A Note on the Mean Square of Riemann Zeta-Function

1. Introduction

Let

$$I={\int }_0^T{\left|\zeta (1/2+it)\right|}^2{\left|A(1/2+it)\right|}^{2}dt$$

where $\zeta \left(s\right)$ is the Riemann zeta-function, and

$$A\left(s\right)=\sum _{1\le m\le M}a\left(m\right)m^{-s}$$

is a finite Dirichlet series. For this type of mean square of $\zeta \left(s\right)$, there are a number of researches, which is related to estimate the number of the zeros of $\zeta \left(s\right)$ on the critical line $Re\left(s\right)=1/2$, see [1,2,3,4,5,6]As before, for two positive integers $h,k$, denote $h^{*}=h/(h,k),k^{*}=k/(h,k)$, and $\overline{h^{*}}$ is the least positive integer such that $\overline{h^{*}}{h}^{*}\equiv 1mod{k}^{*}$. For $s=c+it,c=1+\eta ,0<\eta <1$, denote by

$$\mathcal{M}\left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{d\left(n\right)}{n^s}}\sum _{h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{{\left(hk\right)}^{1-s}}}{(h,k)}^{(1-2s)}e\left({\displaystyle \frac{n\overline{h^{*}}}{k^{*}}}\right)$$

and $V=\underset{\left|t\right|\le M}{sup}\left\{\mathcal{M}(c+it)\right\}$.

Balasubramanian, Conrey, and Heath-Brown proved that

For $a\left(m\right)\ll m^{ϵ}$, and $log\left(m\right)\ll logT$, there is

$$I=T\sum _{h,k\le M}{\displaystyle \frac{a\left(h\right)}{h}}{\displaystyle \frac{\overline{a\left(k\right)}}{k}}(h,k)\left(log{\displaystyle \frac{T{(h,k)}^2}{2\pi hk}}+2\gamma -1\right)+\mathcal{E}.$$

(1.1)

where $\mathcal{E}\ll V{T}^{ϵ}+o\left(T\right)$. In the paper [1], it is applied an auxiliary function $w(t,T_1,T_2)$, which plays an important role of "ferrying", Instead, in this paper, we will make use of a new one, i.e. $\omega (t,T_1,T_2)$ defined in Lemma 2.1. Our result is that

Theorem 1.1.

Suppose that $a\left(m\right)\ll m^{ϵ}$, and $log\left(m\right)\ll logT$, then

$$I=T\sum _{h,k\le M}{\displaystyle \frac{a\left(h\right)}{h}}{\displaystyle \frac{\overline{a\left(k\right)}}{k}}(h,k)\left(log{\displaystyle \frac{T{(h,k)}^2}{2\pi hk}}+2\gamma -1\right)+\tilde{\mathcal{E}}.$$

(1.2)

where $\tilde{\mathcal{E}}\ll V{T}^{-\eta +ϵ}+o\left(T\right)$.

As it is easy to know that $V\ll M^{2+2ϵ+2\eta }$, then with a properly re-define $ϵ,\eta$, it has

$$\tilde{\mathcal{E}},\mathcal{E}\ll M^2{T}^{ϵ}.$$

(1.3)

The main arguments in this paper are most similar to the ones in [1].

2. The Proof of Theorem 1.1

Lemma 2.1.

Suppose that $\lambda =T^{2-ϵ}$, define

$$\omega (t,T_1,T_2)={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_1}^{T_2}\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du.$$

(2.1)

Let $\alpha \ge 10$, denote $\Delta ={(2\alpha \lambda logT)}^{1/2}$, if $t\in [T_1+\Delta ,T_2-\Delta ]$, then

$$\left|\omega (t,T_1,T_2)-1\right|\ll T^{-\alpha }.$$

(2.2)

And if $t\le T_1-\Delta$, or $t\ge T_2+\Delta$, then

$$\left|\omega (t,T_1,T_2)\right|\ll T^{-\alpha }.$$

(2.3)

Proof. 

It is familiar that

$$e^{-\lambda }={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma \left(s\right){\lambda }^{-s}ds.$$

Hence,

$$1-\omega (t,T_1,T_2)=R_1+R_2,$$

where

$$R_1={\displaystyle \frac{e^{\lambda }}{2\pi i}}{\int }_{T_2}^{\infty }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du,$$

$$R_2={\displaystyle \frac{e^{\lambda }}{2\pi i}}{\int }_{-\infty }^{T_1}\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du,$$

By Stirling’s formula, it has

$$\begin{array}{cc}\hfill \mathrm{Re}(log\Gamma (\lambda +(u-t)i\left)\right)=& \left(\lambda -{\displaystyle \frac{1}{2}}\right){\displaystyle \frac{log({\lambda }^2+{(u-t)}^2)}{2}}-\lambda +{\displaystyle \frac{1}{2}}log\left(2\pi \right)\hfill \\ \hfill & -(u-t)arctan\left({\displaystyle \frac{u-t}{\lambda }}\right)+O(1/\lambda )\hfill \end{array}$$

And

$$\begin{array}{cc}\hfill & \mathrm{Re}(log\left(\Gamma (\lambda +(u-t)i)\right))+\mathrm{Re}(log\left({\lambda }^{-(\lambda +(u-t\left)i\right)}\right))+\lambda \hfill \\ \hfill & =-{\displaystyle \frac{{(u-t)}^2}{4{\lambda }^2}}-{\displaystyle \frac{{(u-t)}^2}{2\lambda }}-{\displaystyle \frac{1}{2}}log\lambda +{\displaystyle \frac{1}{2}}log\left(2\pi \right)+O(1/\lambda )\hfill \end{array}$$

Hence, if $t\in [T_1+\Delta ,T_2-\Delta ]$, then

$$\begin{array}{cc}\hfill \left|R_1\right|& \ll {\int }_{T_2}^{\infty }\left|e^{\lambda }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}\right|du\hfill \\ \hfill & \ll {\lambda }^{-1/2}{\int }_{T_2}^{\infty }exp(-{(u-t)}^2/2\lambda )du\hfill \\ \hfill & \ll T^{-\alpha }.\hfill \end{array}$$

and similarly

$$\left|R_2\right|\ll {\lambda }^{-1/2}{\int }_{-\infty }^{T_1}exp(-{(u-t)}^2/2\lambda )du\ll T^{-\alpha }.$$

if $t\le T_1-\Delta$, or $t\ge T_2+\Delta$, then

$$\left|\omega (t,T_1,T_2)\right|\ll {\lambda }^{-1/2}{\int }_{T_1}^{T_2}exp(-{(u-t)}^2/2\lambda )du\ll T^{-\alpha }.$$

□

Lemma 2.2

($\mathrm{Estermann}\left[2\right]$). Let

$$S\left(x,{\displaystyle \frac{h}{k}}\right)=\sum _{n=1}^{\infty }d\left(n\right)e\left({\displaystyle \frac{nh}{k}}\right)e^{2\pi inx}$$

Denote by

$$D\left(s,{\displaystyle \frac{h}{k}}\right)=\sum _{n=1}^{\infty }d\left(n\right)e\left({\displaystyle \frac{nh}{k}}\right)n^{-s}$$

write $z=-2\pi ix$, then for $\mathrm{Im}x>0,\mathrm{Re}s>1,k\ge 1$, then

$$\begin{array}{cc}\hfill S\left(x,{\displaystyle \frac{h}{k}}\right)& ={\displaystyle \frac{1}{z{k}^{*}}}(\gamma -logz-2log{k}^{*})+D\left(0,{\displaystyle \frac{h^{*}}{k^{*}}}\right)\hfill \\ \hfill & -i\underset{\left(c\right)}{\int }{\left(2\pi \right)}^{-s}{\displaystyle \frac{\Gamma \left(s\right)k^{*2s-1}}{sin\pi s}}\left(D\left(s,{\displaystyle \frac{\overline{h^{*}}}{k^{*}}}\right)+\left(cos\pi s\right)D\left(s,-{\displaystyle \frac{\overline{h^{*}}}{k^{*}}}\right)\right)z^{s-1}ds\hfill \end{array}$$

(2.4)

where $1<c<2$, and it has

$$\left|D\left(0,{\displaystyle \frac{h^{*}}{k^{*}}}\right)\right|\le k^{*}{(log2k^{*})}^2$$

As usual, denote

$$\chi (1-s)=2{\left(2\pi \right)}^{-s}\Gamma \left(s\right)cos(\pi s/2).$$

Lemma 2.3.

Suppose that $1<c<2$, let

$$J\left(y\right)={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){\lambda }^{-(s_1-s)}\chi (1-s)y^{-s}ds,$$

then

$$J\left(y\right)={\int }_0^{\infty }{v}^{s_1}{e}^{-\lambda v}(e^{2\pi iyv}+e^{-2\pi iyv}){\displaystyle \frac{dv}{v}}.$$

(2.5)

Proof. 

$$J\left(y\right)={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){\lambda }^{-(s_1-s)}\Gamma \left(s\right)({\left(2\pi iy\right)}^{-s}+{(-2\pi iy)}^{-s})ds=J_1+J_2$$

where

$$\begin{array}{cc}\hfill J_1=& {\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){\lambda }^{-(s_1-s)}\Gamma \left(s\right){\left(2\pi iy\right)}^{-s}ds,\hfill \\ \hfill J_2=& {\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){\lambda }^{-(s_1-s)}\Gamma \left(s\right){(-2\pi iy)}^{-s}ds.\hfill \end{array}$$

By the theory of Mellin Transforms(refer to [7])

$${\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }G\left(s\right)F\left(s\right)x^{-s}ds={\int }_0^{\infty }f\left({\displaystyle \frac{1}{v}}\right)g\left(xv\right){\displaystyle \frac{dv}{v}}$$

where

$$f\left(v\right)={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }F\left(x\right)x^{-s}ds,g\left(v\right)={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }G\left(x\right)x^{-s}ds.$$

We apply this formula to $J_1$ and $J_2$ respectively.

For $J_1$,

$$\begin{array}{cc}\hfill f\left(v\right)=& {\displaystyle \frac{{\lambda }^{-s_1}}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma (s_1-s){(v/\lambda )}^{-s}ds\hfill \\ \hfill =& {\displaystyle \frac{{\lambda }^{-s_1}}{2\pi i}}\underset{\left(\lambda \right)}{\int }\Gamma \left(w\right){(v/\lambda )}^{-(s_1-w)}dw\hfill \\ \hfill =& {\lambda }^{-s_1}{(v/\lambda )}^{-s_1}{e}^{-\lambda /v}=v^{-s_1}{e}^{-\lambda /v},\hfill \end{array}$$

i.e.

$$f(1/v)=v^{s_1}{e}^{-\lambda v}.$$

And

$$g\left(v\right)={\displaystyle \frac{1}{2\pi i}}\underset{\left(c\right)}{\int }\Gamma \left(s\right)v^{-s}ds=e^{-v},$$

i.e.

$$g\left(xv\right)=e^{-xv}.$$

So

$$J_1={\int }_0^{\infty }{v}^{s_1}{e}^{-(\lambda v+2\pi iyv)}{\displaystyle \frac{dv}{v}}$$

Similarly,

$$J_2={\int }_0^{\infty }{v}^{s_1}{e}^{-(\lambda v-2\pi iyv)}{\displaystyle \frac{dv}{v}}$$

□

Denote by $L_{\delta }$ the straight line from 0 to $e^{i\delta }\infty$.

Lemma 2.4.

Let $s_1=\lambda +1/2+iu$, $0\le \delta \le \pi /2$, then

$${\int }_{L_{\delta }}{\displaystyle \frac{v^{s_1}{e}^{-\lambda v}}{v(v-1)}}dv-{\int }_{L_{-\delta }}{\displaystyle \frac{v^{s_1}{e}^{-\lambda v}}{v(v-1)}}dv=-e^{\lambda }2\pi i,$$

(2.6)

And let

$$K={\int }_{L_{\delta }}{\displaystyle \frac{v^{s_1}{e}^{-\lambda v}log(-i(v-1))}{v(v-1)}}dv-{\int }_{L_{-\delta }}{\displaystyle \frac{v^{s_1}{e}^{-\lambda v}log\left(i(v-1)\right)}{v(v-1)}}dv,$$

then

$$K=e^{\lambda }2\pi i(logu+c_0+c_2{u}^{-2}).$$

(2.7)

where $c_0$ and $c_2$ are two constants.

Proof. 

Equality (2.6) is followed by the residue theorem for the two integral paths form a contour with a pole at $v=1$.

To prove (2.7), we change the integral paths $L_{\delta }$ and $L_{-\delta }$ to the positive real axis but with an indentation around $v=1$ with $\mathrm{Im}v>0$ and $\mathrm{Im}v<0$ respectively.

And let $v=e^x$, then

$$\begin{array}{cc}\hfill K=& \underset{{\mathcal{L}}_{+}}{\int }exp(\lambda x+ixu-\lambda e^x){\displaystyle \frac{log(-i(e^x-1))}{2sinh(x/2)}}dx\hfill \\ \hfill & -\underset{{\mathcal{L}}_{-}}{\int }exp(\lambda x+ixu-\lambda e^x){\displaystyle \frac{log\left(i(e^x-1)\right)}{2sinh(x/2)}}dx\hfill \end{array}$$

where the integral path ${\mathcal{L}}_{+}$ is from $-\infty$ to $-ϵ$ then along a upper semicircle $C_{ϵ}^{+}$ to $+ϵ$ and tend to $+\infty$. The integral path ${\mathcal{L}}_{-}$ is same but with a lower semicircle $C_{ϵ}^{-}$. Let $C_{ϵ}$ be the union of $C_{ϵ}^{-}$ and the reversal of $C_{ϵ}^{+}$.

Then

$$\begin{array}{cc}\hfill K=& \pi i{\int }_{-\infty }^{-ϵ}{\displaystyle \frac{exp(\lambda x+ixu-\lambda e^x)}{2sinh(x/2)}}dx-\pi i{\int }_{+ϵ}^{+\infty }{\displaystyle \frac{exp(\lambda x+ixu-\lambda e^x)}{2sinh(x/2)}}dx\hfill \\ \hfill & -{\int }_{C_{ϵ}}exp(\lambda x+ixu-\lambda e^x){\displaystyle \frac{log\left(i(e^x-1)\right)}{2sinh(x/2)}}dx\hfill \\ \hfill & -{\displaystyle \frac{\pi i}{2}}{\int }_{C_{ϵ}^{+}\cup C_{ϵ}^{-}}{\displaystyle \frac{exp(\lambda x+ixu-\lambda e^x)}{2sinh(x/2)}}dx\hfill \end{array}$$

And then

$$\begin{array}{cc}\hfill K=& -{\int }_{C_{ϵ}}{\displaystyle \frac{e^{-\lambda }logx}{x}}dx-R_{ϵ}^{\prime }+\pi i{\int }_{-\infty }^{-ϵ}{\displaystyle \frac{e^{-\lambda }{e}^{iux}}{x}}dx\hfill \\ \hfill & +\pi i{P}_{-}-\pi i{\int }_{+ϵ}^{+\infty }{\displaystyle \frac{e^{-\lambda }{e}^{iux}}{x}}dx-\pi i{P}_{+}-{\displaystyle \frac{\pi i}{2}}{R}_{ϵ}\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill R_{ϵ}=& {\int }_{C_{ϵ}^{+}\cup C_{ϵ}^{-}}{\displaystyle \frac{exp(\lambda x+ixu-\lambda e^x)}{2sinh(x/2)}}dx\hfill \\ \hfill R_{ϵ}^{\prime }=& {\int }_{C_{ϵ}}\left(exp(\lambda x+ixu-\lambda e^x){\displaystyle \frac{log\left(i(e^x-1)\right)}{2sinh(x/2)}}-{\displaystyle \frac{e^{-\lambda }logx}{x}}\right)dx\hfill \\ \hfill P_{-}=& {\int }_{-\infty }^{-ϵ}\left({\displaystyle \frac{e^{iux}exp(\lambda x-\lambda e^x)}{2sinh(x/2)}}-{\displaystyle \frac{e^{-\lambda }{e}^{iux}}{x}}\right)dx\hfill \\ \hfill P_{+}=& {\int }_{+ϵ}^{+\infty }\left({\displaystyle \frac{e^{iux}exp(\lambda x-\lambda e^x)}{2sinh(x/2)}}-{\displaystyle \frac{e^{-\lambda }{e}^{iux}}{x}}\right)dx\hfill \end{array}$$

It is easy to know that $R_{ϵ}$ and $R_{ϵ}^{\prime }\to 0$ as $ϵ\to 0$. Let $P=\underset{ϵ\to 0}{lim}(P_{-}-P_{+})$, then

$$K=e^{-\lambda }2\pi i(logu+c_0+P)$$

where

$$\begin{array}{cc}\hfill c_0=& {\int }_0^1{\displaystyle \frac{1-cosx}{x}}dx-{\int }_1^{\infty }{\displaystyle \frac{cosx}{x}}dx,\hfill \\ \hfill P=& {\displaystyle \frac{1}{2}}{\int }_0^{\infty }cos\left(ux\right)\left({\displaystyle \frac{1}{x}}-{\displaystyle \frac{exp(-\lambda x-\lambda (e^{-x}-1))}{2sinh(x/2)}}\right)dx\hfill \\ \hfill +& {\displaystyle \frac{1}{2}}{\int }_0^{\infty }cos\left(ux\right)\left({\displaystyle \frac{1}{x}}-{\displaystyle \frac{exp(\lambda x-\lambda (e^x-1))}{2sinh(x/2)}}\right)dx\hfill \\ \hfill +& {\displaystyle \frac{i}{2}}{\int }_0^{\infty }sin\left(ux\right)\left({\displaystyle \frac{exp(-\lambda x-\lambda (e^{-x}-1))}{2sinh(x/2)}}-{\displaystyle \frac{1}{x}}\right)dx\hfill \\ \hfill -& {\displaystyle \frac{i}{2}}{\int }_0^{\infty }sin\left(ux\right)\left({\displaystyle \frac{exp(\lambda x-\lambda (e^x-1))}{2sinh(x/2)}}-{\displaystyle \frac{1}{x}}\right)dx\hfill \end{array}$$

By the partial integration, P can be expressed as

$$P=\sum _{2\le n<N}{c}_n{u}^{-n}+O\left(u^{-N}\right).$$

where N is a any positive integer.

□

The following Lemma is direct.

Lemma 2.5.

$${\int }_0^{\infty }{v}^{s_1-1}{e}^{-\lambda x}dx={\lambda }^{-s_1}{\int }_0^{\infty }{x}^{s_1-1}{e}^{-x}dx={\lambda }^{-s_1}\Gamma \left(s_1\right)$$

Lemma 2.6.

Let $c=1+\eta ,0<\eta <1,s_1=\lambda +1/2+iu,T\le u\le 2T,\delta =1/T$, and let

$$W=\underset{L_{\delta }}{\int }\underset{\left(c\right)}{\int }{e}^{\lambda }{v}^{s_1}{e}^{-\lambda v}(1+|s\left|\right){\displaystyle \frac{\Gamma \left(s\right)}{sin\pi s}}(cos\pi s)e^{-\pi is/2}{(v-1)}^{s-1}ds{\displaystyle \frac{dv}{v}}.$$

Then

$$W{\ll }_{ϵ,\eta }{T}^{ϵ}{(logT/T)}^c.$$

and the estimation is also holds in the cases the term $cos\pi s$ is removed, or $L_{\delta }$ is replaced by $L_{-\delta }$ and $e^{-\pi is/2}$ by $e^{\pi is/2}$.

Proof. 

Let $v=x{e}^{i\delta }$, then

$$\left|\Gamma \left(s\right)\right|\ll {(1+|t\left|\right)}^{c-1/2}{e}^{-\pi \left|t\right|/2},$$

$$\left|{\displaystyle \frac{cos\pi s}{sin\pi s}}\right|\ll 1,$$

$$|e^{-\pi s/2}|=e^{\pi t/2},$$

$$|v^{s_1}|\ll x^{\lambda +1/2},$$

$$1/\left|v\right|\ll 1/x,$$

$$|e^{-\lambda v}|=e^{-{\lambda }^{\prime }x},$$

where ${\lambda }^{\prime }=\lambda cos\delta$, and clearly, $\lambda -T^{-ϵ}/2<{\lambda }^{\prime }<\lambda$.

And

$${|(v-1)}^{s-1}{|=|v-1|}^{c-1}{e}^{-targ(v-1)}=a(x,\delta )exp(-tb(x,\delta ))$$

where

$$a(x,\delta )={({(x-a)}^2+2x(1-cos\delta ))}^{(c-1)/2}$$

$$b(x,\delta )=arctan{\displaystyle \frac{xsin\delta }{xcos\delta -1}}$$

Suppose that $\beta \ge 10$, let $\theta ={(2\beta logT/\lambda )}^{1/2}$, there are the following estimates

$$a(x,\delta )\ll \left\{\begin{array}{cc}1,\hfill & ifx\le 1-\theta ,\hfill \\ {\theta }^{c-1},\hfill & if|x-1|\le \theta ,\hfill \\ x^{c-1},\hfill & ifx\ge 1+\theta .\hfill \end{array}\right.$$

and

$$b(x,\delta )\left\{\begin{array}{cc}>\delta ,\hfill & x>0,\hfill \\ \gg {\theta }^{-1}\delta ,\hfill & |x-1|\le \theta ,\hfill \\ \le \pi -C{\theta }^{-1}\delta ,\hfill & |x-1|\le \theta ,\hfill \\ \le \pi -Cx\delta ,\hfill & 0<x\le 1-\theta ,\hfill \\ \le \delta {\theta }^{-1},\hfill & x\ge 1+\theta .\hfill \end{array}\right.$$

where C is a constant. Hence

$${\int }_{-\infty }^{\infty }{(1+|t\left|\right)}^{c+1/2}{e}^{\pi (t-|t\left|\right)/2}{e}^{-tb(x,\delta )}dt\ll \left\{\begin{array}{cc}{\delta }^{-c-3/2},\hfill & x\ge 1+\theta ,\hfill \\ {\left(\theta {\delta }^{-1}\right)}^{c+3/2},\hfill & |x-1|\le \theta ,\hfill \\ {\left(\delta x\right)}^{-c-3/2},\hfill & 0<x\le 1-\theta .\hfill \end{array}\right.$$

Divide the x integral into three pieces $W_1,W_2$ and $W_3$ with $0\le x\le 1-\theta$, $1-\theta \le x\le 1+\theta$ and $1+\theta \le x\le \infty$ respectively, then there are

$$\begin{array}{cc}\hfill & W=W_1+W_2+W_3.\hfill \\ \hfill & W_1\ll {\delta }^{-5/2-\eta }{e}^{\lambda }{\int }_0^{1-\theta }{x}^{\lambda -c-2}{e}^{-{\lambda }^{\prime }x}dx\ll {\delta }^{-5/2-\eta }{T}^{-\beta }\hfill \\ \hfill & W_2\ll T^{ϵ}{\theta }^c\hfill \\ \hfill & W_3\ll {\delta }^{-5/2-\eta }{e}^{\lambda }{\int }_{1+\theta }^{\infty }{x}^{\lambda +c-3/2}{e}^{-{\lambda }^{\prime }x}dx\ll {\delta }^{-5/2-\eta }{T}^{-\beta }\hfill \end{array}$$

□

Lemma 2.7.

Let $s_1=\lambda +1/2+iu$, define

$$\mathfrak{g}\left(u\right)={\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{(1/2)}{\int }\Gamma (s_1-s){\lambda }^{-s_1+s}\zeta \left(s\right)\chi (1-s)\overline{A}\left(s\right)A(1-s)ds.$$

Then for $T\le u\le 2T$, there is

$$\mathfrak{g}\left(u\right)=\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{hk}}(h,k)\left(log{\displaystyle \frac{u{(h,k)}^2}{2\pi hk}}+b_0+O\left(u^{-2}\right)\right)+{\mathcal{E}}_0$$

where ${\mathcal{E}}_0\le V{T}^{-c+ϵ}$, and $b_0$ is a constant.

Proof. 

We move the integral path from $(1/2)$ to $\left(c\right),c=1+\eta ,0<\eta <1$, the residue at $s=1$ is

$$R=e^{\lambda }\Gamma (\lambda -1/2-iu){\lambda }^{-(\lambda -1/2-iu)}\zeta \left(0\right)A\left(1\right)\overline{A}\left(0\right)\ll M^{1+ϵ}{T}^{-\alpha }\ll T^{-\alpha +1}$$

Hence,

$$\mathfrak{g}\left(u\right)=\widehat{\mathfrak{g}}\left(u\right)-R$$

where $\widehat{\mathfrak{g}}\left(u\right)$ is the one of $\mathfrak{g}\left(u\right)$ moved in the new integral path.

$$\begin{array}{cc}\hfill \widehat{\mathfrak{g}}\left(u\right)& =\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{k}}{e}^{\lambda }\sum _{n=1}^{\infty }d\left(n\right)J\left({\displaystyle \frac{nh}{k}}\right)\hfill \\ \hfill & =\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{k}}\sum _{n=1}^{\infty }d\left(n\right)e^{\lambda }{\int }_0^{\infty }{v}^{s_1}{e}^{-\lambda v}(e^{2\pi inh/k}+e^{-2\pi inh/k}){\displaystyle \frac{dv}{v}}\hfill \\ \hfill & =\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{k}}(I_1+I_2)\hfill \end{array}$$

where

$$I_1=M_1+R_1-i{E}_1,I_2=M_2+R_2-i{E}_2.$$

$$\begin{array}{cc}\hfill & M_1=e^{\lambda }\underset{L_{\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{\displaystyle \frac{\gamma -log(-2\pi i(v-1)h/k)-2log{k}^{*}}{-2\pi i{h}^{*}}}{\displaystyle \frac{dv}{v}}\hfill \\ \hfill & R_1=D\left(0,{\displaystyle \frac{h^{*}}{k^{*}}}\right)e^{\lambda }\underset{L_{\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{\displaystyle \frac{dv}{v}}\hfill \\ \hfill & E_1=e^{\lambda }\underset{L_{\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{F}_1\left(v\right){\displaystyle \frac{dv}{v}}\hfill \\ \hfill & F_1\left(v\right)=\underset{\left(c\right)}{\int }{\left(2\pi \right)}^{-s}{\displaystyle \frac{\Gamma \left(s\right)}{sin\pi s}}\left(D\left(s,{\displaystyle \frac{h^{*}}{k^{*}}}\right)+(cos\pi s)D\left(s,-{\displaystyle \frac{\overline{h^{*}}}{k^{*}}}\right)\right)\hfill \\ \hfill & ·{(-2\pi i{h}^{*}{k}^{*}(v-1))}^{s-1}ds\hfill \end{array}$$

And

$$\begin{array}{cc}\hfill & M_2=e^{\lambda }\underset{L_{-\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{\displaystyle \frac{\gamma -log(-2\pi i(v-1)h/k)-2log{k}^{*}}{2\pi i{h}^{*}}}{\displaystyle \frac{dv}{v}}\hfill \\ \hfill & R_2=D\left(0,-{\displaystyle \frac{h^{*}}{k^{*}}}\right)e^{\lambda }\underset{L_{-\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{\displaystyle \frac{dv}{v}}\hfill \\ \hfill & E_2=e^{\lambda }\underset{L_{-\delta }}{\int }{v}^{s_1}{e}^{-\lambda v}{F}_2\left(v\right){\displaystyle \frac{dv}{v}}\hfill \\ \hfill & F_2\left(v\right)=\underset{\left(c\right)}{\int }{\left(2\pi \right)}^{-s}{\displaystyle \frac{\Gamma \left(s\right)}{sin\pi s}}\left(D\left(s,-{\displaystyle \frac{h^{*}}{k^{*}}}\right)+(cos\pi s)D\left(s,{\displaystyle \frac{\overline{h^{*}}}{k^{*}}}\right)\right)\hfill \\ \hfill & ·{\left(2\pi i{h}^{*}{k}^{*}(v-1)\right)}^{s-1}ds\hfill \end{array}$$

By Lemma 2.4,

$$M_1+M_2={\displaystyle \frac{1}{h^{*}}}\left(log{\displaystyle \frac{u{(h,k)}^2}{2\pi hk}}+\gamma +c_0+O\left(u^{-2}\right)\right)$$

and by Lemma 2.5

$$R_1+R_2\ll \mathrm{Re}D\left(0,{\displaystyle \frac{h^{*}}{k^{*}}}\right)T^{-\alpha }\ll M{(logM)}^2{T}^{-\alpha }\ll T^{-\alpha +1}$$

So

$$\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{k}}(R_1+R_2)\ll M^{1+2ϵ}logM{T}^{-\alpha +1}\ll T^{-\alpha +2}$$

The other error term is

$$\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{k}}(E_1+E_2),$$

which may be written as a sum of four terms of a typical one is that

$$Z=\underset{L_{\delta }}{\int }\underset{\left(c\right)}{\int }G(v,s_1,s)\mathcal{M}\left(s\right)dsdv$$

where

$$G(v,s_1,s)=e^{\lambda }{v}^{s_1}{e}^{-\lambda v}{\displaystyle \frac{\Gamma \left(s\right)}{sin\pi s}}{\left(2\pi \right)}^{-2s}{(-2\pi i(v-1))}^{s-1}{v}^{-1},$$

$$\mathcal{M}\left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{d\left(n\right)}{n^s}}\sum _{h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{{\left(hk\right)}^{1-s}}}{(h,k)}^{(1-2s)}e\left({\displaystyle \frac{n\overline{h^{*}}}{k^{*}}}\right).$$

By Lemma 2.6,

$$Z\ll V{T}^{ϵ}{\theta }^c.$$

□

The Proof of Theorem 1.1.

Proof. 

By Lemma 2.1, it has

$$\begin{array}{cc}\hfill & I(T,2T)={\int }_T^{2T}\omega (t,T-\Delta ,2T+\Delta ){\left|\zeta A(1/2+it)\right|}^{2}dt+o\left(1\right)\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_T^{2T}{\int }_{T-\Delta }^{2T+\Delta }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du{\left|\zeta A(1/2+it)\right|}^{2}dt+o\left(1\right)\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T-\Delta }^{2T+\Delta }{\int }_T^{2T}\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}{\left|\zeta A(1/2+it)\right|}^{2}dtdu+o\left(1\right)\hfill \\ \hfill & \le {\int }_{T-\Delta }^{2T+\Delta }\mathfrak{g}\left(u\right)du+o\left(1\right)\hfill \\ \hfill & ={\int }_{T-\Delta }^{2T+\Delta }\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{hk}}(h,k)\left(log{\displaystyle \frac{u{(h,k)}^2}{2\pi hk}}+b_0+O\left(u^{-2}\right)\right)+O\left({\mathcal{E}}_{0}T\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =T\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{hk}}(h,k)\left(log{\displaystyle \frac{T{(h,k)}^2}{2\pi hk}}+b_0-1+2log2\right)+O\left({\mathcal{E}}_{0}T\right)\hfill \\ \hfill & +O\left(\Delta logT\sum _{1\le h,k\le M}{\displaystyle \frac{a\left(h\right)\overline{a\left(k\right)}}{hk}}(h,k)\right)\hfill \end{array}$$

and

$$\sum _{1\le h,k\le M}{\displaystyle \frac{(h,k)}{hk}}\le \sum _{1\le t\le M}t{\left(\sum _{\begin{array}{c}h\le M\\ t|h\end{array}}{h}^{-1}\right)}^2\ll {log}^{3}M$$

that is, the last term is $O\left(\Delta {log}^{4}M{M}^{2ϵ}\right)$.

Then replacing T by $T/2^k,1\le k\le logT$, and summing, it follows

$$I\le T\sum _{h,k\le M}{\displaystyle \frac{a\left(h\right)}{h}}{\displaystyle \frac{\overline{a\left(k\right)}}{k}}(h,k)\left(log{\displaystyle \frac{T{(h,k)}^2}{2\pi hk}}+b_0-1\right)+\tilde{\mathcal{E}}$$

On the other hand, similarly it has

$$\begin{array}{cc}\hfill & I(T,2T)={\int }_T^{2T}\omega (t,T+\Delta ,2T-\Delta ){\left|\zeta A(1/2+it)\right|}^{2}dt+o\left(1\right)\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_T^{2T}{\int }_{T+\Delta }^{2T-\Delta }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}du{\left|\zeta A(1/2+it)\right|}^{2}dt+o\left(1\right)\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T+\Delta }^{2T-\Delta }{\int }_T^{2T}\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}{\left|\zeta A(1/2+it)\right|}^{2}dtdu+o\left(1\right)\hfill \\ \hfill & \ge {\int }_{T+\Delta }^{2T-\Delta }\mathfrak{g}\left(u\right)du+o\left(1\right)\hfill \end{array}$$

Hence, it will lead same bound as the upper bound above. As for the fact $b_0=2\gamma$, it may be followed from the known result of Ingham [5]. □

It should be mentioned that the integrant of $\omega (t,T_1,T_2)$ in general is not as the one of $w(t,T_1,T_2)$ a positive real number, nevertheless, its argument is about ${\displaystyle \frac{t-u}{2\lambda }}-{\displaystyle \frac{{(t-u)}^3}{6{\lambda }^2}}$, which is very small in the context, so can be approximately viewed as a positive real number, and the deduction above is valid as the one in [1].