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 $f$ :
We define
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
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:
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
where
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
$D_{f}(s,\unicode[STIX]{x1D712}_{0})$ 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,
To see this, write
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
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
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
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
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
where
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
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 $c_{f}(n)$ :
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
Then the complete $L$ -function $\unicode[STIX]{x1D6EC}_{f}(s):=\unicode[STIX]{x1D6FE}_{f}^{+}(s)L_{f}(s)$ satisfies
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
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
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)]:
and
where
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
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 $a\in \{0,1\}$ ,
is holomorphic for $\Re (s)>{\textstyle \frac{3}{2}}-M$ except for possible poles for $s\pm \unicode[STIX]{x1D708}\in \mathbb{Z}$ , where
and
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
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
is holomorphic on $\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{3}{2}+m_{0}-M\}$ , where we set
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
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 $\Re (s)\gg 1$ ,
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 $-\Re (s)\gg 1$ ,
Making the substitution $y\mapsto (N(\unicode[STIX]{x1D714}^{2}+1)y)^{-1}$ , this becomes
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 $H_{f}(s,\unicode[STIX]{x1D714}):=G_{f}(s,\unicode[STIX]{x1D714})/\unicode[STIX]{x1D6FE}_{f}^{+}(s)$ :
Next, for $z=x+iy\in \mathbb{H}$ , define
and
where the sum runs over all simple zeros of $\unicode[STIX]{x1D6EC}_{f}(s)$ , and
Lemma 2.1.
Proof. Fix $z=x+iy\in \mathbb{H}$ , and put $\unicode[STIX]{x1D714}=x/y$ . Applying Mellin inversion as in (14), we have
and
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
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
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
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,
Next, writing $\unicode[STIX]{x1D713}_{\mathbb{R}}(s)=(\unicode[STIX]{x1D6E4}_{\mathbb{R}}^{\prime }/\unicode[STIX]{x1D6E4}_{\mathbb{R}})(s)$ , we have
Applying the reflection formula and Legendre duplication formula in the form
we derive
Thus,
Rearranging terms completes the proof. ◻
Lemma 2.2. For any $\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$ ,
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
Hence, writing $\unicode[STIX]{x1D714}=\unicode[STIX]{x1D6FC}/y$ , we have
so that
where
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
Then, applying integration by parts $m$ times, we see that
and
Thus,
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
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
Differentiating $l$ times, we obtain
Using the estimate
we have
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
Proof. Let $z=\unicode[STIX]{x1D6FC}+iy$ , $\unicode[STIX]{x1D6FD}=-1/N\unicode[STIX]{x1D6FC}$ and $u=y/\unicode[STIX]{x1D6FC}$ . Then
so that
By Lemma 2.3, for any $\unicode[STIX]{x1D70E}\geqslant 0$ and any $l_{0}\in \mathbb{Z}_{{\geqslant}0}$ , we have
Similarly, for any $a\in \{0,1\}$ , we have
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
Next, defining
we have
Taking $m_{0}=2l_{0}-j-2l$ and applying Lemma 2.3 with $\unicode[STIX]{x1D70E}$ replaced by $\unicode[STIX]{x1D70E}+j$ , we obtain
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
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
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
we get
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
Put $t=2n+b$ , with $b\in \{0,1\}$ . Then, writing $j=2r+b$ , the above becomes
Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)–(iii)], we get
Turning to $k=1$ , we have
Writing $j=2r-c$ with $c\in \{0,1\}$ , this is
For $b=0$ , applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], the sum over $c$ becomes
For $b=1$ and $c=0$ , the inner sum is
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
For $b=1$ and $c=1$ the inner sum is
and, adding this to the contribution from $c=0$ , for $b=1$ we obtain
Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], this is
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
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
and
Let us assume this for now. Then, since $y\leqslant |\unicode[STIX]{x1D6FC}|/4$ , we have
so that (by the trivial estimate $|\Re (\unicode[STIX]{x1D708})|<{\textstyle \frac{1}{2}}$ ),
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
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
Applying the hypergeometric transformation [Reference Gradshteyn and RyzhikGR15, 9.132(2)] and the defining series (12), this is
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
and
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
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
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
and applying [Reference Gradshteyn and RyzhikGR15, 9.132(2)], this becomes
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
and define
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
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,
is holomorphic on $\unicode[STIX]{x1D6FA}$ . Hence, by Mellin inversion,
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
we find that
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.