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
$f$
:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU1.gif?pub-status=live)
We define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU2.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU3.gif?pub-status=live)
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:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU4.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU5.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU6.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU7.gif?pub-status=live)
$D_{f}(s,\unicode[STIX]{x1D712}_{0})$
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,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn1.gif?pub-status=live)
To see this, write
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU8.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn2.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU9.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU10.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU11.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn3.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn4.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn5.gif?pub-status=live)
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
$c_{f}(n)$
:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU12.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn6.gif?pub-status=live)
Then the complete
$L$
-function
$\unicode[STIX]{x1D6EC}_{f}(s):=\unicode[STIX]{x1D6FE}_{f}^{+}(s)L_{f}(s)$
satisfies
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn7.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn8.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn9.gif?pub-status=live)
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)]:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn10.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn11.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn12.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn13.gif?pub-status=live)
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
$a\in \{0,1\}$
,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU13.gif?pub-status=live)
is holomorphic for
$\Re (s)>{\textstyle \frac{3}{2}}-M$
except for possible poles for
$s\pm \unicode[STIX]{x1D708}\in \mathbb{Z}$
, where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU14.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU15.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU16.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU17.gif?pub-status=live)
is holomorphic on
$\{s\in \unicode[STIX]{x1D6FA}:\Re (s)>\frac{3}{2}+m_{0}-M\}$
, where we set
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU18.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU19.gif?pub-status=live)
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
$\Re (s)\gg 1$
,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn14.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn15.gif?pub-status=live)
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
$-\Re (s)\gg 1$
,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU20.gif?pub-status=live)
Making the substitution
$y\mapsto (N(\unicode[STIX]{x1D714}^{2}+1)y)^{-1}$
, this becomes
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn16.gif?pub-status=live)
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
$H_{f}(s,\unicode[STIX]{x1D714}):=G_{f}(s,\unicode[STIX]{x1D714})/\unicode[STIX]{x1D6FE}_{f}^{+}(s)$
:
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn17.gif?pub-status=live)
Next, for
$z=x+iy\in \mathbb{H}$
, define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU21.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn18.gif?pub-status=live)
where the sum runs over all simple zeros of
$\unicode[STIX]{x1D6EC}_{f}(s)$
, and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU22.gif?pub-status=live)
Lemma 2.1.
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU23.gif?pub-status=live)
Proof. Fix
$z=x+iy\in \mathbb{H}$
, and put
$\unicode[STIX]{x1D714}=x/y$
. Applying Mellin inversion as in (14), we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU24.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU25.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU26.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU27.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU28.gif?pub-status=live)
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,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU29.gif?pub-status=live)
Next, writing
$\unicode[STIX]{x1D713}_{\mathbb{R}}(s)=(\unicode[STIX]{x1D6E4}_{\mathbb{R}}^{\prime }/\unicode[STIX]{x1D6E4}_{\mathbb{R}})(s)$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU30.gif?pub-status=live)
Applying the reflection formula and Legendre duplication formula in the form
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU31.gif?pub-status=live)
we derive
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU32.gif?pub-status=live)
Thus,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU33.gif?pub-status=live)
Rearranging terms completes the proof. ◻
Lemma 2.2. For any
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU34.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU35.gif?pub-status=live)
Hence, writing
$\unicode[STIX]{x1D714}=\unicode[STIX]{x1D6FC}/y$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU36.gif?pub-status=live)
so that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU37.gif?pub-status=live)
where
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn19.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU38.gif?pub-status=live)
Then, applying integration by parts
$m$
times, we see that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU39.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU40.gif?pub-status=live)
Thus,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU41.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU42.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU43.gif?pub-status=live)
Differentiating
$l$
times, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU44.gif?pub-status=live)
Using the estimate
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU45.gif?pub-status=live)
we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU46.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn20.gif?pub-status=live)
Proof. Let
$z=\unicode[STIX]{x1D6FC}+iy$
,
$\unicode[STIX]{x1D6FD}=-1/N\unicode[STIX]{x1D6FC}$
and
$u=y/\unicode[STIX]{x1D6FC}$
. Then
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU47.gif?pub-status=live)
so that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU48.gif?pub-status=live)
By Lemma 2.3, for any
$\unicode[STIX]{x1D70E}\geqslant 0$
and any
$l_{0}\in \mathbb{Z}_{{\geqslant}0}$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU49.gif?pub-status=live)
Similarly, for any
$a\in \{0,1\}$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU50.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU51.gif?pub-status=live)
Next, defining
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU52.gif?pub-status=live)
we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU53.gif?pub-status=live)
Taking
$m_{0}=2l_{0}-j-2l$
and applying Lemma 2.3 with
$\unicode[STIX]{x1D70E}$
replaced by
$\unicode[STIX]{x1D70E}+j$
, we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU54.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU55.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU56.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU57.gif?pub-status=live)
we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU58.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU59.gif?pub-status=live)
Put
$t=2n+b$
, with
$b\in \{0,1\}$
. Then, writing
$j=2r+b$
, the above becomes
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU60.gif?pub-status=live)
Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)–(iii)], we get
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU61.gif?pub-status=live)
Turning to
$k=1$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU62.gif?pub-status=live)
Writing
$j=2r-c$
with
$c\in \{0,1\}$
, this is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU63.gif?pub-status=live)
For
$b=0$
, applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], the sum over
$c$
becomes
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU64.gif?pub-status=live)
For
$b=1$
and
$c=0$
, the inner sum is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU65.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU66.gif?pub-status=live)
For
$b=1$
and
$c=1$
the inner sum is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU67.gif?pub-status=live)
and, adding this to the contribution from
$c=0$
, for
$b=1$
we obtain
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU68.gif?pub-status=live)
Applying [Reference Booker and KrishnamurthyBK11, Lemma A.1(ii)], this is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU69.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn21.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn22.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn23.gif?pub-status=live)
Let us assume this for now. Then, since
$y\leqslant |\unicode[STIX]{x1D6FC}|/4$
, we have
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU70.gif?pub-status=live)
so that (by the trivial estimate
$|\Re (\unicode[STIX]{x1D708})|<{\textstyle \frac{1}{2}}$
),
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn24.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU71.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU72.gif?pub-status=live)
Applying the hypergeometric transformation [Reference Gradshteyn and RyzhikGR15, 9.132(2)] and the defining series (12), this is
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn25.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU73.gif?pub-status=live)
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU74.gif?pub-status=live)
and
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU75.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU76.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU77.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU78.gif?pub-status=live)
and applying [Reference Gradshteyn and RyzhikGR15, 9.132(2)], this becomes
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU79.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU80.gif?pub-status=live)
and define
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU81.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU82.gif?pub-status=live)
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,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU83.gif?pub-status=live)
is holomorphic on
$\unicode[STIX]{x1D6FA}$
. Hence, by Mellin inversion,
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn26.gif?pub-status=live)
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
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqnU84.gif?pub-status=live)
we find that
![](https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20190614081045624-0199:S0010437X19007279:S0010437X19007279_eqn27.gif?pub-status=live)
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.