On moments of $L$ -functions over Dirichlet characters

Avery Bainbridge , Rizwanur Khan and Ze Sen Tang Department of Mathematical Sciences
University of Texas at Dallas
Richardson, TX 75080-3021 [email protected], [email protected], [email protected]

Abstract.

We give a new proof of Heath-Brown’s full asymptotic expansion for the second moment of Dirichlet $L$ -functions and we obtain a corresponding asymptotic expansion for a twisted first moment of Hecke $L$ -functions.

Key words and phrases:

Dirichlet

L

-functions, moments, stationary phase

2020 Mathematics Subject Classification:

11M06

The second and third authors were supported by the National Science Foundation grants DMS-2344044 and DMS-2341239

1. Introduction

The Dirichlet $L$ -functions are ubiquitous objects in analytic number theory. They arise in the study of primes in arithmetic progressions, class numbers of number fields, and many other problems. For a primitive Dirichlet character $\chi$ of modulus $q$ , the associated Dirichlet $L$ -function is defined initially for $\Re(s)>1$ by the absolutely convergent Dirichlet series

L(s,\chi)=\sum_{n\geq 1}\frac{\chi(n)}{n^{s}}

and then by analytic continuation to the rest of $\mathbb{C}$ . The behavior of $L(s,\chi)$ within the critical strip $0\leq s\leq 1$ is the most important and mysterious. An important area of research is to try to understand these $L$ -functions at the center of the critical strip by evaluating the moments

\sum_{\chi\bmod q}|L(\tfrac{1}{2},\chi)|^{2k}

for $k\in\mathbb{N}$ , as $q\to\infty$ . Unfortunately our knowledge of these moments is very limited, with $k=2$ being the largest value for which an asymptotic fornula is known. We specialize to the important case of $q$ an odd prime to describe these results. Heath-Brown [4] proved an asymptotic formula for the fourth moment ( $k=2$ ), as follows:

(1.1)

\displaystyle\sum_{\chi\bmod q}|L(\tfrac{1}{2},\chi)|^{4}=\frac{q-1}{2\pi^{2}}\log^{4}q+O(\log^{3}q).

Subsequently there was great interest in refining this asymptotic. Young [7] obtained an asymptotic formula which additionally includes all lower order main terms up to a much sharper error term: one that saves a power of $q$ over the main term, unlike (1.1) which only saves a power of $\log q$ . This error term was then improved in other works, the best result [1] currently being

\sum_{\chi\bmod q}|L(\tfrac{1}{2},\chi)|^{4}=(q-1)P_{4}(\log q)+O(q^{1-\frac{1}{20}+\epsilon})

for some degree four polynomial $P_{4}(x)$ and any $\epsilon>0$ . For the second moment ( $k=1)$ , Heath-Brown [3] proved an asymptotic formula with arbitrarily high precision. That is, he obtained all main terms up to an error term that saves as large a power of $q$ as desired. Our first contribution is a new proof of this result, using the principle of analytic continuation and stationary phase analysis.

Theorem 1.

[3, cf. Equation (5)] Let $q$ be an odd prime. For any integer $N\geq 0$ we have

(1.2)			$\displaystyle\mathop{{\sum}^{+}}_{\chi\bmod q}\|L(\tfrac{1}{2},\chi)\|^{2}=\frac{q-1}{2}\Big{(}\log\frac{q}{8\pi}+\gamma-\frac{\pi}{2}\Big{)}+q^{\frac{1}{2}}\zeta^{2}(\tfrac{1}{2})+\sum_{n=0}^{N}c_{n}q^{-\frac{n}{2}}+O_{N}\Big{(}q^{\frac{-(N+1)}{2}}\Big{)},$
(1.3)			$\displaystyle\mathop{{\sum}^{-}}_{\chi\bmod q}\|L(\tfrac{1}{2},\chi)\|^{2}=\frac{q-1}{2}\Big{(}\log\frac{q}{8\pi}+\gamma+\frac{\pi}{2}\Big{)}+q^{\frac{1}{2}}\zeta^{2}(\tfrac{1}{2})+\sum_{n=0}^{N}c_{n}^{\prime}q^{-\frac{n}{2}}+O_{N}\Big{(}q^{\frac{-(N+1)}{2}}\Big{)}$

for some constants $c_{n},c_{n}^{\prime}$ , where $\mathop{{\sum}^{+}}$ denotes the sum over the even primitive Dirichlet characters, $\mathop{{\sum}^{-}}$ denotes the sum over the odd primitive Dirichlet characters, and $\gamma$ is Euler’s constant.

One may also include the trivial character in the first sum – this would contribute $|\zeta(\frac{1}{2})(1-q^{-\frac{1}{2}})|^{2}$ .

Our method is robust enough to obtain another, related moment with all main terms. Let $f$ a holomorphic Hecke cusp form for $SL_{2}(\mathbb{Z})$ with weight $k\geq 12$ and Hecke eigenvalues $\lambda_{f}(n)$ . For $\chi$ a primitive Dirichlet character mod $q$ , let $L(s,f\times\chi)$ denote the $L$ -function of $f$ twisted by $\chi$ . For $\Re(s)>1$ , this is given by

L(s,f\times\chi)=\sum_{n\geq 1}\frac{\lambda_{f}(n)\chi(n)}{n^{s}},

and elsewhere by analytic continuation. The following result does not seem to already be in the literature.

Theorem 2.

Let $f$ be a holomorphic Hecke cusp form for $SL_{2}(\mathbb{Z})$ with weight $k\geq 12$ . Let $q$ be an odd prime. For any integer $N\geq 0$ we have

(1.4)			$\displaystyle\mathop{{\sum}^{+}}_{\chi\bmod q}\frac{\tau(\chi)}{q^{\frac{1}{2}}}L(\tfrac{1}{2},f\times\overline{\chi})=q^{\frac{1}{2}}L(\tfrac{1}{2},f)+\sum_{n=0}^{N}c_{n,f}q^{-\frac{n}{2}}+O_{N,f}\Big{(}q^{\frac{-(N+1)}{2}}\Big{)},$
		$\displaystyle\mathop{{\sum}^{-}}_{\chi\bmod q}\frac{\tau(\chi)}{q^{\frac{1}{2}}}L(\tfrac{1}{2},f\times\overline{\chi})=q^{\frac{1}{2}}L(\tfrac{1}{2},f)+\sum_{n=0}^{N}c_{n,f}^{\prime}q^{-\frac{n}{2}}+O_{N,f}\Big{(}q^{\frac{-(N+1)}{2}}\Big{)},$

for some constants $c_{n,f},c_{n,f}^{\prime}$ , where where $\mathop{{\sum}^{+}}$ denotes the sum over the even primitive Dirichlet characters, $\mathop{{\sum}^{-}}$ denotes the sum over the odd primitive Dirichlet characters, and $\tau(\chi)$ denotes the Gauss sum.

We give a detailed proof of Theorem 1. The proof of Theorem 2 is very similar, so we will just sketch the differences to the proof of Theorem 1.

2. sketch

In this section, we describe the main ideas of the proof of (1.2) in Theorem 1. Fix an odd prime $q$ . Define for $s\in\mathbb{C}$ the entire function

F(s,q):=\mathop{{\sum}^{+}}_{\chi\bmod q}L(\tfrac{1}{2}-s,\chi)L(\tfrac{1}{2}+s,\overline{\chi}).

Our strategy is to derive a broader statement of the form

F(s,q)=G(s,q)

where $G(s,q)$ is a function that is holomorphic on some open neighborhood of $s=0$ , and then to recover Theorem 1 as the specialization at $s=0$ . To this end we begin by restricting $F(s,q)$ to $s\in\mathcal{S}_{1}$ , an open set positioned far to the right in $\mathbb{C}$ , as pictured in Figure 1, and calling the restriction $H(s,q)$ . After an application of the functional equation of $L(\tfrac{1}{2}-s,\chi)$ , we obtain

(2.1)

\displaystyle H(s,q)=\frac{\Gamma\left(\tfrac{1}{4}+\tfrac{s}{2}\right)}{\Gamma\left(\tfrac{1}{4}-\tfrac{s}{2}\right)\pi^{s}q^{\frac{1}{2}-s}}\mathop{{\sum}^{+}}_{\chi\bmod q}\tau(\chi)L(\tfrac{1}{2}+s,\overline{\chi})^{2}.

Expanding the $L$ -functions as absolutely convergent Dirichlet series, opening the Gauss sum, and executing the character sum using character orthogonality shows that $H(s,q)$ essentially equals

\sum_{n\geq 1}\frac{\cos(\frac{2\pi n}{q})d(n)}{n^{\frac{1}{2}+s}},

where $d(n)$ is the divisor function. Writing $\cos(\frac{2\pi n}{q})$ as a Mellin transform, we recast this sum in terms of the integral

(2.2)

\displaystyle\frac{1}{2\pi i}\int\limits_{(-\frac{3}{4})}\Gamma(w)\cos(\tfrac{\pi w}{2})\Big{(}\frac{q}{2\pi}\Big{)}^{w}\zeta^{2}(\tfrac{1}{2}+s+w)\ dw.

We then move the line of integration far to the left, picking up residues that contribute to the main terms on the right hand side of (1.2), with the leading main term arising from the double pole of $\zeta^{2}(\tfrac{1}{2}+s+w)$ at $w=\tfrac{1}{2}-s$ . However, the shifted integral (now positioned far to the left at $\Re(w)=-c$ for $c$ large) does not converge absolutely near $s=0$ , and so we need additional ideas to obtain meromorphic continuation near $s=0$ . Keeping in mind that $w$ is far to the left in the shifted integral, we apply the functional equation of $\zeta^{2}(\tfrac{1}{2}+s+w)$ to obtain $\zeta^{2}(\tfrac{1}{2}-s-w)$ , which we can expand into an absolutely convergent Dirichlet series since $-w$ is far to the right. Interchanging the order of integration and summation, we get

(2.3)

\displaystyle\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}-s}}\bigg{(}\frac{1}{2\pi i}\int\limits_{(-c)}\Gamma(w)\cos(\tfrac{\pi w}{2})\Big{(}\frac{nq}{2\pi}\Big{)}^{w}\frac{\pi^{-\frac{1}{2}+s+w}\Gamma^{2}(\frac{1}{4}-\frac{s}{2}-\frac{w}{2})}{\pi^{-\frac{1}{2}-s-w}\Gamma^{2}(\frac{1}{4}+\frac{s}{2}+\frac{w}{2})}dw\bigg{)}.

Evaluating the inner integral is a purely analytic problem (it’s the inverse Mellin transform of a product of special functions). This may be approached via hypergeometric functions, but we prefer to use Stirling’s estimates and techniques of stationary phase analysis. We find that the integrand has oscillation given by

(2.4)

\displaystyle\exp\left(-it\log\frac{|t|}{2\pi enq}\right),

for large $|t|=|\Im(w)|$ . There are stationary points at $t=\pm 2\pi nq$ , and at these points the oscillatory factor (2.4) is independent of $n$ . It is interesting to note that this step is special to the second moment, for such a simplification does not happen if we were trying to apply our method to evaluate higher moments. It turns out that the leading term of the integral is essentially $(nq)^{-2s}$ , which is especially simple, and from this the leading term of the sum (2.3) is

q^{-2s}\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}+s}}=q^{-2s}\zeta^{2}(\tfrac{1}{2}+s).

Thus we recover the Riemann Zeta function and we are able to use its meromorphic continuation from $\mathcal{S}_{1}$ to $\mathcal{S}_{2}$ , and in particular evaluate at $s=0$ . The lower order terms arising from the stationary phase analysis are of a similar shape, their values at $s=0$ decreasing by factors of $q^{-1}$ , and thus contribute to the non-leading main terms on the right hand side of (1.2). We denote by $G(s,q)$ the sum of all these terms plus the residues obtained above, and get that $F(0,q)=G(0,q)$ .

Figure 1.

For the proof of (1.4) in Theorem 2, we observe that the left hand side is the value at $s=0$ of

q^{-\frac{1}{2}}\mathop{{\sum}^{+}}_{\chi\bmod q}\tau(\chi)L(\tfrac{1}{2}+s,f\times\overline{\chi}),

which is analogous to (2.1). We take the Dirichlet series expansion of $L(\tfrac{1}{2}+s,f\times\overline{\chi})=\sum_{n\geq 1}\frac{\lambda_{f}(n)\overline{\chi}(n)}{n^{\frac{1}{2}+s}}$ for $s\in\mathcal{S}_{1}$ , which is analogous to $L(\tfrac{1}{2}+s,\overline{\chi})^{2}=\sum_{n\geq 1}\frac{d(n)\overline{\chi}(n)}{n^{\frac{1}{2}+s}}$ , except that $d(n)$ is replaced with $\lambda_{f}(n)$ . The proof proceeds as for Theorem 1, except for a few changes. In (2.2), we get the entire function $L(\frac{1}{2}+s+w,f)$ instead of $\zeta^{2}(\frac{1}{2}+s+w)$ . As a result, when we shift contours to the left, there is no pole at $s=\frac{1}{2}-w$ , unlike the case was for $\zeta^{2}(\frac{1}{2}+s+w)$ , and therefore the leading main term present in (1.2) does not arise. When the functional equation is applied to $L(\frac{1}{2}+s+w,f)$ , we obtain the ratio of Gamma functions $\frac{\Gamma(\frac{k}{2}-s-w)}{\Gamma(\frac{k}{2}+s+w)}$ instead of what is seen in (2.3). However, the oscillatory behavior (2.4) remains the same after Stirling’s estimates are applied. In the end, we do not get main terms of size $q\log q$ and $q$ in (1.4), and we have the main term $q^{\frac{1}{2}}L(\frac{1}{2},f)$ instead of the $q^{\frac{1}{2}}\zeta^{2}(\frac{1}{2})$ term that is present in (1.2).

3. Preliminaries

We recall here fundamental facts about Dirichlet characters, Dirichlet $L$ -functions, and the Mellin transform. Throughout, $q$ is an odd prime. Recall that a Dirichlet character $\chi$ of modulus $q$ satisfies $\chi(-1)=\pm 1$ , and it is called even if $\chi(-1)=1$ and odd otherwise. Since $q$ is prime, $\chi$ is primitive if and only if $\chi$ is not the principal character (which satisfies $\chi(n)=1$ for all $n$ coprime to $q$ ). We write $e(x)$ for the additive character $e^{2\pi ix}$ .

Lemma 3 (Functional equation).

Let $\chi$ be an even primitive Dirichlet character of modulus $q$ . Then for all $s\in\mathbb{C}$ we have

L(s,\chi)=\tau(\chi)\frac{\pi^{s-\frac{1}{2}}}{q^{s}}\frac{\Gamma(\tfrac{1-s}{2})}{\Gamma(\tfrac{s}{2})}L(1-s,\overline{\chi}),

where $\tau(\chi)=\sum\limits_{k=1}^{q-1}e(\tfrac{k}{q})\chi(k)$ denotes the Gauss sum.

Lemma 4 (An orthogonality relation).

Suppose $(nm,q)=1$ . Then

\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(n)\overline{\chi}(m)=\begin{cases}\frac{q-3}{2}&\textnormal{if }n\equiv m\bmod q\quad\textnormal{or }-n\equiv m\bmod q,\\ -1&\textnormal{otherwise.}\end{cases}

Proof.

The Dirichlet characters of modulus $q$ satisfy the following orthogonality property:

\sum_{\chi\bmod q}\chi(n)\overline{\chi}(m)=\begin{cases}q-1&\textnormal{if }n\equiv m\bmod q,\\ 0&\textnormal{otherwise.}\end{cases}

The result then follows from the equation

\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(n)\overline{\chi}(m)=-1+\sum_{\begin{subarray}{c}\chi\bmod q\\ \chi(-1)=1\end{subarray}}\chi(n)\overline{\chi}(m)=-1+\sum_{\chi\bmod q}\left(\frac{1+\chi(-1)}{2}\right)\chi(n)\overline{\chi}(m),

where the $-1$ term serves to subtract the contribution of the trivial character and the $\frac{1+\chi(-1)}{2}$ factor serves to pick out the condition $\chi(-1)=1$ for the primitive characters. ∎

Corollary 5.

Suppose $(n,q)=1$ . Then

\sum_{k=1}^{q-1}e(\tfrac{kn}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)=(q-1)\cos(\tfrac{2\pi n}{q})+1

Proof.

We separate out the first $(k=1)$ and last ( $k=q-1$ ) terms of the $k$ -sum, so

	$\displaystyle\sum_{k=1}^{q-1}e(\tfrac{kn}{q})$	$\displaystyle\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)$
	$\displaystyle=e(\tfrac{n}{q})$	$\displaystyle\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(n)\overline{\chi}(n)+e(\tfrac{(q-1)n}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi((q-1)n)\overline{\chi}(n)+\sum_{k=2}^{q-2}e(\tfrac{kn}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)$
	$\displaystyle=e(\tfrac{n}{q})$	$\displaystyle\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(n)\overline{\chi}(n)+e(\tfrac{-n}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(-n)\overline{\chi}(n)+\sum_{k=2}^{q-2}e(\tfrac{kn}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)$

where the second equality follows from the fact that $q-1\equiv-1\bmod q$ . Now we see that each sum in the last line corresponds to exactly one of the cases in the previous lemma, so executing the character sums yields

\sum_{k=1}^{q-1}e(\tfrac{kn}{q})\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)=\frac{q-3}{2}\left(e(\tfrac{n}{q})+e(-\tfrac{n}{q})\right)-\sum_{k=2}^{q-2}e(\tfrac{kn}{q})

and we note that

\sum_{k=2}^{q-2}e(\tfrac{kn}{q})=-\left(e(\tfrac{n}{q})+e(-\tfrac{n}{q})\right)+\sum_{k=1}^{q-1}e(\tfrac{kn}{q})=-\left(e(\tfrac{n}{q})+e(-\tfrac{n}{q})\right)-1.\qed

Lemma 6 (Mellin inversion).

[5, Chapter III, Theorem 2] Suppose that $f:(0,\infty)\to\mathbb{R}$ is continuous and the integral

\displaystyle\phi(w)=\int_{0}^{\infty}x^{w-1}f(x)dx

is absolutely convergent for $a<\Re(w)<b$ . Then

f(x)=\frac{1}{2\pi i}\int_{(c)}x^{-w}\phi(w)dw

for $c\in(a,b)$ .

Lemma 7 (An integral representation).

Let $x>0$ . For $d\in(-1,0)$ , we have

(3.1)

\displaystyle\cos(x)=\frac{1}{2\pi i}\int_{(d)}\Gamma(w)\cos\left(\frac{\pi w}{2}\right)x^{w}dw+1.

Further, this representation is absolutely convergent if $d\in(-1,-\frac{1}{2})$ .

Proof.

Let $f(x)=\cos(x)-e^{-x}$ and define

\tilde{f}(w):=\int_{0}^{\infty}x^{w-1}f(x)dx

for $-1<\Re(w)<1$ . This is well defined because we have $f(x)=x+O(x^{2})$ for $|x|\leq\pi$ and $f(x)=\cos(x)+O(e^{-x})$ for $|x|>\pi$ , so that $\tilde{f}(w)$ converges absolutely for $-1<\Re(w)<0$ and converges conditionally for $0\leq\Re(w)<1$ . We claim moreover that $\tilde{f}(w)$ is holomorphic for $-1<\Re(w)<1$ . For $0<\Re(w)<1$ , we can differentiate under the integral sign by absolute convergence. For $0\leq\Re(w)<1$ , we have that $\tilde{f}(w)$ equals the sum of a holomorphic function and $\int_{\pi}^{\infty}x^{w-1}\cos x\ dx$ . But by integration by parts,

\int_{\pi}^{\infty}x^{w-1}\cos x\ dx=\int_{\pi}^{\infty}(w-1)x^{w-2}\sin x\ dx,

which is absolutely convergent, and hence differentiable.

For $0<\Re(w)<1$ , we have by [2, equations 17.43.1, 17.43.3] that

\tilde{f}(w)=\int_{0}^{\infty}x^{w-1}\cos(x)dx-\int_{0}^{\infty}x^{w-1}e^{-x}dx=\Gamma(w)\cos\Big{(}\frac{\pi w}{2}\Big{)}-\Gamma(w).

By analytic continuation, we get that

\tilde{f}(w)=\Gamma(w)\cos\Big{(}\frac{\pi w}{2}\Big{)}-\Gamma(w)

for $-1<\Re(w)<1$ . Then by Lemma 6, we have

(3.2)

\displaystyle f(x)=\cos(x)-e^{-x}=\frac{1}{2\pi i}\int_{(d)}\Gamma(w)\left[\cos\left(\frac{\pi w}{2}\right)-1\right]x^{-w}dw

for $d\in(-1,0)$ . Also by Mellin inversion, since $\int_{0}^{\infty}e^{-x}x^{w-1}dx$ converges absolutely for $0<\Re(w)<1$ , we have

\displaystyle e^{-x}=\frac{1}{2\pi i}\int_{(d)}\Gamma(w)x^{-w}dw

for $d\in(0,1)$ . We move the line of integration to the left, crossing a simple pole of $\Gamma(w)$ at $w=0$ , to obtain

(3.3)

\displaystyle e^{-x}=\frac{1}{2\pi i}\int_{(d)}\Gamma(w)x^{-w}dw+1

for $d\in(-1,0)$ . Adding together (3.2) and (3.3) gives (3.1).

Finally, we have by Stirling’s approximation (equation (3.10)) that

\Gamma(d+it)\cos\left(\frac{\pi(d+it)}{2}\right)\ll|t|^{d-\frac{1}{2}}.

Thus the integral in (3.1) converges absolutely for $d\in(-1,-\frac{1}{2})$ . ∎

Lemma 8 (A residue calculation).

Define

f(s,w,q):=\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi}\right)^{w}\zeta^{2}(\tfrac{1}{2}+s+w)

for $s\in\mathbb{C}$ such that $\tfrac{1}{2}-s\not\in\mathbb{Z}$ . Then $f$ has simple poles at $w=-2n$ for $n\in\mathbb{N}\cup\{0\}$ , and a double pole at $w=\tfrac{1}{2}-s$ . The residues at these poles are

R_{0}(n,s,q):=\underset{w=-2n}{\mathrm{Res}}f(s,w,q)=\frac{(-1)^{n}}{(2n)!}\left(\frac{2\pi}{q}\right)^{2n}\zeta^{2}(\tfrac{1}{2}+s-2n)

and

		$\displaystyle R_{1}(s,q):=\underset{w=\frac{1}{2}-s}{\mathrm{Res}}f(s,w,q)$
(3.4)			$\displaystyle=\Gamma(\tfrac{1}{2}-s)\cos\left(\tfrac{\pi}{4}-\tfrac{\pi s}{2}\right)\left(\tfrac{q}{2\pi}\right)^{\frac{1}{2}-\frac{s}{2}}\Big{(}\tfrac{\Gamma^{\prime}(\frac{1}{2}-s)}{\Gamma(\frac{1}{2}-s)}-\tfrac{\pi}{2}\tan\left(\tfrac{\pi}{4}-\tfrac{\pi s}{2}\right)+\log\left(\tfrac{q}{2\pi}\right)+2\gamma\Big{)}.$

These residues are meromorphic functions of $s$ which are analytic at $s=0$ .

Proof.

The assumption that $\tfrac{1}{2}-s\not\in\mathbb{Z}$ ensures that the poles under discussion do not coincide. Recall that $\Gamma(w)$ has simple poles at the non-positive integers, but $\cos(\tfrac{\pi w}{2})$ vanishes at the negative odd integers, so $\Gamma(w)\cos(\tfrac{\pi w}{2})$ only has simple poles at the non-positive even integers $w=-2n$ . We have the Laurent expansion

\Gamma(w)=\frac{1}{(2n)!}(w+2n)^{-1}+\ldots

about $w=-2n$ , from which we get that

\underset{w=-2n}{\mathrm{Res}}g(w)\Gamma(w)=\frac{g(-2n)}{(2n)!}

for any function $g(w)$ that is holomorphic in a neighborhood of $w=-2n$ . This leads to the calculation of $R_{0}(n,s,q)$ .

Recall that $\zeta^{2}(\tfrac{1}{2}+s+w)$ has a double pole at $w=\tfrac{1}{2}-s$ , and $\cos(\frac{\pi(\frac{1}{2}-s)}{2})\neq 0$ as $\tfrac{1}{2}-s\not\in\mathbb{Z}$ . We have the Laurent expansion

\zeta^{2}(\tfrac{1}{2}+s+w)=(w-\tfrac{1}{2}+s)^{-2}+2\gamma(w-\tfrac{1}{2}+s)^{-1}+\ldots

about $w=\frac{1}{2}-s$ , where $\gamma$ is Euler’s constant. From this we have that if $g(w)$ is holomorphic in a neighborhood of $w=\frac{1}{2}-s$ , then

\underset{w=-2n}{\mathrm{Res}}g(w)\zeta^{2}(\tfrac{1}{2}+s+w)=g^{\prime}(\tfrac{1}{2}-s)+2\gamma g(\tfrac{1}{2}-s).

This leads to the calculation of $R_{1}(s,q)$ . ∎

For future reference, using the values $\Gamma(\frac{1}{2})=\pi^{\frac{1}{2}},\frac{\Gamma^{\prime}(\frac{1}{2})}{\Gamma(\frac{1}{2})}=\log(\frac{1}{4})-\gamma,\cos(\frac{\pi}{4})=2^{-\frac{1}{2}},\tan(\frac{\pi}{4})=1$ , we note that

(3.5)

\displaystyle R_{1}(0,q)=q^{\frac{1}{2}}\Big{(}\log\left(\tfrac{q}{8\pi}\right)+\gamma-\tfrac{\pi}{2}\Big{)}.

Lemma 9 (The convexity bound for the Riemann Zeta function).

[6, Chapter V] Fix $\sigma\in\mathbb{R}$ . For $t\in\mathbb{R}$ with $|t|\geq 1$ , we have

|\zeta(\sigma+it)|\ll|t|^{\alpha},

where

\alpha=\begin{cases}0&\text{ if }\ \sigma>1,\\ \tfrac{1}{2}(1-\sigma)&\text{ if }\ \sigma\in[0,1],\\ \tfrac{1}{2}-\sigma&\text{ if }\ \sigma<0.\end{cases}

Lemma 10 (Stirling’s expansion).

Let $\mathcal{C}\subset\{z:\Re(z)>0\}$ be a fixed compact set. For $z\in\mathcal{C}$ and $t\in\mathbb{R}$ with $|t|>\frac{1}{2}$ , we have

(3.6)

\displaystyle\Gamma(z+it)\cos(\tfrac{\pi}{2}(z+it))=|t|^{(z-\frac{1}{2})}\exp\Big{(}it\log|t|-it\Big{)}\Big{(}\sum_{m=0}^{M}a_{m}(z)t^{-m}+E_{M}(z,t)\Big{)}

for any $M\geq 0$ , where $a_{m}(z)$ and $E_{M}(z,t)$ are holomorphic functions of $z\in\mathcal{C}$ that depend on $\mathrm{sgn}(t)$ such that $a_{m}(z)$ is a polynomial in $z$ of degree at most $2m$ ,

(3.7)

\displaystyle a_{0}(z)=(\tfrac{\pi}{2})^{\frac{1}{2}}e^{-i\frac{\pi}{4}\mathrm{sgn}(t)},

(3.8)

\displaystyle E_{M}(z,t)\ll_{M}|t|^{-M-1}.

Proof.

Throughout the proof, $M$ will denote a natural number, $a_{m}(z)$ and $E_{M}(z,t)$ will denote functions as described in the lemma, but all of these quantities may not be the same from one occurrence to the next, and $a_{m}$ will denote some constants. By [2, equation 8.344], we have

(3.9)

\displaystyle\log\Gamma(z+it)=(z-\tfrac{1}{2}+it)\log(z+it)-(z+it)+\tfrac{1}{2}\ln 2\pi+\sum_{r=1}^{R}\frac{a_{r}}{(z+it)^{r}}+E_{R}(z,t).

Although [2, equation 8.344] just gives a bound for $E_{R}(z,t)$ , we can infer that it is holomorphic for $z\in\mathcal{C}$ since every other term in (3.9) is. Using Taylor’s theorem, we write

\displaystyle\log(z+it)=\log(it)+\log\Big{(}1+\frac{z}{it}\Big{)}=\log|t|+i\frac{\pi}{2}\mathrm{sgn}(t)+\sum_{m=1}^{M}\frac{(-1)^{m+1}}{m}\Big{(}\frac{z}{it}\Big{)}^{m}+E_{M}(z,t)

and

\frac{a_{r}}{(z+it)^{r}}=\frac{a_{r}}{(it)^{r}}\frac{1}{(1+\frac{z}{it})^{r}}=\frac{a_{r}}{(it)^{r}}\bigg{(}1+\sum_{m=1}^{M}\Bigl{(}\begin{array}[]{@{}c@{}}-r\\ m\end{array}\Bigr{)}\Big{(}\frac{z}{it}\Big{)}^{m}+E_{M}(z,t)\bigg{)}.

Plugging back into (3.9), the $-z$ term cancels with $it\cdot\frac{z}{it}$ , and we get

\displaystyle\log\Gamma(z+it)=(z-\tfrac{1}{2}+it)\log|t|-\tfrac{\pi}{2}|t|-it+i(z-\tfrac{1}{2})\tfrac{\pi}{2}\mathrm{sgn}(t)+\tfrac{1}{2}\ln 2\pi+\sum_{m=1}^{M}a_{m}(z)t^{-m}+E_{M}(z,t).

Taking the exponential of both sides we get

	$\displaystyle\Gamma(z+it)$
	$\displaystyle=\sqrt{2\pi}e^{-i\frac{\pi}{4}\mathrm{sgn}(t)}\|t\|^{(z-\frac{1}{2})}\exp\Big{(}-\tfrac{\pi}{2}\|t\|+it\log\|t\|-it+iz\tfrac{\pi}{2}\mathrm{sgn}(t)\Big{)}\prod_{m=1}^{M}\exp(a_{m}(z)t^{-m})\cdot\exp(E_{M}(z,t))$

Next we take a Taylor polynomials for $\exp(x)$ to get

(3.10)			$\displaystyle\Gamma(z+it)$
		$\displaystyle=\sqrt{2\pi}e^{-i\frac{\pi}{4}\mathrm{sgn}(t)}\|t\|^{(z-\frac{1}{2})}\exp\Big{(}-\tfrac{\pi}{2}\|t\|+it\log\|t\|-it+iz\tfrac{\pi}{2}\mathrm{sgn}(t)\Big{)}\Big{(}1+\sum_{m=1}^{M}a_{m}(z)t^{-m}+E_{M}(z,t)\Big{)}.$

To obtain (3.6), we write

(3.11)

\displaystyle\cos(\tfrac{\pi}{2}(z+it))

\displaystyle=\tfrac{1}{2}\big{(}e^{i(\frac{\pi}{2}(z+it))}+e^{-i(\frac{\pi}{2}(z+it))}\big{)}=\tfrac{1}{2}e^{-iz\tfrac{\pi}{2}\mathrm{sgn}(t)+\tfrac{\pi}{2}|t|}\big{(}1+e^{iz\pi\mathrm{sgn}(t)-\pi|t|}\big{)}

and then multiply with (3.10). ∎

4. Obtaining the leading main term in Theorem 1

We focus on (1.2). Fix any constant $c\in\tfrac{1}{2}+\mathbb{Z}$ satisfying $c>10$ . This parameter determines the length of our asymptotic expansion: we will see that the larger $c$ is, the more terms we may take on the right hand side of (1.2). Define the sets

	$\displaystyle\mathcal{S}_{1}:=\{s\in\mathbb{C}\,:\,-1<\Im(s)<1,\ c-\tfrac{3}{2}<\Re(s)<c-\tfrac{1}{2}\},$
	$\displaystyle\mathcal{S}_{2}:=\{s\in\mathbb{C}\,:\,-1<\Im(s)<1,\ -\tfrac{1}{10}<\Re(s)<c-\tfrac{1}{2}\}.$

Define for $s\in\mathbb{C}$ and an odd prime $q$ ,

\displaystyle F(s,q):=\mathop{{\sum}^{+}}_{\chi\bmod q}L(\tfrac{1}{2}-s,\chi)L(\tfrac{1}{2}+s,\overline{\chi})

where the sum is taken over the even primitive Dirichlet characters of modulus $q$ . Let $H(s,q)$ denote the restriction of $F(s,q)$ to the set $\mathcal{S}_{1}$ . By the functional equation given in Lemma 3, we have that

\displaystyle H(s,q)=\frac{\Gamma\left(\tfrac{1}{4}+\tfrac{s}{2}\right)}{\Gamma\left(\tfrac{1}{4}-\tfrac{s}{2}\right)\pi^{s}q^{\frac{1}{2}-s}}\mathop{{\sum}^{+}}_{\chi\bmod q}\tau(\chi)L(\tfrac{1}{2}+s,\overline{\chi})^{2}:=\frac{\Gamma\left(\tfrac{1}{4}+\tfrac{s}{2}\right)}{\Gamma\left(\tfrac{1}{4}-\tfrac{s}{2}\right)\pi^{s}q^{\frac{1}{2}-s}}H_{1}(s,q).

Now, since in $\mathcal{S}_{1}$ we have $\Re(s)>\tfrac{1}{2}$ , the Dirichlet series representation of $L(\tfrac{1}{2}+s,\overline{\chi})$ is absolutely convergent, and hence it follows that

\displaystyle H_{1}(s,q)=\sum_{n,m\geq 1}\frac{\tau(\chi)\overline{\chi}(nm)}{(nm)^{\frac{1}{2}+s}}=\sum_{\begin{subarray}{c}n\geq 1\\ (n,q)=1\end{subarray}}\frac{\tau(\chi)d(n)\overline{\chi}(n)}{n^{\frac{1}{2}+s}}.

The last equality follows by the total multiplicativity of the Dirichlet characters and the fact that $\chi(n)=0$ if $q|n$ . For $(n,q)=1$ , we write the Gauss sum as

\tau(\chi)=\sum_{k=1}^{q-1}e\Big{(}\frac{k}{q}\Big{)}\chi(k)=\sum_{k=1}^{q-1}e\Big{(}\frac{kn}{q}\Big{)}\chi(kn)

and then apply Corollary 5 to get

\displaystyle H_{1}(s,q)=\sum_{\begin{subarray}{c}n\geq 1\\ (n,q)=1\end{subarray}}\sum_{k=1}^{q-1}\frac{e(\tfrac{kn}{q})d(n)}{n^{\frac{1}{2}+s}}\mathop{{\sum}^{+}}_{\chi\bmod q}\chi(kn)\overline{\chi}(n)=\sum_{\begin{subarray}{c}n\geq 1\\ (n,q)=1\end{subarray}}\frac{((q-1)\cos(\tfrac{2\pi n}{q})+1)d(n)}{n^{\frac{1}{2}+s}}.

The sum may be extended to all natural numbers as follows:

	$\displaystyle H_{1}(s,q)$	$\displaystyle=\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\frac{((q-1)\cos(\tfrac{2\pi n}{q})+1)d(n)}{n^{\frac{1}{2}+s}}-\sum_{\begin{subarray}{c}n\geq 1\\ q\|n\end{subarray}}\frac{((q-1)\cos(\tfrac{2\pi n}{q})+1)d(n)}{n^{\frac{1}{2}+s}}$
		$\displaystyle=\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\frac{((q-1)\cos(\tfrac{2\pi n}{q})+1)d(n)}{n^{\frac{1}{2}+s}}-q\sum_{\begin{subarray}{c}m\geq 1\end{subarray}}\frac{d(mq)}{(mq)^{\frac{1}{2}+s}},$

where we wrote $n=qm$ to get the last expression. By the following multiplicative property of the divisor function (which is also satisfied by $\lambda_{f}(n)$ ),

d(mq)=\begin{cases}d(m)d(q)&\text{ if }(m,q)=1,\\ d(m)d(q)-d(\frac{m}{q})&\text{ if }q|m,\end{cases}

and the fact that $d(q)=2$ , we get

(4.1)		$\displaystyle H_{1}(s,q)$	$\displaystyle=(q-1)\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\frac{\cos(\tfrac{2\pi n}{q})d(n)}{n^{\frac{1}{2}+s}}+(1+q^{-2s}-2q^{\frac{1}{2}-s})\zeta^{2}(\tfrac{1}{2}+s)$
		$\displaystyle:=(q-1)H_{2}(s,q)+(1+q^{-2s}-2q^{\frac{1}{2}-s})\zeta^{2}(\tfrac{1}{2}+s).$

Inserting the integral representation of $\cos(\tfrac{2\pi n}{q})$ given in Lemma 7, we have

H_{2}(s,q)=\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}+s}}\frac{1}{2\pi i}\int_{(-\frac{3}{4})}\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi n}\right)^{w}dw+\zeta^{2}(\tfrac{1}{2}+s).

The integral converges absolutely:

(4.2)

\int_{(-\frac{3}{4})}\left|\Gamma(w)\cos\left(\frac{\pi w}{2}\right)\frac{q^{w}}{(2\pi n)^{w}}dw\right|\ll\frac{n^{\frac{3}{4}}}{q^{\frac{3}{4}}},

and $\Re(\tfrac{1}{2}+s-\tfrac{3}{4})>1$ for $s\in\mathcal{S}_{1}$ , so we can interchange summation and integration to get

(4.3)		$\displaystyle H_{2}(s,q)$	$\displaystyle=\frac{1}{2\pi i}\int_{(-\frac{3}{4})}\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi}\right)^{w}\zeta^{2}(\tfrac{1}{2}+s+w)dw+\zeta^{2}(\tfrac{1}{2}+s)$
		$\displaystyle:=H_{3}(s,q)+\zeta^{2}(\tfrac{1}{2}+s).$

Moving the line of integration to $\Re(w)=-c$ , we pick up the residues calculated in Lemma 8, and get

H_{3}(s,q)=R_{1}(s,q)+\sum_{n=1}^{\lfloor c/2\rfloor}R_{0}(n,s,q)+I(s,q),

where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ , and

I(s,q)=\frac{1}{2\pi i}\int_{(-c)}\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi}\right)^{w}\zeta^{2}(\tfrac{1}{2}+s+w)dw.

Note that this integral converges absolutely, since by Lemma 9 we have

|\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)\zeta^{2}(\tfrac{1}{2}+s+w)|\ll|t|^{-c-\frac{1}{2}}|t|^{1-2\Re(\frac{1}{2}+s+w)}\ll|t|^{3}|t|^{-c-\frac{1}{2}}\ll|t|^{-7}

for $w=-c+it$ with $|t|\to\infty$ .

Retracing our steps and inserting the final expressions for $H_{1}(s,q),H_{2}(s,q),H_{3}(s,q)$ back into $H(s,q)$ , we obtain the formula

(4.4)

H(s,q)=\frac{\Gamma(\tfrac{1}{4}+\tfrac{s}{2})}{\Gamma(\tfrac{1}{4}-\tfrac{s}{2})\pi^{s}q^{\frac{1}{2}-s}}\bigg{(}(q-1)R_{1}(s,q)+(q-1)\sum_{n=1}^{\lfloor c/2\rfloor}R_{0}(n,s,q)\\ +(q-1)I(s,q)+(q+q^{-2s}-2q^{\frac{1}{2}-s})\zeta^{2}(\tfrac{1}{2}+s)\bigg{)}.

From this expression, it follows that

H(s,q)-\frac{\Gamma\left(\tfrac{1}{4}+\tfrac{s}{2}\right)(q-1)}{\Gamma\left(\tfrac{1}{4}-\tfrac{s}{2}\right)\pi^{s}q^{\frac{1}{2}-s}}I(s,q),

continues to a meromorphic function on $\mathcal{S}_{2}$ , with value at $s=0$ equal to

(4.5)

\displaystyle\frac{q-1}{2}\left(\gamma+\log(\tfrac{q}{8\pi})-\tfrac{\pi}{2}\right)+\frac{(q-1)}{q^{\frac{1}{2}}}\sum_{n=1}^{\lfloor c/2\rfloor}\frac{(-1)^{n}}{(2n)!}\Big{(}\frac{2\pi}{q}\Big{)}^{2n}\zeta^{2}(\tfrac{1}{2}-2n)+\frac{(q-2q^{\frac{1}{2}})}{q^{\frac{1}{2}}}\zeta^{2}(\tfrac{1}{2}),

using (3.5). We see here the claimed main terms $\frac{q-1}{2}(\log(\frac{q}{8\pi})+\gamma-\frac{\pi}{2})+q^{\frac{1}{2}}\zeta^{2}(\frac{1}{2})$ of Theorem 1, while the rest of the right hand side above is $O(1)$ and contributes to the lower order main terms in Theorem 1. However, it still remains to study the shifted integral $I(s,q)$ .

We will show in the next section that as a function of $s$ , $I(s,q)$ continues to a meromorphic function on $\mathcal{S}_{2}$ , with value at $s=0$ contributing additional lower order main terms of size $O(1)$ .

To frame this in terms of the strategy laid out in the sketch, we let $\mathcal{I}(s,q)$ denote meromorphic continuation of $I(s,q)$ to $\mathcal{S}_{2}$ . Then

G(s,q):=\frac{\Gamma(\tfrac{1}{4}+\tfrac{s}{2})}{\Gamma(\tfrac{1}{4}-\tfrac{s}{2})\pi^{s}q^{\frac{1}{2}-s}}\bigg{(}(q-1)R_{1}(s,q)+(q-1)\sum_{n=1}^{\lfloor c/2\rfloor}R_{0}(n,s,q)\\ +(q-1)\mathcal{I}(s,q)+(q+q^{-2s}-2q^{\frac{1}{2}-s})\zeta^{2}(\tfrac{1}{2}+s)\bigg{)}

is meromorphic for $s\in\mathcal{S}_{2}$ , with $G(s,q)=H(s,q)$ on $\mathcal{S}_{1}$ . Since also $F(s,q)=H(s,q)$ on $\mathcal{S}_{1}$ , we get that $F(s,q)=G(s,q)$ on $\mathcal{S}_{2}$ by the principle of analytic continuation.

The sum over the odd characters. Before moving on we note that we would follow similar calculations for (1.3). A couple of important differences would be as follows. Our main object of study would be

(q-1)\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\frac{\sin(\tfrac{2\pi n}{q})d(n)}{n^{\frac{1}{2}+s}},

instead of $(q-1)\sum\frac{\cos(\frac{2\pi n}{q})d(n)}{n^{\frac{1}{2}+s}}$ as we saw in (4.1). We would use the Mellin inverse transform

(4.6)

\displaystyle\sin(x)=\frac{1}{2\pi i}\int_{(d)}\Gamma(w)\sin\left(\frac{\pi w}{2}\right)x^{w}dw

for $d\in(-1,0)$ instead of (3.1). Note that unlike (3.1), there is no additional $+1$ term here. Consequently, the analogue of (4.3) would not have a $\zeta^{2}(\frac{1}{2}+s)$ term. For (1.2), this is the source of the main term $q^{\frac{1}{2}}\zeta^{2}(\frac{1}{2})$ , as we saw in (4.5). For (1.3), the main term $q^{\frac{1}{2}}\zeta^{2}(\frac{1}{2})$ must then have a different source, which turns out to be the analogue of $\mathcal{I}(0,q)$ . We will explain this in the next section. Using the inverse Mellin transform (4.6), the integral to study would be

\frac{1}{2\pi i}\int_{(-\frac{3}{4})}\Gamma(w)\sin\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi}\right)^{w}\zeta^{2}(\tfrac{1}{2}+s+w)dw,

instead of the integral in (4.3). Moving the line of integration to the left, the residue at $w=\frac{1}{2}-s$ would be

\Gamma(\tfrac{1}{2}-s)\sin\left(\tfrac{\pi}{4}-\tfrac{\pi s}{2}\right)\left(\tfrac{q}{2\pi}\right)^{\frac{1}{2}-\frac{s}{2}}\Big{(}\tfrac{\Gamma^{\prime}(\frac{1}{2}-s)}{\Gamma(\frac{1}{2}-s)}+\tfrac{\pi}{2}\cot\left(\tfrac{\pi}{4}-\tfrac{\pi s}{2}\right)+\log\left(\tfrac{q}{2\pi}\right)+2\gamma\Big{)},

instead of (3.4). Evaluating at $s=0$ yields the leading main term in (1.3) and explains the $+\frac{\pi}{2}$ term there instead of $-\frac{\pi}{2}$ as in (1.2).

5. The shifted integral

The goal of this section is to show that

Lemma 11.

There exists a meromorphic function $\mathcal{I}(s,q)$ defined for $s\in\mathcal{S}_{2}$ such that $\mathcal{I}(s,q)=I(s,q)$ for $s\in\mathcal{S}_{1}$ . At $s=0$ , we have

(5.1)

\displaystyle\mathcal{I}(0,q)=\sum_{n=1}^{N}r_{n}q^{-n}+O_{N}(q^{-N-1}).

for any integer $N\geq 1$ and some constants $r_{n}$ .

Proof.

Let $s\in\mathcal{S}_{1}$ . Applying the reflection formula for $\Gamma(w)$ and the functional equation for $\zeta^{2}(\tfrac{1}{2}+s+w)$ , we get

	$\displaystyle I(s,q)=\frac{1}{2\pi i}\int\limits_{(-c)}\frac{(\frac{q}{2\pi})^{w}(2\pi)^{2(s+w)}}{\Gamma(1-w)\cos(\tfrac{\pi}{2}(1-w))}\cos^{2}(\tfrac{\pi}{2}(\tfrac{1}{2}-s-w))\Gamma^{2}(\tfrac{1}{2}-s-w)\zeta^{2}(\tfrac{1}{2}-s-w)dw$
	$\displaystyle=\int_{-\infty}^{\infty}\frac{(\frac{q}{2\pi})^{-c+it}(2\pi)^{2(s-c+it)-1}}{\Gamma(1+c-it)\cos(\tfrac{\pi}{2}(1+c-it))}\cos^{2}(\tfrac{\pi}{2}(\tfrac{1}{2}-s+c-it))\Gamma^{2}(\tfrac{1}{2}-s+c-it)\zeta^{2}(\tfrac{1}{2}-s+c-it)dt.$

Since $\tfrac{1}{2}-\Re(s)+c>1$ , we can expand

\zeta^{2}(\tfrac{1}{2}-s+c-it)=\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}-s+c-it}}

into an absolutely convergent sum. Exchanging the order of summation and integration, we get

\displaystyle I(s,q)=\sum_{n\geq 1}I(s,q,n),

where

	$\displaystyle I(s,q,n)=$
	$\displaystyle\int_{-\infty}^{\infty}\frac{(\frac{q}{2\pi})^{-c+it}(2\pi)^{2(s-c+it)-1}}{\Gamma(1+c-it)\cos(\tfrac{\pi}{2}(1+c-it))}\cos^{2}(\tfrac{\pi}{2}(\tfrac{1}{2}-s+c-it))\Gamma^{2}(\tfrac{1}{2}-s+c-it)\frac{d(n)}{n^{\frac{1}{2}-s+c-it}}dt.$

The idea now is to understand the oscillatory and stationary behavior of the integrand, and use that information to determine the integral. Such stationary phase analysis can be found in the literature to determine the integral up to a suitable error term. However we cannot simply quote the literature because we need to first establish meromorphic continuation to a neighborhood of $s=0$ , after which we are permitted to determine the value at $s=0$ up to a suitable error term. Therefore, we proceed by revisiting the steps of stationary phase analysis, being careful to obtain meromorphic continuation.

We study the integral $I(s,q,n)$ over finite dyadic intervals, as follows. Note that for each $n\in\mathbb{N}$ , the union

\bigcup\limits_{k>-\log_{2}(2\pi qn)}(-\tfrac{3}{2}(2\pi qn)2^{k},-\tfrac{5}{8}(2\pi qn)2^{k})\ \bigcup\ (-10,10)\bigcup\limits_{k>-\log_{2}(2\pi qn)}(\tfrac{5}{8}(2\pi qn)2^{k},\tfrac{3}{2}(2\pi qn)2^{k}).

forms an open cover of $(-\infty,\infty)$ . Thus there exists a partition of unity of $(-\infty,\infty)$ by a collection of smooth functions of the form $W(\frac{t}{T})$ , where $T=T(n)$ assumes one of the values below, and $W(x)$ is supported on

(5.2)		$\displaystyle\tfrac{5}{8}$	$\displaystyle<x<\tfrac{3}{2}\ \ \ \ \text{ for }\ \ T=(2\pi qn)2^{k},\ k\in\mathbb{Z},\ k\geq-\log_{2}(2\pi qn),$
	$\displaystyle-20$	$\displaystyle<x<20\ \ \ \text{ for }\ \ T=\tfrac{1}{2},$
	$\displaystyle-\tfrac{3}{2}$	$\displaystyle<x<-\tfrac{5}{8}\ \ \text{ for }\ \ T=(2\pi qn)2^{k},\ k\in\mathbb{Z},\ k\geq-\log_{2}(2\pi qn),$

with $W^{(j)}(x)\ll_{j}1$ for any $j\geq 0$ . Then we have that

I(s,q,n)=\sum_{T}I_{T}(s,q),

where

I_{T}(s,q)=\int_{-\infty}^{\infty}W\Big{(}\frac{t}{T}\Big{)}\frac{(\frac{q}{2\pi})^{-c+it}(2\pi)^{2(s-c+it)-1}}{\Gamma(1+c-it)\cos(\tfrac{\pi}{2}(1+c-it))}\\ \times\cos^{2}(\tfrac{\pi}{2}(\tfrac{1}{2}-s+c-it))\Gamma^{2}(\tfrac{1}{2}-s+c-it)\frac{d(n)}{n^{\frac{1}{2}-s+c-it}}dt.

We first consider the contribution of $T=\frac{1}{2}$ to $I(s,q)$ . The integral $I_{\frac{1}{2}}(s,q)$ is of bounded length, the integrand is holomorphic for $s\in\mathcal{S}_{2}$ , and the sum $\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}-\Re(s)+c}}$ is absolutely bounded for $s\in\mathcal{S}_{2}$ . Thus $\sum_{n\geq 1}I_{\frac{1}{2}}(s,q)$ analytically continues to $\mathcal{S}_{2}$ , with

\sum_{n\geq 1}I_{\frac{1}{2}}(0,q)\ll q^{-c}.

Now we restrict to the case $T\neq\frac{1}{2}$ (so that $T\geq 1$ ). We apply Lemma 10 with $M=2c$ for $\Gamma(1+c-it)\cos(\tfrac{\pi}{2}(1+c-it))$ and $\cos(\tfrac{\pi}{2}(\tfrac{1}{2}-s+c-it))\Gamma(\tfrac{1}{2}-s+c-it)$ , which is permissible since $1+c>1$ , $\tfrac{1}{2}-\Re(s)+c>1$ , and $|t|>\frac{5}{8}$ . These conditions hold for $s$ in the wider set $\mathcal{S}_{2}$ too. Thus we get

\displaystyle I_{T}(s,q)=(2\pi q)^{-c}\frac{d(n)}{n^{\frac{1}{2}-s+c}}\int_{-\infty}^{\infty}W\Big{(}\frac{t}{T}\Big{)}\Big{(}\sum_{0\leq m\leq{2c}}a_{m}(s)t^{-m}+E(s,t)\Big{)}|t|^{c-2s-\frac{1}{2}}\exp(i\Phi(t))dt,

where

\Phi(t)=-t\log\left(\frac{|t|}{2\pi eqn}\right),

and $a_{m}(s)$ and $E(s,t)$ are holomorphic functions of $s\in\mathcal{S}_{2}$ such that

a_{m}(s)\ll 1,\ \ \ E(s,t)\ll|t|^{-2c-1}

and

(5.3)

\displaystyle a_{0}(s)=(2\pi)^{2s-1}(\tfrac{\pi}{2})^{\frac{1}{2}}e^{-i\frac{\pi}{4}\mathrm{sgn}(t)}

and

First, we address the contribution of $E(s,t)$ to $I(s,q)$ . This equals

(5.4)

\displaystyle\sum_{n\geq 1}\sum_{T\geq 1}(2\pi q)^{-c}\frac{d(n)}{n^{\frac{1}{2}-s+c}}\int_{-\infty}^{\infty}W\Big{(}\frac{t}{T}\Big{)}E(s,t)|t|^{c-2s-\frac{1}{2}}\exp(i\Phi(t))dt,

which is bounded for $s\in\mathcal{S}_{2}$ by the absolutely convergent sum

\displaystyle\sum_{n\geq 1}\sum_{T\geq 1}q^{-c}\frac{d(n)}{n^{\frac{1}{2}-\Re(s)+c}}T\cdot T^{-2c-1}T^{c-2\Re(s)-\frac{1}{2}}\ll q^{-c}\sum_{n\geq 1}\frac{d(n)}{n^{\frac{1}{2}-\Re(s)+c}}\sum_{T\geq 1}T^{-c}\ll q^{-c}.

Thus (5.4) has analytic continuation to $\mathcal{S}_{2}$ with value at $s=0$ bounded by $O(q^{-c})$ .

It remains to consider

\sum_{n\geq 1}\ \sum_{0\leq m\leq 2c}\ \sum_{T\geq 1}a_{m}(s)I_{T,m}(s,q),

where we define

	$\displaystyle I_{T,m}(s,q)$	$\displaystyle:=(2\pi q)^{-c}\frac{d(n)}{n^{\frac{1}{2}-s+c}}\int_{-\infty}^{\infty}W\Big{(}\frac{t}{T}\Big{)}t^{-m}\|t\|^{c-2s-\frac{1}{2}}\exp(i\Phi(t))dt$
(5.5)			$\displaystyle=(2\pi q)^{-c}\frac{d(n)}{n^{\frac{1}{2}-s+c}}T\int_{-\infty}^{\infty}W(x)(xT)^{-m}\|xT\|^{c-2s-\frac{1}{2}}\exp(i\Phi(xT))dx.$

The last equality follows by the substitution $x=\frac{t}{T}$ . We now assume that $W(x)$ is supported on the positive real axis because the negative case is treated similarly. Thus $W(x)$ is supported on $(\frac{5}{8},\frac{3}{2})$ .

Case I. $T\neq 2\pi qn$ . In this case we have $\frac{T}{2\pi qn}\geq 2$ or $\frac{T}{2\pi qn}\leq\frac{1}{2}$ by definition (5.2). So

\frac{d}{dx}i\Phi(xT)=\frac{d}{dx}\Big{(}-ixT\log\left(\frac{xT}{2\pi eqn}\right)\Big{)}=-iT\log\Big{(}x\frac{T}{2\pi qn}\Big{)}\gg T

for $x\in(\frac{5}{8},\frac{3}{2})$ and

\frac{d^{j}}{dx^{j}}\Big{(}-i\log\Big{(}x\frac{T}{2\pi qn}\Big{)}\Big{)}=i(j-1)!(-x)^{-j}\asymp 1\ \ \text{ for }j\geq 1.

We can write

\displaystyle\int_{-\infty}^{\infty}W(x)x^{-m+c-2s-\frac{1}{2}}\exp(i\Phi(xT))dx=\int_{-\infty}^{\infty}W(x)x^{-m+c-2s-\frac{1}{2}}\frac{-iT\log(x\frac{T}{2\pi qn})}{-iT\log(x\frac{T}{2\pi qn})}\exp(i\Phi(xT))dx

and then repeatedly integrate by parts to get

	$\displaystyle\int_{-\infty}^{\infty}W(x)x^{-m+c-2s-\frac{1}{2}}\exp(i\Phi(xT))dx$	$\displaystyle=\frac{1}{T}\int_{-\infty}^{\infty}\frac{d}{dx}\Big{(}\frac{W(x)x^{-m+c-2s-\frac{1}{2}}}{i\log(x\frac{T}{2\pi qn})}\Big{)}\exp(i\Phi(xT))dx$
		$\displaystyle=\frac{1}{T^{2}}\int_{-\infty}^{\infty}\frac{d}{dx}\Bigg{(}\frac{\frac{d}{dx}\Big{(}\frac{W(x)x^{-m+c-2s-\frac{1}{2}}}{i\log(x\frac{T}{2\pi qn})}\Big{)}}{i\log(x\frac{T}{2\pi qn})}\Bigg{)}\exp(i\Phi(xT))dx,$

and so on, with each of these successive integrals an entire function of $s$ . In this way we see that for $s\in\mathcal{S}_{2}$ , we have

I_{T,m}(s,q)\ll_{k}q^{-c}T^{-k}

for any $k\geq 0$ . This ensures the absolute convergence of the sum

\displaystyle\sum_{n\geq 1}\ \sum_{0\leq m\leq 2c}\sum_{\begin{subarray}{c}T\geq 1\\ T\neq 2\pi qn\end{subarray}}a_{m}(s)I_{T,m}(s,q)

for $s\in\mathcal{S}_{2}$ and therefore its analytic continuation to $\mathcal{S}_{2}$ with value at $s=0$ bounded by $O(q^{-c})$ .

Case II. $T=2\pi nq$ . By construction (5.2), we are considering the unique function $W(\frac{t}{2\pi nq})$ in this partition which contains a neighborhood of $2\pi nq$ in its support. We have, using definition (5.5),

	$\displaystyle I_{2\pi nq,m}(s,q)$	$\displaystyle=(2\pi q)^{-c}\frac{d(n)}{n^{\frac{1}{2}-s+c}}2\pi nq\int_{-\infty}^{\infty}W(x)(x2\pi nq)^{-m+c-2s-\frac{1}{2}}\exp(i\Phi(x2\pi nq))dx$
		$\displaystyle=(2\pi q)^{\frac{1}{2}-m-2s}\frac{d(n)}{n^{s+m}}\int_{-\infty}^{\infty}W(x)x^{-m+c-2s-\frac{1}{2}}\exp(i\phi(x))dx,$

where

\phi(x)=-2\pi nqx\log\Big{(}\frac{x}{e}\Big{)}.

We have

\phi^{\prime}(x)=-2\pi nq\log x,\ \ \ \phi^{(j)}(x)=2\pi nq(j-2)!(-x)^{-j+1}\ \text{ for }j\geq 2.

Note that $\phi^{\prime}(1)=0$ (so that $x=1$ is a stationary point). Taking a Taylor series expansion about $x=1$ , we get

\phi(x)=2\pi nq\Big{(}1-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{6}-\ldots\Big{)}

for $x\in(\frac{5}{8},\frac{3}{2})$ , so that

\exp(i\phi(x))=\exp\Big{(}2\pi inq\Big{(}-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{6}-\ldots\Big{)}\Big{)}.

This uses the nice simplification $e^{2\pi inq}=1$ .

We will evaluate $I_{2\pi nq,m}(s,q)$ by stationary phase analysis. As is the usual technique, we first isolate $x=1$ further. Let $w(x)$ be a smooth function such that

	$\displaystyle w(x)=1\ \ \text{ for }1-(nq)^{-\frac{5}{12}}<x<1+(nq)^{-\frac{5}{12}},$
	$\displaystyle w(x)=0\ \ \text{ for }x<1-2(nq)^{-\frac{5}{12}}\ \text{ or }x>1+2(nq)^{-\frac{5}{12}},$
	$\displaystyle w^{(j)}(1)=0\ \ \text{ for }j\geq 1,$
	$\displaystyle w^{(j)}(x)\ll_{j}((nq)^{\frac{5}{12}})^{j}\ \ \text{ for }j\geq 0.$

We split up the integral as $I_{2\pi nq,m}(s,q)=I_{2\pi nq,m}^{w}(s,q)+I_{2\pi nq,m}^{1-w}(s,q)$ , where

	$\displaystyle I_{2\pi nq,m}^{w}(s,q):=(2\pi q)^{\frac{1}{2}-m-2s}\frac{d(n)}{n^{s+m}}\int_{-\infty}^{\infty}W(x)w(x)x^{-m+c-2s-\frac{1}{2}}\exp(i\phi(x))dx,$
	$\displaystyle I_{2\pi nq,m}^{1-w}(s,q):=(2\pi q)^{\frac{1}{2}-m-2s}\frac{d(n)}{n^{s+m}}\int_{-\infty}^{\infty}W(x)(1-w(x))x^{-m+c-2s-\frac{1}{2}}\exp(i\phi(x))dx.$

Thus the integrand in $I_{2\pi nq,m}^{w}(s,q)$ is supported very close to $x=1$ while the integrand in $I_{2\pi nq,m}^{1-w}(s,q)$ is supported some distance away from $x=1$ .

Repeatedly integrating by parts in $I_{2\pi nq,m}^{1-w}(s,q)$ , we get

	$\displaystyle\int_{-\infty}^{\infty}W(x)(1-w(x))(2\pi x)^{-m+c-2s-\frac{1}{2}}\exp\Big{(}2\pi inq\Big{(}-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{6}-\ldots\Big{)}dx$
	$\displaystyle=\frac{1}{2\pi inq}\int_{-\infty}^{\infty}\frac{d}{dx}\Big{(}\frac{W(x)(1-w(x))(2\pi x)^{-m+c-2s-\frac{1}{2}}}{-(x-1)+\frac{(x-1)^{2}}{2}-\ldots}\Big{)}\exp\Big{(}2\pi inq\Big{(}-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{6}-\ldots\Big{)}dx$
	$\displaystyle=\frac{1}{(2\pi inq)^{2}}\int_{-\infty}^{\infty}\frac{d}{dx}\Bigg{(}\frac{\frac{d}{dx}\Big{(}\frac{W(x)(1-w(x))(2\pi x)^{-m+c-2s-\frac{1}{2}}}{-(x-1)+\frac{(x-1)^{2}}{2}-\ldots}}{{-(x-1)+\frac{(x-1)^{2}}{2}-\ldots}}\Bigg{)}\exp\Big{(}2\pi inq\Big{(}-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{6}-\ldots\Big{)}dx,$

and so on, with each of these successive integrals an entire function of $s$ . Since $(nq)^{-\frac{5}{12}}\ll|x-1|\leq\frac{1}{2}$ in the support of $W(x)(1-w(x))$ , and $w^{(j)}(x)\ll_{j}(nq)^{\frac{5}{12}j}$ for $j\geq 0$ , we see that for $s\in\mathcal{S}_{2}$ , after integrating by parts $k\geq 0$ times we have

\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}a_{m}(s)I_{2\pi nq,m}^{1-w}(s,q)\ll_{k}\sum_{n\geq 1}\ \sum_{0\leq m\leq 2c}q^{\frac{1}{2}-m-2\Re(s)}\frac{d(n)}{n^{\Re(s)+m}}\Big{(}\frac{(nq)^{\frac{5}{6}}}{nq}\Big{)}^{k}\ll q^{-\frac{7c}{6}+1},

where the last bound follows by taking $k=7c$ . This ensures the absolute convergence of the sum above and therefore its analytic continuation to $\mathcal{S}_{2}$ with value at $s=0$ bounded by $O(q^{-c})$ .

It remains to consider $I_{2\pi nq,m}^{w}(s)$ , which we write as

(5.6)

\displaystyle I_{2\pi nq,m}^{w}(s,q)=(2\pi q)^{\frac{1}{2}-m-2s}\frac{d(n)}{n^{s+m}}\int_{-\infty}^{\infty}h(s,x)\exp(-i\pi nq(x-1)^{2})dx,

where

(5.7)

\displaystyle h(s,x)=W(x)w(x)x^{-m+c-2s-\frac{1}{2}}e^{i2\pi nq(\frac{(x-1)^{3}}{3}-\ldots)}.

Let

\hat{h}(s,y)=\int_{-\infty}^{\infty}h(s,x)e^{-2\pi iyx}dx

denote the Fourier transform of $h$ . Integrating by parts $k\geq 0$ times, we have for $s\in\mathcal{S}_{2}$ and $|y|>1$ ,

(5.8)

\displaystyle\hat{h}(s,y)=\hat{h}_{k}(s,y):=\int_{-\infty}^{\infty}\Big{(}\frac{\partial^{k}}{\partial x^{k}}h(s,x)\Big{)}\frac{e^{-2\pi iyx}}{(2\pi iy)^{k}}dx\ll_{k}\Big{(}\frac{(nq)^{\frac{5}{12}}}{|y|}\Big{)}^{k}.

This uses $w^{(j)}(x)\ll_{j}(nq)^{\frac{5}{12}j}$ for $j\geq 0$ and $nq|x-1|^{2}\ll(nq)^{\frac{5}{12}}$ in the support of $w(x)$ . For $|y|\leq 1$ , we have the trivial bound $\hat{h}(s,y)\ll 1$ . Also note that $\hat{h}(s,y)$ (for $|y|\leq 1$ ) and $\hat{h}_{k}(s,y)$ (for $|y|>1$ ) are entire functions of $s$ . By Fourier inversion, we have

	$\displaystyle\int_{-\infty}^{\infty}h(x)e^{-i\pi nq(x-1)^{2}}dx$	$\displaystyle=\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}h(x)e^{-i\pi nq(x-1)^{2}}dx$
		$\displaystyle=\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\Big{(}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi ixy}dy\Big{)}e^{-i\pi nq(x-1)^{2}}dx$
		$\displaystyle=\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\Big{(}\int_{-\alpha}^{\alpha}\hat{h}(s,y)e^{2\pi ixy}dy\Big{)}e^{-i\pi nq(x-1)^{2}}dx,$

where the last line uses the upper bound for $\hat{h}(s,y)$ given in (5.8). We can exchange the order of integration to get

\displaystyle\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\hat{h}(s,y)\int_{-\alpha}^{\alpha}e^{2\pi iyx}e^{-i\pi nq(x-1)^{2}}dx\ dy=\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\hat{h}(s,y)\int_{-\infty}^{\infty}e^{2\pi iyx}e^{-i\pi nq(x-1)^{2}}dx\ dy,

where inner integral was extended to infinity because for $|y|\leq\alpha$ ,

\displaystyle\int_{\alpha}^{\infty}

\displaystyle e^{2\pi iyx-i\pi nq(x-1)^{2}}dx=\lim_{\beta\to\infty}e^{2\pi iy+i\pi\frac{y^{2}}{nq}}\int_{\alpha}^{\beta}e^{-i\pi nq(x-1-\frac{y}{nq})^{2}}dx

by completing the square, and this equals

\displaystyle\lim_{\beta\to\infty}e^{2\pi iy+i\pi\frac{y^{2}}{nq}}\Bigg{(}\frac{e^{-i\pi nq(x-1-\frac{y}{nq})^{2}}}{-2i\pi nq(x-1-\frac{y}{nq})}\Bigg{|}_{x=\alpha}^{x=\beta}-\int_{\alpha}^{\beta}\frac{e^{-i\pi nq(x-1-\frac{y}{nq})^{2}}}{2i\pi nq(x-1-\frac{y}{nq})^{2}}dx\Bigg{)}\ll\alpha^{-1}

by integrating by parts. Now,

	$\displaystyle\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\hat{h}(s,y)\int_{-\infty}^{\infty}e^{2\pi iyx}e^{-i\pi nq(x-1)^{2}}dx\ dy$	$\displaystyle=\lim_{\alpha\to\infty}\int_{-\alpha}^{\alpha}\hat{h}(s,y)\frac{e^{i\pi(2y+\frac{y^{2}}{nq}-\frac{1}{4})}}{\sqrt{nq}}\ dy$
		$\displaystyle=\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi yi}e^{i\pi\frac{y^{2}}{nq}}\ dy.$

by evaluating the Gaussian $x$ -integral using [2, equations 17.23.16, 17.23.17] and then extending the $y$ -integral to infinity using (5.8). Writing

E(y)=\sum_{j=0}^{6c}\frac{1}{j!}\Big{(}\frac{i\pi y^{2}}{nq}\Big{)}^{j},

we have

(5.9)

\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi yi}e^{i\pi\frac{y^{2}}{nq}}\ dy=\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi yi}E(y)dy\\ +\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{-1}^{1}\hat{h}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}dy+\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{|y|>1}\hat{h}_{k}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}\ dy,

where we used (5.8) for the last integral. All the integrals on the right hand side of (5.9) are absolutely convergent, so they are entire functions of $s$ . For the second and third integrals, by Taylor’s theorem applied to the function $e^{ix}$ , we have the following bounds for $s\in\mathcal{S}_{2}$ :

(5.10)			$\displaystyle\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{-1}^{1}\hat{h}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}dy\ll\frac{1}{\sqrt{nq}}\Big{(}\frac{1}{nq}\Big{)}^{6c+1}\ll\Big{(}\frac{1}{nq}\Big{)}^{6c+\frac{3}{2}},$
		$\displaystyle\frac{e^{-\frac{i\pi}{4}}}{\sqrt{nq}}\int_{\|y\|>1}\hat{h}_{k}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}\ dy\ll\frac{1}{\sqrt{nq}}\int_{\|y\|>1}\Big{(}\frac{(nq)^{\frac{5}{12}}}{\|y\|}\Big{)}^{12c+4}\Big{(}\frac{\|y\|^{2}}{nq}\Big{)}^{6c+1}dy\ll\Big{(}\frac{1}{nq}\Big{)}^{c+\frac{1}{6}},$

where we used (5.8) with $k=12c+4$ . Plugging back (5.9) into (5.6), we get that

(5.11)		$\displaystyle\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}a_{m}(s)I_{2\pi nq,m}^{w}(s)$
	$\displaystyle=\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}a_{m}(s)\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}\frac{d(n)}{n^{\frac{1}{2}+s+m}}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi yi}E(y)dy$
	$\displaystyle+\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}a_{m}(s)\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}\frac{d(n)}{n^{\frac{1}{2}+s+m}}\int_{-1}^{1}\hat{h}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}dy$
	$\displaystyle+\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}a_{m}(s)\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}\frac{d(n)}{n^{\frac{1}{2}+s+m}}\int_{\|y\|>1}\hat{h}(s,y)e^{2\pi yi}\Big{(}e^{i\pi\frac{y^{2}}{nq}}-E(y)\Big{)}\ dy.$

In light of the bounds (5.10), the second and third sums on the right hand side of (5.11) analytically continue to $\mathcal{S}_{2}$ , with value at $s=0$ being bounded by $O(q^{-c+1})$ .

It remains to obtain the meromorphic continuation from $\mathcal{S}_{1}$ to $\mathcal{S}_{2}$ of

(5.12)

\displaystyle\sum_{\begin{subarray}{c}n\geq 1\end{subarray}}\ \sum_{0\leq m\leq 2c}\ \sum_{0\leq j\leq 6c}a_{m}(s)I_{2\pi nq,m,j}^{w}(s,q),

where

\displaystyle I_{2\pi nq,m,j}^{w}(s,q):=

\displaystyle\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}\frac{d(n)}{n^{\frac{1}{2}+s+m}}\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi iy}\frac{1}{j!}\Big{(}\frac{i\pi y^{2}}{nq}\Big{)}^{j}dy.

By Fourier inversion, we have

\int_{-\infty}^{\infty}\hat{h}(s,y)e^{2\pi iy}y^{2j}dy=\frac{1}{(2\pi i)^{2j}}\frac{\partial^{2j}}{\partial x^{2j}}h(s,x)\bigg{|}_{x=1}.

Thus keeping in mind definition (5.7), we have

I_{2\pi nq,m,j}^{w}(s,q)=\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}\frac{d(n)}{n^{\frac{1}{2}+s+m}}P_{m,j}\Big{(}s,\frac{1}{nq}\Big{)},

where $P_{m,j}(s,\frac{1}{nq})$ is entire in $s$ and polynomial in $\frac{1}{nq}$ , and the degree of this polynomial is $>1$ for $j>0$ . For $j=0$ , we note that $P_{m,0}(s,\frac{1}{nq})=h(s,1)=1$ . Summing over $n$ , we get that (5.12) equals

\sum_{0\leq m\leq 2c}\ \sum_{0\leq j\leq 6c}a_{m}(s)\sqrt{2\pi}e^{-\frac{i\pi}{4}}(2\pi q)^{-m-2s}P_{m,j}(s,\tfrac{1}{nq})\zeta^{2}(\tfrac{1}{2}+s+m).

This expression continues to a meromorphic function on $\mathcal{S}_{2}$ (because the Riemann Zeta function does) and we have that its value at $s=0$ equals, using (5.3),

(5.13)

\displaystyle a_{0}(0)\sqrt{2\pi}e^{-\frac{i\pi}{4}}\zeta^{2}(\tfrac{1}{2})+\sum_{n=1}^{N}r_{n}^{\prime}q^{-n}+O_{N}(q^{-N-1})=\frac{-i}{2}\zeta^{2}(\tfrac{1}{2})+\sum_{n=1}^{N}r_{n}^{\prime}q^{-n}+O_{N}(q^{-N-1}),

for some constants $r_{n}^{\prime}$ and any $N\geq 1$ , by taking $c$ large enough.

Recall that for this calculation, we have been assuming that $W(x)$ is supported on the positive real axis. In the case that the support of $W(x)$ is on the negative real axis, there is a stationary point at $x=-1$ , and we instead obtain the expression

(5.14)

\displaystyle\frac{i}{2}\zeta^{2}(\tfrac{1}{2})+\sum_{n=1}^{N}r_{n}^{\prime\prime}q^{-n}+O_{N}(q^{-N-1})

for some constants $r_{n}^{\prime\prime}$ . Adding together (5.13) and (5.14), we get (5.1). ∎

The sum over the odd characters. The treatment of the integral

(5.15)

\displaystyle\frac{1}{2\pi i}\int_{(-c)}\Gamma(w)\sin\left(\tfrac{\pi w}{2}\right)\left(\frac{q}{2\pi}\right)^{w}\zeta^{2}(\tfrac{1}{2}+s+w)dw,

required to obtain (1.3), follows similar calculations, but we highlight some important differences with the treatment of the integral $I(s,q)$ . Instead of (3.11), we have

\displaystyle\sin(\tfrac{\pi}{2}(z+it))

\displaystyle=\tfrac{1}{2i}\big{(}e^{i(\frac{\pi}{2}(z+it))}-e^{-i(\frac{\pi}{2}(z+it))}\big{)}=i\mathrm{sgn}(t)\tfrac{1}{2}e^{-iz\tfrac{\pi}{2}\mathrm{sgn}(t)+\tfrac{\pi}{2}|t|}\big{(}1-e^{iz\pi\mathrm{sgn}(t)-\pi|t|}\big{)}.

Thus the leading terms in Stirling’s estimates for $\Gamma(w)\sin\left(\tfrac{\pi w}{2}\right)$ and $\Gamma(w)\cos\left(\tfrac{\pi w}{2}\right)$ differ only by a factor of $i\mathrm{sgn}(t)$ , and this extra factor is carried through the entire calculation. In the end we get that the value at $s=0$ of the leading term in the meromorphic continuation of the integral (5.15) equals

(5.16)

\displaystyle i\cdot\frac{-i}{2}\zeta^{2}(\tfrac{1}{2})+(-i)\cdot\frac{i}{2}\zeta^{2}(\tfrac{1}{2})=\zeta^{2}(\tfrac{1}{2}),

which accounts for the main term $q^{\frac{1}{2}}\zeta^{2}(\tfrac{1}{2})$ in (1.3). We give a few words of explanation for (5.16). In the first term of (5.16), the factor $\frac{-i}{2}\zeta^{2}(\tfrac{1}{2})$ corresponds to (5.13), which arises from the stationary point at $t=2\pi nq$ , and hence this term is also multiplied by the factor $i\mathrm{sgn}(t)=i$ . In the second term of (5.16), the factor $\frac{i}{2}\zeta^{2}(\tfrac{1}{2})$ corresponds to (5.14), which arises from the stationary point at $t=-2\pi nq$ , and hence this term is also multiplied by the factor $i\mathrm{sgn}(t)=-i$ .

References

[1] V. Blomer, É. Fouvry, E. Kowalski, P. Michel, and D. Milićević, Some applications of smooth bilinear forms with Kloosterman sums, Tr. Mat. Inst. Steklova 296 (2017), 24–35, English version published in Proc. Steklov Inst. Math. 296 (2017), no. 1, 18–29.
[2] I. S. Gradshteyn and I. M. Ryzhik, Table of integrals, series, and products, eighth ed., Elsevier/Academic Press, Amsterdam, 2015, Translated from the Russian.
[3] D. R. Heath-Brown, An asymptotic series for the mean value of Dirichlet $L$ -functions, Comment. Math. Helv. 56 (1981), no. 1, 148–161.
[4] by same author, The fourth power mean of Dirichlet’s $L$ -functions, Analysis 1 (1981), no. 1, 25–32.
[5] R.Wong, Asymptotic approximations of integrals, Academic Press, New York, 1989, Reprinted by SIAM,. Philadelphia, 2001.
[6] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., The Clarendon Press, Oxford University Press, New York, 1986, Edited and with a preface by D. R. Heath-Brown.
[7] M. P. Young, The fourth moment of Dirichlet $L$ -functions, Ann. of Math. (2) 173 (2011), no. 1, 1–50.

On moments of LL-functions over Dirichlet characters

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.

Theorem 2.

2. sketch

3. Preliminaries

Lemma 3 (Functional equation).

Lemma 4 (An orthogonality relation).

Proof.

Corollary 5.

Proof.

Lemma 6 (Mellin inversion).

Lemma 7 (An integral representation).

Proof.

Lemma 8 (A residue calculation).

Proof.

Lemma 9 (The convexity bound for the Riemann Zeta function).

Lemma 10 (Stirling’s expansion).

Proof.

4. Obtaining the leading main term in Theorem 1

5. The shifted integral

Lemma 11.

Proof.

References

On moments of $L$ -functions over Dirichlet characters