\Re w=0$, and apply the functional equation (2.4), getting

(3.9)$$\begin{eqnarray}\displaystyle \tilde{{\mathcal{T}}}_{a,b,q}(s,u,v) & = & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\int _{(0)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-s+w}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{(m,q)=1,r,n}\mathop{\sum }_{\unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}am)}e\biggl(\frac{\pm \overline{q}br\unicode[STIX]{x1D708}}{am}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(am))}{n^{s-w}m^{u-w}r^{v+w}}\biggl(\frac{b}{a}\biggr)^{-w}(am)^{s-w-1}\frac{dw}{2\unicode[STIX]{x1D70B}i}.\qquad\end{eqnarray}$$

Note that the possible pole at $w=s$ is cancelled by (3.8). Similarly,

(3.10)$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{q}^{(3)}(s,u,v) & = & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\int _{(0)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{2-s-u-2v}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{r,m,n}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}qam) \\ (\unicode[STIX]{x1D708},q)=1}}e\biggl(\frac{\pm br\unicode[STIX]{x1D708}}{am}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(amq))}{n^{s-w}m^{u-w}r^{v+w}}\biggl(\frac{b}{a}\biggr)^{-w}(amq)^{s-w-1}\frac{dw}{2\unicode[STIX]{x1D70B}i}.\qquad\end{eqnarray}$$

3.5 Voronoi summation

Our next aim is to apply the functional equation for the $r$-sum. This requires some preparation because $b\unicode[STIX]{x1D708}$ is not necessarily coprime to $am$. Therefore we introduce various new variables. We write $(m,b)=\unicode[STIX]{x1D6FD}_{1}$ and $b=\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}$, $m=\unicode[STIX]{x1D6FD}_{1}m^{\prime }$, $(m^{\prime },\unicode[STIX]{x1D6FD}_{2})=1$. Next, we write $(\unicode[STIX]{x1D708},am^{\prime })=m_{1}$, $(m_{1},a)=\unicode[STIX]{x1D6FC}_{1}$, and $a=\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}$, $m_{1}=\unicode[STIX]{x1D6FC}_{1}m_{1}^{\prime }$, $(m_{1}^{\prime },\unicode[STIX]{x1D6FC}_{2})=1$, and further $m^{\prime }=m_{1}^{\prime }m_{2}$, $\unicode[STIX]{x1D708}=\unicode[STIX]{x1D6FC}_{1}m_{1}^{\prime }\unicode[STIX]{x1D708}^{\prime }$, $(\unicode[STIX]{x1D708}^{\prime },\unicode[STIX]{x1D6FC}_{2}m_{2})=1$. Dropping the primes for notational simplicity, we recast the second line in (3.9) as

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\mathop{\sum }_{r,n}\mathop{\sum }_{\substack{ (m_{1},\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)=1 \\ (m_{2},q\unicode[STIX]{x1D6FD}_{2})=1}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}) \\ (\unicode[STIX]{x1D708},\unicode[STIX]{x1D6FC}_{2}m_{2})=1}}e\biggl(\frac{\pm \overline{q}\unicode[STIX]{x1D6FD}_{2}r\unicode[STIX]{x1D708}}{\unicode[STIX]{x1D6FC}_{2}m_{2}}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}))}{n^{s-w}(\unicode[STIX]{x1D6FD}_{1}m_{1}m_{2})^{u-s+1}r^{v+w}}\frac{a^{s-1}}{b^{w}}\end{eqnarray}$$

and the second line in (3.10) as

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\mathop{\sum }_{r,n}\mathop{\sum }_{\substack{ (m_{1},\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)=1 \\ (m_{2},\unicode[STIX]{x1D6FD}_{2})=1}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\,(q\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}) \\ (\unicode[STIX]{x1D708},\unicode[STIX]{x1D6FC}_{2}m_{2}q)=1}}e\biggl(\frac{\pm \unicode[STIX]{x1D6FD}_{2}r\unicode[STIX]{x1D708}}{\unicode[STIX]{x1D6FC}_{2}m_{2}}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}q))}{n^{s-w}(\unicode[STIX]{x1D6FD}_{1}m_{1}m_{2})^{u-s+1}r^{v+w}}\frac{a^{s-1}}{b^{w}}.\end{eqnarray}$$

Note that in both cases the $m_{1}$-sum is $\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)$. Both terms are now in shape to apply Voronoi summation. We express the $r$-sum in terms of the twisted $L$-function (2.5), shift the $w$-contour to $\Re w=-4$ (picking up a possible residue at $w=1-v$), and apply the functional equation (2.7). In this way we see that $\tilde{{\mathcal{T}}}_{a,b,q}(s,u,v)$ equals

(3.11)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}\int _{(-4)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-s+w}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,G_{f}^{-\unicode[STIX]{x1D70F}}(1-v-w)\mathop{\sum }_{r,n}\mathop{\sum }_{(m_{2},q\unicode[STIX]{x1D6FD}_{2})=1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}) \\ (\unicode[STIX]{x1D708},\unicode[STIX]{x1D6FC}_{2}m_{2})=1}}e\biggl(\frac{\pm \unicode[STIX]{x1D70F}q\overline{\unicode[STIX]{x1D6FD}_{2}\unicode[STIX]{x1D708}}r}{\unicode[STIX]{x1D6FC}_{2}m_{2}}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}))}{n^{s-w}(\unicode[STIX]{x1D6FD}_{1}m_{2})^{u-s+1}r^{1-v-w}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{a^{s-1}}{(\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2})^{w}}(\unicode[STIX]{x1D6FC}_{2}m_{2})^{1-2v-2w}\frac{dw}{2\unicode[STIX]{x1D70B}i}+{\mathcal{P}}_{a,b,q}^{(1)}(s,u,v),\end{eqnarray}$$

where the polar term ${\mathcal{P}}_{a,b,q}^{(1)}(s,u,v)$ vanishes unless $f$ is Eisenstein, in which case it equals

(3.12)$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{a,b,q}^{(1)}(s,u,v) & = & \displaystyle \mathop{\{}_{w=1-v}\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-s+w}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\mid n}\mathop{\sum }_{(m_{2},q\unicode[STIX]{x1D6FD}_{2})=1}\frac{a^{s-1}\unicode[STIX]{x1D6FD}_{1}r_{\unicode[STIX]{x1D6FC}_{2}m_{2}}(n/\unicode[STIX]{x1D6FD}_{1})}{(\unicode[STIX]{x1D6FD}_{1}m_{2})^{u-s+1}n^{s-w}b^{w}}L\biggl(v+w,\frac{\ast }{\unicode[STIX]{x1D6FC}_{2}m_{2}},f\biggr),\qquad\end{eqnarray}$$

where $r_{c}(n)$ denotes the Ramanujan sum. (Recall that by (2.6) the residue is independent of the numerator $\ast$ in the twist of the $L$-function.) Similarly,

(3.13)$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{q}^{(3)}(s,u,v) & = & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}\int _{(-4)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-w-u-2v}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)G_{f}^{-\unicode[STIX]{x1D70F}}(1-v-w)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{r,n}\mathop{\sum }_{(m_{2},\unicode[STIX]{x1D6FD}_{2})=1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}) \\ (\unicode[STIX]{x1D708},\unicode[STIX]{x1D6FC}_{2}m_{2}q)=1}}e\biggl(\frac{\pm \unicode[STIX]{x1D70F}\overline{\unicode[STIX]{x1D6FD}_{2}\unicode[STIX]{x1D708}}r}{\unicode[STIX]{x1D6FC}_{2}m_{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}q))}{n^{s-w}(\unicode[STIX]{x1D6FD}_{1}m_{2})^{u-s+1}r^{1-v-w}}\frac{a^{s-1}}{b^{w}}(\unicode[STIX]{x1D6FC}_{2}m_{2})^{1-2v-2w}\frac{dw}{2\unicode[STIX]{x1D70B}i}+{\mathcal{P}}_{a,b,q}^{(2)}(s,u,v),\qquad\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{a,b,q}^{(2)}(s,u,v) & = & \displaystyle \mathop{\{}_{w=1-v}\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-w-u-2v}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\mid n}\mathop{\sum }_{(m_{2},\unicode[STIX]{x1D6FD}_{2})=1}\frac{a^{s-1}\unicode[STIX]{x1D6FD}_{1}r_{\unicode[STIX]{x1D6FC}_{2}m_{2}q}(n/\unicode[STIX]{x1D6FD}_{1})}{(\unicode[STIX]{x1D6FD}_{1}m_{2})^{u-s+1}n^{s-w}b^{w}}L\biggl(v+w,\frac{\ast }{\unicode[STIX]{x1D6FC}_{2}m_{2}},f\biggr).\nonumber\end{eqnarray}$$

We will compute the two polar terms in a moment, but we observe already at this point that now the time has come to combine the two main terms. Indeed, the main term in (3.13) simply counteracts the condition $(m_{2},q)=1$ of the main term in (3.11) and supplies the missing terms $q\mid m$. Combining the two, we see that

$$\begin{eqnarray}\tilde{{\mathcal{T}}}_{a,b,q}(s,u,v)+{\mathcal{T}}_{q}^{(3)}(s,u,v)={\mathcal{T}}_{a,b,q}^{\ast }(s,u,v)+\mathop{\sum }_{j=1}^{2}{\mathcal{P}}_{a,b,q}^{(j)}(s,u,v),\end{eqnarray}$$

where

(3.14)$$\begin{eqnarray}\displaystyle {\mathcal{T}}_{a,b,q}^{\ast }(s,u,v) & = & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}\int _{(-4)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)G_{f}^{-\unicode[STIX]{x1D70F}}(1-v-w)q^{1-w-u-2v}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\frac{\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{s-2v-2w}}{\unicode[STIX]{x1D6FD}_{1}^{u-s+1+w}\unicode[STIX]{x1D6FD}_{2}^{w}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{r,n}\mathop{\sum }_{(m_{2},\unicode[STIX]{x1D6FD}_{2})=1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D708}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}) \\ (\unicode[STIX]{x1D708},\unicode[STIX]{x1D6FC}_{2}m_{2})=1}}e\biggl(\frac{\pm \unicode[STIX]{x1D70F}q\overline{\unicode[STIX]{x1D6FD}_{2}\unicode[STIX]{x1D708}}r}{\unicode[STIX]{x1D6FC}_{2}m_{2}}\biggr)\frac{\unicode[STIX]{x1D706}(r)e(\unicode[STIX]{x1D70E}n\unicode[STIX]{x1D708}/(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{1}m_{2}))}{n^{s-w}m_{2}^{u-s+2v+2w}r^{1-v-w}}\frac{dw}{2\unicode[STIX]{x1D70B}i}.\qquad\end{eqnarray}$$

3.6 Computation of polar terms

In this subsection we compute ${\mathcal{P}}_{a,b,q}^{(j)}(s,u,v)$ for $j=1,2$. We consider first

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\mathop{\sum }_{n}\mathop{\sum }_{(m_{2},q\unicode[STIX]{x1D6FD}_{2})=1}\frac{a^{s-1}\unicode[STIX]{x1D6FD}_{1}r_{\unicode[STIX]{x1D6FC}_{2}m_{2}}(n)}{(\unicode[STIX]{x1D6FD}_{1}m_{2})^{u-s+1}(\unicode[STIX]{x1D6FD}_{1}n)^{s-w}b^{w}}L\biggl(v+w,\frac{\ast }{\unicode[STIX]{x1D6FC}_{2}m_{2}},f\biggr)\end{eqnarray}$$

corresponding to the last four sums in (3.12) for $w$ in a neighbourhood of $1-v$. Substituting

$$\begin{eqnarray}r_{\unicode[STIX]{x1D6FC}_{2}m_{2}}(n)=\mathop{\sum }_{\substack{ d_{1}d_{2}=\unicode[STIX]{x1D6FC}_{2}m \\ d_{1}\mid n}}d_{1}\unicode[STIX]{x1D707}(d_{2}),\end{eqnarray}$$

we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D701}(s-w)\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\frac{a^{s-1}}{\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{w}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{(m_{2},q\unicode[STIX]{x1D6FD}_{2})=1}\mathop{\sum }_{d_{1}d_{2}=\unicode[STIX]{x1D6FC}_{2}m_{2}}\frac{\unicode[STIX]{x1D707}(d_{2})}{d_{1}^{s-w-1}}\frac{1}{m_{2}^{u-s+1}}L\biggl(v+w,\frac{\ast }{\unicode[STIX]{x1D6FC}_{2}m_{2}},f\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D701}(s-w)\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\frac{a^{s-1}}{\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{w}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FC}_{2}\mid d_{1}d_{2} \\ (d_{1}d_{2}/\unicode[STIX]{x1D6FC}_{2},q\unicode[STIX]{x1D6FD}_{2})=1}}\frac{\unicode[STIX]{x1D707}(d_{2})}{d_{1}^{s-w-1}}\frac{\unicode[STIX]{x1D6FC}_{2}^{u-s+1}}{(d_{1}d_{2})^{u-s+1}}L\biggl(v+w,\frac{\ast }{d_{1}d_{2}},f\biggr).\nonumber\end{eqnarray}$$

We write $(d_{1},\unicode[STIX]{x1D6FC}_{2})=A_{1}$, $A_{1}A_{2}=\unicode[STIX]{x1D6FC}_{2}$, $A_{2}\mid d_{2}$, eventually getting

(3.15)$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{a,b,q}^{(1)}(s,u,v) & = & \displaystyle \mathop{\{}_{w=1-v}\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-s+w}\unicode[STIX]{x1D701}(s-w)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}A_{1}A_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\frac{\unicode[STIX]{x1D701}^{(A_{1}A_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\unicode[STIX]{x1D707}(A_{2})A_{2}^{w}}{\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{w}(\unicode[STIX]{x1D6FC}_{1}A_{2})^{1-s}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{(d_{1}d_{2},qA_{2}\unicode[STIX]{x1D6FD}_{2})=1}\frac{\unicode[STIX]{x1D707}(d_{2})L(v+w,\ast /A_{1}A_{2}d_{1}d_{2},f)}{d_{1}^{u-w}d_{2}^{u-s+1}}.\end{eqnarray}$$

By (2.6), this is a linear combination of

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}A_{1}A_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D6F7}_{a,b,q}^{\pm }(1-v)G^{-\unicode[STIX]{x1D70E}}(s+v-1)q^{2-v-s}\unicode[STIX]{x1D701}(s+v-1)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D701}^{(A_{1}A_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1-s+u)\unicode[STIX]{x1D701}^{(qA_{2}\unicode[STIX]{x1D6FD}_{2})}(u+v)}{\unicode[STIX]{x1D701}^{(qA_{2}\unicode[STIX]{x1D6FD}_{2})}(2+u-s)\unicode[STIX]{x1D6FC}_{1}^{1-s}A_{2}^{1+v-s}A_{1}\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{1-v}}\nonumber\end{eqnarray}$$

and derivatives thereof. The same computation shows that

(3.16)$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{a,b,q}^{(2)}(s,u,v) & = & \displaystyle \mathop{\{}_{w=1-v}\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E}\in \{\pm \}}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G^{-\unicode[STIX]{x1D70E}}(s-w)q^{1-w-u-2v}\unicode[STIX]{x1D701}(s-w)\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{q_{1}q_{2}=q}\frac{1}{q_{1}^{s-w-1}}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}A_{1}A_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\frac{\unicode[STIX]{x1D701}^{(A_{1}A_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1-s+u)\unicode[STIX]{x1D707}(A_{2}q_{2})A_{1}^{w}}{\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{w}(\unicode[STIX]{x1D6FC}_{1}A_{2})^{1-s}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\substack{ (d_{1},A_{2}\unicode[STIX]{x1D6FD}_{2}q_{2})=1 \\ (d_{2},A_{2}\unicode[STIX]{x1D6FD}_{2})=1}}\frac{\unicode[STIX]{x1D707}(d_{2})L(v+w,\ast /A_{1}A_{2}d_{1}d_{2},f)}{d_{1}^{u-w}d_{2}^{u-s+1}},\end{eqnarray}$$

which is a linear combination of

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{q_{1}q_{2}=q}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}A_{1}A_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D6F7}_{a,b,q}^{\pm }(1-v)G^{-\unicode[STIX]{x1D70E}}(s+v-1)\unicode[STIX]{x1D701}(s+v-1)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{q_{1}^{2-s-u-2v}}{q_{2}^{u+v}}\frac{\unicode[STIX]{x1D701}^{(A_{1}A_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1-s+u)\unicode[STIX]{x1D701}^{(A_{2}\unicode[STIX]{x1D6FD}_{2}q_{2})}(u+v-1)}{\unicode[STIX]{x1D701}^{(A_{2}\unicode[STIX]{x1D6FD}_{2})}(2+u-s)\unicode[STIX]{x1D6FC}_{1}^{1-s}A_{2}^{2-s}A_{1}^{v}\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{1-v}}\nonumber\end{eqnarray}$$

and derivatives thereof.

3.7 Application of the Kuznetsov formula

We return to (3.14) and recognize the $\unicode[STIX]{x1D708}$-sum as a Kloosterman sum. More precisely, the $\unicode[STIX]{x1D708}$-sum vanishes unless $\unicode[STIX]{x1D6FD}_{1}\mid n$, so that the second and third lines of (3.14) equal

(3.17)$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}\unicode[STIX]{x1D6FD}_{2}=b}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}\unicode[STIX]{x1D6FD}_{2}q)}(1+u-s)\frac{\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{s-2v-2w}}{\unicode[STIX]{x1D6FD}_{1}^{u}\unicode[STIX]{x1D6FD}_{2}^{w}}\mathop{\sum }_{r,n}\mathop{\sum }_{(m_{2},\unicode[STIX]{x1D6FD}_{2})=1}\frac{\unicode[STIX]{x1D706}(r)S(\pm \unicode[STIX]{x1D70F}q\overline{\unicode[STIX]{x1D6FD}_{2}}r,\unicode[STIX]{x1D70E}n,\unicode[STIX]{x1D6FC}_{2}m_{2})}{n^{s-w}m_{2}^{u-s+2v+2w}r^{1-v-w}}.\end{eqnarray}$$

For $(\unicode[STIX]{x1D6FD}_{2},q\unicode[STIX]{x1D6FC}_{2}m)=1$, we have, by the twisted multiplicativity of Kloosterman sums,

$$\begin{eqnarray}S(\pm \unicode[STIX]{x1D70F}qr,\unicode[STIX]{x1D70E}\unicode[STIX]{x1D6FD}_{2}n,\unicode[STIX]{x1D6FD}_{2}\unicode[STIX]{x1D6FC}_{2}m)=r_{\unicode[STIX]{x1D6FD}_{2}}(r)S(\pm \unicode[STIX]{x1D70F}qr\overline{\unicode[STIX]{x1D6FD}_{2}},\unicode[STIX]{x1D70E}n,\unicode[STIX]{x1D6FC}_{2}m).\end{eqnarray}$$

At this point, we use the fact that $b$ is square-free; in particular, the Ramanujan sum $r_{\unicode[STIX]{x1D6FD}_{2}}(r)$ does not vanish. Write $B_{1}=(\unicode[STIX]{x1D6FD}_{2},r)$, $B_{2}B_{1}=\unicode[STIX]{x1D6FD}_{2}$, $r=r^{\prime }B_{1}$, $(r^{\prime },B_{2})=1$. Then

$$\begin{eqnarray}\displaystyle S(\pm \unicode[STIX]{x1D70F}qr\overline{\unicode[STIX]{x1D6FD}_{2}},\unicode[STIX]{x1D70E}n,\unicode[STIX]{x1D6FC}_{2}m) & = & \displaystyle \frac{1}{r_{\unicode[STIX]{x1D6FD}_{2}}(r)}S(\pm \unicode[STIX]{x1D70F}qr^{\prime }B_{1},\unicode[STIX]{x1D70E}B_{1}B_{2}n,B_{1}B_{2}\unicode[STIX]{x1D6FC}_{2}m)\nonumber\\ \displaystyle & = & \displaystyle \frac{\unicode[STIX]{x1D719}(B_{1})}{\unicode[STIX]{x1D719}(B_{1})\unicode[STIX]{x1D707}(B_{2})}S(\pm \unicode[STIX]{x1D70F}qr^{\prime },\unicode[STIX]{x1D70E}B_{2}n,B_{2}\unicode[STIX]{x1D6FC}_{2}m),\nonumber\end{eqnarray}$$

so that (3.17) is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{1}B_{2}=b}\frac{\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}B_{1}B_{2}q)}(1+u-s)\unicode[STIX]{x1D707}(B_{2})\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{s-2v-2w}}{\unicode[STIX]{x1D6FD}_{1}^{u}B_{2}^{w}B_{1}^{1-v}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{(r,B_{2})=1,n}\mathop{\sum }_{(m_{2},\unicode[STIX]{x1D6FD}_{2})=1}\frac{\unicode[STIX]{x1D706}(rB_{1})S(\pm \unicode[STIX]{x1D70F}qr,\unicode[STIX]{x1D70E}B_{2}n,B_{2}\unicode[STIX]{x1D6FC}_{2}m_{2})}{n^{s-w}m_{2}^{u-s+2v+2w}r^{1-v-w}}.\nonumber\end{eqnarray}$$

Here we can drop the condition $(m_{2},B_{2})=1$, since otherwise the Kloosterman sum vanishes (since $(r,B_{2})=1$). We remove the remaining condition $(m_{2},B_{1})=1$ by Möbius inversion, getting

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{3}B_{4}B_{2}=b}\frac{\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}B_{2}B_{3}B_{4}q)}(1+u-s)\unicode[STIX]{x1D707}(B_{2})\unicode[STIX]{x1D707}(B_{3})\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{s-2v-2w}}{\unicode[STIX]{x1D6FD}_{1}^{u}B_{2}^{w}B_{3}^{1+u-s+v+2w}B_{4}^{1-v}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \mathop{\sum }_{(r,B_{2})=1,n}\mathop{\sum }_{m_{2}}\frac{\unicode[STIX]{x1D706}(rB_{3}B_{4})S(\pm \unicode[STIX]{x1D70F}qr,\unicode[STIX]{x1D70E}B_{2}n,B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}m_{2})}{n^{s-w}m_{2}^{u-s+2v+2w}r^{1-v-w}}.\nonumber\end{eqnarray}$$

Rearranging, we obtain the final expression

(3.18)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}q^{(3-s-u-2v)/2}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{3}B_{4}B_{2}=b}\frac{\unicode[STIX]{x1D707}(B_{2})\unicode[STIX]{x1D707}(B_{3})\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{u}\unicode[STIX]{x1D701}^{(\unicode[STIX]{x1D6FC}_{2}B_{2}B_{3}B_{4}q)}(1-s+u)}{\unicode[STIX]{x1D6FD}_{1}^{u}B_{2}^{(s-u-1-2v)/2}B_{3}^{1-v}B_{4}^{1-v}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{(r,B_{2})=1,n}\frac{\unicode[STIX]{x1D706}(rB_{3}B_{4})}{n^{(s+u-1+2v)/2}r^{(1-s+u)/2}}\mathop{\sum }_{B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}\mid m_{2}}\frac{S(\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}qr,B_{2}n,m_{2})}{m_{2}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}\biggl(\frac{\sqrt{qrB_{2}n}}{m_{2}}\biggr),\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

for ${\mathcal{T}}_{a,b,q}^{\ast }(s,u,v)$, where

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F9}(x) & = & \displaystyle \unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(x)=x^{u-s+2v-1}\int _{(-4)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }(w)G_{f}^{-\unicode[STIX]{x1D70F}}(1-v-w)G^{-\unicode[STIX]{x1D70E}}(s-w)x^{2w}\frac{dw}{2\unicode[STIX]{x1D70B}i}\nonumber\\ \displaystyle & = & \displaystyle \int _{(0)}\unicode[STIX]{x1D6F7}_{a,b,s}^{\pm }\biggl(\frac{1+s-u-2v-w}{2}\biggr)G_{f}^{-\unicode[STIX]{x1D70F}}\biggl(\frac{1-s+u+w}{2}\biggr)G^{-\unicode[STIX]{x1D70E}}\biggl(\frac{s+u+2v-1+w}{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,x^{-w}\frac{dw}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

In the region (3.1), the integrand is holomorphic in $2\unicode[STIX]{x1D703}-8<\Re w<34$ (recalling (2.8) and (3.7)) and rapidly decaying on vertical lines; in particular, the assumption $x^{j}\unicode[STIX]{x1D6F9}^{(j)}(x)\ll \min (x,x^{-3/2})$ for $0\leqslant j\leqslant 3$ of the Kuznetsov formula (2.14) is satisfied. By (2.14), the $m_{2}$-sum equals

$$\begin{eqnarray}{\mathcal{A}}_{B}^{\text{Maa}\unicode[STIX]{x00DF}}(\unicode[STIX]{x1D716}qr,B_{2}r;\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9})+{\mathcal{A}}_{B}^{\text{Eis}}(\unicode[STIX]{x1D716}qr,B_{2}r;\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9})+{\mathcal{A}}_{B}^{\text{hol}}(\unicode[STIX]{x1D716}qr,B_{2}r;\mathscr{L}^{\text{hol}}\unicode[STIX]{x1D6F9})\end{eqnarray}$$

with $B=B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}$ and $\unicode[STIX]{x1D716}=\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}$. In the larger region

(3.19)$$\begin{eqnarray}1/2\leqslant \Re s,\Re v\leqslant 3,\quad \Re s\leqslant \Re u\leqslant 11,\end{eqnarray}$$

the integrand of $\unicode[STIX]{x1D6F9}$ is holomorphic in $2\unicode[STIX]{x1D703}-1<\Re w<32$ (and meromorphic in $|\Re w|<32$) and rapidly decaying on vertical lines. By [Reference Blomer and KhanBK19b, Lemma 3a] and (3.7), we conclude that, uniformly in this region,

(3.20)$$\begin{eqnarray}\mathscr{L}^{\pm }\unicode[STIX]{x1D6F9}(t)\ll (1+|t|)^{-30},\quad \mathscr{L}^{\text{hol}}\unicode[STIX]{x1D6F9}(k)\ll k^{-30}\end{eqnarray}$$

as long as $a\asymp b$ for $t\in \mathbb{R}\cup [-7i/64,7i/64]$ and $k\in 2\mathbb{N}$. Recall again that $\unicode[STIX]{x1D6F9}$ depends on $s,u,v$, which are currently restricted to (3.1).

3.8 The cuspidal contribution

We start with the analysis of the Maaß spectrum. Inserting definitions (2.15) and (2.9) and using the notation and conventions of §2.4, we obtain

$$\begin{eqnarray}\displaystyle {\mathcal{A}}_{B}^{\text{Maa}\unicode[STIX]{x00DF}}(\unicode[STIX]{x1D716}qr,B_{2}r;\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}) & = & \displaystyle \mathop{\sum }_{B_{0}\mid B}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(B_{0})}\mathop{\sum }_{M\mid (B/B_{0})}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,B}(\unicode[STIX]{x1D716}qr)\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,B}(B_{2}n)\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}(t_{\unicode[STIX]{x1D713}})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{B_{0}\mid B}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(B_{0})}\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}^{(1-\unicode[STIX]{x1D716})/2}\mathop{\sum }_{M\mid (B/B_{0})}\frac{\mathop{\prod }_{p\mid B_{0}}(1-p^{-2})}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})B\unicode[STIX]{x1D708}(B)}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{d_{1},d_{2}\mid M}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{1})\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{2})\frac{d_{1}d_{2}}{M}\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}\biggl(\frac{qr}{d_{1}}\biggr)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}\biggl(\frac{B_{2}n}{d_{2}}\biggr)\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}(t_{\unicode[STIX]{x1D713}}).\nonumber\end{eqnarray}$$

Summing over $n$ and $r$ as in (3.18), we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{B_{0}\mid B}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(B_{0})}\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}^{(1-\unicode[STIX]{x1D716})/2}\mathop{\sum }_{M\mid (B/B_{0})}\frac{\mathop{\prod }_{p\mid B_{0}}(1-p^{-2})}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})B\unicode[STIX]{x1D708}(B)}\mathop{\sum }_{d_{1},d_{2}\mid M}\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{1})\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{2})\frac{d_{1}d_{2}}{M}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{(r,B_{2})=1}\frac{\unicode[STIX]{x1D706}(rB_{3}B_{4})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(qr/d_{1})}{r^{(1-s+u)/2}}\mathop{\sum }_{n}\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2}n/d_{2})}{n^{(s+u-1+2v)/2}}\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}(t_{\unicode[STIX]{x1D713}}).\nonumber\end{eqnarray}$$

Since $(q,B)=1$, we have $(d_{1},q)=1$, and so by (2.1), the $r$-sum equals

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x1D6FF}_{(d_{1},B_{2})=1}}{d_{1}^{(1-s+u)/2}}\mathop{\sum }_{(r,B_{2})=1}\frac{\unicode[STIX]{x1D706}(rd_{1}B_{3}B_{4})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(qr)}{r^{(1-s+u)/2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D6FF}_{(d_{1},B_{2})=1}}{d_{1}^{(1-s+u)/2}}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FF}_{1}\mid d_{1}B_{3}B_{4} \\ (\unicode[STIX]{x1D6FF}_{1},B_{0})=1}}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1})\unicode[STIX]{x1D706}(d_{1}B_{3}B_{4}/\unicode[STIX]{x1D6FF}_{1})}{\unicode[STIX]{x1D6FF}_{1}^{(1-s+u)/2}}\mathop{\sum }_{\unicode[STIX]{x1D6FF}_{2}\mid q}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{2})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q/\unicode[STIX]{x1D6FF})}{\unicode[STIX]{x1D6FF}_{2}^{(1-s+u)/2}}\frac{L^{(B_{2})}(f\times \unicode[STIX]{x1D713},(1-s+u)/2)}{\unicode[STIX]{x1D701}^{(B_{2})}(1-s+u)}.\nonumber\end{eqnarray}$$

Similarly, the $n$-sum equals

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\frac{(B_{2},d_{2})}{d_{2}}\biggr)^{(s+u-1+2v)/2}\mathop{\sum }_{n}\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2}n/(d_{2},B_{2}))}{n^{(s+u-1+2v)/2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl(\frac{(B_{2},d_{2})}{d_{2}}\biggr)^{(s+u-1+2v)/2}\mathop{\sum }_{\substack{ B^{\ast }\mid B_{2}/(d_{2},B_{2}) \\ (B^{\ast },B_{0})=1}}\frac{\unicode[STIX]{x1D707}(B^{\ast })\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2}/((d_{2},B_{2})B^{\ast }))}{(B^{\ast })^{(s+u-1+2v)/2}}L(\unicode[STIX]{x1D713},(s+u-1+2v)/2).\nonumber\end{eqnarray}$$

Putting everything together, the Maaß contribution to (3.18) equals

$$\begin{eqnarray}\mathop{\sum }_{A\mid ab}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(A)}\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Maa}\unicode[STIX]{x00DF}}(s,u,v,\unicode[STIX]{x1D713})\frac{L(\frac{s+u-1+2v}{2},\unicode[STIX]{x1D713})L(\frac{1-s+u}{2},f\times \unicode[STIX]{x1D713})}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})},\end{eqnarray}$$

where

(3.21)$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Maa}\unicode[STIX]{x00DF}}(s,u,v,\unicode[STIX]{x1D713})\nonumber\\ \displaystyle & & \displaystyle \quad :=\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}^{(1\mp \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F})/2}q^{(3-s-u-2v)/2}\underset{A\mid B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}{\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{3}B_{4}B_{2}=b}}\frac{\unicode[STIX]{x1D707}(B_{2})\unicode[STIX]{x1D707}(B_{3})\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{u}}{\unicode[STIX]{x1D6FD}_{1}^{u}B_{2}^{(s-u-1-2v)/2}B_{3}^{1-v}B_{4}^{1-v}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{M\mid (B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}/A)}\mathop{\sum }_{\substack{ d_{1},d_{2}\mid M \\ (d_{1},B_{2})=1}}\frac{\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{1})\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M,d_{2})d_{1}d_{2}((B_{2},d_{2})/d_{2})^{(s+u-1+2v)/2}}{Md_{1}^{(1-s+u)/2}L_{B_{2}}(f\times \unicode[STIX]{x1D713},(1-s+u)/2)\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}_{2}B_{3}B_{4}q}(1-s+u)}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FF}_{1}\mid d_{1}B_{3}B_{4} \\ (\unicode[STIX]{x1D6FF}_{1},A)=1}}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1})\unicode[STIX]{x1D706}(d_{1}B_{3}B_{4}/\unicode[STIX]{x1D6FF}_{1})}{\unicode[STIX]{x1D6FF}_{1}^{(1-s+u)/2}}\mathop{\sum }_{\unicode[STIX]{x1D6FF}_{2}\mid q}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{2})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q/\unicode[STIX]{x1D6FF})}{\unicode[STIX]{x1D6FF}_{2}^{(1-s+u)/2}}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\sum }_{\substack{ B^{\ast }\mid (B_{2}/(d_{2},B_{2})) \\ (B^{\ast },A)=1}}\frac{\unicode[STIX]{x1D707}(B^{\ast })\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2}/((d_{2},B_{2})B^{\ast }))}{(B^{\ast })^{(s+u-1+2v)/2}}\mathscr{L}^{\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(t_{\unicode[STIX]{x1D713}})\end{eqnarray}$$

for $\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(A)$ of spectral parameter $t_{\unicode[STIX]{x1D713}}$ and parity $\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}$. Clearly this expression is holomorphic in the region (3.19). We proceed to confirm the bound (1.11) for $\Re s=\Re u=\Re v=1/2$. This requires a little more than a trivial bound of (3.21). The critical variable is $B_{2}$. In order to get enough saving, we need to exploit some cancellation. To this end, we write $M=M_{1}M_{2}$, where $(M_{1},B_{2})=1$ and $M_{2}\mid B_{2}$. (Recall that $ab$ is square-free.) Since $(d_{1},B_{2})=1$, we have $d_{1}\mid M_{1}$, and we write $d_{2}=d_{2}^{\prime }d_{2}^{\prime \prime }$ with $d_{2}^{\prime }\mid M_{1}$, $d_{2}^{\prime \prime }\mid M_{2}$. In this way, the $M_{2}$-sum becomes

(3.22)$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{d_{2}^{\prime \prime }\mid M_{2}\mid (B_{2}/(A,B_{2}))}\frac{\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M_{2},1)\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(M_{2},d_{2}^{\prime \prime })d_{2}^{\prime \prime }((B_{2},d_{2}^{\prime \prime })/d_{2}^{\prime \prime })^{(s+u-1+2v)/2}}{M_{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{\substack{ B^{\ast }\mid (B_{2}/(d_{2}^{\prime \prime },B_{2})) \\ (B^{\ast },A)=1}}\frac{\unicode[STIX]{x1D707}(B^{\ast })\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2}/((d_{2}^{\prime \prime },B_{2})B^{\ast }))}{(B^{\ast })^{(s+u-1+2v)/2}}.\end{eqnarray}$$

If $B_{2}\mid A$, this is equal to $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(B_{2})\ll B_{2}^{-1/2}$ by (2.3). If $B_{2}\nmid A$, this is equal to

$$\begin{eqnarray}\mathop{\prod }_{p\mid B_{2}}\biggl(\biggl(\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)-\frac{1}{p^{(s+u-1+2v)/2}}\biggr)\biggl(1+\frac{\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(p,1)^{2}}{p}\biggr)+\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(p,1)\unicode[STIX]{x1D709}_{\unicode[STIX]{x1D713}}(p,p)\biggr).\end{eqnarray}$$

By (2.10), the leading term $\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)$ cancels (to first-order approximation), and each $p$-factor in the preceding display is bounded by $p^{-1/2}+p^{3\unicode[STIX]{x1D703}-1}\ll p^{-1/2}$ for $\Re s=\Re u=\Re v=1/2$. Hence in all cases the $M_{2}$-sum is $\ll B_{2}^{-1/2+\unicode[STIX]{x1D700}}$. Combining with (3.20) and (2.11), we obtain

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Maa}\unicode[STIX]{x00DF}}(s,u,v,\unicode[STIX]{x1D713})\ll \frac{(abq)^{\unicode[STIX]{x1D700}}q^{1/2}}{(1+|t_{\unicode[STIX]{x1D713}}|)^{30}}\underset{A\mid B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}{\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{3}B_{4}B_{2}=b}}\mathop{\sum }_{\substack{ d_{1},d_{2}\mid M\mid (B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}/A) \\ (M,B_{2})=1}}\frac{d_{1}^{1/2}d_{2}^{1/2-\unicode[STIX]{x1D703}}(1+|\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q)|)}{\unicode[STIX]{x1D6FC}_{2}^{1/2}B_{3}^{3/2-\unicode[STIX]{x1D703}}M^{1-2\unicode[STIX]{x1D703}}B_{2}^{1/2}}\end{eqnarray}$$

for $\Re s=\Re u=\Re v=1/2$. This is increasing in $d_{1},d_{2}$, and the result is increasing in $M$, so that one easily confirms (1.11).

The same formula holds for the holomorphic contribution to (3.18), except that the transform $\mathscr{L}^{\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(t_{\unicode[STIX]{x1D713}})$ has to be replaced with $\mathscr{L}^{\text{hol}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(k_{\unicode[STIX]{x1D713}})$ and $\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}=0$ if $\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}=-1$. The corresponding bound (1.12) is even simpler to obtain because $\unicode[STIX]{x1D703}=0$ in the holomorphic case.

3.9 The Eisenstein contribution

By (2.12), we have

(3.23)$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{A}}_{B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}^{\text{Eis}}(\unicode[STIX]{x1D716}qr,B_{2}r;\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9})\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{\mathbb{R}}\frac{1}{B|\unicode[STIX]{x1D701}^{(B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2})}(1+2it)|^{2}}\mathop{\sum }_{v\mid B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}\frac{1}{v}\mathop{\sum }_{b_{1},b_{2}\mid v}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}\mid B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}/v}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\unicode[STIX]{x1D707}(b_{1}\unicode[STIX]{x1D6FE}_{1})\unicode[STIX]{x1D707}(b_{2}\unicode[STIX]{x1D6FE}_{2})b_{1}b_{2}\biggl(\frac{b_{1}\unicode[STIX]{x1D6FE}_{2}}{b_{2}\unicode[STIX]{x1D6FE}_{1}}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{qr}{b_{1}\unicode[STIX]{x1D6FE}_{1}},t\biggr)\unicode[STIX]{x1D702}\biggl(\frac{B_{2}n}{b_{2}\unicode[STIX]{x1D6FE}_{2}},-t\biggr)\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}(t)\frac{dt}{2\unicode[STIX]{x1D70B}}.\end{eqnarray}$$

We saw in the previous subsection that the $B_{2}$-variable was the most critical variable, and we finally used the strong bound (2.3) to get a sufficient saving. We do not have a direct analogue of this bound in the Eisenstein case when we make the choice of basis arising from (2.12), but luckily we can obtain additional cancellation by summing non-trivially over the cusps $v$. This again requires some subtle manipulations.

Since $(B_{2},B_{3}\unicode[STIX]{x1D6FC}_{2})=1$, we write $v=v_{1}v_{2}$ with $v_{1}\mid B_{3}\unicode[STIX]{x1D6FC}_{2}$, $v_{2}\mid B_{3}\unicode[STIX]{x1D6FC}_{2}$, $b_{j}=b_{j}^{\prime }b_{j}^{\prime \prime }$, where $b_{j}^{\prime }\mid v_{1}$, $b_{j}^{\prime \prime }\mid v_{2}$, and $\unicode[STIX]{x1D6FE}_{j}=\unicode[STIX]{x1D6FE}_{j}^{\prime }\unicode[STIX]{x1D6FE}_{j}^{\prime \prime }$, where $\unicode[STIX]{x1D6FE}_{j}^{\prime }\mid B_{2}/v_{1}$, $\unicode[STIX]{x1D6FE}_{j}^{\prime \prime }\mid B_{3}\unicode[STIX]{x1D6FC}_{2}/v_{2}$. The key observation is that $(qr,B_{2})=1$ in our application, so that $b_{1}^{\prime }=\unicode[STIX]{x1D6FE}_{1}^{\prime }=1$. In this way, we can recast the previous $v,b_{1},b_{2},\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}$-sum as

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\substack{ b_{2}^{\prime }\mid v_{1}\mid B_{2} \\ b_{1}^{\prime \prime },b_{2}^{\prime \prime }\mid v_{2}\mid B_{3}\unicode[STIX]{x1D6FC}_{2}}}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{2}^{\prime }\mid (B_{2}/v_{1})}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{1}^{\prime \prime },\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }\mid (B_{3}\unicode[STIX]{x1D6FC}_{2}/v_{2})}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D707}(b_{1}^{\prime \prime }\unicode[STIX]{x1D6FE}_{1}^{\prime \prime })\unicode[STIX]{x1D707}(b_{2}^{\prime }b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime })b_{1}^{\prime \prime }b_{2}^{\prime }b_{2}^{\prime \prime }}{v_{1}v_{2}}\biggl(\frac{b_{1}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }}{b_{2}^{\prime }b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{1}^{\prime \prime }}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{qr}{b_{1}^{\prime \prime }\unicode[STIX]{x1D6FE}_{1}^{\prime \prime }},t\biggr)\unicode[STIX]{x1D702}\biggl(\frac{B_{2}n}{b_{2}^{\prime }b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }},-t\biggr).\nonumber\end{eqnarray}$$

We consider only the $B_{2}$-part,

(3.24)$$\begin{eqnarray}\mathop{\sum }_{b_{2}^{\prime }\mid v_{1}\mid B_{2}}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{2}^{\prime }\mid (B_{2}/v_{1})}\frac{1}{v_{1}}\unicode[STIX]{x1D707}(b_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime })b_{2}^{\prime }\biggl(\frac{\unicode[STIX]{x1D6FE}_{2}^{\prime }}{b_{2}^{\prime }}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{B_{2}n}{b_{2}^{\prime }b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }},-t\biggr),\end{eqnarray}$$

for fixed $b_{2}^{\prime \prime },\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }$, where we parametrize $v_{1}=b_{2}^{\prime }b^{\ast }$, $B_{2}/v_{1}=\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}^{\ast }$, getting

$$\begin{eqnarray}\mathop{\sum }_{b_{2}^{\prime }b^{\ast }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}^{\ast }=B_{2}}\frac{1}{b^{\ast }}\unicode[STIX]{x1D707}(b_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime })\biggl(\frac{\unicode[STIX]{x1D6FE}_{2}^{\prime }}{b_{2}^{\prime }}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(b^{\ast }\unicode[STIX]{x1D6FE}^{\ast }\frac{n}{b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }},-t\biggr).\end{eqnarray}$$

We must have $b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }\mid n$, so we write $n=b_{2}^{\prime \prime }\unicode[STIX]{x1D6FE}_{2}^{\prime \prime }n^{\ast }$. Applying the Hecke relation (2.1) for $\unicode[STIX]{x1D702}(n,t)$, we obtain

$$\begin{eqnarray}\mathop{\sum }_{\substack{ b_{2}^{\prime }b^{\ast }\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FE}^{\ast }=B_{2} \\ \unicode[STIX]{x1D6FF}\mid (n^{\ast },b^{\ast }\unicode[STIX]{x1D6FE}^{\ast })}}\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF})\frac{1}{b^{\ast }}\unicode[STIX]{x1D707}(b_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime })\biggl(\frac{\unicode[STIX]{x1D6FE}_{2}^{\prime }}{b_{2}^{\prime }}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{b^{\ast }\unicode[STIX]{x1D6FE}^{\ast }}{\unicode[STIX]{x1D6FF}},-t\biggr)\unicode[STIX]{x1D702}\biggl(\frac{n^{\ast }}{\unicode[STIX]{x1D6FF}},-t\biggr).\end{eqnarray}$$

We parametrize $\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}$, $b^{\ast }=\unicode[STIX]{x1D6FF}_{1}b_{0}$, $\unicode[STIX]{x1D6FE}^{\ast }=\unicode[STIX]{x1D6FF}_{2}\unicode[STIX]{x1D6FE}_{0}$, so that the previous line is equal to

$$\begin{eqnarray}\mathop{\sum }_{\substack{ b_{2}^{\prime }\unicode[STIX]{x1D6FF}_{1}b_{0}\unicode[STIX]{x1D6FE}_{2}^{\prime }\unicode[STIX]{x1D6FF}_{2}\unicode[STIX]{x1D6FE}_{0}=B_{2} \\ \unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}\mid n^{\ast }}}\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})\frac{1}{\unicode[STIX]{x1D6FF}_{1}b}\unicode[STIX]{x1D707}(b_{2}^{\prime }\unicode[STIX]{x1D6FE}_{2}^{\prime })\biggl(\frac{\unicode[STIX]{x1D6FE}_{2}^{\prime }}{b_{2}^{\prime }}\biggr)^{it}\unicode[STIX]{x1D702}(b_{0},-t)\unicode[STIX]{x1D702}(\unicode[STIX]{x1D6FE}_{0},-t)\unicode[STIX]{x1D702}\biggl(\frac{n^{\ast }}{\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}},-t\biggr).\end{eqnarray}$$

The key point is now that, by Möbius inversion, the $b_{2}^{\prime },\unicode[STIX]{x1D6FE}_{2}^{\prime },\unicode[STIX]{x1D6FE}_{0}$-sum disappears, so that (3.24) is equal to

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FF}_{1}b_{0}\unicode[STIX]{x1D6FF}_{2}=B_{2} \\ \unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}\mid n^{\ast }}}\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})\frac{1}{\unicode[STIX]{x1D6FF}_{1}b}\unicode[STIX]{x1D702}(b_{0},-t)\unicode[STIX]{x1D702}\biggl(\frac{n^{\ast }}{\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}},-t\biggr),\end{eqnarray}$$

and hence (3.23) is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\mathbb{R}}\underset{b_{2}\unicode[STIX]{x1D6FE}_{2}\mid n}{\mathop{\sum }_{b_{1},b_{2}\mid v\mid B_{3}\unicode[STIX]{x1D6FC}_{2}}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}\mid B_{3}\unicode[STIX]{x1D6FC}_{2}/v}}\frac{\unicode[STIX]{x1D707}(b_{1}\unicode[STIX]{x1D6FE}_{1})\unicode[STIX]{x1D707}(b_{2}\unicode[STIX]{x1D6FE}_{2})b_{1}b_{2}}{vB_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}|\unicode[STIX]{x1D701}^{(B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2})}(1+2it)|^{2}}\biggl(\frac{b_{1}\unicode[STIX]{x1D6FE}_{2}}{b_{2}\unicode[STIX]{x1D6FE}_{1}}\biggr)^{it}\unicode[STIX]{x1D702}\biggl(\frac{qr}{b_{1}\unicode[STIX]{x1D6FE}_{1}},t\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FF}_{1}b_{0}\unicode[STIX]{x1D6FF}_{2}=B_{2} \\ \unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}\mid n}}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})\unicode[STIX]{x1D702}(b_{0},-t)}{\unicode[STIX]{x1D6FF}_{1}b_{0}}\unicode[STIX]{x1D702}\biggl(\frac{n}{b_{2}\unicode[STIX]{x1D6FE}_{2}\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}},-t\biggr)\mathscr{L}^{\unicode[STIX]{x1D716}}\unicode[STIX]{x1D6F9}(t)\frac{dt}{2\unicode[STIX]{x1D70B}}.\nonumber\end{eqnarray}$$

After this manoeuvre, we are now in shape to sum over $r$ and $n$ as in (3.18). This gives

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\mathbb{R}}\mathop{\sum }_{b_{1},b_{2}\mid v\mid B_{3}\unicode[STIX]{x1D6FC}_{2}}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}\mid (B_{3}\unicode[STIX]{x1D6FC}_{2}/v)}\frac{\unicode[STIX]{x1D707}(b_{1}\unicode[STIX]{x1D6FE}_{1})\unicode[STIX]{x1D707}(b_{2}\unicode[STIX]{x1D6FE}_{2})b_{1}b_{2}}{vB_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}|\unicode[STIX]{x1D701}^{(B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2})}(1+2it)|^{2}}\biggl(\frac{b_{1}\unicode[STIX]{x1D6FE}_{2}}{b_{2}\unicode[STIX]{x1D6FE}_{1}}\biggr)^{it}\mathop{\sum }_{\unicode[STIX]{x1D6FF}_{1}b_{0}\unicode[STIX]{x1D6FF}_{2}=B_{2}}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})\unicode[STIX]{x1D702}(b_{0},-t)}{\unicode[STIX]{x1D6FF}_{1}b_{0}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{(r,B_{2})=1}\frac{\unicode[STIX]{x1D706}(rb_{1}\unicode[STIX]{x1D6FE}_{1}B_{3}B_{4})\unicode[STIX]{x1D702}(qr,t)}{(b_{1}\unicode[STIX]{x1D6FE}_{1}r)^{(1-s+u)/2}}\mathop{\sum }_{n}\frac{\unicode[STIX]{x1D702}(n,-t)}{(b_{2}\unicode[STIX]{x1D6FE}_{2}\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}n)^{(s+u-1+2v)/2}}\mathscr{L}^{\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}}\unicode[STIX]{x1D6F9}(t)\frac{dt}{2\unicode[STIX]{x1D70B}}.\nonumber\end{eqnarray}$$

The $n$-sum can be easily evaluated in terms of the Riemann zeta function. The $r$-sum requires multiple applications of (2.1). Checking local factors, we confirm that, for $(B,q)=(B_{2},Bq)=1$, $B$ square-free, and $q$ prime, we have

$$\begin{eqnarray}\mathop{\sum }_{(B_{2},r)=1}\frac{\unicode[STIX]{x1D706}(rB)\unicode[STIX]{x1D702}(qr,t)}{r^{z}}=\frac{\unicode[STIX]{x1D702}(q,t)-\unicode[STIX]{x1D706}(q)q^{-z}}{1-q^{-2z}}\mathop{\prod }_{p\mid B}\frac{\unicode[STIX]{x1D706}(p)-\unicode[STIX]{x1D702}(p,t)p^{-z}}{1-p^{-2z}}\mathop{\sum }_{(B_{2},r)=1}\frac{\unicode[STIX]{x1D706}(r)\unicode[STIX]{x1D702}(r,t)}{r^{z}}.\end{eqnarray}$$

Putting everything together, the Eisenstein contribution to (3.18) is equal to

(3.25)$$\begin{eqnarray}\int _{\mathbb{R}}\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Eis}}(s,u,v,t)\frac{\unicode[STIX]{x1D701}(\frac{s+u-1+2v}{2}+it)L(\frac{s+u-1+2v}{2}-it)\unicode[STIX]{x1D701}(\frac{1-s+u}{2}+it,f)L(\frac{1-s+u}{2}-it,f)}{\unicode[STIX]{x1D701}(1+2it)\unicode[STIX]{x1D701}(1-2it)}\frac{dt}{2\unicode[STIX]{x1D70B}},\end{eqnarray}$$

where

(3.26)$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Eis}}(s,u,v,t):=\mathop{\sum }_{\pm }\mathop{\sum }_{\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\in \{\pm \}}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2}=a}\mathop{\sum }_{\unicode[STIX]{x1D6FD}_{1}B_{3}B_{4}B_{2}=b}\frac{q^{(3-s-u-2v)/2}\unicode[STIX]{x1D707}(B_{2})\unicode[STIX]{x1D707}(B_{3})\unicode[STIX]{x1D6FC}_{1}^{s-1}\unicode[STIX]{x1D6FC}_{2}^{u}}{\unicode[STIX]{x1D6FD}_{1}^{u}B_{2}^{(s-u-1-2v)/2}B_{3}^{1-v}B_{4}^{1-v}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{b_{1},b_{2}\mid v\mid B_{3}\unicode[STIX]{x1D6FC}_{2}}\mathop{\sum }_{\unicode[STIX]{x1D6FE}_{1},\unicode[STIX]{x1D6FE}_{2}\mid (B_{3}\unicode[STIX]{x1D6FC}_{2}/v)}\frac{\unicode[STIX]{x1D707}(b_{1}\unicode[STIX]{x1D6FE}_{1})\unicode[STIX]{x1D707}(b_{2}\unicode[STIX]{x1D6FE}_{2})b_{1}b_{2}}{vB_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}|\unicode[STIX]{x1D701}^{(B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2})}(1+2it)|^{2}}\biggl(\frac{b_{1}\unicode[STIX]{x1D6FE}_{2}}{b_{2}\unicode[STIX]{x1D6FE}_{1}}\biggr)^{it}\frac{\unicode[STIX]{x1D702}(q,t)-\unicode[STIX]{x1D706}(q)q^{-(1-s+u)/2}}{1-q^{-(1-s+u)}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{\unicode[STIX]{x1D6FF}_{1}b_{0}\unicode[STIX]{x1D6FF}_{2}=B_{2}}\frac{\unicode[STIX]{x1D707}(\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})\unicode[STIX]{x1D702}(b_{0},-t)}{\unicode[STIX]{x1D6FF}_{1}b_{0}(b_{1}\unicode[STIX]{x1D6FE}_{1})^{(1-s+u)/2}(b_{2}\unicode[STIX]{x1D6FE}_{2}\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2})^{(s+u-1+2v)/2}}\mathop{\prod }_{p\mid b_{1}\unicode[STIX]{x1D6FE}_{1}B_{3}B_{4}}\frac{\unicode[STIX]{x1D706}(p)-\unicode[STIX]{x1D702}(p,t)p^{-(1-s+u)/2}}{1-p^{-(1-s+u)}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D701}_{B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}(1+2it)\unicode[STIX]{x1D701}_{B_{2}B_{3}\unicode[STIX]{x1D6FC}_{2}}(1-2it)}{\unicode[STIX]{x1D701}_{\unicode[STIX]{x1D6FC}_{2}B_{3}B_{4}q}(1-s+u)L_{B_{2}}(\frac{1-s+u}{2}+it,f)L_{B_{2}}(\frac{1-s+u}{2}-it,f)}\mathscr{L}^{\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(t).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

The term (3.25) is clearly holomorphic in the range (3.1) and it can easily be extended as long as $\Re (u-s)>1$ and $\Re (s+u+2v)>3$. To pass these two hyperplanes, we observe that the presence of the Riemann zeta function in the numerator contributes residues, and so we apply the argument of [Reference Blomer and KhanBK19b, Lemma 16] to show that the meromorphic continuation of (3.25) in the region $\Re (u-s)<1$ and $\Re (s+u+2v)<3$ is given by the same expression plus the polar term

(3.27)$$\begin{eqnarray}\displaystyle {\mathcal{P}}_{a,b,q}^{(3)}(s,u,v) & := & \displaystyle \sum \mathop{\{}_{\substack{ t=\pm i(1+s-u)/2 \\ t=\pm i(3-s-u-2v)/2}}(\pm i)\frac{\mathop{\unicode[STIX]{x1D6E9}}_{a,b,q}^{\text{Eis}}(s,u,v,t)\unicode[STIX]{x1D701}(\frac{s+u-1+2v}{2}+it)}{\unicode[STIX]{x1D701}(1+2it)\unicode[STIX]{x1D701}(1-2it)}\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D701}\biggl(\frac{s+u-1+2v}{2}-it\biggr)L\biggl(\frac{1-s+u}{2}+it,f\biggr)L\biggl(\frac{1-s+u}{2}-it,f\biggr).\qquad\end{eqnarray}$$

A trivial estimation confirms (1.13) for the term on the right-hand side of (3.25) with $\Re s=\Re u=\Re v=1/2$, $t\in \mathbb{R}$, $a\asymp b$ (which differs from the meromorphic continuation of (3.25) to this region by (3.27)).

It remains to meromorphically continue and bound the joint polar term

(3.28)$$\begin{eqnarray}{\mathcal{P}}_{a,b,q}(s,u,v):=\mathop{\sum }_{j=1}^{3}{\mathcal{P}}_{a,b,q}^{(j)}(s,u,v),\end{eqnarray}$$

where we recall (3.15) and (3.16) for $j=1,2$. In these cases, it is easily seen that ${\mathcal{P}}_{a,b,q}^{(j)}(s,u,v)$ continues meromorphically to a neighbourhood of (3.19), and for $1/2-\unicode[STIX]{x1D700}<\Re s=\Re u=\Re v<1/2+\unicode[STIX]{x1D700}$, $a\asymp b$, we have the bound

$$\begin{eqnarray}|{\mathcal{P}}_{a,b,q}^{(1)}(s,u,v)|+|{\mathcal{P}}_{a,b,q}^{(2)}(s,u,v)|\ll q(ab)^{-1/2}(abq)^{\unicode[STIX]{x1D700}}\end{eqnarray}$$

away from poles. The treatment of ${\mathcal{P}}_{a,b,q}^{(3)}(s,u,v)$ requires slightly more effort, because we need to analyze $\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Eis}}(s,u,v,t)$ for $|\Im t|\leqslant 1/2$. The meromorphic continuation of $\mathscr{L}^{\pm \unicode[STIX]{x1D70E}\unicode[STIX]{x1D70F}}\unicode[STIX]{x1D6F9}_{a,b,s,u,v}^{\pm ,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}}(t)$ with at most finitely many poles (and hence of $\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Eis}}(s,u,v,t)$) to that region follows from [Reference Blomer and KhanBK19b, Lemma 3b]. Again, a trivial upper bound yields

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{a,b,q}^{\text{Eis}}(s,u,v,t)\ll (abq)^{\unicode[STIX]{x1D700}}\frac{q}{(ab)^{1/2-\unicode[STIX]{x1D703}}}\end{eqnarray}$$

for fixed $s,u,v,t$ with $1/2-\unicode[STIX]{x1D700}<\Re s=\Re u=\Re v<1/2+\unicode[STIX]{x1D700}$, $|\Im t|<1/2+\unicode[STIX]{x1D700}$, $a\asymp b$ away from possible poles, so that also

$$\begin{eqnarray}{\mathcal{P}}_{a,b,q}^{(3)}(s,u,v)\ll (abq)^{\unicode[STIX]{x1D700}}\frac{q}{(ab)^{1/2-\unicode[STIX]{x1D703}}}\end{eqnarray}$$

in the region $1/2-\unicode[STIX]{x1D700}<\Re s=\Re u=\Re v<1/2+\unicode[STIX]{x1D700}$, away from possible poles. We have established (1.14) as an equality of meromorphic functions, but since all terms except possibly ${\mathcal{P}}_{a,b,q}(s,u,v)$ are holomorphic for $\Re s=\Re u=\Re v=1/2$, ${\mathcal{P}}_{a,b,q}(s,u,v)$ must also be holomorphic for $\Re s=\Re u=\Re v=1/2$, and the general bound (1.15) then follows by Cauchy’s integral theorem in the same way as at the end of [Reference Blomer and KhanBK19b, §10].

4.1 Initial manipulations

Let $P=TQ$. By ‘negligible’, we mean a quantity that is $O(P^{-100})$. By a dyadic decomposition, we may replace the conditions $q\leqslant Q$, $|t_{\unicode[STIX]{x1D713}}|\leqslant T$ with $\frac{1}{2}Q\leqslant q\leqslant Q$, $\frac{1}{2}T\leqslant t_{\unicode[STIX]{x1D713}}\leqslant T$ or $t_{\unicode[STIX]{x1D713}}\in [0,1]\cup [-i\unicode[STIX]{x1D703},i\unicode[STIX]{x1D703}]$ where in the last case we formally put $T=1$.

Let $E_{3}$ denote the standard minimal Eisenstein series for $\text{SL}_{3}(\mathbb{Z})$ with Fourier coefficients

$$\begin{eqnarray}A(n,m)=\mathop{\sum }_{d\mid (n,m)}\unicode[STIX]{x1D707}(d)\unicode[STIX]{x1D70F}_{3}(n/d)\unicode[STIX]{x1D70F}_{3}(m/d).\end{eqnarray}$$

Then, for $\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)$ with $|t_{\unicode[STIX]{x1D713}}|\leqslant T$, we have

$$\begin{eqnarray}L(s,\unicode[STIX]{x1D713})^{3}=L(s,\unicode[STIX]{x1D713}\times E_{3})=\mathop{\sum }_{n}\mathop{\sum }_{(m,q)=1}\frac{A(n,m)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n)}{n^{s}m^{2s}}=P_{q}(s)\mathop{\sum }_{n,m}\frac{A(n,m)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n)}{n^{s}m^{2s}}\end{eqnarray}$$

for $\Re s>1$, where

$$\begin{eqnarray}P_{q}(s)=\mathop{\prod }_{p\mid q}\biggl(1-\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)}{p^{s}}\biggr)^{-3}\biggl(1-\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(p)}{p^{s}}+\frac{1}{p^{2s}}\biggr)^{3}\end{eqnarray}$$

is holomorphic and uniformly bounded in $\Re s\geqslant 1/2$. By a standard approximate functional equation, we have

$$\begin{eqnarray}|L(1/2,\unicode[STIX]{x1D713})^{3}|\leqslant 2\biggl|\mathop{\sum }_{n,m}\frac{A(n,m)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n)}{n^{1/2}m}V_{\unicode[STIX]{x1D713}}\biggl(\frac{nm^{2}}{q^{3/2}}\biggr)\biggr|,\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle V_{\unicode[STIX]{x1D713}}(y) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{(2)}P({\textstyle \frac{1}{2}}+u)\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+u+\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}+it_{\unicode[STIX]{x1D713}}))^{3}\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+u+\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}-it_{\unicode[STIX]{x1D713}}))^{3}}{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}+it_{\unicode[STIX]{x1D713}}))^{3}\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D713}}-it_{\unicode[STIX]{x1D713}}))^{3}}\unicode[STIX]{x1D70B}^{-3u}e^{u^{2}}y^{-u}\frac{du}{u}.\nonumber\end{eqnarray}$$

Shifting the contour to the far right, we see that $V_{\unicode[STIX]{x1D713}}(y)$ is negligible if $y\geqslant T^{3}P^{\unicode[STIX]{x1D700}}$. Remembering this, we shift the contour to $\Re u=\unicode[STIX]{x1D700}$. There we may truncate the integral at $|\Im u|\leqslant P^{\unicode[STIX]{x1D700}}$ at the cost of a negligible error. Applying a smooth dyadic decomposition, we have shown that

$$\begin{eqnarray}L(1/2,\unicode[STIX]{x1D713})^{3}\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}\int _{-P^{\unicode[STIX]{x1D700}}}^{P^{\unicode[STIX]{x1D700}}}\mathop{\sum }_{2^{\unicode[STIX]{x1D708}}=N\leqslant Q^{3/2}T^{3}P^{\unicode[STIX]{x1D700}}}\biggl|\mathop{\sum }_{n,m}\frac{A(n,m)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n)}{(nm^{2})^{1/2+iv}}V\biggl(\frac{nm^{2}}{N}\biggr)\biggr|\,dv,\end{eqnarray}$$

where $V$ has support in $[1,2]$, is independent of $\unicode[STIX]{x1D713}$, and satisfies $V^{(j)}(y)\;\ll _{j}\;1$ for all $j\in \mathbb{N}_{0}$. Multiplying two such expressions together and using the Cauchy–Schwarz inequality, we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle L(1/2,\unicode[STIX]{x1D713})^{6}\nonumber\\ \displaystyle & & \displaystyle \quad \;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}\max _{|v|\leqslant P^{\unicode[STIX]{x1D700}}}\max _{N\leqslant Q^{3/2}T^{3}P^{\unicode[STIX]{x1D700}}}\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n_{1})\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}.\nonumber\end{eqnarray}$$

For $\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)$, we have

$$\begin{eqnarray}\frac{\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(n)\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(m)}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}=q\mathop{\prod }_{p\mid q}\biggl(1-\frac{1}{p}\biggr)^{-1}\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},1,q}(n)\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},1,q}(m)\end{eqnarray}$$

by (2.9). For the purpose of Theorem 2, it therefore suffices to bound

(4.1)$$\begin{eqnarray}\displaystyle {\mathcal{S}}_{v}(Q,T,N) & := & \displaystyle \mathop{\sum }_{q}W\biggl(\frac{q}{Q}\biggr)Q\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)}h_{T}(t_{\unicode[STIX]{x1D713}})\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},1,q}(n_{1})\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},1,q}(n_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)},\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $N\leqslant Q^{3/2}T^{3}P^{\unicode[STIX]{x1D700}}$, $|v|\leqslant P^{\unicode[STIX]{x1D700}}$, and

$$\begin{eqnarray}h_{T}(t)=e^{-(t/T)^{2}}\mathop{\prod }_{n=1}^{\lfloor \unicode[STIX]{x1D700}^{-1}\rfloor }\biggl(\frac{1}{T^{2}}\biggl(t^{2}+\frac{(2n-1)^{2}}{4}\biggr)\biggr).\end{eqnarray}$$

Note that this function satisfies the assumptions of Lemmas 1 and 2.

4.2 The Eisenstein contribution associated with the trivial character

The $\unicode[STIX]{x1D713}$-sum in (4.1) can be evaluated by the Kuznetsov formula (2.16). To this end, we need to add, using positivity, the contribution from the oldforms and the continuous spectrum. As mentioned in the introduction, this manoeuvre is costly, and we single out the contribution of the continuous spectrum associated with the trivial character:

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{S}}_{v}^{\ast }(Q,T,N):=\mathop{\sum }_{q}W\biggl(\frac{q}{Q}\biggr)Q\mathop{\sum }_{M\mid q}\int _{\mathbb{R}}h_{T}(t)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{1},t)\overline{\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{2},t)}}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\frac{dt}{2\unicode[STIX]{x1D70B}},\nonumber\end{eqnarray}$$

which we rewrite in more compact form as

(4.2)$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(2)}\int _{(2)}\int _{(1)}\int _{\mathbb{R}}\widehat{V}(z_{1})\widehat{V}(z_{2})\widehat{W}(s)Q^{1+s}N^{z_{1}+z_{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{{\mathcal{D}}_{t}(s,1/2+iv+z_{1},1/2-iv+z_{2};0,0)}{|\unicode[STIX]{x1D701}(1+2it)|^{2}}h_{T}(t)\frac{dt}{2\unicode[STIX]{x1D70B}}\frac{ds\,dz_{1}\,dz_{2}}{(2\unicode[STIX]{x1D70B}i)^{3}},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{D}}_{t}(s,z_{1},z_{2};w_{1},w_{2})\nonumber\\ \displaystyle & & \displaystyle \quad =|\unicode[STIX]{x1D701}(1+2it)|^{2}\mathop{\sum }_{q,n_{1},n_{2},m_{1},m_{2}}\mathop{\sum }_{M\mid q}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{1},t)\overline{\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{2},t)}}{q^{s}(n_{1}m_{1}^{2})^{z_{1}}(n_{2}m_{2}^{2})^{z_{2}}m_{1}^{2w_{1}}m_{2}^{2w_{2}}}.\nonumber\end{eqnarray}$$

Recalling the definition (cf. (2.13))

$$\begin{eqnarray}\displaystyle & & \displaystyle |\unicode[STIX]{x1D701}(1+2it)|^{2}\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{1},t)\overline{\unicode[STIX]{x1D70C}_{\text{triv},M,q}(n_{2},t)}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{|\unicode[STIX]{x1D701}^{(q)}(1+2it)|^{2}(n_{1}/n_{2})^{it}}{q\unicode[STIX]{x1D708}(q)\tilde{\mathfrak{n}}_{q}(M)^{2}}\mathop{\sum }_{\unicode[STIX]{x1D6FF}_{1},\unicode[STIX]{x1D6FF}_{2}\mid M}\frac{\unicode[STIX]{x1D6FF}_{1}\unicode[STIX]{x1D6FF}_{2}\unicode[STIX]{x1D707}(M/\unicode[STIX]{x1D6FF}_{1})\unicode[STIX]{x1D707}(M/\unicode[STIX]{x1D6FF}_{2})}{M}\mathop{\sum }_{\substack{ c_{1}\unicode[STIX]{x1D6FF}_{1}f_{1}=n_{1} \\ (c_{1},q/M)=1}}\mathop{\sum }_{\substack{ c_{2}\unicode[STIX]{x1D6FF}_{2}f_{2}=n_{2} \\ (c_{2},q/M)=1}}\biggl(\frac{c_{2}}{c_{1}}\biggr)^{2it},\nonumber\end{eqnarray}$$

we see that (for $t\in \mathbb{R}$) the series ${\mathcal{D}}_{t}(s,z_{1},z_{2};w_{1},w_{2})$ is absolutely convergent in $\Re s>0$, $\Re z_{1},\Re z_{2}>1$, $\Re (z_{1}+w_{1}),\Re (z_{2}+w_{2})>1/2$, and admits an Euler product of the shape

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{p} (1+\frac{1}{p^{s+1}}+\mathop{\sum }_{j=1,2}\mathop{\sum }_{\pm }\frac{3}{p^{z_{j}\pm it}}\nonumber\\ \displaystyle & & \displaystyle \quad +\,O\biggl(\frac{1}{p^{2\min (\Re z_{1},\Re z_{2},\Re (z_{1}+w_{1}),\Re (z_{2}+w_{2}))}}+\frac{1}{p^{\Re s+\min (\Re z_{1},1)+\min (\Re z_{2},1)}}\biggr)\!),\nonumber\end{eqnarray}$$

where the bounds in the error term hold uniformly in

$$\begin{eqnarray}\Re z_{1},\quad \Re z_{2},\quad \Re (z_{1}+w_{1}),\quad \Re (z_{2}+w_{2}),\quad \Re s+\min (\Re z_{1},1)+\min (\Re z_{2},1)>0.\end{eqnarray}$$

In particular, we have

(4.3)$$\begin{eqnarray}{\mathcal{D}}_{t}(s,z_{1},z_{2};w_{1},w_{2})=\unicode[STIX]{x1D701}(s+1)\unicode[STIX]{x1D701}(z_{1}+it)^{3}\unicode[STIX]{x1D701}(z_{1}-it)^{3}\unicode[STIX]{x1D701}(z_{2}+it)^{3}\unicode[STIX]{x1D701}(z_{2}-it)^{3}{\mathcal{E}}_{t}(s,z_{1},z_{2};w_{1},w_{2}),\end{eqnarray}$$

where ${\mathcal{E}}_{t}(s,z_{1},z_{2};w_{1},w_{2})$ is holomorphic and uniformly bounded in

(4.4)$$\begin{eqnarray}\Re z_{1},~\Re z_{2},~\Re (z_{1}+w_{1}),~\Re (z_{2}+w_{2})\geqslant 1/2+\unicode[STIX]{x1D700},\quad \Re s+\min (\Re z_{1},1)+\min (\Re z_{2},1)\geqslant 1+\unicode[STIX]{x1D700},\end{eqnarray}$$

as long as $\Im t=0$. Hence in (4.2), we may shift the contours to $\Re s=-1+\unicode[STIX]{x1D700}$ (picking up a residue at $s=0$), and in the remaining integral we shift the $z_{1},z_{2}$-contours to $\Re z_{1}=\Re z_{2}=1/2+\unicode[STIX]{x1D700}$, getting

(4.5)$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{S}}_{v}^{\ast }(Q,T,N)=\widehat{W}(0)Q\int _{(2)}\int _{(2)}\int _{\mathbb{R}}\mathop{\prod }_{\pm }\frac{\unicode[STIX]{x1D701}(1/2+iv+z_{1}\pm it)^{3}\unicode[STIX]{x1D701}(1/2-iv+z_{2}\pm it)^{3}}{\unicode[STIX]{x1D701}(1\pm 2it)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,{\mathcal{E}}_{t}(0,1/2+iv+z_{1},1/2-iv+z_{2};0,0)\widehat{V}(z_{1})\widehat{V}(z_{2})N^{z_{1}+z_{2}}h_{T}(t)\frac{dz_{1}\,dz_{2}}{(2\unicode[STIX]{x1D70B}i)^{2}}\frac{dt}{2\unicode[STIX]{x1D70B}}+O(TNP^{\unicode[STIX]{x1D700}}).\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

4.3 Applying the Kuznetsov formula twice

By the Kuznetsov formula (and positivity), we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{S}}_{v}(Q,T,N)+{\mathcal{S}}_{v}^{\ast }(Q,T,N)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant Q\mathop{\sum }_{q}W\biggl(\frac{q}{Q}\biggr)\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl(\unicode[STIX]{x1D6FF}_{n_{1},n_{2}}\int _{-\infty }^{\infty }h_{T}(t)\frac{t\tanh (\unicode[STIX]{x1D70B}t)\,dt}{2\unicode[STIX]{x1D70B}^{2}}+\mathop{\sum }_{c}\frac{S(n_{1},n_{2},qc)}{qc}\mathscr{K}h_{T}\biggl(\frac{\sqrt{n_{1}n_{2}}}{qc}\biggr)\biggr).\nonumber\end{eqnarray}$$

The diagonal term is easy to deal with and is trivially bounded by

(4.6)$$\begin{eqnarray}O_{\unicode[STIX]{x1D700}}(P^{\unicode[STIX]{x1D700}}Q^{2}T^{2}).\end{eqnarray}$$

By Mellin inversion, we can recast the off-diagonal term as

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(-12)}Q^{s+1}\widehat{W}(s)\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}(n_{1}n_{2})^{s/2}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{c,q}c^{s}\frac{S(n_{1},n_{2},qc)}{qc}\mathscr{K}_{s}h_{T}\biggl(\frac{\sqrt{n_{1}n_{2}}}{qc}\biggr)\frac{ds}{2\unicode[STIX]{x1D70B}i}\nonumber\end{eqnarray}$$

with $\mathscr{K}_{s}h_{T}(x)=x^{s}\mathscr{K}h_{T}(x)$ as in §2.6. Applying the Kuznetsov formula immediately in the other direction (which we may do by Lemma 1, but this time the summation is over $q$ instead of $c$), we obtain by (2.14) that the previous expression is equal to

(4.7)$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(-12)}Q^{s+1}\widehat{W}(s)\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}(n_{1}n_{2})^{s/2}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{c}c^{s}({\mathcal{A}}_{c}^{\text{Maa}\unicode[STIX]{x00DF}}(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})+{\mathcal{A}}_{c}^{\text{Eis}}(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})+{\mathcal{A}}_{c}^{\text{hol}}(n_{1},n_{2};\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h_{T}))\frac{ds}{2\unicode[STIX]{x1D70B}i}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Lemma 2(b) implies that $\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t)$ has analytic continuation to $\Re s<1$, and we proceed to derive a uniform bound. If $|t|\geqslant 10|\Im s|$ (so that $t\pm \frac{1}{2}|\Im s|\asymp t$), we have

$$\begin{eqnarray}\displaystyle \mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t) & \ll & \displaystyle \int _{\mathbb{R}}\frac{e^{-|\unicode[STIX]{x1D70F}|/T}(1+|\unicode[STIX]{x1D70F}|)}{(1+|t|+|\unicode[STIX]{x1D70F}|)^{2-2\Re s}}\,d\unicode[STIX]{x1D70F}+\frac{e^{-|t|/T}(1+|t|)}{(1+|t|)^{1-\Re s}}\int _{0}^{1+|t|}\frac{1}{(1+|\unicode[STIX]{x1D70F}|)^{1-\Re s}}\,d\unicode[STIX]{x1D70F}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{T^{2}}{(1+|t|)^{2-2\Re s}}+e^{-|t|/T}((1+|t|)^{\Re s}+(1+|t|)^{2\Re s})\ll \frac{T^{2+\max (0,-\Re s)}}{(1+|t|)^{2-2\Re s}}.\nonumber\end{eqnarray}$$

If $|t|\leqslant 10|\Im s|$, we have trivially $\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t)\ll T^{2}$, so that altogether we obtain the uniform bound

(4.8)$$\begin{eqnarray}\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t)\;\ll _{\Re s}\;(1+|\Im s|)^{2-2\Re s}\frac{T^{2+\max (0,-\Re s)}}{(1+|t|)^{2-2\Re s}}.\end{eqnarray}$$

The problematic expression in (4.7) is the part of ${\mathcal{A}}_{c}^{\text{Eis}}(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})$ that is associated with the trivial character. We spell this out explicitly as

$$\begin{eqnarray}\displaystyle {\mathcal{S}}_{v}^{\ast \ast }(Q,T,N) & = & \displaystyle \int _{(20)}\int _{(20)}\int _{(-12)}\int _{\mathbb{R}}\widehat{V}(z_{1})\widehat{V}(z_{2})\widehat{W}(s)Q^{1+s}N^{z_{1}+z_{2}}\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{{\mathcal{D}}_{t}(-s,1/2+iv+s/2+z_{1},1/2-iv+s/2+z_{2};-s/2,-s/2)}{|\unicode[STIX]{x1D701}(1+2it)|^{2}}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t)\frac{dt}{2\unicode[STIX]{x1D70B}}\frac{ds\,dz_{1}\,dz_{2}}{(2\unicode[STIX]{x1D70B}i)^{3}}.\nonumber\end{eqnarray}$$

Shifting the $s$-contour to the far left and simultaneously the $z_{1},z_{2}$-contours to $\Re z_{1}=\frac{1}{2}(1-\Re s)+\unicode[STIX]{x1D700}$, we see from (4.8) that the $t$-integral is negligible for $|t|\geqslant \sqrt{NT/Q}P^{\unicode[STIX]{x1D700}}$. In particular, we may truncate at $|t|\leqslant (T+\sqrt{NT/Q})P^{\unicode[STIX]{x1D700}}$.

Next we shift the $s$-contour to $\Re s=\unicode[STIX]{x1D700}$, past the pole at $s=0$. By Lemma 2(a), the residue matches exactly the main term in (4.5) except for the truncation of the $t$-integral, but by the rapid decay of $\mathscr{L}^{+}\mathscr{K}_{0}h_{T}=h_{T}$ for $|t|\geqslant T$, we may reinsert the tail at the cost of a negligible error.

To estimate the remaining integral, we shift the $z_{1},z_{2}$-contours to left, past the triple poles at $z_{1}=1/2-iv-s/2\pm it$, $z_{2}=1/2-iv-s/2\pm it$ to $\Re z_{1},\Re z_{2}=\unicode[STIX]{x1D700}$. Thus we need to bound the contributions from the remaining integral and the two residues. The remaining multiple integral contains a $t$-integral that can be bounded by

$$\begin{eqnarray}\;\ll _{\unicode[STIX]{x1D700}}\;\int _{|t|\leqslant (T+\sqrt{NT/Q})P^{\unicode[STIX]{x1D700}}}\biggl|\unicode[STIX]{x1D701}\biggl(\frac{1}{2}+\unicode[STIX]{x1D700}+it+i\unicode[STIX]{x1D70F}\biggr)\biggr|^{12}\displaystyle \frac{T^{2}}{(1+|t|)^{2-\unicode[STIX]{x1D700}}}\,dt\;\ll _{\unicode[STIX]{x1D700}}\;T^{2}(1+|\unicode[STIX]{x1D70F}|)^{2}P^{\unicode[STIX]{x1D700}},\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}=\pm v+\Im z_{j}+\frac{1}{2}\Im s_{j}$ and we used Heath-Brown’s twelfth moment bound [Reference Heath-BrownHea78]. Thus the total contribution of the remaining integral is $O_{\unicode[STIX]{x1D700}}(QT^{2}P^{\unicode[STIX]{x1D700}})$. It remains to deal with the two residues. Here the rapid decay of $\widehat{W}$ and $\widehat{V}_{1,2}$ and their derivatives at $z=1/2\pm iv-s/2\pm it$ makes the $t$-integral rapidly convergent regardless of the real part of $s$, so we may shift the contour to $\Re s=1-\unicode[STIX]{x1D700}$ (so that $\Re z_{j}=\unicode[STIX]{x1D700}$), getting a contribution of $O_{\unicode[STIX]{x1D700}}(Q^{2}T^{2}P^{\unicode[STIX]{x1D700}})$.

Combining (4.6) and the error term in (4.5) with the previous two error terms, we have accomplished so far the bound

(4.9)$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{S}}_{v}(Q,T,N)\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}(Q^{2}T^{2}+NT)\nonumber\\ \displaystyle & & \displaystyle \quad +\, |\int _{(-12)}Q^{s+1}\widehat{W}(s)\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}(n_{1}n_{2})^{s/2}}V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{c}c^{s}({\mathcal{A}}_{c}^{\text{Maa}\unicode[STIX]{x00DF}}(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})+{\mathcal{A}}_{c}^{\text{Eis},\ast }(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})+{\mathcal{A}}_{c}^{\text{hol}}(n_{1},n_{2};\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h_{T}))\frac{ds}{2\unicode[STIX]{x1D70B}i} |,\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where ${\mathcal{A}}_{c}^{\text{Eis},\ast }$ denotes the contribution of level-$c$ Eisenstein series without the trivial character.

4.4 The endgame

We consider the Maaß contribution in (4.9) given by

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(-12)}Q^{s+1}\widehat{W}(s)\mathop{\sum }_{c}c^{s}\mathop{\sum }_{c_{0}M\mid c}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(c_{0})}\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(n)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,N}(m)}}{(n_{1}m_{1}^{2})^{1/2+iv}(n_{2}m_{2}^{2})^{1/2-iv}(n_{1}n_{2})^{s/2}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,V\biggl(\frac{n_{1}m_{1}^{2}}{N}\biggr)\overline{V\biggl(\frac{n_{2}m_{2}^{2}}{N}\biggr)}\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t_{\unicode[STIX]{x1D713}})\frac{ds}{2\unicode[STIX]{x1D70B}i}.\nonumber\end{eqnarray}$$

Shifting the $s$-contour to the far left, we see that we can truncate both the $c$-sum and the $\unicode[STIX]{x1D713}$-sum at $c(1+|t_{\unicode[STIX]{x1D713}}|^{2})\leqslant P^{\unicode[STIX]{x1D700}}NT/Q$ at the cost of a negligible error (recall (4.8) and the rapid decay of $\widehat{V}$). Having done this, we shift the $s$-contour back to $\Re s=0$. By Mellin inversion, we obtain

(4.10)$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(0)}Q^{s+1}\widehat{W}(s)\underset{c(1+|t_{\unicode[STIX]{x1D713}}|^{2})\leqslant P^{\unicode[STIX]{x1D700}}NT/Q}{\mathop{\sum }_{c}\mathop{\sum }_{c_{0}\mid c}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(c_{0})}}\frac{c^{s}}{c\unicode[STIX]{x1D708}(c)}\mathop{\prod }_{p\mid c_{0}}(1-p^{-2})\int _{(\unicode[STIX]{x1D700})}\int _{(\unicode[STIX]{x1D700})}N^{z_{1}+z_{2}}\widehat{V}(z_{1})\widehat{V}(z_{2})\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{{\mathcal{D}}_{\unicode[STIX]{x1D713},c}^{\text{Maa}\unicode[STIX]{x00DF}}(1/2+iv+s/2+z_{1},1/2-iv+s/2+z_{2},-s/2,-s/2)}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathscr{L}^{+}\mathscr{K}_{s}h_{T}(t_{\unicode[STIX]{x1D713}})\frac{dz_{1}\,dz_{2}}{(2\unicode[STIX]{x1D70B}i)^{2}}\frac{ds}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

where (recalling the notation in (2.9))

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{D}}_{\unicode[STIX]{x1D713},c}^{\text{Maa}\unicode[STIX]{x00DF}}(z_{1},z_{2},w_{1},w_{2})\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})c\unicode[STIX]{x1D708}(c)}{\mathop{\prod }_{p\mid c_{0}}(1-p^{-2})}\mathop{\sum }_{M\mid (c/c_{0})}\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,c}(n)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D713},M,c}(m)}}{(n_{1}m_{1}^{2})^{z_{1}}(n_{2}m_{2}^{2})^{z_{2}}m_{1}^{2w_{1}}m_{2}^{2w_{2}}}\nonumber\end{eqnarray}$$

for $\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(c_{0})$ with $c_{0}\mid c$. Using (2.9) and (2.11) (with $\unicode[STIX]{x1D703}\leqslant 1/2$), we see as in (4.3) that

(4.11)$$\begin{eqnarray}{\mathcal{D}}_{\unicode[STIX]{x1D713},c}^{\text{Maa}\unicode[STIX]{x00DF}}(z_{1},z_{2},w_{1},w_{2})=L(z_{1},\unicode[STIX]{x1D713})^{3}L(z_{2},\unicode[STIX]{x1D713})^{3}{\mathcal{E}}_{\unicode[STIX]{x1D713},c}^{\text{Maa}\unicode[STIX]{x00DF}}(z_{1},z_{2},w_{1},w_{2}),\end{eqnarray}$$

where

$$\begin{eqnarray}{\mathcal{E}}_{\unicode[STIX]{x1D713},c}^{\text{Maa}\unicode[STIX]{x00DF}}(z_{1},z_{2},w_{1},w_{2})\;\ll _{\unicode[STIX]{x1D700}}\;c^{\unicode[STIX]{x1D700}}\end{eqnarray}$$

uniformly in $\Re z_{1},\Re z_{2},\Re (z_{1}+w_{1}),\Re (z_{2}+w_{2})\geqslant 1/2+\unicode[STIX]{x1D700}$. The convexity bound for $L(z,\unicode[STIX]{x1D713})$ is

$$\begin{eqnarray}L(z,\unicode[STIX]{x1D713})\;\ll _{\unicode[STIX]{x1D700}}\;(c_{0}(1+|t_{\unicode[STIX]{x1D713}}|+|\Im z|)^{2})^{1/4+\unicode[STIX]{x1D700}},\quad \Re z\geqslant 1/2.\end{eqnarray}$$

We can afford to use the convexity bound on four of the six $L$-functions in (4.11). We may then truncate the $s,z_{1},z_{2}$-contours at height $P^{\unicode[STIX]{x1D700}}$, and after a trivial estimation, we bound (4.10) by

(4.12)$$\begin{eqnarray}\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}Q\frac{NT}{Q}\max _{|\unicode[STIX]{x1D709}|\leqslant P^{\unicode[STIX]{x1D700}}}\underset{c(1+|t_{\unicode[STIX]{x1D713}}|^{2})\leqslant P^{\unicode[STIX]{x1D700}}NT/Q}{\mathop{\sum }_{c}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(c)}}\frac{1}{c}\biggl|L\biggl(\frac{1}{2}+\unicode[STIX]{x1D700}+i\unicode[STIX]{x1D709},\unicode[STIX]{x1D713}\biggr)\biggr|^{2}\frac{T^{2}}{(1+|t_{\unicode[STIX]{x1D713}}|^{2})}.\end{eqnarray}$$

It is an easy exercise with the Kuznetsov formula or the spectral large sieve to obtain a Lindelöf on average bound for the second moment, which can safely be left to the reader: the length of the approximate functional equation in each factor is $O_{\unicode[STIX]{x1D700}}(P^{\unicode[STIX]{x1D700}}c^{1/2}(1+|t_{\unicode[STIX]{x1D713}}|))$, so the Kloosterman term in the Kuznetsov formula is essentially invisible. Thus by Weyl’s law, the total contribution of the previous expression is

$$\begin{eqnarray}\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}N^{2}T^{2}Q^{-1}\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}Q^{2}T^{8}\end{eqnarray}$$

for $N\leqslant Q^{3/2}T^{3}P^{\unicode[STIX]{x1D700}}$, and this majorizes all preceding error terms.

The contribution of ${\mathcal{A}}_{c}^{\text{hol}}(n_{1},n_{2};\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h_{T})$ can be bounded in same way using the analogous bound for $\mathscr{L}^{\text{hol}}\mathscr{K}_{s}h_{T}$ in Lemma 2(b).

Finally, for the contribution ${\mathcal{A}}_{c}^{\text{Eis},\ast }(n_{1},n_{2};\mathscr{L}^{+}\mathscr{K}_{s}h_{T})$, we observe that after removing the trivial character, the analogously defined function

$$\begin{eqnarray}\displaystyle & & \displaystyle {\mathcal{D}}_{(\unicode[STIX]{x1D712},t),c}^{\text{Eis}}(z_{1},z_{2},w_{1},w_{2})\nonumber\\ \displaystyle & & \displaystyle \quad =|L(1+2it,\unicode[STIX]{x1D712}^{2})|^{2}c\unicode[STIX]{x1D708}(c)\mathop{\sum }_{c_{\unicode[STIX]{x1D712}}^{2}\mid M\mid c}\mathop{\sum }_{n_{1},n_{2},m_{1},m_{2}}\frac{A(n_{1},m_{1})A(n_{2},m_{2})\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D712},M,c}(n)\overline{\unicode[STIX]{x1D70C}_{\unicode[STIX]{x1D712},M,c}(m)}}{(n_{1}m_{1}^{2})^{z_{1}}(n_{2}m_{2}^{2})^{z_{2}}m_{1}^{2w_{1}}m_{2}^{2w_{2}}}\nonumber\end{eqnarray}$$

is pole-free in $\Re z_{1},\Re z_{2},\Re (z_{1}+w_{1}),\Re (z_{2}+w_{2})\geqslant 1/2+\unicode[STIX]{x1D700}$ since $\unicode[STIX]{x1D712}$ is primitive of conductor ${>}1$, and it can be approximated by $L(z_{1}+it,\unicode[STIX]{x1D712})^{3}L(z_{1}-it,\overline{\unicode[STIX]{x1D712}})^{3}L(z_{2}+it,\unicode[STIX]{x1D712})^{3}L(z_{2}-it,\overline{\unicode[STIX]{x1D712}})^{3}$ in this region up to a holomorphic factor bounded by $O_{\unicode[STIX]{x1D700}}(c^{\unicode[STIX]{x1D700}})$. Here we can even afford to apply the convexity bound for all 12 Dirichlet $L$-functions. The quantity corresponding to (4.12) is then

$$\begin{eqnarray}P^{\unicode[STIX]{x1D700}}Q\biggl(\frac{NT}{Q}\biggr)^{3/2}\mathop{\sum }_{c}\int _{c(1+|t|)^{2}\leqslant P^{\unicode[STIX]{x1D700}}NT/Q}\frac{\#\{\unicode[STIX]{x1D712}:c_{\unicode[STIX]{x1D712}}^{2}\mid c\}T^{2}}{c(1+|t|^{2})}\,dt\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}Q\biggl(\frac{NT}{Q}\biggr)^{3/2}\;\ll _{\unicode[STIX]{x1D700}}\;P^{\unicode[STIX]{x1D700}}Q^{7/4}T^{6}.\end{eqnarray}$$

This completes the proof of Theorem 2.

5.1 Proof of Corollary 3

From [Reference Blomer and KhanBK19b, §12.1], we quote

(5.1)$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)}L(1/2,\unicode[STIX]{x1D713})^{5}e^{-t_{\unicode[STIX]{x1D713}}^{2}}\;\ll _{\unicode[STIX]{x1D700}}\;q^{\unicode[STIX]{x1D700}}\max _{|\unicode[STIX]{x1D70F}|\leqslant (\log q)^{2}}\mathop{\sum }_{\ell \leqslant q^{1/2+\unicode[STIX]{x1D700}}}\frac{1}{\ell ^{1/2}}\biggl|\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(q)}\frac{L(1/2,f)^{4}}{L(1,\text{Ad}^{2}f)}\unicode[STIX]{x1D706}_{f}(\ell )h_{\unicode[STIX]{x1D70F}}(t_{f})\biggr|+q^{-10},\end{eqnarray}$$

where

$$\begin{eqnarray}h_{\unicode[STIX]{x1D70F}}(t_{f})=\frac{L_{\infty }(1/2+\unicode[STIX]{x1D700}+i\unicode[STIX]{x1D70F},f)}{L_{\infty }(1/2,f)}\frac{G_{f}(\unicode[STIX]{x1D700}+i\unicode[STIX]{x1D70F})}{G_{f}(0)}e^{-t_{f}^{2}}(1+|t_{f}|)^{\unicode[STIX]{x1D700}}\end{eqnarray}$$

with

$$\begin{eqnarray}G_{f}(s)=\mathop{\prod }_{j=0}^{1000}\mathop{\prod }_{\unicode[STIX]{x1D716}_{1},\unicode[STIX]{x1D716}_{2}\in \{\pm 1\}}\biggl(\frac{1}{2}+\unicode[STIX]{x1D716}_{1}s+i\unicode[STIX]{x1D716}_{2}t_{f}+j\biggr).\end{eqnarray}$$

This is an application of a carefully designed approximate functional equation. Now the formula two displays below [Reference Blomer and KhanBK19b, (11.4)], together with [Reference Blomer and KhanBK19b, Lemma 1], shows that for $q$ prime,

$$\begin{eqnarray}\frac{\unicode[STIX]{x1D719}(q)}{q^{2}}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(q)}\frac{L(1/2,f)^{4}}{L(1,\text{Ad}^{2}f)}\unicode[STIX]{x1D706}_{f}(\ell )h_{\unicode[STIX]{x1D70F}}(t_{f})=\mathop{\sum }_{ab=\ell }{\mathcal{M}}_{q,a}^{+}(1/2,1/2,\mathfrak{h}_{\unicode[STIX]{x1D70F}})\biggl(\frac{a}{b}\biggr)^{1/2}+O(\ell ^{\unicode[STIX]{x1D703}+\unicode[STIX]{x1D700}}q^{-1})\end{eqnarray}$$

with $\mathfrak{h}_{\unicode[STIX]{x1D70F}}=(h_{\unicode[STIX]{x1D70F}},0)$ in the notation of [Reference Blomer and KhanBK19b, (1.3), (1.7)]. Here the error term also includes the oldforms of level 1. On the other hand, [Reference Blomer and KhanBK19b, (11.4)] states that

$$\begin{eqnarray}\mathop{\sum }_{ab=\ell }{\mathcal{M}}_{q,a}^{+}(1/2,1/2,\mathfrak{h}_{\unicode[STIX]{x1D70F}})\biggl(\frac{a}{b}\biggr)^{1/2}\;\ll _{\unicode[STIX]{x1D700}}\;(\ell q)^{\unicode[STIX]{x1D700}}\mathop{\sum }_{ab=\ell }\biggl(\frac{a}{b}\biggr)^{1/2}\biggl(\frac{1}{a}+\frac{1}{q}+\mathop{\sum }_{\pm }\big|{\mathcal{M}}_{a,q}^{\pm }(1/2,1/2;\mathscr{T}_{1/2,1/2}^{\pm }\mathfrak{h}_{\unicode[STIX]{x1D70F}})\big|\biggr)\end{eqnarray}$$

uniformly in $|\unicode[STIX]{x1D70F}|\leqslant (\log q)^{2}$, and the analysis of [Reference Blomer and KhanBK19b, §11] shows

$$\begin{eqnarray}{\mathcal{M}}_{a,q}^{\pm }(1/2,1/2;\mathscr{T}_{1/2,1/2}^{\pm }\mathfrak{h}_{\unicode[STIX]{x1D70F}})\;\ll _{\unicode[STIX]{x1D700}}\;(aq)^{\unicode[STIX]{x1D700}}\biggl(\frac{1}{q^{1/2}}+\frac{1}{aq^{1/2}}\mathop{\sum }_{a_{0}\mid a}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a_{0})}\frac{|L(1/2,f)|^{4}}{L(1,\text{Ad}^{2}f)}\frac{|\unicode[STIX]{x1D6EC}_{f}(q,1/2)|}{(1+|t_{f}|)^{15}}\biggr)\end{eqnarray}$$

(again uniformly in $|\unicode[STIX]{x1D70F}|\leqslant (\log q)^{2}$) where $\unicode[STIX]{x1D6EC}_{f}(q,1/2):=\unicode[STIX]{x1D706}_{f}(q)-q^{-1/2}$ for $q$ prime. Combining these estimates, we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x1D719}(q)}{q^{2}}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(q)}\frac{L(1/2,f)^{4}}{L(1,\text{Ad}^{2}f)}\unicode[STIX]{x1D706}_{f}(\ell )h_{\unicode[STIX]{x1D70F}}(t_{f})\nonumber\\ \displaystyle & & \displaystyle \quad \;\ll _{\unicode[STIX]{x1D700}}\;(\ell q)^{\unicode[STIX]{x1D700}}\biggl(\frac{1}{\ell ^{1/2}}+\frac{\ell ^{1/2}}{q^{1/2}}+\frac{1}{(\ell q)^{1/2}}\mathop{\sum }_{a_{0}\mid \ell }\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a_{0})}\frac{|L(1/2,f)|^{4}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\frac{(1+|\unicode[STIX]{x1D706}_{f}(q)|)}{(1+|t_{f}|)^{15}}\biggr).\nonumber\end{eqnarray}$$

Substituting back into (5.1), this yields

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)}L(1/2,\unicode[STIX]{x1D713})^{5}e^{-t_{\unicode[STIX]{x1D713}}^{2}} & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle q^{1+\unicode[STIX]{x1D700}}\mathop{\sum }_{\ell \leqslant q^{1/2+\unicode[STIX]{x1D700}}}\biggl(\frac{1}{\ell }+\frac{1}{q^{1/2}}+\frac{1}{\ell q^{1/2}}\mathop{\sum }_{a\mid \ell }\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a)}\frac{|L(1/2,f)|^{4}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\frac{(1+|\unicode[STIX]{x1D706}_{f}(q)|)}{(1+|t_{f}|)^{15}}\biggr)\nonumber\\ \displaystyle & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle q^{1+\unicode[STIX]{x1D700}}+q^{1/2+\unicode[STIX]{x1D700}}\mathop{\sum }_{a\leqslant q^{1/2+\unicode[STIX]{x1D700}}}\frac{1}{a}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a)}\frac{|L(1/2,f)|^{4}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\frac{(1+|\unicode[STIX]{x1D706}_{f}(q)|)}{(1+|t_{f}|)^{15}}\nonumber\\ \displaystyle & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle q^{1+\unicode[STIX]{x1D700}}+q^{1/2+\unicode[STIX]{x1D700}}\max _{A\leqslant q^{1/2+\unicode[STIX]{x1D700}}}\frac{1}{A}\mathop{\sum }_{a\asymp A}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a)}\frac{|L(1/2,f)|^{4}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\frac{(1+|\unicode[STIX]{x1D706}_{f}(q)|)}{(1+|t_{f}|)^{15}}.\nonumber\end{eqnarray}$$

So far this is essentially a restatement of the analysis in [Reference Blomer and KhanBK19b], but now we insert an additional application of Hölder’s inequality. In this way, we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)}L(1/2,\unicode[STIX]{x1D713})^{5}e^{-t_{\unicode[STIX]{x1D713}}^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad \;\ll _{\unicode[STIX]{x1D700}}\;q^{1+\unicode[STIX]{x1D700}}+q^{1/2+\unicode[STIX]{x1D700}}\max _{A\leqslant q^{1/2+\unicode[STIX]{x1D700}}}\frac{1}{A}\biggl(\mathop{\sum }_{a\asymp A}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a)}\frac{|L(1/2,f)|^{6}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})(1+|t_{f}|)^{15}}\biggr)^{2/3}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl(\mathop{\sum }_{a\asymp A}\mathop{\sum }_{f\in {\mathcal{B}}^{\ast }(a)}\frac{(1+|\unicode[STIX]{x1D706}_{f}(q)|)^{3}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})(1+|t_{f}|)^{15}}\biggr)^{1/3}.\nonumber\end{eqnarray}$$

By Theorem 2 and Lemma 3, we obtain

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(q)}L(1/2,\unicode[STIX]{x1D713})^{5}e^{-t_{\unicode[STIX]{x1D713}}^{2}}\;\ll _{\unicode[STIX]{x1D700}}\;q^{\unicode[STIX]{x1D700}}\biggl(q+q^{1/2}\max _{A\leqslant q^{1/2+\unicode[STIX]{x1D700}}}\frac{1}{A}A^{4/3}(A^{2}+q^{1/2})^{1/3}q^{\unicode[STIX]{x1D703}/3}\biggr)\;\ll _{\unicode[STIX]{x1D700}}\;q^{1+\unicode[STIX]{x1D703}/3+\unicode[STIX]{x1D700}}.\end{eqnarray}$$

5.2 Proof of Theorem 4

By a dyadic decomposition, we can replace the summation condition $m\leqslant M$ by $m\asymp M$. Let us also assume without loss of generality that $\Vert \mathbf{a}\Vert _{\infty }\leqslant 1$. In order to apply Theorem 1, we would like to bound $L(1/2,\unicode[STIX]{x1D712})$ by a small integral over the imaginary axis. This can be done by a standard argument based on the functional equation and the residue theorem (which seems to have been first applied by Heath-Brown [Reference Heath-BrownHea78, Lemma 3]) as follows. Fix $0<\unicode[STIX]{x1D700}<1/10$ and suppose that $\unicode[STIX]{x1D712}$ is a primitive character modulo $q$. We have

$$\begin{eqnarray}\displaystyle L(1/2,\unicode[STIX]{x1D712})^{4} & = & \displaystyle \int _{(\unicode[STIX]{x1D700})}L(1/2+s,\unicode[STIX]{x1D712})^{4}\frac{e^{2s^{2}}}{s}\frac{ds}{2\unicode[STIX]{x1D70B}i}+\int _{(\unicode[STIX]{x1D700})}L(1/2-s,\unicode[STIX]{x1D712})^{4}\frac{e^{2s^{2}}}{s}\frac{ds}{2\unicode[STIX]{x1D70B}i}\nonumber\\ \displaystyle & = & \displaystyle \int _{(\unicode[STIX]{x1D700})}L(1/2+s,\unicode[STIX]{x1D712})^{4}f(s)\frac{ds}{2\unicode[STIX]{x1D70B}i},\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}f(s)=\biggl(1+\frac{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+s+\mathfrak{a}))^{4}}{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}-s+\mathfrak{a}))^{4}}\biggl(\frac{q}{\unicode[STIX]{x1D70B}}\biggr)^{4s}\biggr)\frac{e^{2s^{2}}}{s}\end{eqnarray}$$

with $\mathfrak{a}=0$ if $\unicode[STIX]{x1D712}$ is even and $\mathfrak{a}=1$ if $\unicode[STIX]{x1D712}$ is odd. Applying the same argument again, we have

$$\begin{eqnarray}L(1/2+s,\unicode[STIX]{x1D712})^{4}=\int _{(-\unicode[STIX]{x1D700})}L(1/2+s+u,\unicode[STIX]{x1D712})^{4}g_{s}(u)\frac{ds}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

where

$$\begin{eqnarray}g_{s}(u)=-\biggl(1+\frac{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+s+u+\mathfrak{a}))^{4}}{\unicode[STIX]{x1D6E4}({\textstyle \frac{1}{2}}({\textstyle \frac{1}{2}}+s-u+\mathfrak{a}))^{4}}\biggl(\frac{q}{\unicode[STIX]{x1D70B}}\biggr)^{4u}\biggr)\frac{e^{2u^{2}}}{u}.\end{eqnarray}$$

Inserting and changing variables, we obtain

$$\begin{eqnarray}L(1/2,\unicode[STIX]{x1D712})^{4}=\int _{(0)}L(1/2+v,\unicode[STIX]{x1D712})^{4}h(v)\frac{dv}{2\unicode[STIX]{x1D70B}i},\end{eqnarray}$$

where

$$\begin{eqnarray}h(v)=\int _{(\unicode[STIX]{x1D700})}g_{s}(v-s)f(s)\frac{ds}{2\unicode[STIX]{x1D70B}i}\ll e^{-|v|^{2}}q^{4\unicode[STIX]{x1D700}}.\end{eqnarray}$$

Now choosing $F$ as in (1.7), we get

$$\begin{eqnarray}|L(1/2,\unicode[STIX]{x1D712})|^{4}\;\ll _{\unicode[STIX]{x1D700}}\;q^{\unicode[STIX]{x1D700}}\int _{(0)}L(1/2+z,\unicode[STIX]{x1D712})^{2}L(1/2-z,\overline{\unicode[STIX]{x1D712}})^{2}F(z)\frac{dz}{2\unicode[STIX]{x1D70B}i}.\end{eqnarray}$$

Opening the square, we have

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D712}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)}\biggl|\mathop{\sum }_{m\asymp M}a(m)\unicode[STIX]{x1D712}(m)\biggr|^{2}|L(1/2,\unicode[STIX]{x1D712})|^{4}\;\ll _{\unicode[STIX]{x1D700}}\;M^{2}+q^{\unicode[STIX]{x1D700}}\mathop{\sum }_{d}\mathop{\sum }_{\substack{ m_{1},m_{2}\asymp M/d \\ (m_{1},m_{2})=1}}|{\mathcal{T}}_{m_{1},m_{2},q}(1/2,1/2,1/2)|\end{eqnarray}$$

with the notation as in (1.6) in the special case where $f$ is the standard Eisenstein series with $\unicode[STIX]{x1D703}=0$. By Theorem 1 and (1.8) we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D712}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)}\biggl|\mathop{\sum }_{m\asymp M}a(m)\unicode[STIX]{x1D712}(m)\biggr|^{2}|L(1/2,\unicode[STIX]{x1D712})|^{4}\nonumber\\ \displaystyle & & \displaystyle \quad \;\ll _{\unicode[STIX]{x1D700}}\;M^{2}+(Mq)^{1+\unicode[STIX]{x1D700}}+q^{\unicode[STIX]{x1D700}}\mathop{\sum }_{d}\mathop{\sum }_{\substack{ m_{1},m_{2}\asymp M/d \\ (m_{1},m_{2})=1}}\mathop{\sum }_{\ast \in \{\text{Maa}\unicode[STIX]{x00DF},\text{hol},\text{Eis}\}}|{\mathcal{M}}_{m_{1},m_{2},q}^{\ast }(1/2,1/2,1/2)|.\nonumber\end{eqnarray}$$

We only deal with the Maaß case; the other two cases are similar but easier. By (1.10) and (1.11), we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{d}\mathop{\sum }_{\substack{ m_{1},m_{2}\asymp M/d \\ (m_{1},m_{2})=1}}|{\mathcal{M}}_{m_{1},m_{2},q}^{\text{Maa}\unicode[STIX]{x00DF}}(1/2,1/2,1/2)|\nonumber\\ \displaystyle & & \displaystyle \quad \;\ll _{\unicode[STIX]{x1D700}}\;(qM)^{\unicode[STIX]{x1D700}}\mathop{\sum }_{d}\mathop{\sum }_{\substack{ m_{1},m_{2}\asymp M/d \\ (m_{1},m_{2})=1}}\mathop{\sum }_{N\mid m_{1}m_{2}}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N)}\frac{q^{1/2}}{N^{1/2}}\frac{(1+|\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q)|)}{(1+|t_{\unicode[STIX]{x1D713}}|)^{30}}\frac{L(1/2,\unicode[STIX]{x1D713})^{3}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}.\nonumber\end{eqnarray}$$

We drop the condition $(m_{1},m_{2})=1$ and write $m_{1}m_{2}=m=NK$, obtaining by the standard divisor bound that the previous display is bounded by

$$\begin{eqnarray}\displaystyle & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle (qM)^{\unicode[STIX]{x1D700}}\mathop{\sum }_{d}\mathop{\sum }_{NK\asymp M^{2}/d^{2}}\frac{q^{1/2}}{N^{1/2}}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N)}\frac{(1+|\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q)|)}{(1+|t_{\unicode[STIX]{x1D713}}|)^{30}}\frac{L(1/2,\unicode[STIX]{x1D713})^{3}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\nonumber\\ \displaystyle & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle (qM)^{\unicode[STIX]{x1D700}}\mathop{\sum }_{d}\mathop{\sum }_{N\ll M^{2}/d^{2}}\frac{q^{1/2}M^{2}}{d^{2}N^{3/2}}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N)}\frac{(1+|\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q)|)}{(1+|t_{\unicode[STIX]{x1D713}}|)^{30}}\frac{L(1/2,\unicode[STIX]{x1D713})^{3}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}\nonumber\\ \displaystyle & \;\ll _{\unicode[STIX]{x1D700}}\; & \displaystyle (q^{1/2}M^{2})^{1+\unicode[STIX]{x1D700}}\max _{{\mathcal{N}}\ll M^{2}}\frac{1}{{\mathcal{N}}^{3/2}}\mathop{\sum }_{N\asymp {\mathcal{N}}}\mathop{\sum }_{\unicode[STIX]{x1D713}\in {\mathcal{B}}^{\ast }(N)}\frac{(1+|\unicode[STIX]{x1D706}_{\unicode[STIX]{x1D713}}(q)|)}{(1+|t_{\unicode[STIX]{x1D713}}|)^{30}}\frac{L(1/2,\unicode[STIX]{x1D713})^{3}}{L(1,\text{Ad}^{2}\unicode[STIX]{x1D713})}.\nonumber\end{eqnarray}$$

By the Cauchy–Schwarz inequality, Theorem 2, and Lemma 3, this is

$$\begin{eqnarray}\;\ll _{\unicode[STIX]{x1D700}}\;(q^{1/2}M^{2})^{1+\unicode[STIX]{x1D700}}\max _{{\mathcal{N}}\ll M^{2}}\frac{1}{{\mathcal{N}}^{3/2}}({\mathcal{N}}+q^{1/4}{\mathcal{N}}^{1/2}){\mathcal{N}}\;\ll _{\unicode[STIX]{x1D700}}\;(q^{1/2}M^{2})^{1+\unicode[STIX]{x1D700}}(M+q^{1/4}).\end{eqnarray}$$

For $M\leqslant q^{1/4}$, we obtain altogether

$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D712}\hspace{-7.11317pt}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)}\biggl|\mathop{\sum }_{m\asymp M}a(m)\unicode[STIX]{x1D712}(m)\biggr|^{2}|L(1/2,\unicode[STIX]{x1D712})|^{4}\;\ll _{\unicode[STIX]{x1D700}}\;(Mq)^{1+\unicode[STIX]{x1D700}},\end{eqnarray}$$

as desired.