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Simple zeros of automorphic $L$-functions

Published online by Cambridge University Press:  30 May 2019

Andrew R. Booker
Affiliation:
School of Mathematics, University of Bristol, University Walk, Bristol BS8 1TW, UK email [email protected]
Peter J. Cho
Affiliation:
Department of Mathematical Sciences, Ulsan National Institute of Science and Technology, Ulsan, Korea email [email protected]
Myoungil Kim
Affiliation:
Department of Mathematical Sciences, Ulsan National Institute of Science and Technology, Ulsan, Korea email [email protected]
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Abstract

We prove that the complete $L$-function associated to any cuspidal automorphic representation of $\operatorname{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$ has infinitely many simple zeros.

Type
Research Article
Copyright
© The Authors 2019 

1 Introduction

In [Reference BookerBoo16], the first author showed that the complete $L$ -functions associated to classical holomorphic newforms have infinitely many simple zeros. The purpose of this paper is to extend that result to the remaining degree- $2$ automorphic $L$ -functions over $\mathbb{Q}$ , that is, those associated to cuspidal Maass newforms. This also extends work of the second author [Reference ChoCho13] which established a quantitative estimate for the first few Maass forms of level $1$ . When combined with the holomorphic case from [Reference BookerBoo16], we obtain the following theorem.

Theorem 1.1. Let $\mathbb{A}_{\mathbb{Q}}$ denote the adèle ring of $\mathbb{Q}$ , and let $\unicode[STIX]{x1D70B}$ be a cuspidal automorphic representation of $\operatorname{GL}_{2}(\mathbb{A}_{\mathbb{Q}})$ . Then the associated complete $L$ -function $\unicode[STIX]{x1D6EC}(s,\unicode[STIX]{x1D70B})$ has infinitely many simple zeros.

The basic idea of the proof is the same as in [Reference BookerBoo16], which is in turn based on the method of Conrey and Ghosh [Reference Conrey and GhoshCG88]. Let $f$ be a primitive Maass cuspform of weight $k\in \{0,1\}$ for $\unicode[STIX]{x1D6E4}_{0}(N)$ with nebentypus character $\unicode[STIX]{x1D709}$ , and let $L_{f}(s)$ be the finite $L$ -function attached to $f$ :

$$\begin{eqnarray}L_{f}(s)=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)n^{-s}.\end{eqnarray}$$

We define

$$\begin{eqnarray}D_{f}(s)=L_{f}(s)\frac{d^{2}}{ds^{2}}\log L_{f}(s)=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)n^{-s}.\end{eqnarray}$$

Then it is easy to see that $D_{f}(s)$ has a pole at some point if and only if $L_{f}(s)$ has a simple zero there.

For $\unicode[STIX]{x1D6FC}\in \mathbb{Q}$ and $j\geqslant 0$ we define the additive twists

$$\begin{eqnarray}L_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(j)})=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)\cos ^{(j)}(2\unicode[STIX]{x1D70B}n\unicode[STIX]{x1D6FC})n^{-s},\quad D_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(j)})=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)\cos ^{(j)}(2\unicode[STIX]{x1D70B}n\unicode[STIX]{x1D6FC})n^{-s},\end{eqnarray}$$

where $\cos ^{(j)}$ denotes the $j$ th derivative of the cosine function. Let $q\nmid N$ be a prime and $\unicode[STIX]{x1D712}_{0}$ the principal character mod $q$ . Then we have the following expansions of the trigonometric functions in terms of Dirichlet characters:

$$\begin{eqnarray}\displaystyle \cos \biggl(\frac{2\unicode[STIX]{x1D70B}n}{q}\biggr) & = & \displaystyle 1-\frac{q}{q-1}\unicode[STIX]{x1D712}_{0}(n)+\frac{\sqrt{q}}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}(modq) \\ \unicode[STIX]{x1D712}(-1)=1 \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\overline{\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D712}}}\unicode[STIX]{x1D712}(n),\nonumber\\ \displaystyle \sin \biggl(\frac{2\unicode[STIX]{x1D70B}n}{q}\biggr) & = & \displaystyle \frac{\sqrt{q}}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}(modq) \\ \unicode[STIX]{x1D712}(-1)=-1}}\overline{\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D712}}}\unicode[STIX]{x1D712}(n),\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D712}}$ denotes the root number of the Dirichlet $L$ -function $L(s,\unicode[STIX]{x1D712})$ . In particular, we have

$$\begin{eqnarray}D_{f}\biggl(s,\frac{1}{q},\cos \biggr)=D_{f}(s)-\frac{q}{q-1}D_{f}(s,\unicode[STIX]{x1D712}_{0})+\frac{\sqrt{q}}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}(modq) \\ \unicode[STIX]{x1D712}(-1)=1 \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\overline{\unicode[STIX]{x1D716}_{\unicode[STIX]{x1D712}}}D_{f}(s,\unicode[STIX]{x1D712}),\end{eqnarray}$$

where

$$\begin{eqnarray}D_{f}(s,\unicode[STIX]{x1D712})=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)\unicode[STIX]{x1D712}(n)n^{-s}\end{eqnarray}$$

is the corresponding multiplicative twist.

By the non-vanishing results for automorphic $L$ -functions [Reference Jacquet and ShalikaJS77], all non-trivial poles of $D_{f}(s)$ and $D_{f}(s,\unicode[STIX]{x1D712})$ for $\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}$ are located in the critical strip $\{s\in \mathbb{C}:0<\Re (s)<1\}$ . However, for the case of the principal character, since

$$\begin{eqnarray}L_{f}(s,\unicode[STIX]{x1D712}_{0})=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)\unicode[STIX]{x1D712}_{0}(n)n^{-s}=(1-\unicode[STIX]{x1D706}_{f}(q)q^{-s}+\unicode[STIX]{x1D709}(q)q^{-2s})L_{f}(s),\end{eqnarray}$$

$D_{f}(s,\unicode[STIX]{x1D712}_{0})$ has a pole at every simple zero of the local Euler factor polynomial, $1-\unicode[STIX]{x1D706}_{f}(q)q^{-s}+\unicode[STIX]{x1D709}(q)q^{-2s}$ , at which $L_{f}(s)$ does not vanish.

Since $f$ is cuspidal, the Rankin–Selberg method implies that the average of $|\unicode[STIX]{x1D706}_{f}(q)|^{2}$ over primes $q$ is $1$ , that is,

(1) $$\begin{eqnarray}\lim _{x\rightarrow \infty }\frac{\mathop{\sum }_{\substack{ q\text{ prime},\,q\leqslant x}}|\unicode[STIX]{x1D706}_{f}(q)|^{2}}{\#\{q\text{ prime}:q\leqslant x\}}=1.\end{eqnarray}$$

To see this, write

$$\begin{eqnarray}-\frac{L_{f}^{\prime }}{L_{f}}(s)=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D6EC}(n)a_{n}n^{-s},\end{eqnarray}$$

where $\unicode[STIX]{x1D6EC}$ is the von Mangoldt function and $a_{n}=0$ unless $n$ is prime or a prime power. Then, by [Reference Liu and YeLY07, Lemma 5.2], we have

(2) $$\begin{eqnarray}\mathop{\sum }_{n\leqslant x}\unicode[STIX]{x1D6EC}(n)|a_{n}|^{2}\sim x\quad \text{as }x\rightarrow \infty .\end{eqnarray}$$

By the estimate of Kim and Sarnak [Reference KimKim03], we have $|a_{n}|\leqslant n^{7/64}+n^{-7/64}$ , so the contribution of composite $n$ to (2) is $O(x^{23/32})$ . Since $a_{q}=\unicode[STIX]{x1D706}_{f}(q)$ for primes $q$ , this implies that

$$\begin{eqnarray}\mathop{\sum }_{\substack{ q\text{ prime} \\ q\leqslant x}}(\log q)|\unicode[STIX]{x1D706}_{f}(q)|^{2}\sim x,\end{eqnarray}$$

and (1) follows by partial summation and the prime number theorem.

In particular, there are infinitely many $q\nmid N$ such that $|\unicode[STIX]{x1D706}_{f}(q)|<2$ . For any such $q$ , it follows that $D_{f}(s,\unicode[STIX]{x1D712}_{0})$ has infinitely many poles on the line $\Re (s)=0$ . In view of the above, $D_{f}(s,1/q,\cos )$ inherits these poles when they occur. On the other hand, under the assumption that $L_{f}(s)$ has at most finitely many non-trivial simple zeros, we will show that $D_{f}(s,1/q,\cos )$ is holomorphic apart from possible poles along two horizontal lines. The contradiction between these two implies the main theorem.

1.1 Overview

We begin with an overview of the proof. First, by [Reference Duke, Friedlander and IwaniecDFI02, (4.36)], $f$ has the Fourier–Whittaker expansion

$$\begin{eqnarray}f(x+iy)=\mathop{\sum }_{n=1}^{\infty }(\unicode[STIX]{x1D70C}(n)W_{k/2,\unicode[STIX]{x1D708}}(4\unicode[STIX]{x1D70B}ny)e(nx)+\unicode[STIX]{x1D70C}(-n)W_{-k/2,\unicode[STIX]{x1D708}}(4\unicode[STIX]{x1D70B}ny)e(-nx)),\end{eqnarray}$$

where $W_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}}$ is the Whittaker function defined in [Reference Duke, Friedlander and IwaniecDFI02, (4.20)] and $\unicode[STIX]{x1D708}=\sqrt{{\textstyle \frac{1}{4}}-\unicode[STIX]{x1D706}}$ , where $\unicode[STIX]{x1D706}$ is the eigenvalue of $f$ with respect to the weight- $k$ Laplace operator. When $k=1$ , the Selberg eigenvalue conjecture holds, so that $\unicode[STIX]{x1D708}\in i[0,\infty )$ . When $k=0$ the conjecture remains open, but we have the partial result of Kim and Sarnak [Reference KimKim03] that $\unicode[STIX]{x1D708}\in (0,{\textstyle \frac{7}{64}}]\cup i[0,\infty )$ .

Since $f$ is primitive, it is an eigenfunction of the operator $Q_{sk}$ defined in [Reference Duke, Friedlander and IwaniecDFI02, (4.65)], so that

$$\begin{eqnarray}\unicode[STIX]{x1D70C}(-n)=\unicode[STIX]{x1D716}\frac{\unicode[STIX]{x1D6E4}((1+k)/2+\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}((1-k)/2+\unicode[STIX]{x1D708})}\unicode[STIX]{x1D70C}(n)=\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}^{k}\unicode[STIX]{x1D70C}(n)\end{eqnarray}$$

for some $\unicode[STIX]{x1D716}\in \{\pm 1\}$ . Further, we have $\unicode[STIX]{x1D70C}(n)=\unicode[STIX]{x1D70C}(1)\unicode[STIX]{x1D706}_{f}(n)/\sqrt{n}$ . Choosing the normalization $\unicode[STIX]{x1D70C}(1)=\unicode[STIX]{x1D70B}^{-k/2}$ and writing $e(\pm nx)=\cos (2\unicode[STIX]{x1D70B}nx)\pm i\sin (2\unicode[STIX]{x1D70B}nx)$ , we obtain the expansion

(3) $$\begin{eqnarray}f(x+iy)=\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{\sqrt{n}}(V_{f}^{+}(ny)\cos (2\unicode[STIX]{x1D70B}nx)+iV_{f}^{-}(ny)\sin (2\unicode[STIX]{x1D70B}nx)),\end{eqnarray}$$

where

(4) $$\begin{eqnarray}V_{f}^{\pm }(y)=\unicode[STIX]{x1D70B}^{-k/2}(W_{k/2,\unicode[STIX]{x1D708}}(4\unicode[STIX]{x1D70B}y)\pm \unicode[STIX]{x1D716}\unicode[STIX]{x1D708}^{k}W_{-k/2,\unicode[STIX]{x1D708}}(4\unicode[STIX]{x1D70B}y))=\left\{\begin{array}{@{}ll@{}}4\sqrt{y}K_{\unicode[STIX]{x1D708}}(2\unicode[STIX]{x1D70B}y)\quad & \text{if }k=0\text{ and }\unicode[STIX]{x1D716}=\pm 1,\\ 0\quad & \text{if }k=0\text{ and }\unicode[STIX]{x1D716}=\mp 1,\\ 4yK_{\unicode[STIX]{x1D708}\pm \unicode[STIX]{x1D716}/2}(2\unicode[STIX]{x1D70B}y)\quad & \text{if }k=1.\end{array}\right.\end{eqnarray}$$

Let $\bar{f}(z):=\overline{f(-\bar{z})}$ denote the dual of $f$ . Since $f$ is primitive, it is also an eigenfunction of the operator $\overline{W}_{k}$ defined in [Reference Duke, Friedlander and IwaniecDFI02, (6.10)], so we have

(5) $$\begin{eqnarray}f(z)=\unicode[STIX]{x1D702}\biggl(i\frac{|z|}{z}\biggr)^{k}\bar{f}\biggl(-\frac{1}{Nz}\biggr)\end{eqnarray}$$

for some $\unicode[STIX]{x1D702}\in \mathbb{C}$ with $|\unicode[STIX]{x1D702}|=1$ .

Next we define a formal Fourier series $F(z)$ associated to $D_{f}(s)$ by replacing $\unicode[STIX]{x1D706}_{f}(n)$ in the above by $c_{f}(n)$ :

$$\begin{eqnarray}\displaystyle F(x+iy)=\mathop{\sum }_{n=1}^{\infty }\frac{c_{f}(n)}{\sqrt{n}}(V_{f}^{+}(ny)\cos (2\unicode[STIX]{x1D70B}nx)+iV_{f}^{-}(ny)\sin (2\unicode[STIX]{x1D70B}nx)). & & \displaystyle \nonumber\end{eqnarray}$$

We expect $F(z)$ to satisfy a relation similar to the modularity relation (5). To make this precise, we first recall the functional equation for $L_{f}(s)$ . Define

(6) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{f}^{\pm }(s)=\unicode[STIX]{x1D6E4}_{\mathbb{ R}}\biggl(s+\frac{1\mp (-1)^{k}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708}\biggr)\unicode[STIX]{x1D6E4}_{\mathbb{R}}\biggl(s+\frac{1\mp \unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708}\biggr).\end{eqnarray}$$

Then the complete $L$ -function $\unicode[STIX]{x1D6EC}_{f}(s):=\unicode[STIX]{x1D6FE}_{f}^{+}(s)L_{f}(s)$ satisfies

(7) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}(s)=\unicode[STIX]{x1D702}\unicode[STIX]{x1D716}^{1-k}N^{1/2-s}\unicode[STIX]{x1D6EC}_{\bar{f}}(1-s),\end{eqnarray}$$

with $\unicode[STIX]{x1D702}$ as above.

We define a completed version of $D_{f}(s)$ by multiplying by the same $\unicode[STIX]{x1D6E4}$ -factor: $\unicode[STIX]{x1D6E5}_{f}(s):=\unicode[STIX]{x1D6FE}_{f}^{+}(s)D_{f}(s)$ . Then, differentiating the functional equation (7), we obtain

(8) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}(s)+(\unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s))\unicode[STIX]{x1D6EC}_{f}(s)=\unicode[STIX]{x1D702}\unicode[STIX]{x1D716}^{1-k}N^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s),\end{eqnarray}$$

where $\unicode[STIX]{x1D713}_{f}(s):=(d/ds)\log \unicode[STIX]{x1D6FE}_{f}^{+}(s)$ . In §2, we take a suitable inverse Mellin transform of (8). Under the assumption that $\unicode[STIX]{x1D6EC}_{f}(s)$ has at most finitely many simple zeros, this yields a pseudo-modularity relation for $F$ of the form

(9) $$\begin{eqnarray}F(z)+A(z)=\unicode[STIX]{x1D702}\biggl(i\frac{|z|}{z}\biggr)^{k}\overline{F}\biggl(-\frac{1}{Nz}\biggr)+B(z),\end{eqnarray}$$

for certain auxiliary functions $A$ and $B$ , where $\overline{F}(z):=\overline{F(-\bar{z})}$ . Roughly speaking, $A$ is the contribution from the correction term $(\unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s))\unicode[STIX]{x1D6EC}_{f}(s)$ in (8), and $B$ comes from the non-trivial poles of $\unicode[STIX]{x1D6E5}_{f}(s)$ .

The main technical ingredient needed to carry this out is the following pair of Mellin transforms involving the $K$ -Bessel function and trigonometric functions [Reference Gradshteyn and RyzhikGR15, 6.699(3) and 6.699(4)]:

(10) $$\begin{eqnarray}\displaystyle \int _{0}^{\infty }x^{\unicode[STIX]{x1D706}+1}K_{\unicode[STIX]{x1D707}}(ax)\sin (bx)\frac{dx}{x} & = & \displaystyle 2^{\unicode[STIX]{x1D706}}b\unicode[STIX]{x1D6E4}\biggl(\frac{2+\unicode[STIX]{x1D706}+\unicode[STIX]{x1D707}}{2}\biggr)\unicode[STIX]{x1D6E4}\biggl(\frac{2+\unicode[STIX]{x1D706}-\unicode[STIX]{x1D707}}{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \cdot \,\operatorname{2F1}\biggl(\frac{2+\unicode[STIX]{x1D706}+\unicode[STIX]{x1D707}}{2},\frac{2+\unicode[STIX]{x1D706}-\unicode[STIX]{x1D707}}{2};\frac{3}{2};-\frac{b^{2}}{a^{2}}\biggr)\end{eqnarray}$$

and

(11) $$\begin{eqnarray}\displaystyle \int _{0}^{\infty }x^{\unicode[STIX]{x1D706}+1}K_{\unicode[STIX]{x1D707}}(ax)\cos (bx)\frac{dx}{x} & = & \displaystyle \frac{2^{\unicode[STIX]{x1D706}-1}}{a^{\unicode[STIX]{x1D706}+1}}\unicode[STIX]{x1D6E4}\biggl(\frac{1+\unicode[STIX]{x1D706}+\unicode[STIX]{x1D707}}{2}\biggr)\unicode[STIX]{x1D6E4}\biggl(\frac{1+\unicode[STIX]{x1D706}-\unicode[STIX]{x1D707}}{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \cdot \,\operatorname{2F1}\biggl(\frac{1+\unicode[STIX]{x1D706}+\unicode[STIX]{x1D707}}{2},\frac{1+\unicode[STIX]{x1D706}-\unicode[STIX]{x1D707}}{2};\frac{1}{2};-\frac{b^{2}}{a^{2}}\biggr),\end{eqnarray}$$

where

(12) $$\begin{eqnarray}\operatorname{2F1}(a,b;c;z)=\mathop{\sum }_{j=1}^{\infty }\frac{a(a+1)\cdots (a+j-1)\cdot b(b+1)\cdots (b+j-1)}{c(c+1)\cdots (c+j-1)}\frac{z^{j}}{j!}\end{eqnarray}$$

is the Gauss hypergeometric function. The origin of these hypergeometric factors is explained in the introduction to [Reference Booker and ThenBT18], and the need to analyze them is the main difference between this paper and the holomorphic case from [Reference BookerBoo16] (for which the corresponding factors are elementary functions).

Specializing (9) to $z=\unicode[STIX]{x1D6FC}+iy$ for $\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$ , we have

(13) $$\begin{eqnarray}F(\unicode[STIX]{x1D6FC}+iy)+A(\unicode[STIX]{x1D6FC}+iy)=\unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D6FC}+iy|}{\unicode[STIX]{x1D6FC}+iy}\biggr)^{k}\overline{F}\biggl(-\frac{1}{N(\unicode[STIX]{x1D6FC}+iy)}\biggr)+B(\unicode[STIX]{x1D6FC}+iy).\end{eqnarray}$$

We will take the Mellin transform of (13). Without difficulty the reader can guess that the transform of $F(\unicode[STIX]{x1D6FC}+iy)$ will be a combination of $D_{f}(s,\unicode[STIX]{x1D6FC},\cos )$ and $D_{f}(s,\unicode[STIX]{x1D6FC},\sin )$ . The calculation of the other terms is non-trivial, but ultimately we obtain the following proposition, which will play the role of [Reference BookerBoo16, Proposition 2.1].

Proposition 1.2. Suppose that $\unicode[STIX]{x1D6EC}_{f}(s)$ has at most finitely many simple zeros. Then, for every $M\in \mathbb{Z}_{{\geqslant}0}$ and $a\in \{0,1\}$ ,

$$\begin{eqnarray}\displaystyle & & \displaystyle P_{f}(s;a,0)\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(a+k)})\nonumber\\ \displaystyle & & \displaystyle \quad -\,\unicode[STIX]{x1D702}(-\!\operatorname{sgn}\unicode[STIX]{x1D6FC})^{k}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}\mathop{\sum }_{m=0}^{M-1}\frac{(2\unicode[STIX]{x1D70B}N\unicode[STIX]{x1D6FC})^{m}}{m!}P_{f}(s;a,m)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+m,-\frac{1}{N\unicode[STIX]{x1D6FC}},\cos ^{(a+m)}\biggr)\nonumber\end{eqnarray}$$

is holomorphic for $\Re (s)>{\textstyle \frac{3}{2}}-M$ except for possible poles for $s\pm \unicode[STIX]{x1D708}\in \mathbb{Z}$ , where

$$\begin{eqnarray}P_{f}(s;a,m)=\frac{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a}}(1-s)}{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a}}(1-s-2\lfloor m/2\rfloor )}\left\{\begin{array}{@{}ll@{}}{\displaystyle \frac{s+2\lfloor m/2\rfloor -(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}}{2\unicode[STIX]{x1D70B}}}\quad & \text{if }k=1\text{ and }2\nmid m,\\ 0\quad & \text{if }k=0\text{ and }(-1)^{a}=-\unicode[STIX]{x1D716},\\ 1\quad & \text{otherwise,}\end{array}\right.\end{eqnarray}$$

and

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(a)})=\unicode[STIX]{x1D6FE}_{f}^{(-)^{a}}(s)D_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(a)}).\end{eqnarray}$$

1.2 Proof of Theorem 1.1

Assuming Proposition 1.2 for the moment, we can complete the proof of Theorem 1.1 for the case of $\unicode[STIX]{x1D70B}$ corresponding to a Maass cusp form, $f$ . First, as noted above, we may choose a prime $q\nmid N$ for which $D_{f}(s,1/q,\cos )$ has infinitely many poles on the line $\Re (s)=0$ . Then, by Dirichlet’s theorem on primes in an arithmetic progression, for any $M\in \mathbb{Z}_{{>}0}$ there are distinct primes $q_{0},q_{1},\ldots ,q_{M-1}$ such that $q_{j}\equiv q(modN)$ and $D_{\bar{f}}(s,-q_{j}/N,\cos ^{(a)})=D_{\bar{f}}(s,-q/N,\cos ^{(a)})$ for all $j$ , $a$ .

Let $m_{0}$ be an integer with $0\leqslant m_{0}\leqslant M-1$ . By the Vandermonde determinant, there exist rational numbers $c_{0},c_{1},\ldots ,c_{M-1}$ such that

$$\begin{eqnarray}\mathop{\sum }_{j=0}^{M-1}c_{j}q_{j}^{-m}=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }m=m_{0}\\ 0\quad & \text{if }m\neq m_{0}\end{array}\right.\quad \text{for all }m\in \{0,1,\ldots ,M-1\}.\end{eqnarray}$$

We fix $\unicode[STIX]{x1D6FF}\in \{0,1\}$ and apply Proposition 1.2 with $a\equiv \unicode[STIX]{x1D6FF}+m_{0}(mod2)$ and $\unicode[STIX]{x1D6FC}=1/q_{j}$ for $j=0,1,\ldots ,M-1$ . Multiplying by $(-1)^{k}c_{j}(q_{j}^{2}/N)^{s-1/2}$ , summing over $j$ and replacing $s$ by $s-m_{0}$ , we find that

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=0}^{M-1}(-1)^{k}c_{j}\biggl(\frac{q_{j}^{2}}{N}\biggr)^{s-m_{0}-1/2}P_{f}(s-m_{0};\unicode[STIX]{x1D6FF}+m_{0},0)\unicode[STIX]{x1D6E5}_{f}\biggl(s-m_{0},\frac{1}{q_{j}},\cos ^{(\unicode[STIX]{x1D6FF}+m_{0}+k)}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad -\,\unicode[STIX]{x1D702}\frac{(-2\unicode[STIX]{x1D70B}N)^{m_{0}}}{m_{0}!}P_{f}(s-m_{0};\unicode[STIX]{x1D6FF}+m_{0},m_{0})\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s,-\frac{q}{N},\cos ^{(\unicode[STIX]{x1D6FF})}\biggr)\nonumber\end{eqnarray}$$

is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{3}{2}+m_{0}-M\}$ , where we set

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}=\{s\in \mathbb{C}:s\pm \unicode[STIX]{x1D708}\notin \mathbb{Z}\}.\end{eqnarray}$$

Since $D_{f}(s-m_{0},1/q_{j},\cos ^{(\unicode[STIX]{x1D6FF}+m_{0}+k)})$ is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)<m_{0}-\frac{1}{2}\}$ , choosing $m_{0}=2+\unicode[STIX]{x1D6FF}+(1-\unicode[STIX]{x1D716})/2$ and $M$ arbitrarily large, we conclude that $D_{\bar{f}}(s,-q/N,\cos ^{(\unicode[STIX]{x1D6FF})})$ is holomorphic on $\unicode[STIX]{x1D6FA}$ .

Next we apply Proposition 1.2 again with $a=k$ , $\unicode[STIX]{x1D6FC}=1/q$ and $M=2$ . When $k=1$ or $k=0$ and $\unicode[STIX]{x1D716}=1$ , we see that $D_{f}(s,1/q,\cos )$ is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)=0\}$ . This is a contradiction, and Theorem 1.1 follows in these cases.

The remaining case is that of odd Maass forms of weight $0$ . The above argument with $\unicode[STIX]{x1D6FF}=1$ shows that $D_{f}(s,-q/N,\sin )$ is entire apart from possible poles for $s\pm \unicode[STIX]{x1D708}\in \mathbb{Z}$ . Applying Proposition 1.2 with $a=1$ , $\unicode[STIX]{x1D6FC}=-q/N$ and $M=3$ , we find that

$$\begin{eqnarray}\displaystyle & & \displaystyle -\unicode[STIX]{x1D6E5}_{f}\biggl(s,-\frac{q}{N},\sin \biggr)+\unicode[STIX]{x1D702}\biggl(\frac{q^{2}}{N}\biggr)^{s-1/2} [\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s,\frac{1}{q},\sin \biggr)-2\unicode[STIX]{x1D70B}q\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+1,\frac{1}{q},\cos \biggr)\nonumber\\ \displaystyle & & \displaystyle \quad -\,\frac{(2\unicode[STIX]{x1D70B}q)^{2}}{2!}P_{f}(s;1,2)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+2,\frac{1}{q},\sin \biggr)\!]\nonumber\end{eqnarray}$$

is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>-\frac{5}{2}\}$ . Since $D_{\bar{f}}(s,1/q,\sin )$ is holomorphic on the lines $\Re (s)=-1$ and $\Re (s)=1$ , we see that $D_{\bar{f}}(s,1/q,\cos )$ is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)=0\}$ . This is again a contradiction, and concludes the proof.

2 Proof of Proposition 1.2

Using expansion (3), we take the Mellin transform of (5) along the line $z=(\unicode[STIX]{x1D714}+i)y$ . First, the left-hand side becomes, for $\Re (s)\gg 1$ ,

(14) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }f(\unicode[STIX]{x1D714}y+iy)y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{\sqrt{n}}\int _{0}^{\infty }(V_{f}^{+}(ny)\cos (2\unicode[STIX]{x1D70B}n\unicode[STIX]{x1D714}y)+iV_{f}^{-}(ny)\sin (2\unicode[STIX]{x1D70B}n\unicode[STIX]{x1D714}y))y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =G_{f}(s,\unicode[STIX]{x1D714})L_{f}(s),\end{eqnarray}$$

where, by (4), (10) and (11),

(15) $$\begin{eqnarray}\displaystyle & & \displaystyle G_{f}(s,\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{0}^{\infty }(V_{f}^{+}(y)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}y)+iV_{f}^{-}(y)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}y))y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\left\{\begin{array}{@{}ll@{}}(2\unicode[STIX]{x1D70B}i\unicode[STIX]{x1D714})^{(1-\unicode[STIX]{x1D716})/2}\unicode[STIX]{x1D6FE}_{f}^{+}(s)\operatorname{2F1}\biggl({\displaystyle \frac{s+(1-\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708}}{2}},{\displaystyle \frac{s+(1-\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708}}{2}};1-{\displaystyle \frac{\unicode[STIX]{x1D716}}{2}};-\unicode[STIX]{x1D714}^{2}\biggr)\quad & \text{if }k=0,\\ \unicode[STIX]{x1D6FE}_{f}^{+}(s)\operatorname{2F1}\biggl({\displaystyle \frac{s+(1+\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708}}{2}},{\displaystyle \frac{s+(1-\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708}}{2}};{\displaystyle \frac{1}{2}};-\unicode[STIX]{x1D714}^{2}\biggr)\quad \\ \quad +\,2\unicode[STIX]{x1D70B}i\unicode[STIX]{x1D714}\unicode[STIX]{x1D6FE}_{f}^{-}(s+1)\operatorname{2F1}\biggl({\displaystyle \frac{s+(3-\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708}}{2}},{\displaystyle \frac{s+(3+\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708}}{2}};{\displaystyle \frac{3}{2}};-\unicode[STIX]{x1D714}^{2}\biggr)\quad & \text{if }k=1.\end{array}\right.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Note that we have $G_{\bar{f}}(s,\unicode[STIX]{x1D714})=\overline{G_{f}(\bar{s},-\unicode[STIX]{x1D714})}$ .

On the other hand, the Mellin transform of the right-hand side of (5) is, for $-\Re (s)\gg 1$ ,

$$\begin{eqnarray}\unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}\int _{0}^{\infty }\bar{f}\biggl(-\frac{\unicode[STIX]{x1D714}}{N(\unicode[STIX]{x1D714}^{2}+1)y}+\frac{i}{N(\unicode[STIX]{x1D714}^{2}+1)y}\biggr)y^{s-1/2}\frac{dy}{y}.\end{eqnarray}$$

Making the substitution $y\mapsto (N(\unicode[STIX]{x1D714}^{2}+1)y)^{-1}$ , this becomes

(16) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}(N(1+\unicode[STIX]{x1D714}^{2}))^{1/2-s}\int _{0}^{\infty }\bar{f}(-\unicode[STIX]{x1D714}y+iy)y^{1/2-s}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}(N(1+\unicode[STIX]{x1D714}^{2}))^{1/2-s}G_{\bar{f}}(1-s,-\unicode[STIX]{x1D714})L_{\bar{f}}(1-s).\end{eqnarray}$$

By (5), (14) and (16) must continue to entire functions and equal each other. In particular, taking $\unicode[STIX]{x1D714}\rightarrow 0$ , we recover the functional equation (7). Equating (14) with (16) and dividing by (7), we discover the functional equation for the hypergeometric factor $H_{f}(s,\unicode[STIX]{x1D714}):=G_{f}(s,\unicode[STIX]{x1D714})/\unicode[STIX]{x1D6FE}_{f}^{+}(s)$ :

(17) $$\begin{eqnarray}H_{f}(s,\unicode[STIX]{x1D714})=\unicode[STIX]{x1D716}^{1-k}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}(1+\unicode[STIX]{x1D714}^{2})^{1/2-s}H_{\bar{f}}(1-s,-\unicode[STIX]{x1D714}).\end{eqnarray}$$

Next, for $z=x+iy\in \mathbb{H}$ , define

$$\begin{eqnarray}A(z)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}(\unicode[STIX]{x1D713}^{\prime }(s+\unicode[STIX]{x1D708})+\unicode[STIX]{x1D713}^{\prime }(s-\unicode[STIX]{x1D708}))H_{f}(s,x/y)\unicode[STIX]{x1D6EC}_{f}(s)y^{1/2-s}\,ds\end{eqnarray}$$

and

(18) $$\begin{eqnarray}B(z)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}X_{f}(s)\unicode[STIX]{x1D6EC}_{f}(s)H_{f}(s,x/y)y^{1/2-s}\,ds-\mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})H_{f}(\unicode[STIX]{x1D70C},x/y)y^{1/2-\unicode[STIX]{x1D70C}},\end{eqnarray}$$

where the sum runs over all simple zeros of $\unicode[STIX]{x1D6EC}_{f}(s)$ , and

$$\begin{eqnarray}X_{f}(s)=\frac{\unicode[STIX]{x1D70B}^{2}}{4}\biggl[\csc ^{2}\biggl(\frac{\unicode[STIX]{x1D70B}}{2}\biggl[s+\frac{1+(-1)^{k}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708}\biggr]\biggr)+\csc ^{2}\biggl(\frac{\unicode[STIX]{x1D70B}}{2}\biggl[s+\frac{1+\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708}\biggr]\biggr)\biggr].\end{eqnarray}$$

Lemma 2.1.

$$\begin{eqnarray}F(z)+A(z)=\unicode[STIX]{x1D702}\biggl(i\frac{|z|}{z}\biggr)^{k}\overline{F}\biggl(-\frac{1}{Nz}\biggr)+B(z)\quad \text{for all }z\in \mathbb{H}.\end{eqnarray}$$

Proof. Fix $z=x+iy\in \mathbb{H}$ , and put $\unicode[STIX]{x1D714}=x/y$ . Applying Mellin inversion as in (14), we have

$$\begin{eqnarray}F(z)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}D_{f}(s)G_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s}\,ds\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D702}\biggl(i\frac{|z|}{z}\biggr)^{k}\overline{F}\biggl(-\frac{1}{Nz}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}\cdot \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}G_{\bar{f}}(s,-\unicode[STIX]{x1D714})D_{\bar{f}}(s)(N(1+\unicode[STIX]{x1D714}^{2})y)^{s-1/2}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D702}\biggl(i\frac{|\unicode[STIX]{x1D714}+i|}{\unicode[STIX]{x1D714}+i}\biggr)^{k}\cdot \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=-1}H_{\bar{f}}(1-s,-\unicode[STIX]{x1D714})\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s)(N(1+\unicode[STIX]{x1D714}^{2})y)^{1/2-s}\,ds.\nonumber\end{eqnarray}$$

Applying (17) and (8), and using the fact that $\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s)$ is holomorphic for $\Re (s)\leqslant \frac{1}{2}$ , the last line becomes

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=-1}\unicode[STIX]{x1D702}\unicode[STIX]{x1D716}^{1-k}H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s)(Ny)^{1/2-s}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=-1}H_{f}(s,\unicode[STIX]{x1D714})[\unicode[STIX]{x1D6E5}_{f}(s)+(\unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s))\unicode[STIX]{x1D6EC}_{f}(s)]y^{1/2-s}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=-1}H_{f}(s,\unicode[STIX]{x1D714})[\unicode[STIX]{x1D6E5}_{f}(s)+\unicode[STIX]{x1D713}_{f}^{\prime }(s)\unicode[STIX]{x1D6EC}_{f}(s)]y^{1/2-s}\,ds\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s)\unicode[STIX]{x1D6EC}_{f}(s)y^{1/2-s}\,ds.\nonumber\end{eqnarray}$$

Shifting the contour of the first integral to the right and using that $\unicode[STIX]{x1D713}_{f}^{\prime }(s)$ is holomorphic for $\Re (s)\geqslant \frac{1}{2}$ , we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D6E5}_{f}(s)y^{1/2-s}\,ds-\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{{\mathcal{C}}}H_{f}(s,\unicode[STIX]{x1D714})(\unicode[STIX]{x1D6E5}_{f}(s)+\unicode[STIX]{x1D713}_{f}^{\prime }(s)\unicode[STIX]{x1D6EC}_{f}(s))y^{1/2-s}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad +\,\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}(\unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s))H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D6EC}_{f}(s)y^{1/2-s}\,ds,\nonumber\end{eqnarray}$$

where ${\mathcal{C}}$ is the contour running from $2-i\infty$ to $2+i\infty$ and from $-1+i\infty$ to $-1-i\infty$ . Note that

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}(s)+\unicode[STIX]{x1D713}_{f}^{\prime }(s)\unicode[STIX]{x1D6EC}_{f}(s)=\unicode[STIX]{x1D6EC}_{f}(s)\frac{d^{2}}{ds^{2}}\log \unicode[STIX]{x1D6EC}_{f}(s),\end{eqnarray}$$

which has a pole at every simple zero $\unicode[STIX]{x1D70C}$ of $\unicode[STIX]{x1D6EC}_{f}(s)$ , with residue $-\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})$ . Hence,

$$\begin{eqnarray}-\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{{\mathcal{C}}}H_{f}(s,\unicode[STIX]{x1D714})(\unicode[STIX]{x1D6E5}_{f}(s)+\unicode[STIX]{x1D713}_{f}^{\prime }(s)\unicode[STIX]{x1D6EC}_{f}(s))y^{1/2-s}\,ds=\mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})H_{f}(\unicode[STIX]{x1D70C},\unicode[STIX]{x1D714})y^{1/2-\unicode[STIX]{x1D70C}}.\end{eqnarray}$$

Next, writing $\unicode[STIX]{x1D713}_{\mathbb{R}}(s)=(\unicode[STIX]{x1D6E4}_{\mathbb{R}}^{\prime }/\unicode[STIX]{x1D6E4}_{\mathbb{R}})(s)$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{f}(s)=\unicode[STIX]{x1D713}_{\mathbb{R}}\biggl(s+\frac{1-(-1)^{k}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708}\biggr)+\unicode[STIX]{x1D713}_{\mathbb{R}}\biggl(s+\frac{1-\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708}\biggr).\end{eqnarray}$$

Applying the reflection formula and Legendre duplication formula in the form

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{\mathbb{R}}^{\prime }(s)=\frac{\unicode[STIX]{x1D70B}^{2}}{4}\csc ^{2}\biggl(\frac{\unicode[STIX]{x1D70B}s}{2}\biggr)-\unicode[STIX]{x1D713}_{\mathbb{R}}^{\prime }(2-s)\quad \text{and}\quad \unicode[STIX]{x1D713}_{\mathbb{ R}}^{\prime }(s)+\unicode[STIX]{x1D713}_{\mathbb{ R}}^{\prime }(s+1)=\unicode[STIX]{x1D713}^{\prime }(s),\end{eqnarray}$$

we derive

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s)=\unicode[STIX]{x1D713}^{\prime }(s+\unicode[STIX]{x1D708})+\unicode[STIX]{x1D713}^{\prime }(s-\unicode[STIX]{x1D708})-X_{f}(s). & & \displaystyle \nonumber\end{eqnarray}$$

Thus,

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}(\unicode[STIX]{x1D713}_{f}^{\prime }(s)-\unicode[STIX]{x1D713}_{\bar{f}}^{\prime }(1-s))H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D6EC}_{f}(s)y^{1/2-s}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad =A(z)-\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=1/2}X_{f}(s)H_{f}(s,\unicode[STIX]{x1D714})\unicode[STIX]{x1D6EC}_{f}(s)y^{1/2-s}\,ds.\nonumber\end{eqnarray}$$

Rearranging terms completes the proof. ◻

Lemma 2.2. For any $\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$ ,

$$\begin{eqnarray}\frac{1}{\unicode[STIX]{x1D6E4}(s+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}(s-\unicode[STIX]{x1D708})}\int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+iy)y^{s-1/2}\frac{dy}{y}\end{eqnarray}$$

continues to an entire function of $s$ .

Proof. Define $\unicode[STIX]{x1D6F7}(s)=\unicode[STIX]{x1D713}^{\prime }(s+\unicode[STIX]{x1D708})+\unicode[STIX]{x1D713}^{\prime }(s-\unicode[STIX]{x1D708})$ . Then we have $\unicode[STIX]{x1D6F7}(s)=\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)x^{1/2-s}\,dx$ , where $\unicode[STIX]{x1D719}(x)=\cosh (\unicode[STIX]{x1D708}\log x)\log x/\text{sinh}((1/2)\log x)$ . Applying (15) and the change of variables $y\mapsto xt$ , we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(s)G_{f}(s,\unicode[STIX]{x1D714}) & = & \displaystyle \int _{1}^{\infty }\int _{0}^{\infty }\unicode[STIX]{x1D719}(x)(V_{f}^{+}(y)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}y)+iV_{f}^{-}(y)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}y))\biggl(\frac{y}{x}\biggr)^{s-1/2}\frac{dy}{y}\,dx\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }\biggl(\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)(V_{f}^{+}(tx)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}tx)+iV_{f}^{-}(tx)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D714}tx))\,dx\biggr)t^{s-1/2}\frac{dt}{t}.\nonumber\end{eqnarray}$$

Hence, writing $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D6FC}/y$ , we have

$$\begin{eqnarray}\displaystyle A(\unicode[STIX]{x1D6FC}+iy) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}\unicode[STIX]{x1D6EC}_{f}(s)\unicode[STIX]{x1D6F7}(s)H_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s}\,ds\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{\sqrt{n}}\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}\unicode[STIX]{x1D6F7}(s)G_{f}(s,\unicode[STIX]{x1D714})(ny)^{1/2-s}\,ds\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{\sqrt{n}}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)(V_{f}^{+}(nxy)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)+iV_{f}^{-}(nxy)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx))\,dx,\nonumber\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+iy)y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{\sqrt{n}}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\int _{0}^{\infty }(V_{f}^{+}(nxy)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)+iV_{f}^{-}(nxy)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx))y^{s-1/2}\frac{dy}{y}\,dx\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)n^{-s}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)x^{1/2-s}(\widetilde{V}_{f}^{+}(s)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)+i\widetilde{V}_{f}^{-}(s)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx))\,dx,\nonumber\end{eqnarray}$$

where

(19) $$\begin{eqnarray}\widetilde{V}_{f}^{\pm }(s)=\int _{0}^{\infty }V_{f}^{\pm }(y)y^{s-1/2}\frac{dy}{y}=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D6FE}_{f}^{\pm }(s)\quad & \text{if }k=1\text{ or }\unicode[STIX]{x1D716}=\pm 1,\\ 0\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

A case-by-case inspection of (6) shows that $\widetilde{V}_{f}^{\pm }(s)/(\unicode[STIX]{x1D6E4}(s+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}(s-\unicode[STIX]{x1D708}))$ is entire for both choices of sign.

Define $\unicode[STIX]{x1D719}_{j}=\unicode[STIX]{x1D719}_{j}(x,s)$ for $j\geqslant 0$ by

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{0}=\unicode[STIX]{x1D719}\quad \text{and}\quad \unicode[STIX]{x1D719}_{j+1}=x\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{j}}{\unicode[STIX]{x2202}x}-\biggl(s+j-\frac{1}{2}\biggr)\unicode[STIX]{x1D719}_{j}.\end{eqnarray}$$

Then, applying integration by parts $m$ times, we see that

$$\begin{eqnarray}\displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\cos (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)x^{1/2-s}\,dx & = & \displaystyle \mathop{\sum }_{j=0}^{m-1}\frac{\cos ^{(j+1)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)^{j+1}}\unicode[STIX]{x1D719}_{j}(1,s)\nonumber\\ \displaystyle & & \displaystyle +\,\int _{1}^{\infty }\frac{\cos ^{(m)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)^{m}}\unicode[STIX]{x1D719}_{k}(x,s)x^{1/2-m-s}\,dx\nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\sin (2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)x^{1/2-s}\,dx & = & \displaystyle \mathop{\sum }_{j=0}^{m-1}\frac{\sin ^{(j+1)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)^{j+1}}\unicode[STIX]{x1D719}_{j}(1,s)\nonumber\\ \displaystyle & & \displaystyle +\,\int _{1}^{\infty }\frac{\sin ^{(m)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}n)^{m}}\unicode[STIX]{x1D719}_{k}(x,s)x^{1/2-m-s}\,dx.\nonumber\end{eqnarray}$$

Thus,

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+iy)y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\widetilde{V}_{f}^{+}(s)(\mathop{\sum }_{j=0}^{m-1}\frac{\unicode[STIX]{x1D719}_{j}(1,s)L(f,s+j+1,\unicode[STIX]{x1D6FC},\cos ^{(j+1)})}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC})^{j+1}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC})^{m}}\mathop{\sum }_{n=1}^{\infty }\frac{a_{f}(n)}{n^{s+m}}\int _{1}^{\infty }\cos ^{(m)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)\unicode[STIX]{x1D719}_{m}(x,s)x^{1/2-m-s}\,dx\!)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,i\widetilde{V}_{f}^{-}(s)(\mathop{\sum }_{j=0}^{m-1}\frac{\unicode[STIX]{x1D719}_{j}(1,s)L(f,s+j+1,\unicode[STIX]{x1D6FC},\sin ^{(j+1)})}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC})^{j+1}}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\frac{1}{(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC})^{m}}\mathop{\sum }_{n=1}^{\infty }\frac{a_{f}(n)}{n^{s+m}}\int _{1}^{\infty }\sin ^{(m)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FC}nx)\unicode[STIX]{x1D719}_{m}(x,s)x^{1/2-m-s}\,dx\!).\nonumber\end{eqnarray}$$

It follows from [Reference Booker and KrishnamurthyBK11, Proposition 3.1] that $L_{f}(s,\unicode[STIX]{x1D6FC},\cos )$ and $L_{f}(s,\unicode[STIX]{x1D6FC},\sin )$ continue to entire functions. We see by induction that $\unicode[STIX]{x1D719}_{m}(x,s)\ll _{m}((1+|s|)(1+|\unicode[STIX]{x1D708}|))^{m}x^{-1}$ uniformly for $x\geqslant 1$ , and thus the integral terms above are holomorphic for $\Re (s)>\frac{1}{2}-m$ . Choosing $m$ arbitrarily large, the lemma follows.◻

Lemma 2.3. For any $\unicode[STIX]{x1D70E}\geqslant 0$ and any $l\in \mathbb{Z}_{{\geqslant}0}$ , we have

$$\begin{eqnarray}\frac{y^{l}}{l!}(V_{\bar{f}}^{\pm })^{(l)}(y)\ll _{\unicode[STIX]{x1D70E}}2^{l}y^{-\unicode[STIX]{x1D70E}}\quad \text{for }y>0.\end{eqnarray}$$

Proof. In view of (19), since $|\Re (\unicode[STIX]{x1D708})|<\frac{1}{2}$ , for any $\unicode[STIX]{x1D70E}\geqslant 0$ we have the integral representation

$$\begin{eqnarray}V_{\bar{f}}^{\pm }(y)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=\unicode[STIX]{x1D70E}+1/2}\widetilde{V}_{\bar{f}}^{\pm }(s)y^{1/2-s}\,ds.\end{eqnarray}$$

Differentiating $l$ times, we obtain

$$\begin{eqnarray}\frac{y^{l}}{l!}(V_{\bar{f}}^{\pm })^{(l)}(y)=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=\unicode[STIX]{x1D70E}+1/2}\binom{\frac{1}{2}-s}{l}\widetilde{V}_{\bar{f}}^{\pm }(s)y^{1/2-s}\,ds.\end{eqnarray}$$

Using the estimate

$$\begin{eqnarray}\biggl|\binom{\frac{1}{2}-s}{l}\biggr|=\biggl|\binom{s-\frac{1}{2}+l}{l}\biggr|\leqslant 2^{|s-1/2|+l},\end{eqnarray}$$

we have

$$\begin{eqnarray}\frac{y^{l}}{l!}(V_{\bar{f}}^{\pm })^{(l)}(y)\leqslant 2^{l}y^{-\unicode[STIX]{x1D70E}}\cdot \frac{1}{2\unicode[STIX]{x1D70B}}\int _{\Re (s)=\unicode[STIX]{x1D70E}+1/2}2^{|s-1/2|}|\widetilde{V}_{\bar{f}}^{\pm }(s)\,ds|\ll _{\unicode[STIX]{x1D70E}}2^{l}y^{-\unicode[STIX]{x1D70E}},\end{eqnarray}$$

where the last inequality is justified by Stirling’s formula. ◻

Lemma 2.4. Let $\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$ and $z=\unicode[STIX]{x1D6FC}+iy$ for some $y\in (0,|\unicode[STIX]{x1D6FC}|/2]$ . Then, for any integer $T\geqslant 0$ , we have

(20) $$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(i\frac{|z|}{z}\biggr)^{k}\overline{F}\biggl(-\frac{1}{Nz}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =O_{\unicode[STIX]{x1D6FC},T}(y^{T-1})+(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{t=0}^{T-1}\frac{(2\unicode[STIX]{x1D70B}iN\unicode[STIX]{x1D6FC})^{t}}{t!}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{a\in \{0,1\}}\frac{i^{-a}}{2\unicode[STIX]{x1D70B}i}\int _{\Re (s)=2}P_{f}(s;a+t,t)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+t,-\frac{1}{N\unicode[STIX]{x1D6FC}},\cos ^{(a)}\biggr)\biggl(\frac{y}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)^{1/2-s}\,ds.\end{eqnarray}$$

Proof. Let $z=\unicode[STIX]{x1D6FC}+iy$ , $\unicode[STIX]{x1D6FD}=-1/N\unicode[STIX]{x1D6FC}$ and $u=y/\unicode[STIX]{x1D6FC}$ . Then

$$\begin{eqnarray}-\frac{1}{Nz}=\frac{\unicode[STIX]{x1D6FD}}{1+u^{2}}+i\frac{|\unicode[STIX]{x1D6FD}u|}{1+u^{2}},\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle \biggl(i\frac{|z|}{z}\biggr)^{k}\overline{F}\biggl(-\frac{1}{Nz}\biggr) & = & \displaystyle \biggl(i\operatorname{sgn}(\unicode[STIX]{x1D6FC})\frac{|1+iu|}{1+iu}\biggr)^{k}\overline{F}\biggl(\frac{\unicode[STIX]{x1D6FD}}{1+u^{2}}+i\frac{|\unicode[STIX]{x1D6FD}u|}{1+u^{2}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \biggl(i\operatorname{sgn}(\unicode[STIX]{x1D6FC})\frac{|1+iu|}{1+iu}\biggr)^{k}\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)}{\sqrt{n}}\biggl(V_{\bar{f}}^{+}\biggl(\frac{|\unicode[STIX]{x1D6FD}nu|}{1+u^{2}}\biggr)\cos \biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+u^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle +\,iV_{\bar{f}}^{-}\biggl(\frac{|\unicode[STIX]{x1D6FD}nu|}{1+u^{2}}\biggr)\sin \biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+u^{2}}\biggr)\!\biggr).\nonumber\end{eqnarray}$$

By Lemma 2.3, for any $\unicode[STIX]{x1D70E}\geqslant 0$ and any $l_{0}\in \mathbb{Z}_{{\geqslant}0}$ , we have

$$\begin{eqnarray}\displaystyle V_{\bar{f}}^{\pm }\biggl(\frac{|\unicode[STIX]{x1D6FD}nu|}{1+u^{2}}\biggr) & = & \displaystyle \mathop{\sum }_{l=0}^{\infty }\frac{1}{l!}(V_{\bar{f}}^{\pm })^{(l)}(|\unicode[STIX]{x1D6FD}nu|)\biggl(\frac{\unicode[STIX]{x1D6FD}nu^{3}}{1+u^{2}}\biggr)^{l}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{l=0}^{l_{0}-1}\frac{1}{l!}(V_{\bar{f}}^{\pm })^{(l)}(|\unicode[STIX]{x1D6FD}nu|)\biggl(\frac{\unicode[STIX]{x1D6FD}nu^{3}}{1+u^{2}}\biggr)^{l}+O_{\unicode[STIX]{x1D70E}}\biggl(|\unicode[STIX]{x1D6FD}nu|^{-\unicode[STIX]{x1D70E}}\mathop{\sum }_{l=l_{0}}^{\infty }\biggl(\frac{2u^{2}}{1+u^{2}}\biggr)^{l}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{l=0}^{l_{0}-1}\frac{1}{l!}(V_{\bar{f}}^{\pm })^{(l)}(|\unicode[STIX]{x1D6FD}nu|)\biggl(\frac{\unicode[STIX]{x1D6FD}nu^{3}}{1+u^{2}}\biggr)^{l}+O_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70E},l_{0}}(|nu|^{-\unicode[STIX]{x1D70E}}u^{2l_{0}}).\nonumber\end{eqnarray}$$

Similarly, for any $a\in \{0,1\}$ , we have

$$\begin{eqnarray}\displaystyle \cos ^{(a)}\biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+u^{2}}\biggr) & = & \displaystyle \mathop{\sum }_{j=0}^{\infty }\frac{1}{j!}\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(-\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}nu^{2}}{1+u^{2}}\biggr)^{j}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{j=0}^{j_{0}-1}\frac{1}{j!}\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(-\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}nu^{2}}{1+u^{2}}\biggr)^{j}+O\biggl(\frac{1}{j_{0}!}\biggl|\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}nu^{2}}{1+u^{2}}\biggr|^{j_{0}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{j=0}^{j_{0}-1}\frac{1}{j!}\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(-\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}nu^{2}}{1+u^{2}}\biggr)^{j}+O_{\unicode[STIX]{x1D6FC},j_{0}}((nu^{2})^{j_{0}}),\nonumber\end{eqnarray}$$

by the Lagrange form of the error in Taylor’s theorem. Taking $j_{0}=2(l_{0}-l)$ and applying Lemma 2.3 with $\unicode[STIX]{x1D70E}$ replaced by $\unicode[STIX]{x1D70E}+2(l_{0}-l)$ , we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle V_{\bar{f}}^{(-)^{a}}\biggl(\frac{|\unicode[STIX]{x1D6FD}nu|}{1+u^{2}}\biggr)\cos ^{(a)}\biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+u^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j+2l<2l_{0}}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!l!}(V_{\bar{f}}^{(-)^{a}})^{(l)}(|\unicode[STIX]{x1D6FD}nu|)\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)u^{l}\biggl(\frac{\unicode[STIX]{x1D6FD}nu^{2}}{1+u^{2}}\biggr)^{j+l}+O_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70E},l_{0}}(|nu|^{-\unicode[STIX]{x1D70E}}u^{2l_{0}}).\nonumber\end{eqnarray}$$

Next, defining

$$\begin{eqnarray}b_{j,k,l,m}=\left\{\begin{array}{@{}ll@{}}\left(\begin{array}{@{}c@{}}j+l-1+\biggl\lfloor{\displaystyle \frac{m}{2}}\biggr\rfloor+{\displaystyle \frac{k}{2}}\\ \biggl\lfloor{\displaystyle \frac{m}{2}}\biggr\rfloor\end{array}\right)\quad & \text{if }k=1\text{ or }k=0\text{ and }2\mid m,\\ 0\quad & \text{otherwise},\end{array}\right.\end{eqnarray}$$

we have

$$\begin{eqnarray}\displaystyle \biggl(\frac{|1+iu|}{1+iu}\biggr)^{k}(1+u^{2})^{-j-l} & = & \displaystyle (1-iu)^{k}(1+u^{2})^{-j-l-k/2}=\mathop{\sum }_{m=0}^{\infty }b_{j,k,l,m}(-iu)^{m}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{m=0}^{m_{0}-1}b_{j,k,l,m}(-iu)^{m}+O\biggl(\mathop{\sum }_{m=m_{0}}^{\infty }2^{j+l+m/2}|u|^{m}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{m=0}^{m_{0}-1}b_{j,k,l,m}(-iu)^{m}+O_{j,l,m_{0}}(|u|^{m_{0}}).\nonumber\end{eqnarray}$$

Taking $m_{0}=2l_{0}-j-2l$ and applying Lemma 2.3 with $\unicode[STIX]{x1D70E}$ replaced by $\unicode[STIX]{x1D70E}+j$ , we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(i\operatorname{sgn}(\unicode[STIX]{x1D6FC})\frac{|1+iu|}{1+iu}\biggr)^{k}V_{\bar{f}}^{(-)^{a}}\biggl(\frac{|\unicode[STIX]{x1D6FD}nu|}{1+u^{2}}\biggr)\cos ^{(a)}\biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+u^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{j+2l+m<2l_{0}}\frac{(-2\unicode[STIX]{x1D70B})^{j}(-i)^{m}}{j!l!}b_{j,k,l,m}(\unicode[STIX]{x1D6FD}nu)^{j+l}(V_{\bar{f}}^{(-)^{a}})^{(l)}(|\unicode[STIX]{x1D6FD}nu|)\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)u^{j+2l+m}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,O_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70E},l_{0}}(|nu|^{-\unicode[STIX]{x1D70E}}u^{2l_{0}}).\nonumber\end{eqnarray}$$

Recalling the definition of $u$ , multiplying by $c_{\bar{f}}(n)/\sqrt{n}$ and summing over $n$ and both choices of $a$ , the error term converges if $\unicode[STIX]{x1D70E}\geqslant 1$ , to give

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{a\in \{0,1\}}i^{-a}\biggl(i\frac{|\unicode[STIX]{x1D6FC}+iy|}{\unicode[STIX]{x1D6FC}+iy}\biggr)^{k}\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)}{\sqrt{n}}V_{\bar{f}}^{(-)^{a}}\biggl(\frac{ny}{N(\unicode[STIX]{x1D6FC}^{2}+y^{2})}\biggr)\cos ^{(a)}\biggl(\frac{2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n}{1+(y/\unicode[STIX]{x1D6FC})^{2}}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j+2l+m<2l_{0}}(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)}{\sqrt{n}}\frac{(2\unicode[STIX]{x1D70B}i)^{j}}{j!l!}b_{j,k,l,m}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)^{j+l}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,(V_{\bar{f}}^{(-)^{a}})^{(l)}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)\cos ^{(j+a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(\frac{y}{i\unicode[STIX]{x1D6FC}}\biggr)^{j+2l+m}+O_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70E},l_{0}}(y^{2l_{0}-\unicode[STIX]{x1D70E}})\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j+2l+m<2l_{0}}(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)}{\sqrt{n}}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!l!}b_{j,k,l,m}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)^{j+l}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,(V_{\bar{f}}^{(-)^{a+j}})^{(l)}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)\cos ^{(a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(\frac{y}{i\unicode[STIX]{x1D6FC}}\biggr)^{j+2l+m}+O_{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D70E},l_{0}}(y^{2l_{0}-\unicode[STIX]{x1D70E}}).\nonumber\end{eqnarray}$$

Taking the Mellin transform of a single term of the sum over $j,l,m$ and making the change of variables $y\mapsto N\unicode[STIX]{x1D6FC}^{2}y/n$ , we get

$$\begin{eqnarray}\displaystyle & & \displaystyle (i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\int _{0}^{\infty }\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)}{\sqrt{n}}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!l!}b_{j,k,l,m}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)^{j+l}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,(V_{\bar{f}}^{(-)^{a+j}})^{(l)}\biggl(\frac{ny}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)\cos ^{(a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)\biggl(\frac{y}{i\unicode[STIX]{x1D6FC}}\biggr)^{j+2l+m}y^{s-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{j+2l+m}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}b_{j,k,l,m}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{n=1}^{\infty }\frac{c_{\bar{f}}(n)\cos ^{(a)}(2\unicode[STIX]{x1D70B}\unicode[STIX]{x1D6FD}n)}{n^{s+j+2l+m}}\int _{0}^{\infty }\frac{y^{l}}{l!}(V_{\bar{f}}^{(-)^{a+j}})^{(l)}(y)y^{s+2j+2l+m-1/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}b_{j,k,l,m}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\binom{\frac{1}{2}-s-t-j}{l}\widetilde{V}_{\bar{f}}^{(-)^{a+j}}(s+t+j),\nonumber\end{eqnarray}$$

where $t=j+2l+m$ .

Next we fix $t\in \mathbb{Z}_{{\geqslant}0}$ and sum over all $(j,l,m)$ satisfying $j+2l+m=t$ . When $k=0$ , $b_{j,k,l,m}$ vanishes unless $m$ is even. Hence, defining

$$\begin{eqnarray}I_{k}(m)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }k=1\text{ or }2\mid m,\\ 0\quad & \text{otherwise},\end{array}\right.\end{eqnarray}$$

we get

$$\begin{eqnarray}\displaystyle & & \displaystyle (i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{j+2l+m=t}I_{k}(t-j)\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}\binom{j+l-1+\lfloor \frac{m}{2}\rfloor +\frac{k}{2}}{\lfloor \frac{m}{2}\rfloor }\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\binom{\frac{1}{2}-s-t-j}{l}\widetilde{V}_{\bar{f}}^{(-)^{a+j}}(s+t+j)\nonumber\\ \displaystyle & & \displaystyle \quad =(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{j=0}^{t}I_{k}(t-j)\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\widetilde{V}_{\bar{f}}^{(-)^{a+j}}(s+t+j)\mathop{\sum }_{l=0}^{\lfloor (t-j)/2\rfloor }\binom{j+\lfloor \frac{t-j}{2}\rfloor +\frac{k}{2}-1}{\lfloor \frac{t-j}{2}\rfloor -l}\binom{\frac{1}{2}-s-t-j}{l}\nonumber\\ \displaystyle & & \displaystyle \quad =(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{a\in \{0,1\}}i^{-a}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{j=0}^{t}I_{k}(t-j)\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\widetilde{V}_{\bar{f}}^{(-)^{a+j}}(s+t+j)\binom{\lfloor \frac{t-j}{2}\rfloor +\frac{k-1}{2}-s-t}{\lfloor \frac{t-j}{2}\rfloor },\nonumber\end{eqnarray}$$

by the Chu–Vandermonde identity.

We now consider two cases according to the weight, $k$ . When $k=0$ , the inner sum vanishes identically when $(-1)^{a+t}=-\unicode[STIX]{x1D716}$ , so we may assume that $(-1)^{a+t}=\unicode[STIX]{x1D716}$ . Thus, in this case, we have

$$\begin{eqnarray}\displaystyle (N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(iN\unicode[STIX]{x1D6FC})^{t}i^{-a}D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\mathop{\sum }_{\substack{ j\leqslant t \\ j\equiv t(mod2)}}\frac{(2\unicode[STIX]{x1D70B})^{j}}{j!}\unicode[STIX]{x1D6FE}_{\bar{f}}^{(-)^{a+t}}(s+t+j)\binom{\frac{t-j}{2}-\frac{1}{2}-s-t}{\frac{t-j}{2}}. & & \displaystyle \nonumber\end{eqnarray}$$

Put $t=2n+b$ , with $b\in \{0,1\}$ . Then, writing $j=2r+b$ , the above becomes

$$\begin{eqnarray}\displaystyle & & \displaystyle (N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(iN\unicode[STIX]{x1D6FC})^{t}i^{-a}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{r=0}^{n}\frac{(2\unicode[STIX]{x1D70B})^{2r+b}}{(2r+b)!}\frac{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+2r+b+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+2r+b-\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+b+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+b-\unicode[STIX]{x1D708})}\binom{n-r-\frac{1}{2}-s-t}{n-r}\nonumber\\ \displaystyle & & \displaystyle \quad =(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(iN\unicode[STIX]{x1D6FC})^{t}i^{-a}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})(-1)^{n}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{r=0}^{n}\biggl(\frac{2\unicode[STIX]{x1D70B}}{2r+1}\biggr)^{b}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t+b+\unicode[STIX]{x1D708})/2}{r}\binom{-(s+t+b-\unicode[STIX]{x1D708})/2}{r}\binom{s+t-\frac{1}{2}}{n-r}.\nonumber\end{eqnarray}$$

Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)–(iii)], we get

$$\begin{eqnarray}\displaystyle & & \displaystyle (N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(iN\unicode[STIX]{x1D6FC})^{t}i^{-a}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\biggl(\frac{2\unicode[STIX]{x1D70B}}{2n+1}\biggr)^{b}\frac{4^{n}n!^{2}}{(2n)!}\binom{(s+t-1-b+\unicode[STIX]{x1D708})/2}{n}\binom{(s+t-1-b-\unicode[STIX]{x1D708})/2}{n}\nonumber\\ \displaystyle & & \displaystyle \quad =(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}\frac{(2\unicode[STIX]{x1D70B}iN\unicode[STIX]{x1D6FC})^{t}}{t!}i^{-a}\frac{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s)}{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s-2n)}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)}).\nonumber\end{eqnarray}$$

Turning to $k=1$ , we have

$$\begin{eqnarray}\displaystyle & & \displaystyle i\operatorname{sgn}(\unicode[STIX]{x1D6FC})(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\mathop{\sum }_{j=0}^{t}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\unicode[STIX]{x1D6FE}_{\bar{f}}^{(-)^{a+j}}(s+t+j)\binom{\lfloor \frac{t-j}{2}\rfloor -s-t}{\lfloor \frac{t-j}{2}\rfloor }\nonumber\\ \displaystyle & & \displaystyle \quad =i\operatorname{sgn}(\unicode[STIX]{x1D6FC})(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{a\in \{0,1\}}i^{-a}D_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{j=0}^{t}\frac{(-2\unicode[STIX]{x1D70B})^{j}}{j!}\unicode[STIX]{x1D6FE}_{\bar{f}}^{(-1)^{a+j}}(s+t+j)\binom{\lfloor \frac{t-j}{2}\rfloor -s-t}{\lfloor \frac{t-j}{2}\rfloor }.\nonumber\end{eqnarray}$$

Writing $j=2r-c$ with $c\in \{0,1\}$ , this is

$$\begin{eqnarray}\displaystyle & & \displaystyle i\operatorname{sgn}(\unicode[STIX]{x1D6FC})(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{c\in \{0,1\}}\mathop{\sum }_{2r-c\leqslant t}\frac{(-2\unicode[STIX]{x1D70B})^{2r-c}}{(2r-c)!}\binom{n-r+\lfloor \frac{b+c}{2}\rfloor -s-t}{n-r+\lfloor \frac{b+c}{2}\rfloor }\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\frac{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+2r-c+(1-(-1)^{a+c}\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}_{\mathbb{ R}}(s+t+2r-c+(1+(-1)^{a+c}\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+(1-(-1)^{a}\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708})\unicode[STIX]{x1D6E4}_{\mathbb{R}}(s+t+(1+(-1)^{a}\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708})}\nonumber\\ \displaystyle & & \displaystyle \quad =i\operatorname{sgn}(\unicode[STIX]{x1D6FC})(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}(-iN\unicode[STIX]{x1D6FC})^{t}\mathop{\sum }_{a\in \{0,1\}}i^{-a}\unicode[STIX]{x1D6E5}_{\bar{f}}(s+t,\unicode[STIX]{x1D6FD},\cos ^{(a)})\mathop{\sum }_{c\in \{0,1\}}(-1)^{n+bc}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\mathop{\sum }_{2r-c\leqslant t}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{r-c\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}}\binom{-(s+t+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{r-c\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}}\binom{s+t-1}{n+bc-r}.\nonumber\end{eqnarray}$$

For $b=0$ , applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], the sum over $c$ becomes

$$\begin{eqnarray}\displaystyle & & \displaystyle (-1)^{n}\mathop{\sum }_{r=0}^{n}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t-1+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{r}\!\binom{-(s+t-1+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{r}\!\binom{s+t-1}{n-r}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{4^{n}n!^{2}}{(2n)!}\binom{(s+2n-2+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{n}\binom{(s+2n-2+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{n}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{(-2\unicode[STIX]{x1D70B})^{2n}}{(2n)!}\frac{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(1-s+(1+(-1)^{a}\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(1-s-2n+(1+(-1)^{a}\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708})}\frac{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(1-s+(1-(-1)^{a}\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}_{\mathbb{R}}(1-s-2n+(1-(-1)^{a}\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708})}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{(-2\unicode[STIX]{x1D70B})^{t}}{t!}\frac{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a}}(1-s)}{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a}}(1-s-2n)}=\frac{(-2\unicode[STIX]{x1D70B})^{t}}{t!}\frac{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s)}{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s-2\lfloor t/2\rfloor )}.\nonumber\end{eqnarray}$$

For $b=1$ and $c=0$ , the inner sum is

$$\begin{eqnarray}(-1)^{n}\mathop{\sum }_{r=0}^{n}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{r}\binom{-(s+t+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{r}\binom{s+t-1}{n-r}.\end{eqnarray}$$

Writing $\binom{s+t-1}{n-r}=\binom{s+t}{n-r+1}-\binom{s+t-1}{n-r+1}$ and applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], we get

$$\begin{eqnarray}\displaystyle & & \displaystyle (-1)^{n} [\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{(s+t-\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{n+1}\binom{(s+t-\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{-(s+t+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{n+1}\binom{-(s+t+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{n+1}\! ]\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{n+1} [\mathop{\sum }_{r=0}^{n+1}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{r}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\binom{-(s+t+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{r}\binom{s+t-1}{n-r+1}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{-(s+t+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{n+1}\binom{-(s+t+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{n+1}\! ]\nonumber\\ \displaystyle & & \displaystyle \quad =(-1)^{n}\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{(s+t-1+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\binom{(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{n+1}\mathop{\sum }_{r=0}^{n+1}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}+1)/2}{r}\binom{-(s+t+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{r}\binom{s+t-1}{n-r+1}.\nonumber\end{eqnarray}$$

For $b=1$ and $c=1$ the inner sum is

$$\begin{eqnarray}\displaystyle (-1)^{n+1}\mathop{\sum }_{r=1}^{n+1}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{-(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}+1)/2}{r-1}\binom{-(s+t+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{r}\binom{s+t-1}{n+1-r}, & & \displaystyle \nonumber\end{eqnarray}$$

and, adding this to the contribution from $c=0$ , for $b=1$ we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle (-1)^{n}\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{(s+t-1+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\binom{(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}+(-1)^{n+1}\binom{s+t-1}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{n+1}\mathop{\sum }_{r=1}^{n+1}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{1-(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}+1)/2}{r}\binom{-(s+t+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{r}\binom{s+t-1}{n-r+1}\nonumber\\ \displaystyle & & \displaystyle \quad =(-1)^{n}\frac{(-4)^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{(s+t-1+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\binom{(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-1)^{n+1}\mathop{\sum }_{r=0}^{n+1}\frac{(-4)^{r}r!^{2}}{(2r)!}\binom{1-(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}+1)/2}{r}\!\binom{-(s+t+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{r}\!\binom{s+t-1}{n-r+1}.\nonumber\end{eqnarray}$$

Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], this is

$$\begin{eqnarray}\displaystyle & & \displaystyle -\frac{4^{n+1}(n+1)!^{2}}{(2n+2)!}\binom{(s+t-1+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\biggl[\binom{(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2}{n+1}-\binom{(s+t-(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})/2-1}{n+1}\biggr]\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{4^{n+1}(n+1)!^{2}}{(2n+2)!}\frac{(s+(-1)^{a}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}+2n)/2}{n+1}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\binom{(s+2n-2+\frac{1+(-1)^{a}\unicode[STIX]{x1D716}}{2}-\unicode[STIX]{x1D708})/2}{n}\binom{(s+2n-2+\frac{1-(-1)^{a}\unicode[STIX]{x1D716}}{2}+\unicode[STIX]{x1D708})/2}{n}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{s+2\lfloor t/2\rfloor -(-1)^{a+t}\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}}{2\unicode[STIX]{x1D70B}}\frac{(-2\unicode[STIX]{x1D70B})^{t}}{t!}\frac{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s)}{\unicode[STIX]{x1D6FE}_{f}^{(-)^{a+t}}(1-s-2\lfloor t/2\rfloor )}.\nonumber\end{eqnarray}$$

In all cases, the result matches the formula for $P_{f}(s;a+t,t)$ . Taking $l_{0}=\lceil T/2\rceil$ , $\unicode[STIX]{x1D70E}=1$ and applying Mellin inversion, we get (20), with $T+1$ in place of $T$ when $T$ is odd. In that case, we estimate the final term of the sum by shifting the contour to $\Re (s)=\frac{3}{2}-T$ , which yields $O(y^{T-1})$ .◻

Lemma 2.5. Assume that $\unicode[STIX]{x1D6EC}_{f}(s)$ has at most finitely many simple zeros, and let $\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$ and $z=\unicode[STIX]{x1D6FC}+iy$ for some $y\in (0,|\unicode[STIX]{x1D6FC}|/4]$ . Then there are numbers $a_{j}(\unicode[STIX]{x1D6FC}),b_{j}(\unicode[STIX]{x1D6FC})\in \mathbb{C}$ such that, for any integer $M\geqslant 0$ , we have

(21) $$\begin{eqnarray}B(\unicode[STIX]{x1D6FC}+iy)=O_{\unicode[STIX]{x1D6FC},f,M}(y^{M})+\mathop{\sum }_{j=0}^{M-1}y^{j+1/2}\left\{\begin{array}{@{}ll@{}}a_{j}(\unicode[STIX]{x1D6FC})+b_{j}(\unicode[STIX]{x1D6FC})\log y\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC})y^{\unicode[STIX]{x1D708}}+b_{j}(\unicode[STIX]{x1D6FC})y^{-\unicode[STIX]{x1D708}}\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

Proof. Let $s\in \mathbb{C}$ with $\Re (s)\in (0,1)$ , and set $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D6FC}/y$ . We will show that there are numbers $a_{j}(\unicode[STIX]{x1D6FC},s),b_{j}(\unicode[STIX]{x1D6FC},s)\in \mathbb{C}$ satisfying

(22) $$\begin{eqnarray}H_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s}=\mathop{\sum }_{j=0}^{\infty }y^{j+1/2}\left\{\begin{array}{@{}ll@{}}a_{j}(\unicode[STIX]{x1D6FC},s)+b_{j}(\unicode[STIX]{x1D6FC},s)\log y\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC},s)y^{\unicode[STIX]{x1D708}}+b_{j}(\unicode[STIX]{x1D6FC},s)y^{-\unicode[STIX]{x1D708}}\quad & \text{otherwise,}\end{array}\right.\end{eqnarray}$$

and

(23) $$\begin{eqnarray}a_{j}(\unicode[STIX]{x1D6FC},s),b_{j}(\unicode[STIX]{x1D6FC},s)\ll _{f,\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D700}}(2e^{\unicode[STIX]{x1D70B}/2})^{(1+\unicode[STIX]{x1D700})|s|}|2/\unicode[STIX]{x1D6FC}|^{j+1/2}\sqrt{j+1}\quad \text{for all }\unicode[STIX]{x1D700}>0.\end{eqnarray}$$

Let us assume this for now. Then, since $y\leqslant |\unicode[STIX]{x1D6FC}|/4$ , we have

$$\begin{eqnarray}\mathop{\sum }_{j=M}^{\infty }\biggl(\frac{2y}{|\unicode[STIX]{x1D6FC}|}\biggr)^{j+1/2}\sqrt{j+1}\ll _{\unicode[STIX]{x1D6FC},M}y^{M+1/2},\end{eqnarray}$$

so that (by the trivial estimate $|\Re (\unicode[STIX]{x1D708})|<{\textstyle \frac{1}{2}}$ ),

(24) $$\begin{eqnarray}\displaystyle H_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s} & = & \displaystyle O_{f,\unicode[STIX]{x1D6FC},M,\unicode[STIX]{x1D700}}((2e^{\unicode[STIX]{x1D70B}/2})^{(1+\unicode[STIX]{x1D700})|s|}y^{M})\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{j=0}^{M-1}y^{j+1/2}\left\{\begin{array}{@{}ll@{}}a_{j}(\unicode[STIX]{x1D6FC},s)+b_{j}(\unicode[STIX]{x1D6FC},s)\log y\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC},s)y^{\unicode[STIX]{x1D708}}+b_{j}(\unicode[STIX]{x1D6FC},s)y^{-\unicode[STIX]{x1D708}}\quad & \text{otherwise}.\end{array}\right.\end{eqnarray}$$

We substitute this expansion into (18). By hypothesis, $\unicode[STIX]{x1D6EC}_{f}(s)$ has at most finitely many simple zeros, so the sum over $\unicode[STIX]{x1D70C}$ in (18) is a finite linear combination of the series (24) with $s=\unicode[STIX]{x1D70C}$ , which yields an expansion of the shape (21). As for the integral term in (18), by the convexity bound and Stirling’s formula, we have

$$\begin{eqnarray}X_{f}(s)\unicode[STIX]{x1D6EC}_{f}(s)\ll _{f,\unicode[STIX]{x1D700}}e^{-(3\unicode[STIX]{x1D70B}/2-\unicode[STIX]{x1D700})|s|}\quad \text{for }\Re (s)={\textstyle \frac{1}{2}},~\unicode[STIX]{x1D700}>0.\end{eqnarray}$$

Since $2<e^{\unicode[STIX]{x1D70B}}$ , the integral converges absolutely and again yields something of the shape (21).

It remains to show (22) and (23). First suppose that $k=0$ . Then, by (15), we have

$$\begin{eqnarray}H_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s}=|\unicode[STIX]{x1D6FC}/\unicode[STIX]{x1D714}|^{1/2-s}(2\unicode[STIX]{x1D70B}i\unicode[STIX]{x1D714})^{(1-\unicode[STIX]{x1D716})/2}\operatorname{ 2F1}\biggl(\frac{s+(1-\unicode[STIX]{x1D716})/2+\unicode[STIX]{x1D708}}{2},\frac{s+(1-\unicode[STIX]{x1D716})/2-\unicode[STIX]{x1D708}}{2};1-\frac{\unicode[STIX]{x1D716}}{2};-\unicode[STIX]{x1D714}^{2}\biggr).\end{eqnarray}$$

Applying the hypergeometric transformation [Reference Gradshteyn and RyzhikGR15, 9.132(2)] and the defining series (12), this is

(25) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D70B}i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{(1-\unicode[STIX]{x1D716})/2}|\unicode[STIX]{x1D6FC}|^{1/2-s}\unicode[STIX]{x1D70B}^{1/2}\mathop{\sum }_{\pm }\frac{|y/\unicode[STIX]{x1D6FC}|^{1/2\pm \unicode[STIX]{x1D708}}\unicode[STIX]{x1D6E4}(\mp \unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}(1-(s+(1+\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708})/2)\unicode[STIX]{x1D6E4}((s+(1-\unicode[STIX]{x1D716})/2\mp \unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\operatorname{2F1}\biggl(\frac{s+(1-\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708}}{2},\frac{s+(1+\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708}}{2};1\pm \unicode[STIX]{x1D708};-\biggl(\frac{y}{\unicode[STIX]{x1D6FC}}\biggr)^{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =(\unicode[STIX]{x1D70B}i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{(1-\unicode[STIX]{x1D716})/2}|\unicode[STIX]{x1D6FC}|^{1/2-s}\unicode[STIX]{x1D70B}^{1/2}\mathop{\sum }_{j=0}^{\infty }\mathop{\sum }_{\pm }\frac{\unicode[STIX]{x1D6E4}(\mp \unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}(1-(s+(1+\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708})/2)\unicode[STIX]{x1D6E4}((s+(1-\unicode[STIX]{x1D716})/2\mp \unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\frac{\binom{-\frac{s+(1-\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708}}{2}}{j}\binom{-\frac{s+(1+\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708}}{2}}{j}}{\binom{-1\mp \unicode[STIX]{x1D708}}{j}}\biggl|\frac{y}{\unicode[STIX]{x1D6FC}}\biggr|^{2j+1/2\pm \unicode[STIX]{x1D708}}.\end{eqnarray}$$

To pass from this to (22), we replace $2j$ by $j$ and set $a_{j}=b_{j}=0$ when $j$ is odd.

When $\unicode[STIX]{x1D708}\neq 0$ we use the estimates

$$\begin{eqnarray}\biggl|\binom{-\frac{s+a\pm \unicode[STIX]{x1D708}}{2}}{j}\biggr|=\biggl|\binom{\frac{s+a\pm \unicode[STIX]{x1D708}}{2}+j-1}{j}\biggr|\leqslant 2^{|s+a\pm \unicode[STIX]{x1D708}|/2+j}\ll _{f}2^{|s|/2+j}\quad \text{for }a\in \{0,1\},\end{eqnarray}$$
$$\begin{eqnarray}\biggl|\binom{-1\mp \unicode[STIX]{x1D708}}{j}\biggr|=\mathop{\prod }_{l=1}^{j}\biggl|1\pm \frac{\unicode[STIX]{x1D708}}{l}\biggr|\geqslant \mathop{\prod }_{l=1}^{j}\biggl|1-\frac{1}{2l}\biggr|=\biggl|\binom{-\frac{1}{2}}{j}\biggr|\gg \frac{1}{\sqrt{2j+1}}\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D70B}i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{(1-\unicode[STIX]{x1D716})/2}|\unicode[STIX]{x1D6FC}|^{1/2-s}\unicode[STIX]{x1D70B}^{1/2}\frac{\unicode[STIX]{x1D6E4}(\mp \unicode[STIX]{x1D708})}{\unicode[STIX]{x1D6E4}(1-(s+(1+\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708})/2)\unicode[STIX]{x1D6E4}((s+(1-\unicode[STIX]{x1D716})/2\mp \unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \quad \ll _{f,\unicode[STIX]{x1D700}}e^{(\unicode[STIX]{x1D70B}/2+\unicode[STIX]{x1D700})|s|}\quad \text{for all }\unicode[STIX]{x1D700}>0\nonumber\end{eqnarray}$$

to obtain (23).

When $\unicode[STIX]{x1D708}=0$ , (25) has a singularity arising from the $\unicode[STIX]{x1D6E4}(\pm \unicode[STIX]{x1D708})$ factors, but we can still understand the formula by analytic continuation. To remove the singularity, we replace $y^{\pm \unicode[STIX]{x1D708}}$ by $(y^{\pm \unicode[STIX]{x1D708}}-1)+1$ . Since

$$\begin{eqnarray}\lim _{\unicode[STIX]{x1D708}\rightarrow 0}\unicode[STIX]{x1D6E4}(\pm \unicode[STIX]{x1D708})(y^{\pm \unicode[STIX]{x1D708}}-1)=\log y,\end{eqnarray}$$

in the terms with $y^{\pm \unicode[STIX]{x1D708}}-1$ we can simply take the limit and estimate the remaining factors as before; this gives the $b_{j}$ terms in (22) and (23). The terms with $1$ can be written in the form $y^{2j+1/2}(h_{j}(\unicode[STIX]{x1D708})+h_{j}(-\unicode[STIX]{x1D708}))$ , where $h_{j}$ is meromorphic with a simple pole at $\unicode[STIX]{x1D708}=0$ , and independent of $y$ . Then $h_{j}(\unicode[STIX]{x1D708})+h_{j}(-\unicode[STIX]{x1D708})$ is even, so it has a removable singularity at $\unicode[STIX]{x1D708}=0$ . By the Cauchy integral formula, we have

$$\begin{eqnarray}\lim _{\unicode[STIX]{x1D708}\rightarrow 0}(h_{j}(\unicode[STIX]{x1D708})+h_{j}(-\unicode[STIX]{x1D708}))=\frac{1}{2\unicode[STIX]{x1D70B}i}\int _{|\unicode[STIX]{x1D708}|=1/2}\frac{h_{j}(\unicode[STIX]{x1D708})+h_{j}(-\unicode[STIX]{x1D708})}{\unicode[STIX]{x1D708}}\,d\unicode[STIX]{x1D708}.\end{eqnarray}$$

Since the above estimates hold uniformly for $\unicode[STIX]{x1D708}\in \mathbb{C}$ with $|\unicode[STIX]{x1D708}|=\frac{1}{2}$ , they also hold for $\lim _{\unicode[STIX]{x1D708}\rightarrow 0}(h_{j}(\unicode[STIX]{x1D708})+h_{j}(-\unicode[STIX]{x1D708}))$ . This concludes the proof of (22) and (23) when $k=0$ .

Turning to $k=1$ , by (15) we have

$$\begin{eqnarray}\displaystyle & & \displaystyle H_{f}(s,\unicode[STIX]{x1D714})y^{1/2-s}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\unicode[STIX]{x1D6FF}\in \{0,1\}}\biggl|\frac{\unicode[STIX]{x1D6FC}}{\unicode[STIX]{x1D714}}\biggr|^{1/2-s}(i\unicode[STIX]{x1D714}(s-\unicode[STIX]{x1D716}\unicode[STIX]{x1D708}))^{\unicode[STIX]{x1D6FF}}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\operatorname{2F1}\biggl(\frac{s+(-1)^{\unicode[STIX]{x1D6FF}}((1+\unicode[STIX]{x1D716})/2)+\unicode[STIX]{x1D708}}{2}+\unicode[STIX]{x1D6FF},\frac{s+(-1)^{\unicode[STIX]{x1D6FF}}((1-\unicode[STIX]{x1D716})/2)-\unicode[STIX]{x1D708}}{2}+\unicode[STIX]{x1D6FF};\frac{1}{2}+\unicode[STIX]{x1D6FF};-\unicode[STIX]{x1D714}^{2}\biggr),\nonumber\end{eqnarray}$$

and applying [Reference Gradshteyn and RyzhikGR15, 9.132(2)], this becomes

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D70B}^{1/2}|\unicode[STIX]{x1D6FC}|^{1/2-s}\mathop{\sum }_{\unicode[STIX]{x1D6FF}\in \{0,1\}}\biggl(\frac{i\operatorname{sgn}(\unicode[STIX]{x1D6FC})(s-\unicode[STIX]{x1D716}\unicode[STIX]{x1D708})}{2}\biggr)^{\unicode[STIX]{x1D6FF}}\mathop{\sum }_{\pm }\biggl|\frac{y}{\unicode[STIX]{x1D6FC}}\biggr|^{1/2+(1\pm (-1)^{\unicode[STIX]{x1D6FF}}\unicode[STIX]{x1D716})/2\pm \unicode[STIX]{x1D708}}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\unicode[STIX]{x1D6E4}(\mp (\unicode[STIX]{x1D708}+(-1)^{\unicode[STIX]{x1D6FF}}(\unicode[STIX]{x1D716}/2)))}{\unicode[STIX]{x1D6E4}((s+(-1)^{\unicode[STIX]{x1D6FF}}((1\mp \unicode[STIX]{x1D716})/2)\mp \unicode[STIX]{x1D708})/2+\unicode[STIX]{x1D6FF})\unicode[STIX]{x1D6E4}(1/2-(s+(-1)^{\unicode[STIX]{x1D6FF}}((1\pm \unicode[STIX]{x1D716})/2)\pm \unicode[STIX]{x1D708})/2)}\nonumber\\ \displaystyle & & \displaystyle \quad \cdot \,\operatorname{2F1} (\frac{s+(-1)^{\unicode[STIX]{x1D6FF}}((1\pm \unicode[STIX]{x1D716})/2)\pm \unicode[STIX]{x1D708}}{2}+\unicode[STIX]{x1D6FF},\frac{s+(-1)^{\unicode[STIX]{x1D6FF}}((1\pm \unicode[STIX]{x1D716})/2)\pm \unicode[STIX]{x1D708}}{2}+\frac{1}{2};\nonumber\\ \displaystyle & & \displaystyle \qquad 1\pm \biggl(\unicode[STIX]{x1D708}+(-1)^{\unicode[STIX]{x1D6FF}}\frac{\unicode[STIX]{x1D716}}{2}\biggr);-\biggl(\frac{y}{\unicode[STIX]{x1D6FC}}\biggr)^{2} ).\nonumber\end{eqnarray}$$

In this case no singularity arises from the $\unicode[STIX]{x1D6E4}$ -factor in the numerator, so expanding the final $\operatorname{2F1}$ as a series and applying a similar analysis to the above, we arrive at (22) and (23).◻

With the lemmas in place, we can now complete the proof of Proposition 1.2. Let

$$\begin{eqnarray}\unicode[STIX]{x1D712}_{(0,|\unicode[STIX]{x1D6FC}|/4]}(y)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }y\leqslant {\displaystyle \frac{|\unicode[STIX]{x1D6FC}|}{4}},\\ 0\quad & \text{if }y>{\displaystyle \frac{|\unicode[STIX]{x1D6FC}|}{4}},\end{array}\right.\end{eqnarray}$$

and define

$$\begin{eqnarray}\displaystyle g(y) & = & \displaystyle F(\unicode[STIX]{x1D6FC}+iy)+A(\unicode[STIX]{x1D6FC}+iy)-\unicode[STIX]{x1D712}_{(0,|\unicode[STIX]{x1D6FC}|/4]}(y)\mathop{\sum }_{j=0}^{M-1}y^{j+1/2}\left\{\begin{array}{@{}ll@{}}a_{j}(\unicode[STIX]{x1D6FC})+b_{j}(\unicode[STIX]{x1D6FC})\log y\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC})y^{\unicode[STIX]{x1D708}}+b_{j}(\unicode[STIX]{x1D6FC})y^{-\unicode[STIX]{x1D708}}\quad & \text{otherwise,}\end{array}\right.\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D702}(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}\mathop{\sum }_{t=0}^{M-1}\frac{(2\unicode[STIX]{x1D70B}iN\unicode[STIX]{x1D6FC})^{t}}{t!}\mathop{\sum }_{a\in \{0,1\}}\frac{i^{-a}}{2\unicode[STIX]{x1D70B}i}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\int _{\Re (s)=2}P_{f}(s;a+t,t)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+t,-\frac{1}{N\unicode[STIX]{x1D6FC}},\cos ^{(a)}\biggr)\biggl(\frac{y}{N\unicode[STIX]{x1D6FC}^{2}}\biggr)^{1/2-s}\,ds.\nonumber\end{eqnarray}$$

By Lemmas 2.1, 2.4 and 2.5, we have $g(y)=O_{\unicode[STIX]{x1D6FC},M}(y^{M-1})$ for $y\leqslant |\unicode[STIX]{x1D6FC}|/4$ . On the other hand, shifting the contour of the above to the right, we see that $g$ decays rapidly as $y\rightarrow \infty$ . Hence, $\int _{0}^{\infty }g(y)y^{s-1/2}(dy/y)$ converges absolutely and defines a holomorphic function for $\Re (s)>\frac{5}{2}-M$ .

We have

$$\begin{eqnarray}\displaystyle \int _{0}^{\infty }F(\unicode[STIX]{x1D6FC}+iy)y^{s-1/2}\frac{dy}{y}=\mathop{\sum }_{a\in \{0,1\}}i^{-a}\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(a)})\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }k=1\text{ or }(-1)^{a}=\unicode[STIX]{x1D716},\\ 0\quad & \text{otherwise.}\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

By Lemma 2.2, $\int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+iy)y^{s-1/2}(dy/y)$ continues to a holomorphic function on $\unicode[STIX]{x1D6FA}$ . Similarly,

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }y^{s-1/2}\frac{dy}{y}\cdot \unicode[STIX]{x1D712}_{(0,|\unicode[STIX]{x1D6FC}|/4]}(y)\mathop{\sum }_{j=0}^{M-1}y^{j+1/2}\left\{\begin{array}{@{}ll@{}}a_{j}(\unicode[STIX]{x1D6FC})+b_{j}(\unicode[STIX]{x1D6FC})\log y\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC})y^{\unicode[STIX]{x1D708}}+b_{j}(\unicode[STIX]{x1D6FC})y^{-\unicode[STIX]{x1D708}}\quad & \text{otherwise,}\end{array}\right.\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j=0}^{M-1}\left\{\begin{array}{@{}ll@{}}{\displaystyle \frac{|\unicode[STIX]{x1D6FC}/4|^{s+j}}{s+j}}\biggl[a_{j}(\unicode[STIX]{x1D6FC})+b_{j}(\unicode[STIX]{x1D6FC})\biggl(\log |\unicode[STIX]{x1D6FC}/4|-{\displaystyle \frac{1}{s+j}}\biggr)\biggr]\quad & \text{if }\unicode[STIX]{x1D708}=k=0,\\ a_{j}(\unicode[STIX]{x1D6FC}){\displaystyle \frac{|\unicode[STIX]{x1D6FC}/4|^{s+j+\unicode[STIX]{x1D708}}}{s+j+\unicode[STIX]{x1D708}}}+b_{j}(\unicode[STIX]{x1D6FC}){\displaystyle \frac{|\unicode[STIX]{x1D6FC}/4|^{s+j-\unicode[STIX]{x1D708}}}{s+j-\unicode[STIX]{x1D708}}}\quad & \text{otherwise},\end{array}\right.\nonumber\end{eqnarray}$$

is holomorphic on $\unicode[STIX]{x1D6FA}$ . Hence, by Mellin inversion,

(26) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{a\in \{0,1\}}i^{-a}\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(a)})\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }k=1\text{ or }(-1)^{a}=\unicode[STIX]{x1D716},\\ 0\quad & \text{otherwise,}\end{array}\right.\nonumber\\ \displaystyle & & \displaystyle \quad -\unicode[STIX]{x1D702}(i\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}\mathop{\sum }_{t=0}^{M-1}\frac{(2\unicode[STIX]{x1D70B}iN\unicode[STIX]{x1D6FC})^{t}}{t!}\mathop{\sum }_{a\in \{0,1\}}i^{-a}P_{f}(s;a+t,t)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+t,-\frac{1}{N\unicode[STIX]{x1D6FC}},\cos ^{(a)}\biggr)\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{5}{2}-M\}$ .

Denoting (26) by $h(\unicode[STIX]{x1D6FC})$ , we consider the combination $\frac{1}{2}(i^{k+a_{0}}h(\unicode[STIX]{x1D6FC})+i^{-k-a_{0}}h(-\unicode[STIX]{x1D6FC}))$ for some $a_{0}\in \{0,1\}$ . This picks out the term with $a\equiv k+a_{0}(mod2)$ in the first sum over $a$ , and $a\equiv t+a_{0}(mod2)$ in the second. Therefore, since

$$\begin{eqnarray}P_{f}(s;a_{0},0)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if }k=1\text{ or }(-1)^{a_{0}}=\unicode[STIX]{x1D716},\\ 0\quad & \text{otherwise},\end{array}\right.\end{eqnarray}$$

we find that

(27) $$\begin{eqnarray}\displaystyle & & \displaystyle P_{f}(s;a_{0},0)\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D6FC},\cos ^{(k+a_{0})})\nonumber\\ \displaystyle & & \displaystyle \quad -\,\unicode[STIX]{x1D702}(-\!\operatorname{sgn}(\unicode[STIX]{x1D6FC}))^{k}(N\unicode[STIX]{x1D6FC}^{2})^{s-1/2}\mathop{\sum }_{t=0}^{M-1}\frac{(2\unicode[STIX]{x1D70B}N\unicode[STIX]{x1D6FC})^{t}}{t!}P_{f}(s;a_{0},t)\unicode[STIX]{x1D6E5}_{\bar{f}}\biggl(s+t,-\frac{1}{N\unicode[STIX]{x1D6FC}},\cos ^{(t+a_{0})}\biggr)\end{eqnarray}$$

is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{5}{2}-M\}$ . Finally, replacing $M$ by $M+1$ and discarding the final term of the sum, we see that (27) is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{3}{2}-M\}$ , as required.

Footnotes

A. R. Booker was partially supported by EPSRC grant EP/K034383/1. P. J. Cho was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (2016R1D1A1B03935186). No data were created in the course of this study.

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