2 Preliminaries and notation

Throughout the paper, $\varepsilon $ will denote a sufficiently small positive number, different at each occurrence, and all implied constants are allowed to depend on $\varepsilon $ .

We follow the notation of [Reference BlomerBlo]. For integers $m,n$ , we denote by $\chi _m(n)$ the Kronecker symbol

$$ \begin{align*}\chi_m(n)=\left(\frac{m}{n}\right).\end{align*} $$

Assume that m is odd and write it as $m=m_0m_1^2$ with $m_0$ squarefree. Then $\chi _m$ is a character of conductor $|m_0|$ if $m\equiv 1\mod 4$ and $|4m_0|$ if $m\equiv 3\mod 4.$ We denote by $\psi _1,\psi _{-1},\psi _2,\psi _{-2}$ the four Dirichlet characters modulo 8 given by the Kronecker symbol $\psi _j(n)=\left (\frac {j}{n}\right ).$ We also let

$$ \begin{align*}\tilde\chi_m=\begin{cases}\chi_m,&{\text{ if }m\equiv1\ \ \mod 4,}\\\chi_{-m},&{\text{ if }m\equiv3\ \ \mod 4.}\end{cases}\end{align*} $$

With this notation, quadratic reciprocity tells us that for odd positive integers $m,n$ ,

(2.1) $$ \begin{align} \chi_m(n)=\tilde\chi_n(m). \end{align} $$

The fundamental discriminants m correspond to primitive real characters of conductor $|m|$ . In such cases, the completed L-function is

$$ \begin{align*}\Lambda(s,\chi_{m})=\left(\frac{|m|}{\pi}\right) ^{\frac{s+a}2}\Gamma\left(\frac{s+a}{2}\right)L(s,\chi_{m}),\end{align*} $$

where $a=0$ or $1$ depending on whether the character is even or odd, i.e., whether $\chi _m(-1)=1$ or $-1$ , and we have the functional equation

(2.2) $$ \begin{align} \Lambda(s,\chi_m)=\Lambda(1-s,\chi_m). \end{align} $$

All primitive real characters can be uniquely written as $\chi _{m_0}\psi _j$ for some positive odd squarefree integer $m_0$ and $j\in \{\pm 1,\pm 2\}$ .

If m is not a fundamental discriminant, then $\chi _m$ is a character of conductor $m_0\mid 4m$ , and we have

(2.3) $$ \begin{align} L(s,\chi_m)=L(s,\chi_{m_0})\cdot\prod_{p\mid\frac{|m|}{m_0}}\left(1-\frac{\chi_{m_0}(p)}{p^s}\right). \end{align} $$

A subscript $2$ of an L-function means that the Euler factor at $2$ is removed, so, in particular,

(2.4) $$ \begin{align} L_2(s,\chi)=\sum_{n\mathrm{\ odd}}\frac{\chi(n)}{n^s}. \end{align} $$

We now record two estimates that will be used later.

The first estimate holds for any s with $\mathrm {Re}(s)\geq 1/2$ :

(2.5) $$ \begin{align} \sum_{\substack{m\leq X,\\m\mathrm{\ odd}}}|L_2(s,\chi_m\psi_j)|\ll_{\varepsilon} X^{1+\varepsilon}|s|^{\frac14+\varepsilon}. \end{align} $$

It follows after applying Hölder’s inequality on the bound for the fourth moment, proved by Heath-Brown [Reference Heath-BrownHea, Theorem 2].

The second is conditional under RH, and it says that for any fixed $\sigma>1/2$ , we have

(2.6) $$ \begin{align} \left|\frac1{\zeta(\sigma+it)}\right|\ll_\varepsilon(1+|t|)^\varepsilon. \end{align} $$

It follows from [Reference Carneiro and ChandeeCC, Theorem 2].

For a function $f(x)$ , we denote by $\hat f(s)$ its Mellin transform, which is defined as

(2.7) $$ \begin{align} \hat f(s)=\int_{0}^\infty f(x)x^{s-1}dx, \end{align} $$

when the integral converges. If $\hat f$ is analytic in the strip $a<\mathrm {Re}(s)<b$ , then the inverse Mellin transform is given by

(2.8) $$ \begin{align} f(x)=\frac1{2\pi i}\int\limits_{(c)}x^{-s}\hat f(s)ds, \end{align} $$

where the integral is over the vertical line $\mathrm {Re}(s)=c$ , and $a<c<b$ is arbitrary.

We will use the following estimate for the Gamma function, which is a consequence of Stirling’s formula: for a fixed $\sigma \in \mathbb {R}$ and $|t|\geq 1$ , we have

(2.9) $$ \begin{align} |\Gamma(\sigma+it)|\asymp e^{-|t|\frac\pi2}|t|^{\sigma-1/2}. \end{align} $$

We will also use the formula

(2.10) $$ \begin{align} \frac{\Gamma\left(\frac{1-s}{2}\right)}{\Gamma\left(\frac{s}{2}\right)}=\frac{2^s\sin (\pi s/2)\Gamma(1-s)}{\sqrt\pi}. \end{align} $$

We write the functional equation for the Riemann zeta function as

(2.11) $$ \begin{align} \zeta(s)=\chi(s)\zeta(1-s), \end{align} $$

where

(2.12) $$ \begin{align} |\chi(\sigma+it)|\ll_{\sigma}(1+|t|)^{1/2-\sigma}. \end{align} $$

We will also use the estimate

(2.13) $$ \begin{align} \int_{-T}^T|\zeta(\sigma+it)|^2dt\ll T^{1+\varepsilon}, \end{align} $$

which is true for any $\sigma \geq 1/2.$

3 Outline of the proof and double Dirichlet series

Applying Mellin inversion to $S(X,Y;\varphi ,\psi )$ twice, we obtain

(3.1) $$ \begin{align} S(X,Y;\varphi,\psi)=\left(\frac{1}{2\pi i}\right)^2\int\limits_{(\sigma)}\int\limits_{(\omega)}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds, \end{align} $$

where for $\mathrm {Re}(s)=\sigma $ and $\mathrm {Re}(w)=\omega $ large enough, we have the absolutely convergent double Dirichlet series

(3.2) $$ \begin{align} A(s,w)=\sum_{m\mathrm{\ odd}}\sum_{n\mathrm{\ odd}}\frac{\left(\frac{m}{n}\right)}{m^wn^s}=\sum_{m\mathrm{\ odd}}\frac{L_2(s,\chi_m)}{m^w}. \end{align} $$

We use the results of Blomer to meromorphically continue $A(s,w)$ to the whole $\mathbb {C}^2$ , shift the two integrals to the left and compute the contribution of the crossed polar lines.

We now cite and sketch the proof of Lemma 2 in [Reference BlomerBlo]. For two characters $\psi ,\psi '$ of conductor dividing $8$ , we define

(3.3) $$ \begin{align} Z(s,w;\psi,\psi'):=\zeta_2(2s+2w-1)\sum_{m,n\mathrm{\ odd}}\frac{\chi_m(n)\psi(n)\psi'(m)}{m^wn^s}, \end{align} $$

which converges absolutely if $\mathrm {Re}(s)$ and $\mathrm {Re}(w)$ are large enough, and we let

(3.4) $$ \begin{align} Z(s,w):=Z(s,w;\psi_1,\psi_1)=\zeta_2(2s+2w-1)A(s,w). \end{align} $$

We also denote

$$ \begin{align*}\mathbf{Z}(s,w;\psi)=\left( \begin{array}{c} Z(s,w;\psi,\psi_1)\\ Z(s,w;\psi,\psi_{-1})\\ Z(s,w;\psi,\psi_{2})\\ Z(s,w;\psi,\psi_{-2})\\ \end{array} \right),\hspace{10pt} \mathbf{Z}(s,w)=\left( \begin{array}{c} \mathbf{Z}(s,w,\psi_1)\\ \mathbf{Z}(s,w,\psi_{-1})\\ \mathbf{Z}(s,w,\psi_{2})\\ \mathbf{Z}(s,w,\psi_{-2})\\ \end{array} \right).\end{align*} $$

Proof sketch

We write the Dirichlet series for $Z(s,w;\psi ,\psi ')$ in two ways.

First, writing $m=m_0m_1^2$ with $\mu ^2(m_0)=1$ , we have

(3.5) $$ \begin{align} \begin{aligned} Z(s,w;\psi,\psi')&=\zeta_2(2s+2w-1)\sum_{\substack{m_0\mathrm{\ odd},\\ \mu^2(m_0)=1}}\frac{L_2(s,\chi_{m_0}\psi)\psi'(m_0)}{m_0^w}\\[5pt]&\hspace{30pt}\times\sum_{m_1\mathrm{\ odd}}\frac1{m_1^{2w}}\prod_{p|m_1}\left(1-\frac{\chi_{m_0}\psi(p)}{p^s}\right)\\&=\zeta_2(2s+2w-1)\sum_{\substack{m_0\mathrm{\ odd},\\\mu^2(m_0)=1}}\frac{L_2(s,\chi_{m_0}\psi)\psi'(m_0)\zeta_2(2w)}{m_0^wL_2(s+2w,\chi_{m_0}\psi)}. \end{aligned} \end{align} $$

If $\psi $ is nontrivial, the right-hand side converges absolutely in the region

$$ \begin{align*}\{(s,w):\mathrm{Re}(w)>1\text{ and }\mathrm{Re}(s+w)>3/2\},\end{align*} $$

the second condition comes from using the functional equation in the numerator when $\mathrm {Re}(s)<1/2$ . When $\psi $ is the trivial character, the summand corresponding to $m_0=1$ is $\frac {\zeta _2(s)\zeta _2(2w)}{\zeta _2(s+2w)}$ , so there is a pole at $s=1$ with residue $\zeta _2(2w)/2.$ Note that the other potential polar lines coming from $\frac {\zeta _2(2w)}{L_2(s+2w,\chi _{d_0}\psi )}$ are outside of the considered region.

The second way to write $Z(s,w;\psi ,\psi ')$ is by exchanging summations and using the quadratic reciprocity. We obtain

(3.6) $$ \begin{align}\begin{aligned} Z(s,w;\psi,\psi')&=\zeta_2(2s+2w-1)\sum_{m,n\mathrm{\ odd}}\frac{\chi_m(n)\psi(n)\psi'(m)}{m^wn^s}\\&=\zeta_2(2s+2w-1)\sum_{n\mathrm{\ odd}}\frac{L_2(w,\tilde\chi_n\psi')\psi(n)}{n^s}. \end{aligned} \end{align} $$

We can again write $n=n_0n_1^2$ with $\mu ^2(n_0)=1$ and obtain a series that is absolutely convergent in the region

$$ \begin{align*}\{(s,w):\mathrm{Re}(s)>1\text{ and }\mathrm{Re}(s+w)>3/2\},\end{align*} $$

unless $\psi '$ is the trivial character, in which case there is a pole at $w=1$ coming from the summands when n is a square, and the residue is $\zeta _2(2s)/2.$

Note that (3.6) gives a link between $Z(s,w;\psi ,\psi ')$ and $Z(w,s;\psi ',\psi )$ , which gives us a functional equation relating $\mathbf {Z}(s,w)$ with $\mathbf {Z}(w,s)$ . To finish the proof and obtain the meromorphic continuation to the whole $\mathbb {C}^2$ , we use the functional equation in the numerator of (3.5), which gives a functional equation relating $\mathbf {Z}(s,w)$ and $\mathbf {Z}(1-s,s+w-1/2)$ , where the change in the second coordinate comes from the conductor in the functional equation for $L(s,\chi _m)$ . Notice that this change of variables interchanges $2s+2w-1$ and $2w$ , leaves $s+2w$ fixed, and maps the line $w=1$ to $s+w=3/2$ , which becomes a new polar line.

We can iterate the two transformations coming from (3.5) and (3.6) and obtain a function meromorphic on a tube region of the form $\{(s,w):\mathrm {Re}(s)^2+\mathrm {Re}(w)^2>c\}$ for some c. During this process, we obtain some additional potential polar lines, but these will be canceled by the gamma factors coming from the functional equations. To obtain a continuation to the region $\{(s,w):\mathrm {Re}(s)^2+\mathrm {Re}(w)^2\leq c\}$ , we use Bochner’s Tube theorem from multivariable complex analysis, which states that a function that is holomorphic on a tube region can be continued to its convex hull (see [Reference BochnerBoc]).

The proof that the function is polynomially bounded in vertical strips is similar to the proof of Proposition 4.11 in [Reference Diaconu, Goldfeld and HoffsteinDGH].

We will also use the following estimate, which is Theorem 2 in [Reference BlomerBlo].

Theorem 3.2 For any $Y_1,Y_2\geq 1$ and characters $\psi ,\psi '$ modulo 8, we have

(3.7) $$ \begin{align} \int_{-Y_1}^{Y_1}\int_{-Y_2}^{Y_2}|Z(1/2+it,1/2+iu;\psi,\psi')|^2dudt\ll(Y_1Y_2)^{1+\varepsilon}. \end{align} $$

In the next two sections, we are going to shift the two integrals in (3.1) to the left and compute the contribution of the crossed polar lines. By Theorem 3.1, the polar lines of $A(s,w)=\frac {Z(s,w)}{\zeta _2(2s+2w-1)}$ are the following:

• • The polar lines of $Z(s,w)$ , which give us the main term in Theorem 1.2:

  1. – the line $s=1$ with residue $\mathrm {res}_{(1,w)}A(s,w)=\frac {\zeta _2(2w)}{2\zeta _2(2w+1)}$ ,

  2. – the line $w=1$ with residue $\mathrm {res}_{(s,1)}A(s,w)=\frac {\zeta _2(2s)}{2\zeta _2(2s+1)}$ ,

  3. – the line $s+w=3/2$ , whose residue will be computed in Lemma 5.1.

• • Zeros of $\zeta _2(2s+2w-1)=\zeta (2s+2w-1)\left (1-2^{1-2s-2w}\right )$ , which are the lines $s+w=\frac {\rho +1}{2},$ where $\rho $ is such that $\zeta (\rho )=0$ , or $s+w=\frac {k\pi i}{\log 2}+\frac 12$ for some $k\in \mathbb {Z}$ . All these satisfy $\mathrm {Re}(s+w)<1$ , and even $\mathrm {Re}(s+w)\leq \frac 34$ if we assume RH.

We will see that the main term comes from the polar lines of $Z(s,w)$ , whereas the polar lines coming from the zeros of $\zeta _2(2s+2w-1)$ determine how far to the left we will be able to shift the integrals, so they give us our error term.

4 Contribution of the polar lines $s=1$ and $w=1$

In this section, we shift the integrals to the left and compute the contribution of the polar lines $s=1$ and $w=1$ . We begin with $\sigma =2$ and $\omega =2$ in (3.1), where everything converges absolutely.

Then we move the inner integral to the line $\mathrm {Re}(w)=3/4+\varepsilon $ , so we obtain

(4.1) $$ \begin{align}\begin{aligned} S(X,Y;\varphi,\psi)=\left( \frac1{2\pi i}\right)^2&\int\limits_{(2)}\int\limits_{(3/4+\varepsilon)}A(s,w)X^wY^s\hat \varphi(w)\hat\psi(s)dwds\\&+\frac1{2\pi i}\int\limits_{(2)}XY^s\hat\varphi(1)\hat\psi(s)\mathrm{res}_{(s,1)}A(s,w)ds. \end{aligned}\end{align} $$

This shift of integrals is justified by the fast decay of the Mellin transform and polynomial boundedness of $A(s,w)$ in vertical strips.

Now we compute the second integral in (4.1), which equals

(4.2) $$ \begin{align} \frac{\hat\varphi(1)X}{2\pi i}\int\limits_{(2)}\frac{Y^s\hat\psi(s)\zeta_2(2s)}{2\zeta_2(2s+1)}ds. \end{align} $$

We again estimate this integral using the residue theorem. The integrand has the following poles:

• • At $s=1/2$ with residue

  $$ \begin{align*}\frac{Y^{1/2}\hat\psi\left(\frac{1}{2}\right)}{8\zeta_2(2)}=\frac{Y^{1/2}\hat\psi\left(\frac{1}{2}\right)}{\pi^2}.\end{align*} $$

• • Zeros of

  $$ \begin{align*}\zeta_2(2s+1)=\left(1-\frac{1}{2^{2s+1}}\right)\zeta(2s+1).\end{align*} $$
  These are at the points $s=\frac {\rho -1}{2},$ where $\zeta (\rho )=0$ , and
  $$ \begin{align*}s=\frac{k\pi i}{\log 2}-\frac12,\ k\in\mathbb{Z}.\end{align*} $$
  These poles have $\mathrm {Re}(s)<0$ and if we assume RH, they all have $\mathrm {Re}(s)\leq -1/4$ .

Therefore, we have the following:

(4.3) $$ \begin{align} \begin{aligned} \frac{\hat\varphi(1)X}{2\pi i}&\int\limits_{(2)}\frac{Y^s\hat\psi(s)\zeta_2(2s)}{2\zeta_2(2s+1)}ds=\frac{\hat\varphi(1)\hat\psi\left(\frac{1}{2}\right)XY^{1/2}}{\pi^2}+\frac{\hat\varphi(1)X}{2\pi i}\int\limits_{(\delta)}\frac{Y^s\hat\psi(s)\zeta_2(2s)}{2\zeta_2(2s+1)}ds. \end{aligned} \end{align} $$

Depending whether we assume RH or not, we take $\delta =-\frac 14+\varepsilon $ or $\delta =\varepsilon $ , bound the integral trivially (we use (2.6) when $\delta =-1/4+\varepsilon $ ) and get

(4.4) $$ \begin{align} \frac{\hat\varphi(1)X}{2\pi i}\int\limits_{(2)}\frac{Y^s\hat\psi(s)\zeta_2(2s)}{2\zeta_2(2s+1)}ds=\frac{\hat\varphi(1)\hat\psi\left(\frac{1}{2}\right)XY^{1/2}}{\pi^2}+O\left( XY^{\delta}\right). \end{align} $$

Using this in (4.1), we obtain

(4.5) $$ \begin{align} \begin{aligned} S(X,Y;\varphi,\psi)=&\frac{\hat\varphi(1)\hat\psi\left(\frac{1}{2}\right)XY^{1/2}}{\pi^2}\\&+\left(\frac{1}{2\pi i}\right)^2\int\limits_{(2)}\int\limits_{(3/4+\varepsilon)}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds+O(XY^{\delta}), \end{aligned} \end{align} $$

Note that when $\varphi =\psi =1_{[0,1]}$ , the first term is $\frac {2}{\pi ^{2}}XY^{1/2}$ and corresponds to the contribution when n is a square.

Next, we exchange the integrals and shift the integral over $\mathrm {Re}(s)=2$ to $\mathrm {Re}(s)=3/4$ , crossing the polar line at $s=1$ . The computation of the residues coming from this polar line is completely analogous to the previous case, and the result is stated in the following theorem.

Theorem 4.1 Let $\varepsilon>0$ . Then we have

(4.6) $$ \begin{align}\begin{aligned} \sum_{m\mathrm{\ odd}}\sum_{n\mathrm{\ odd}}\left(\frac{m}{n}\right)\varphi\left(\frac{m}{X}\right)\psi\left(\frac{n}{Y}\right)&=\frac{\hat\varphi(1)\hat\psi\left(\frac{1}{2}\right)XY^{1/2}+\hat\psi(1)\hat\varphi\left(\frac{1}{2}\right)YX^{1/2}}{\pi^2}\\&\hspace{-100pt}+\left(\frac{1}{2\pi i}\right)^2\int\limits_{(3/4)}\int\limits_{(3/4+\varepsilon)}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds+O_\varepsilon\left( YX^\delta+XY^\delta\right), \end{aligned}\end{align} $$

where $\delta =\varepsilon $ . If we assume the Riemann Hypothesis, then we can take $\delta =-1/4+\varepsilon $ .

5 Contribution of the polar line $s+w=3/2$

Before further shifting the integrals, we need to compute the residues on the polar line $s+w=3/2$ , which is done in the following lemma.

Lemma 5.1 For all $s\in \mathbb {C}$ ,

(5.1) $$ \begin{align} \mathrm{res}_{\left( s,\frac32-s\right)}Z(s,w)=\frac{\sqrt\pi\sin\left(\frac{\pi s}{2}\right)\Gamma\left( s-\frac12\right)\zeta(2s-1)}{2(2\pi)^s}. \end{align} $$

Proof We use the functional equation (28) in [Reference BlomerBlo], from which it follows that

(5.2) $$ \begin{align}\begin{aligned} &Z(1-u,u+v-1/2)=\frac{\pi^{-u+\frac12}\Gamma\left(\frac u2\right)}{\left( 4^{1-u}-4\right)\Gamma\left(\frac{1-u}2\right)}\cdot\Bigg(-4^uZ(u,v;\psi_1,\psi_1)\\&+\left( 4^u-2\right) Z(u,v;\psi_1,\psi_{-1})+\left( 2^u-2^{1-u}\right) \left( Z(u,v;\psi_1,\psi_2)+ Z(u,v;\psi_1,\psi_{-2})\right)\Bigg). \end{aligned}\end{align} $$

Under the change of variables $(s,w)=(1-u,u+v-1/2)$ , the line $v=1$ transforms to the line $s+w=3/2$ . Since $v=1$ is a polar line of $Z(u,v;\psi ,\psi ')$ if and only if $\psi '=\psi _1$ , the residue comes only from the first term in the parenthesis on the right-hand side of (5.2), and is given by

$$ \begin{align*}\mathrm{res}_{\left(1-u,u+\frac12\right)}Z(u,v;\psi_1,\psi_1)=\frac{\pi^{-u+\frac12}\Gamma\left(\frac{u}{2}\right)\left(-4^u\right)\zeta_2(2u)}{2\left(4^{1-u}-4\right)\Gamma\left(\frac{1-u}{2}\right)}=\frac{\pi^{-u+\frac12}\Gamma\left(\frac{u}{2}\right)\zeta(2u)}{2\cdot4^{1-u}\Gamma\left(\frac{1-u}{2}\right)},\end{align*} $$

so we have

$$ \begin{align*}\begin{aligned} \mathrm{res}_{\left( s,\frac32-s\right)}Z(s,w)&=\frac{\pi^{s-\frac12}\Gamma\left(\frac {1-s}2\right)\zeta(2-2s)}{2\cdot4^s\Gamma\left(\frac{s}{2}\right)}=\frac{\sqrt\pi\sin\left(\frac{\pi s}{2}\right)\Gamma\left( s-\frac12\right)\zeta(2s-1)}{2(2\pi)^s}, \end{aligned}\end{align*} $$

where the last equality follows after using the functional equation for $\zeta (s)$ and the formula (2.10).

We are now ready to prove Theorem 1.1.

Proof of Theorem 1.1

We move the integral from (4.6) further to the left. According to the discussion at the end of Section 3, we know that except the line $s+w=3/2$ , the integrand has no poles with $\mathrm {Re}(s+w)\geq 1$ , or with $\mathrm {Re}(s+w)>3/4$ if we assume RH. Hence we have

(5.3) $$ \begin{align}\begin{aligned} \left(\frac{1}{2\pi i}\right)^2&\int\limits_{(3/4)}\int\limits_{(3/4+\varepsilon)}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds\\[5pt]&= \left(\frac{1}{2\pi i}\right)^2\int\limits_{(3/4)}\int\limits_{(\delta')}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds\\[5pt]&\quad+\frac1{2\pi i}\int\limits_{(3/4)}X^{\frac32-s}Y^{s}\hat\varphi\left(\frac32-s\right)\hat\psi\left( s\right)\mathrm{res}_{\left( s,\frac32-s\right)}A(s,w)ds, \end{aligned}\end{align} $$

where $\delta '=1/4+\varepsilon $ , or $\varepsilon $ under RH. Using Lemma 5.1 and (3.4), we have

$$ \begin{align*}\mathrm{res}_{\left( s,\frac32-s\right)}A(s,w)=\frac{\mathrm{res}_{\left( s,\frac32-s\right)}Z(s,w)}{\zeta_2(2)}=\frac{\sqrt\pi\sin\left(\frac{\pi s}{2}\right)\Gamma\left( s-\frac12\right)\zeta(2s-1)}{2\zeta_2(2)(2\pi)^s}.\end{align*} $$

Therefore, the second integral on the right-hand side of (5.3) is

(5.4) $$ \begin{align} \begin{aligned} \frac{2X^{\frac 32}}{i\pi^{\frac52}}\int\limits_{(3/4)}\left(\frac{Y}{2\pi X}\right)^s\hat\varphi\left(\frac32-s\right)\hat\psi(s)\Gamma\left( s-\frac12\right)\sin\left(\frac{\pi s}{2}\right)\zeta(2s-1)ds. \end{aligned} \end{align} $$

Hence, we have

(5.5) $$ \begin{align} \begin{aligned} S(X,Y;\varphi,\psi)&=\frac{\hat\varphi(1)\hat\psi\left(\frac{1}{2}\right)XY^{1/2}+\hat\psi(1)\hat\varphi\left(\frac{1}{2}\right)YX^{1/2}}{\pi^2}\\&\quad+\frac{2X^{\frac32}}{i\pi^{\frac52}}\int\limits_{(3/4)}\left(\frac{Y}{2\pi X}\right)^s\hat\varphi\left(\frac32-s\right)\hat\psi(s)\Gamma\left( s-\frac12\right)\sin\left(\frac{\pi s}{2}\right)\zeta(2s-1)ds\\&\quad+\left(\frac{1}{2\pi i}\right)^2\int\limits_{(3/4)}\int\limits_{(\delta')}A(s,w)X^wY^s\hat\varphi(w)\hat\psi(s)dwds+O\left( XY^\delta+YX^\delta\right)\\[5pt]&=\frac2{\pi^2}\cdot X^{3/2}\cdot D\left(\frac YX;\varphi,\psi\right)+O\left( XY^\delta+YX^\delta\right), \end{aligned} \end{align} $$

where the last equality follows after trivially bounding the second integral, and using $X^{\delta '}Y^{\frac 34}\ll XY^\delta +YX^\delta $ .

6 Removing the smooth weights

In this section, we show how to remove the smooth weights from Theorem 1.1 and prove Theorem 1.2. We choose the weights $\varphi =\psi $ to be a smooth function which is $1$ on the interval $ \left [\frac 1U,1-\frac 1U\right ]$ for some U to be chosen later, and 0 outside of $(0,1)$ , and which satisfies

(6.1) $$ \begin{align} |\hat\varphi(\sigma+it)|\ll_{j,\sigma}\frac{U^{j-1}}{1+|t|^j} \end{align} $$

for all $j\geq 1$ . Then using the Pólya–Vinogradov inequality (see (3.1) in [Reference Chinta, Friedberg and HoffsteinCFS]), we have

(6.2) $$ \begin{align} |S(X,Y)-S(X,Y;\varphi,\psi)|\ll\frac{X^{3/2}+Y^{3/2}}{U}\log(XY). \end{align} $$

Now we need to estimate the dependence on U of the error term in the computation of $S(X,Y;\varphi ,\psi )$ . These come from the following:

• • The error from the polar lines $s=1$ and $w=1$ .

• • The error from the shifted integral.

• • The difference of the main terms.

The error from the polar lines $s=1$ is

(6.3) $$ \begin{align} \ll XY^{\sigma}\left|\int_{(\delta)}\frac{\hat\psi(s)\zeta_2(2s)}{\zeta_2(2s+1)}ds\right|, \end{align} $$

where $\delta =\varepsilon $ , or $-1/4+\varepsilon $ , and similarly for the error from the polar line $w=1$ . We have $\frac {1}{\zeta _2(2\delta +2it+1)}\ll 1$ in both cases by (2.6).

We can shift the integral (6.3) to any vertical line $\mathrm {Re}(s)=\sigma $ with $1/4\geq \sigma \geq \delta $ , and bound the integral using the Cauchy–Schwarz inequality as

(6.4) $$ \begin{align} \begin{aligned} &\ll\int_{(\sigma)}|\hat\psi(s)\zeta(2s)|ds\\ &\ll\left(\int_{(\sigma)}\frac{|\zeta(1-2s)|^2}{(1+|s|)^{1+\varepsilon}}ds\right)^{\frac12}\left(\int_{(\sigma)}|\hat\psi(s)\chi(2s)|^2(1+|s|)^{1+\varepsilon}ds\right)^{\frac12}\\ &\ll U^{j-1}\int_{(\sigma)}(1+|s|)^{2-4\sigma-2j+\varepsilon}ds\\ &\ll U^{1/2-2\sigma+\varepsilon}, \end{aligned} \end{align} $$

where the first integral on the second line converges by (2.13), and we took $j=3/2-2\sigma +\varepsilon .$

A similar computation for the polar line $w=1$ gives the same result with $X,Y$ interchanged, so the error from these terms is

(6.5) $$ \begin{align} (XY^\sigma+YX^{\sigma})U^{1/2-2\sigma+\varepsilon}. \end{align} $$

For the error coming from the shifted integral, we need to estimate

(6.6) $$ \begin{align} \left|\int_{(\sigma)}\int_{(\omega)}A(s,w)\hat\varphi(w)\hat\psi(s)X^wY^sdwds\right|. \end{align} $$

We have

(6.7) $$ \begin{align} A(s,w)=\frac{Z(s,w)}{\zeta_2(2s+2w-1)}, \end{align} $$

so the integral is

(6.8) $$ \begin{align} \ll X^\omega Y^\sigma\int_{(\sigma)}\int_{(\omega)}\left| Z(s,w)\hat\varphi(w)\hat\psi(s)\right| dwds \end{align} $$

provided $\sigma +\omega>1+\varepsilon $ or $>3/4+\varepsilon $ if we assume RH. We take $\sigma =\omega =1/2+\varepsilon $ and estimate the double integral using the Cauchy–Schwarz inequality as follows:

(6.9) $$ \begin{align} &\int_{(\frac12+\varepsilon)}\int_{(\frac12+\varepsilon)}\left| Z(s,w)\hat\varphi(w)\hat\psi(s)\right| dwds\nonumber\\ &\ll\left(\,\int\limits_{(\frac12+\varepsilon)}\int\limits_{(\frac12+\varepsilon)}\frac{\left| Z(s,w)\right|{}^2}{(1+|s|)^{1+\varepsilon}(1+|w|)^{1+\varepsilon}}dwds\right)^{\frac12}\nonumber \\&\quad\times\left(\,\int\limits_{(\frac12+\varepsilon)}\int\limits_{(\frac12+\varepsilon)}\left|\hat\varphi(w)\hat\psi(s)\right|{}^2(1+|s|)^{1+\varepsilon}(1+|w|)^{1+\varepsilon} dwds\right)^{\frac12}\\ &\ll U^{j_1+j_2-2}\left(\,\int\limits_{(\frac12+\varepsilon)}\int\limits_{(\frac12+\varepsilon)}(1+|s|)^{1+\varepsilon-2j_1}(1+|w|)^{1+\varepsilon-2j_2} dwds\right)^{1/2}\nonumber\\ &\ll U^{\varepsilon},\nonumber \end{align} $$

where the integral on the second line converges by (3.7), and we took $j_1=j_2=1+\varepsilon .$ It follows that the error from the shifted integral is

(6.10) $$ \begin{align} \ll (XY)^{1/2+\varepsilon}U^{\varepsilon}. \end{align} $$

The error from the difference of the main terms is

(6.11) $$ \begin{align} \begin{aligned} E(X,Y;\varphi,\psi)&=\frac{2}{\pi^2}X^{3/2}\left( D\left(\frac YX;\varphi,\psi\right)-D\left(\frac{Y}{X}\right)\right) \\\ &\hspace{-50pt}\ll XY^{1/2}\left| \hat\varphi(1)\hat\psi(1/2)-2\right|+YX^{1/2}\left|\hat \psi(1)\hat\varphi(1/2)-2\right|\\ &\hspace{-38pt}+X^{3/2}\int_{(3/4)}\left|\left(\frac{Y}{2\pi X}\right)^s\Gamma\left( s-\frac12\right)\sin\left(\frac{\pi s}{2}\right)\zeta(2s-1)\right|\\ &\hspace{-38pt}\times\left|\hat\varphi(3/2-s)\hat\psi(s)-\frac1{s(3/2-s)}\right| ds\\ &\hspace{-50pt}\ll XY^{1/2}\left| \hat\varphi(1)\hat\psi(1/2)-2\right|+YX^{1/2}\left|\hat \psi(1)\hat\varphi(1/2)-2\right|\\ &\hspace{-38pt}+(XY)^{3/4}\int_{(3/4)}\left(1+|t|\right)^{-\frac14}\left|\zeta(2s-1)\right|\left|\hat\varphi\left(\frac32-s\right)\hat\psi(s)-\frac1{s\left(\frac32-s\right)}\right| ds. \end{aligned} \end{align} $$

We estimate the Mellin transforms in the following lemma.

Lemma 6.1 For $s=\sigma +it, 0<\sigma <1$ and $\varphi ,\psi $ as above, we have

(6.12) $$ \begin{align} \hat\varphi(1)\hat\psi(1/2)=2+O\left( U^{-1/2}\right), \end{align} $$

and

(6.13) $$ \begin{align} \left|\hat\varphi(3/2-s)\hat\psi(s)-\frac1{s(3/2-s)}\right|\ll\frac{\left|\hat\varphi(3/2-s)\right|}{U^\sigma}+\frac{1}{|s|U^{3/2-\sigma}}. \end{align} $$

Proof For (6.12), we have

(6.14) $$ \begin{align} &\hat\varphi(1)\hat\psi(1/2)=\int_{0}^\infty \varphi(u)du\int_0^\infty\psi(v)v^{-1/2}dv\nonumber\\ &=\left(\left(\int\limits_{0}^{\frac 1U}+\int\limits_{1-\frac 1U}^{1}\right)\varphi(u)du+1-\frac2U\right)\left(\left(\int\limits_{0}^{\frac 1U}+\int\limits_{1-\frac1U}^{1}\right)\frac{\psi(v)}{\sqrt v}dv+\int\limits_{\frac1U}^{1-\frac1U}\frac{1}{\sqrt v}dv\right)\\ &=\left(1+O\left(\frac 1U\right)\right)\left( O\left(\frac1{\sqrt U}\right)+2\left(\sqrt{1-\frac{1}{U}}-\frac1{\sqrt U}\right)\right)\nonumber\\ &= 2+O\left( U^{-1/2}\right).\nonumber \end{align} $$

For (6.13), we first use the triangle inequality to get

(6.15) $$ \begin{align} \begin{aligned} &\left|\hat\varphi(3/2-s)\hat\psi(s)-\frac1{s(3/2-s)}\right|\\ &\leq\left|\hat\varphi(3/2-s)\right|\left|\hat\psi(s)-\frac1s\right|+\left|\frac1s\right|\left|\hat\varphi(3/2-s)-\frac{1}{3/2-s}\right|. \end{aligned} \end{align} $$

Now we have

(6.16) $$ \begin{align} \begin{aligned} \left|\hat\psi(s)-\frac1s\right|&\leq\int_{0}^\infty\left|\psi(u)-1\right| u^{\sigma-1}du\\ &=\int_{0}^{\frac1U}\left|\psi(u)-1\right| u^{\sigma-1}du+\int_{1-\frac1U}^{1}\left|\psi(u)-1\right| u^{\sigma-1}du\\ &\ll \frac1{U^\sigma}, \end{aligned} \end{align} $$

and similarly

(6.17) $$ \begin{align} \left|\hat\varphi(3/2-s)-\frac1{3/2-s}\right|\leq\int_{0}^\infty\left|\varphi(u)-1\right| u^{1/2-\sigma}du\ll\frac{1}{U^{3/2-\sigma}}.\\[-47pt]\nonumber \end{align} $$

Using this Lemma in (6.11), we get

(6.18) $$ \begin{align} E(X,Y;\varphi,\psi)\ll\frac{XY^{1/2}+YX^{1/2}}{\sqrt U}+\left(\frac{XY}{U}\right)^{3/4}. \end{align} $$

Putting everything together, the error in both cases is

(6.19) $$ \begin{align} \frac{X^{\frac32+\varepsilon}+Y^{\frac32+\varepsilon}}{U}+\frac{XY^{\frac12}+YX^{\frac12}}{\sqrt U}+\left(\frac{XY}{U}\right)^{\frac34}+(XY^{\sigma}+YX^{\sigma})U^{\frac12-2\sigma+\varepsilon}+(XY)^{\frac12}U^{\varepsilon}, \end{align} $$

where $1/4\geq \sigma \geq \delta $ . In both cases, the best choice is $\sigma =1/4$ . Then we can take $U=\min \{X,Y\}$ and obtain the error

(6.20) $$ \begin{align} XY^{1/4+\varepsilon}+YX^{1/4+\varepsilon} \end{align} $$

in the range $Y^{2/5}\leq X\leq Y^{5/2}.$ In the remaining range, the result follows from (1.2) and the asymptotic expansion of $C(\alpha )$ (1.5) and (1.6), together with the fact that $C(\alpha )=D(\alpha )$ as will be proved in the next section.

7 Proving that $C(\alpha )=D(\alpha )$

In this section, we show that $C(\alpha )=D(\alpha )$ . Recall that

(7.1) $$ \begin{align} \begin{aligned} D(\alpha)=\sqrt\alpha+\alpha-\frac{2\sqrt\pi}{2\pi i}\int\limits_{(3/4)}\left(\frac{\alpha}{2\pi}\right)^s\cdot\frac{\Gamma\left( s-\frac32\right)\sin\left(\frac{\pi s}{2}\right)\zeta(2s-1)}{s}ds. \end{aligned} \end{align} $$

We shift the integral to the left, capturing the pole at $s=\frac 12$ , which contributes

$$ \begin{align*}2\sqrt\pi\left(\frac{\alpha}{2\pi}\right)^{1/2}\cdot\frac{-\sin(\pi/4)\zeta(0)}{1/2}=\alpha^{1/2}.\end{align*} $$

The horizontal integrals vanish by (2.9) and a convexity estimate for $\zeta (s)$ , so

(7.2) $$ \begin{align} \begin{aligned} D(\alpha)&=\alpha-\frac{2\sqrt\pi}{2\pi i}\int\limits_{(-1/4)}\left(\frac{\alpha}{2\pi}\right)^s\cdot\frac{\Gamma\left( s-\frac32\right)\sin\left(\frac{\pi s}{2}\right)\zeta(2s-1)}{s}ds\\&=\alpha-\frac{2\sqrt\pi}{2\pi i}\int\limits_{(1/4)}\alpha^{-s}\cdot\frac{(2\pi)^s\Gamma\left(-s-\frac32\right)\sin\left(\frac{\pi s}{2}\right)\zeta(-2s-1)}{s}ds. \end{aligned} \end{align} $$

This integral is an inverse Mellin transform, so we rewrite $C(\alpha )$ using Mellin inversion. We have

(7.3) $$ \begin{align} \begin{aligned} C(\alpha)&=\alpha+\alpha^{3/2}\frac{2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2}\int_{0}^{1/\alpha}\sqrt y\sin\left(\frac{\pi k^2}{2y}\right)dy\\&=\alpha+\frac2\pi\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2\alpha}{2u}\right)du\\&=\alpha+\frac2{\pi}\cdot\frac{1}{2\pi i}\int\limits_{(c)}\alpha^{-s}\hat f(s)ds, \end{aligned} \end{align} $$

where

(7.4) $$ \begin{align} f(x)=\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du. \end{align} $$

For $0<\mathrm {Re}(s)<1$ , we have

(7.5) $$ \begin{align} \begin{aligned} \hat f(s)&=\int_0^\infty\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du\ x^{s-1}dx\\&=\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\int_0^\infty \sin\left(\frac{\pi k^2 x}{2u}\right) x^{s-1}dx\ du,\end{aligned} \end{align} $$

which isn’t obvious as the double integral doesn’t converge absolutely, but we will justify the interchange of summation and integrals in Lemma 7.1. We can now make a change of variables $y=\frac {\pi k^2x}{2u}$ and obtain

(7.6) $$ \begin{align} \begin{aligned} \hat f(s)&=\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\int_0^\infty \sin(y) y^{s-1}dy\ \left(\frac{2u}{\pi k^2}\right)^sdu\\&=\left(\frac{2}{\pi}\right)^s\zeta(2+2s)\int_0^1u^{s+1/2}du\int_0^\infty\sin(y)y^{s-1}dy\\&=\frac{2^s\zeta(2+2s)\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}{\pi^s(s+3/2)}, \end{aligned} \end{align} $$

which holds for $0<\mathrm {Re}(s)<1,$ so we can take $c=1/4$ in (7.3). It therefore suffices to show that

(7.7) $$ \begin{align} -2\sqrt\pi\frac{(2\pi)^s\Gamma\left(-s-\frac32\right)\sin\left(\frac{\pi s}{2}\right)\zeta(-2s-1)}{s}=\frac2\pi\cdot\frac{2^s\zeta(2+2s)\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}{\pi^s(s+3/2)}. \end{align} $$

The functional equation for the zeta function gives

(7.8) $$ \begin{align} \zeta(2s+2)=\pi^{2s+3/2}\cdot\frac{\Gamma\left(-s-\frac12\right)}{\Gamma(s+1)}\zeta(-2s-1), \end{align} $$

so using $s\Gamma (s)=\Gamma (s+1)$ and $(-s-3/2)\Gamma (-s-3/2)=\Gamma (-s-1/2)$ gives the result.

It remains to justify the interchange of the order of summation and integrations in (7.5).

Lemma 7.1 If $0<\mathrm {Re}(s)<1$ , it holds that

(7.9) $$ \begin{align} \begin{aligned} &\int_0^\infty\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du\ x^{s-1}dx\\&=\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\int_0^\infty \sin\left(\frac{\pi k^2 x}{2u}\right) x^{s-1}dx\ du. \end{aligned} \end{align} $$

Proof We have

(7.10) $$ \begin{align} \begin{aligned} &\int_0^\infty\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du\ x^{s-1}dx\\&=\lim_{A\rightarrow\infty}\int_{1/A}^A\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du\ x^{s-1}dx. \end{aligned} \end{align} $$

We can now interchange the integrals and summation, because

(7.11) $$ \begin{align} \begin{aligned} \sum_{k=1}^\infty\frac1{k^2}\int_0^1\int_{1/A}^A\left|\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)x^{s-1}\right| dx\ du\ll A, \end{aligned} \end{align} $$

so

(7.12) $$ \begin{align} \begin{aligned} &\lim_{A\rightarrow\infty}\int_{1/A}^A\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\sin\left(\frac{\pi k^2 x}{2u}\right)du\ x^{s-1}dx\\&=\lim_{A\rightarrow\infty}\sum_{k=1}^\infty\frac1{k^2}\int_0^1\sqrt u\int_{1/A}^A\sin\left(\frac{\pi k^2 x}{2u}\right)x^{s-1}dx\ du. \end{aligned} \end{align} $$

To insert the limit inside the sum and integral, we use the Dominated convergence theorem with the bound $\sqrt u|\int _{1/A}^A\sin \left (\frac {\pi k^2 x}{2u}\right )x^{s-1}dx|\leq K\sqrt u$ for an absolute constant K independent of u and A in our range, which holds because $\int _0^\infty \sin (y)y^{s-1}dy$ converges.