An Approximation to Riemann Hypothesis

1. Introduction

Riemann zeta-function $\zeta \left(s\right)$ is originally defined as

$$\zeta \left(s\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^s}},\mathrm{for}\mathrm{Re}s>1.$$

and it also can be expressed as the product form

$$\zeta \left(s\right)=\prod _p{\displaystyle \frac{1}{1-1/p^s}},\mathrm{for}\mathrm{Re}s>1.$$

This formula is called Euler’s product formula, which indicates the relation between $\zeta \left(s\right)$ and prime numbers. About $\zeta \left(s\right)$ there is a well-known Riemann hypothesis, states that all the non-trivial zeros of $\zeta \left(s\right)$ are on the critical line $\mathrm{Re}s=1/2$. The researches on the conjecture are no doubt a most time-consuming one in mathematics, refer to see the survey paper [3].

The so-called trivial zeros of $\zeta \left(s\right)$ are $s=-2,-4,\dots$, and nontrivial zeros of $\zeta \left(s\right)$ are known all in the critical strip $0\le \mathrm{Re}s\le 1$.

Denote by $N\left(T\right)$ the number of zeros of $\zeta (\sigma +it)$ in the region $0\le \sigma \le 1$, $0\le t\le T$, and by $N_0\left(T\right)$ the number of zeros on the critical line $\sigma =1/2$, $0\le t\le T$. Riemann hypothesis is that

$$N_0\left(T\right)=N\left(T\right).$$

For $N\left(T\right)$, it is known that

$$N\left(T\right)={\displaystyle \frac{T}{2\pi }}log{\displaystyle \frac{T}{2\pi }}-{\displaystyle \frac{T}{2\pi }}+O(logT).$$

(1.1)

And for $N_0\left(T\right)$, Hardy firstly shown that there are infinity many zeros on the critical line, and then he and Littlewood [5] and Selberg [8] proved that

$$\kappa ={\displaystyle \frac{N_0\left(T\right)}{N\left(T\right)}}>0.$$

Levinson [6] proved

$$\kappa \ge {\displaystyle \frac{1}{3}}.$$

and then this result has been improved successively. Conrey [2], Feng [4] proved respectively

$$\kappa \ge 0.407,\kappa \ge 0.412.$$

In this paper, we will prove that

Theorem 1.1. 

$$N\left(T\right)=N_0\left(T\right)+\mathcal{E}.$$

(1.2)

where $\mathcal{E}\ll T^{1/3}{(logT)}^{7/3}$.

The main arguments in this paper are based the papers [1,6,7], but instead of using Riemann-Siegel formula, it will be applied an auxiliary function $\omega (s,T_1,T_2)$ defined in Lemma 2.1, which will play a role of mollifier and ferry, it firstly used in [7] but here with a small modification.

2. Some Lemmas

Lemma 2.1. 

Suppose that $T\le T_1\le T_2\le 2T$, $\lambda =T^{2/3}{(logT)}^{5/3}$, $s_1=\lambda +c+iu,(c\ge 0)$, $s=v+it$,define

$$\omega (s,T_1,T_2)={\displaystyle \frac{e^{\lambda }}{2\pi i}}{\int }_{T_1}^{T_2}\Gamma (s_1-s){\lambda }^{s-s_1}d{s}_1.$$

(2.1)

Let $\sigma =\lambda +c-v$, $\eta =min\left\{|x-t||x\in [T_1,T_2]\right\}$, there is

$$|\omega (s,T_1,T_2)|\ll exp(-({(c-v)}^2+{\eta }^2)/2\sigma )$$

(2.2)

Let $\Delta ={(20\lambda logT)}^{1/2}$, if $t\in [T_1+\Delta ,T_2-\Delta ]$, then

$$|\omega (c+it,T_1,T_2)-1|\ll T^{-10}.$$

(2.3)

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

$$|\omega (c+it,T_1,T_2)|\ll T^{-10}.$$

(2.4)

If $|t-u|=o\left({\sigma }^{2/3}\right)$, then

$$arg\left(\Gamma (s_1-s){\lambda }^{s-s_1}\right)={\displaystyle \frac{t-u}{2\sigma }}-{\displaystyle \frac{{(t-u)}^3}{6{\sigma }^2}}+ϵ.$$

(2.5)

Proof. 

By Stirling’s formula, it has

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

And

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

Hence,

$$|\omega (s,T_1,T_2)|\le e^{-{(c-v)}^2/2\sigma }{\displaystyle \frac{{\sigma }^{-1/2}}{\sqrt{2\pi }}}{\int }_{T_1}^{T_2}{e}^{-{(u-t)}^2/2\sigma }du\ll e^{-({(c-v)}^2+{\eta }^2)/2\sigma }.$$

Besides, it is familiar that

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

Hence,

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

where

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

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

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

$$\begin{array}{cc}\hfill \left|R_1\right|& \ll {\int }_{T_2}^{\infty }\left|e^{\lambda }\Gamma (\lambda +(u-t)i){\lambda }^{-(\lambda +(u-t\left)i\right)}\right|du\hfill \\ \hfill & \ll {\lambda }^{-1/2}{\int }_{T_2}^{\infty }exp(-{(u-t)}^2/2\lambda )du\hfill \\ \hfill & \ll T^{-10}.\hfill \end{array}$$

and similarly

$$\left|R_2\right|\ll {\lambda }^{-1/2}{\int }_{-\infty }^{T_1}exp(-{(u-t)}^2/2\lambda )du\ll T^{-10}.$$

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

$$\left|\omega (c+it,T_1,T_2)\right|\ll {\lambda }^{-1/2}{\int }_{T_1}^{T_2}exp(-{(u-t)}^2/2\lambda )du\ll T^{-10}.$$

If $|t-u|=o\left({\sigma }^{2/3}\right)$, then

$$\begin{array}{c}\hfill \mathrm{Im}\left(log\left(\Gamma (s_1-s){\lambda }^{s-s_1}\right)\right)={\displaystyle \frac{t-u}{2\sigma }}+{\displaystyle \frac{{(t-u)}^3}{3{\sigma }^2}}-{\displaystyle \frac{{(t-u)}^3}{2{\sigma }^2}}\\ \hfill & -{\displaystyle \frac{{(t-u)}^5}{5{\sigma }^4}}+{\displaystyle \frac{{(t-u)}^5}{4{\sigma }^4}}+ϵ\hfill \\ \hfill & ={\displaystyle \frac{t-u}{2\sigma }}-{\displaystyle \frac{{(t-u)}^3}{6{\sigma }^2}}+{\displaystyle \frac{{(t-u)}^5}{20{\sigma }^4}}+ϵ.\hfill \end{array}$$

□

Corollary 2.1. 

Divide $\omega (s,T_1,T_2)$ into two parts,

$$\omega (s,T_1,T_2)={\omega }_1(s,T_1,T_2)+{\omega }_2(s,T_1,T_2),$$

$${\omega }_1(s,T_1,T_2)={\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{|u-t|\le \Delta }{\int }\Gamma (s_1-s){\lambda }^{s-s_1}d{s}_1,$$

$${\omega }_2(s,T_1,T_2)={\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{|u-t|\ge \Delta }{\int }\Gamma (s_1-s){\lambda }^{s-s_1}d{s}_1.$$

Then

$$|{\omega }_2(s,T_1,T_2)|\le O\left(T^{-10}\right),$$

and

$$|arg\left({\omega }_1(s,T_1,T_2)\right)|\le O({\Delta }^3/{\lambda }^2).$$

This indicates ${\omega }_1(s,T_1,T_2)$ is the domination of $\omega (s,T_1,T_2)$, and its argument is small.

Lemma 2.2. 

Let ${\zeta }_1\left(s\right),{\zeta }_2\left(s\right)$ be $\zeta \left(s\right)$ or ${\zeta }^{\prime }\left(s\right)$ and

$${\zeta }_1\left(s\right)=\sum _n{a}_n{n}^{-s},for\mathrm{Re}s>1,$$

$${\zeta }_2\left(s\right)=\sum _n{b}_n{n}^{-s},for\mathrm{Re}s>1.$$

λ same as in Lemma 2.1, $0\le a\le 1/2$, $s_1=\lambda +c+iu$, $c\ge 0$, define

$$g\left(u\right)={\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{\left(a\right)}{\int }\Gamma (s_1-s){\lambda }^{-s_1+s}{\zeta }_1\left(s\right){\overline{\zeta }}_2\left(s\right)ds$$

(2.9)

Then for $T\le u\le 2T$, $1<\beta <\lambda +c$, there is

$$g\left(u\right)=e^{\lambda }\sum _{h=1}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{a_k{\overline{b}}_h}{h^{2\beta }}}{\left({\displaystyle \frac{h}{k}}\right)}^{s_1}{e}^{-\lambda h/k}+O\left(T^{-10}\right).$$

(2.10)

Proof. 

We move the integral path from $\left(a\right)$ to $\left(\beta \right)$, the residue at the pole $s=1$ is

$$R\ll e^{\lambda }\Gamma (\lambda +c-1+iu){\lambda }^{-(\lambda +c-1+iu)}\ll T^{-10}$$

Hence,

$$\begin{array}{cc}\hfill g\left(u\right)=& \sum _{h=1}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{a_k{\overline{b}}_h}{k^{\beta }{h}^{\beta }}}{\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{\left(\beta \right)}{\int }\Gamma (s_1-s){\lambda }^{-s_1+s}{\left({\displaystyle \frac{h}{k}}\right)}^{it}ds+O\left(T^{-10}\right)\hfill \\ \hfill =& \sum _{h=1}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{a_k{\overline{b}}_h}{h^{2\beta }}}{\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{\left(\beta \right)}{\int }\Gamma (s_1-s){\lambda }^{-s_1+s}{\left({\displaystyle \frac{h}{k}}\right)}^{s}ds+O\left(T^{-10}\right)\hfill \\ \hfill =& \sum _{h=1}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{a_k{\overline{b}}_h}{h^{2\beta }}}{\displaystyle \frac{e^{\lambda }}{2\pi i}}\underset{(\lambda +c-\beta )}{\int }\Gamma \left(w\right){\lambda }^{-w}{\left({\displaystyle \frac{h}{k}}\right)}^{s_1-w}dw+O\left(T^{-10}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill =& e^{\lambda }\sum _{h=1}^{\infty }\sum _{k=1}^{\infty }{\displaystyle \frac{a_k{\overline{b}}_h}{h^{2\beta }}}{\left({\displaystyle \frac{h}{k}}\right)}^{s_1}{e}^{-\lambda h/k}+O\left(T^{-10}\right).\hfill \end{array}$$

□

Lemma 2.3. 

Let $\theta ={(20logT/\lambda )}^{1/2}$, λ as before, $0\le c\le \Delta /10,$ then

$$\begin{array}{cc}\hfill J_1=& {\int }_{1+\theta }^{\infty }{e}^{\lambda }{v}^{\lambda +c}{e}^{-\lambda v}dv\ll T^{-8}{\lambda }^{-1/2},\hfill \end{array}$$

(2.13)

$$\begin{array}{cc}\hfill J_2=& {\int }_0^{1-\theta }{e}^{\lambda }{v}^{\lambda +c}{e}^{-\lambda v}dv\ll T^{-10}.\hfill \end{array}$$

(2.13)

$$\begin{array}{cc}\hfill J_3=& {\int }_0^{1-\theta }{e}^{\lambda }log(j/v)v^{\lambda +c}{e}^{-\lambda v}dv\ll (logj+\theta )T^{-10},(j\ge 1).\hfill \end{array}$$

(2.13)

Proof. 

For (2.11), by the integration by parts

$$\begin{array}{cc}\hfill J_1=& {\left.-{\displaystyle \frac{e^{\lambda }{v}^{\lambda +c}{e}^{-\lambda v}}{\lambda }}\right|}_{1+\theta }^{\infty }+{\displaystyle \frac{\lambda +c}{\lambda }}{\int }_{1+\theta }^{\infty }{e}^{\lambda }{v}^{\lambda +c-1}{e}^{-\lambda v}dv\hfill \\ \hfill \le & {\displaystyle \frac{exp\left(-2{\theta }^2\lambda /5\right)}{\lambda }}+(1+\theta /10){\displaystyle \frac{J_1}{1+\theta }}\hfill \end{array}$$

That is,

$${\displaystyle \frac{9}{10}}{\displaystyle \frac{\theta \lambda }{1+\theta }}{J}_1\le exp(-2{\theta }^2\lambda /5)\le T^{-8}$$

and

$$J_1\le T^{-8}{\lambda }^{-1/2}.$$

For (2.12),

$$J_2\le e^{\lambda }{(1-\theta )}^{\lambda }{e}^{-\lambda (1-\theta )}\le exp(-{\theta }^2\lambda /2)\le T^{-10}.$$

For (2.13),

$$J_3\le e^{\lambda }log(j/(1-\theta )){(1-\theta )}^{\lambda +c}{e}^{-\lambda (1-\theta )}\ll (logj+\theta )T^{-10}.$$

□

3. The Proof of Theorem 1.1

Proof. 

Let $h\left(s\right)={\pi }^{-s/2}\Gamma (s/2)$, then the functional equation of $\zeta \left(s\right)$ can be written as

$$h\left(s\right)\zeta \left(s\right)=h(1-s)\zeta (1-s)$$

(3.1)

By Stirling’s formula, it has

$$logh\left(s\right)={\displaystyle \frac{1}{2}}(s-1)log{\displaystyle \frac{s}{2\pi }}-{\displaystyle \frac{s}{2}}+C_0+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.2)

Let $f\left(s\right)=logh\left(s\right)$, then

$$f^{\prime }\left(s\right)={\displaystyle \frac{h^{\prime }\left(s\right)}{h\left(s\right)}}={\displaystyle \frac{1}{2}}log{\displaystyle \frac{s}{2\pi }}+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.3)

and for larger t

$$f^{\prime }\left(s\right)+f^{\prime }(1-s)=log{\displaystyle \frac{t}{2\pi }}+O\left({\displaystyle \frac{1}{s}}\right)$$

(3.4)

Taking logarithm of equation (3.1), and then derivative, it follows

$$h\left(s\right)\zeta \left(s\right)(f^{\prime }\left(s\right)+f^{\prime }(1-s))=-h\left(s\right){\zeta }^{\prime }\left(s\right)-h(1-s){\zeta }^{\prime }(1-s)$$

(3.5)

We note that the right side of (3.5) is a sum of two conjugative complex numbers as $s=1/2+it$, so the zeros of the right side of (3.5) occur if and only if

$$arg\left(h\left(s\right){\zeta }^{\prime }\left(s\right)\right)\equiv \pi /2mod\pi$$

(3.6)

On the left side of (3.5), clearly, $h\left(s\right)$ is never zero, and by (3.4), so these zeros are just the zeros of $\zeta (1/2+it)$.

Moreover, let $\chi \left(s\right)=h(1-s)/h\left(s\right)$, then $\zeta \left(s\right)=\chi \left(s\right)\zeta (1-s)$, and

$${\zeta }^{\prime }\left(s\right)=-\chi \left(s\right)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta (1-s)+{\zeta }^{\prime }(1-s)\}$$

(3.7)

By (3.6), the zeros of $\zeta (1/2+it)$ are the ones

$$arg\left(h(1-s)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta (1-s)+{\zeta }^{\prime }(1-s)\}\right)\equiv \pi /2mod\pi$$

on $\sigma =1/2$, equivalently,

$$arg\left(h\left(s\right)\{(f^{\prime }\left(s\right)+f^{\prime }(1-s))\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)\}\right)\equiv \pi /2mod\pi$$

(3.8)

on $\sigma =1/2$. Write $\mathcal{L}\left(s\right)=f^{\prime }\left(s\right)+f^{\prime }(1-s)$, and denote by

$$G\left(s\right)=\zeta \left(s\right)+{\zeta }^{\prime }\left(s\right)/\mathcal{L}\left(s\right)$$

(3.9)

The investigation above means

$$N_0\left(T\right)={\displaystyle \frac{1}{\pi }}{\Delta }_0^{T}arg\left(hG(1/2+it)\right)$$

(3.10)

By(3.2), it can be known that

$${\Delta }_0^{T}arg\left(h(1/2+it)\right)={\displaystyle \frac{T}{2}}log{\displaystyle \frac{T}{2\pi }}-{\displaystyle \frac{T}{2}}+O(logT)$$

(3.11)

So, the main task to determine $N_0\left(T\right)$ is to calculate ${\Delta }_0^{T}arg\left(G(1/2+it)\right)$.

Let $L=log(T/2\pi )$, $U\le T$, and let D be the rectangle with the vertices $1/2+iT$, $c+iT$, $c+i(T+U)$, $1/2+i(T+U)$,$(c\ge 3)$. First of all, we might as well assume there are no zeros of $G\left(s\right)$ on the boundary of D, then by the principle of argument, the change of $argG\left(s\right)$ around D is equal to $2\pi$ times $N_G\left(D\right)$, the number of zeros of $G\left(s\right)$ in D.

On the right side of D

$$|G(c+it)-1|\le \sum _{n\ge 2}{n}^{-c}+O(1/L)\le 1/3$$

so, $argG\left(s\right)$ change less than $\pi$. On the lower side and the upper side of D, by a known result [9, $\S 9.4$], a extension of Jessen’s theorem, taking account on the order of $G\left(s\right)$, we can know that $argG\left(s\right)=O\left(L\right)$ as $0<\sigma \le 3$, and $argG(\sigma +it)=O\left(2^{-\sigma }\right)$ as $\sigma \ge 3$, hence, for any $0\le b\le c$, it has

$${\int }_b^{c}argG(\sigma +iT)d\sigma ,{\int }_b^{c}argG(\sigma +i(T+U))d\sigma \ll O\left(L\right)$$

(3.12)

So,

$${\Delta }_T^{T+U}arg\left(G(1/2+it)\right)=-2\pi N_G\left(D\right)+O(logT)$$

(3.13)

Now the work is turned into to evaluate $N_G\left(D\right)$.

Take $a=1/2-1/L$, and let $\mathcal{D}$ be the rectangle with vertices $a+iT$, $c+iT$, $c+i(T+U)$, $a+i(T+U)$. Taking the integral $\underset{\mathcal{D}}{\int }logG\left(s\right)ds$, by the Littlewood’s Lemma [9, $\S 9.9$], it has

$$\begin{array}{cc}\hfill & {\int }_T^{T+U}log\left|G(a+it)\right|dt-{\int }_T^{T+U}log\left|G(c+it)\right|dt+{\int }_a^{c}argG(\sigma +i(T+U))d\sigma \hfill \\ \hfill & -{\int }_a^{c}argG(\sigma +iT)d\sigma =2\pi \sum \mathrm{dist}\hfill \end{array}$$

(3.14)

where $\sum \mathrm{dist}$ is the sum of the distances of the zeros of $G\left(s\right)$ from the left.

By (3.9), it is easy to know

$${\int }_T^{T+U}logG(c+it)dt={\int }_T^{T+U}log\zeta (c+it)dt+O(1/L)$$

and it is familiar that

$$log\zeta \left(s\right)=\sum _n{\displaystyle \frac{-\Lambda \left(n\right)}{n^{s}logn}}$$

So

$${\int }_T^{T+U}log\left|G(c+it)\right|dt\ll 1.$$

With (3.12), the rest is to calculate the first integral of (3.14).

By the concavity of logarithm, it has

$$\begin{array}{cc}\hfill {\int }_T^{T+U}log\left|G(a+it)\right|dt& ={\displaystyle \frac{1}{2}}{\int }_T^{T+U}log{\left|G(a+it)\right|}^{2}dt\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}Ulog\left({\displaystyle \frac{1}{U}}{\int }_T^{T+U}{\left|G(a+it)\right|}^{2}dt\right)\hfill \end{array}$$

(3.16)

At first, we simplify $G\left(s\right)$ as

$$G_0\left(s\right)=\zeta \left(s\right)+{\displaystyle \frac{{\zeta }^{\prime }\left(s\right)}{L}}.$$

Then

$$G\left(s\right)=G_0\left(s\right)+E\left(s\right).$$

$$E\left(s\right)=\left({\displaystyle \frac{1}{\mathcal{L}\left(s\right)}}-{\displaystyle \frac{1}{L}}\right){\zeta }^{\prime }\left(s\right)\ll {\displaystyle \frac{1}{L^3}}{\zeta }^{\prime }\left(s\right).$$

And

$$\begin{array}{cc}\hfill {\int }_{T_1}^{T_2}{\left|G(a+it)\right|}^{2}dt=& {\int }_{T_1}^{T_2}{\left|G_0(a+it)\right|}^{2}dt+2\mathrm{Re}{\int }_{T_1}^{T_2}{G}_0(a+it)E(a-it)dt\hfill \\ \hfill & +{\int }_{T_1}^{T_2}{\left|E(a+it)\right|}^{2}dt\hfill \end{array}$$

By Cauchy inequality

$${\int }_{T_1}^{T_2}{G}_0(a+it)E(a-it)dt\le {\left({\int }_{T_1}^{T_2}|G_0{(a+it)|}^{2}dt{\int }_{T_1}^{T_2}{\left|E(a+it)\right|}^{2}dt\right)}^{1/2}$$

The third integral in the right side of (3.16) is much smaller than the first one, which will be actually calculated later, hence

$${\int }_{T_1}^{T_2}{\left|G(a+it)\right|}^{2}dt=(1+ϵ){\int }_{T_1}^{T_2}{\left|G_0(a+it)\right|}^{2}dt.$$

Moreover, let

$$\varphi (s,T_1-\Delta ,T_2+\Delta )={\omega }_1^{1/2}(s,T_1-\Delta ,T_2+\Delta ).$$

(3.17)

By (2.8), we can know that on the upper side and the lower side of $\mathcal{D}$, there is

$$arg\left(\varphi (s,T_1-\Delta ,T_2+\Delta )\right)\le O({\Delta }^3/{\lambda }^2)$$

(3.18)

and

$$\begin{array}{cc}\hfill {\int }_a^{c}arg\left(\varphi (v+T_{j}i,T_1-\Delta ,T_2+\Delta )\right)dv& \le O(c{\Delta }^3/{\lambda }^2),(j=1,2).\hfill \end{array}$$

(3.19)

It is assumed that $c\le \Delta /10$.

Moreover, by Lemma 2.1, there is

$${\int }_{T_1}^{T_2}log\left|\varphi (c+it,T_1-\Delta ,T_2+\Delta )\right|dt\ll T^{-9}.$$

(3.20)

(3.19) and (3.20) indicate that function $\varphi (s,T_1-\Delta ,T_2+\Delta )$ may be used as a mollifier. Let

$$\mathcal{G}\left(s\right)=G_0\left(s\right)\varphi (s,T_1-\Delta ,T_2+\Delta ).$$

(3.21)

By (2.8), for $\mathrm{Re}s\le c$, it has

$$\begin{array}{cc}\hfill & {\int }_{T_1}^{T_2}{\left|\mathcal{G}\left(s\right)\right|}^{2}dt={\int }_{T_1}^{T_2}|G_0{\left(s\right)|}^2{\left|\varphi (s,T_1-\Delta ,T_2+\Delta )\right|}^{2}dt\hfill \\ \hfill & ={\int }_{T_1}^{T_2}|G_0{\left(s\right)|}^2\left|{\omega }_1(s,T_1-\Delta ,T_2+\Delta )\right|dt\hfill \\ \hfill & \le (1+O(L^3/\lambda )){\int }_{T_1}^{T_2}{\left|G_0\left(s\right)\right|}^2{\omega }_1(s,T_1-\Delta ,T_2+\Delta )dt\hfill \\ \hfill & \le (1+O(L^3/\lambda )){\int }_{T_1}^{T_2}{\left|G_0\left(s\right)\right|}^2\omega (s,T_1-\Delta ,T_2+\Delta )dt+O\left(T^{-8}\right)\hfill \end{array}$$

(3.22)

In the next is mainly to calculate the last integral.

By Lemma 2.1, it has

$$\begin{array}{cc}\hfill & {\int }_{T_1}^{T_2}\omega (a+it,T_1-\Delta ,T_2+\Delta ){\left|G_0(a+it)\right|}^{2}dt\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_1}^{T_2}{\int }_{T_1-\Delta }^{T_2+\Delta }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}du{\left|G_0(a+it)\right|}^{2}dt\hfill \\ \hfill & ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{T_1}^{T_2}\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}{\left|G_0(a+it)\right|}^{2}dtdu\hfill \\ \hfill & \le {\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{-\infty }^{\infty }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}{\left|G_0(a+it)\right|}^{2}dtdu+ϵ\hfill \\ \hfill & =I_{11}+I_{12}+I_{21}+I_{22}+ϵ.\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill I_{11}& ={\displaystyle \frac{e^{\lambda }}{2\pi }}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{-\infty }^{\infty }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}{\left|\zeta (a+it)\right|}^{2}dtdu\hfill \\ \hfill I_{12}& ={\displaystyle \frac{e^{\lambda }}{2\pi L}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{-\infty }^{\infty }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}\zeta (a+it){\zeta }^{\prime }(a-it)dtdu\hfill \\ \hfill I_{21}& ={\displaystyle \frac{e^{\lambda }}{2\pi L}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{-\infty }^{\infty }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}\zeta (a-it){\zeta }^{\prime }(a+it)dtdu\hfill \\ \hfill I_{22}& ={\displaystyle \frac{e^{\lambda }}{2\pi L^2}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\int }_{-\infty }^{\infty }\Gamma (s_1-(a+it)){\lambda }^{-(s_1-(a+it))}{\left|{\zeta }^{\prime }(a+it)\right|}^{2}dtdu.\hfill \end{array}$$

In the following specify $T_1=T,T_2=T+U.$

We first calculate $I_{11}$, by Lemma 2.2,

$$\begin{array}{cc}\hfill I_{11}& =e^{\lambda }{\int }_{T_1-\Delta }^{T_2+\Delta }\sum _{j_1,j_2}{\displaystyle \frac{1}{j_2^{2\beta }}}{\rho }^{s_1}exp(-\lambda \rho )du,\rho =j_2/j_1.\hfill \\ \hfill & =I_{11,0}+I_{11,1}+I_{11,2}+I_{11,3}.\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill I_{11,0}=& \sum _{j_1=j_2}{\displaystyle \frac{e^{\lambda }}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du,\hfill \\ \hfill I_{11,1}=& \sum _{\rho \ge 1+\theta }{\displaystyle \frac{e^{\lambda }}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{11,2}=& \sum _{\rho \le 1-\theta }{\displaystyle \frac{e^{\lambda }}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{11,3}=& \sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{\displaystyle \frac{e^{\lambda }}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du.\hfill \end{array}$$

Clearly

$$I_{11,0}=(U+2\Delta )\sum _{j_2}{\displaystyle \frac{1}{j_2^{2\beta }}}=c_{11}(U+2\Delta ).$$

By Lemma 2.3,

$$\begin{array}{cc}\hfill I_{11,1}& \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}\sum _{j_2/j_1\ge 1+\theta }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}{\int }_1^{j_2/(1+\theta )}{e}^{\lambda }{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{\int }_{1+\theta }^{j_2}{e}^{\lambda }{v}^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{T}^{-8}{\lambda }^{-1/2}\ll T^{-8}.\hfill \end{array}$$

And,

$$\begin{array}{cc}\hfill I_{11,2}& \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}\sum _{j_2/j_1\le 1-\theta }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}{\int }_{j_2/(1-\theta )}^{\infty }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{\int }_0^{1-\theta }{e}^{\lambda }{v}^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{T}^{-10}\ll T^{-9}\hfill \end{array}$$

And

$$\begin{array}{cc}\hfill I_{11,3}& \ll (U+2\Delta )\sum _{j_2}{\displaystyle \frac{1}{j_2^{2\beta }}}\sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll (U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{1}{j_2^{2\beta }}}\left({\displaystyle \frac{j_2}{1-\theta }}-{\displaystyle \frac{j_2}{1+\theta }}\right)T^{1/10}\hfill \\ \hfill & \ll (U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{2\theta }{j_2^{2\beta -1}(1-{\theta }^2)}}{T}^{1/10}\hfill \\ \hfill & \ll (U+2\Delta ){\theta }^{2\beta -1}{T}^{1/10}\ll T^{-10},(\beta \ge 10logT).\hfill \end{array}$$

For $I_{12}$, by Lemma 2.2,

$$\begin{array}{cc}\hfill I_{12}=& {\displaystyle \frac{1}{L}}{e}^{\lambda }{\int }_{T_1-\Delta }^{T_2+\Delta }\sum _{j_1,j_2}{\displaystyle \frac{-log{j}_2}{j_2^{2\beta }}}{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill =& I_{12,0}+I_{12,1}+I_{12,2}+I_{12,3}.\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill I_{12,0}={\displaystyle \frac{1}{L}}& \sum _{j_1=j_2}{\displaystyle \frac{-e^{\lambda }log{j}_2}{j_2^{2c}}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du,\hfill \\ \hfill I_{12,1}={\displaystyle \frac{1}{L}}& \sum _{\rho \ge 1+\theta }{\displaystyle \frac{-e^{\lambda }log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{12,2}={\displaystyle \frac{1}{L}}& \sum _{\rho \le 1-\theta }{\displaystyle \frac{-e^{\lambda }log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{12,3}={\displaystyle \frac{1}{L}}& \sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{\displaystyle \frac{-e^{\lambda }log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du.\hfill \end{array}$$

Clearly,

$$I_{12,0}=(U+2\Delta ){\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{-log{j}_2}{j_2^{2\beta }}}=c_{12}(U+2\Delta ).$$

By Lemma 2.3,

$$\begin{array}{cc}\hfill I_{12,1}& \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}\sum _{j_2/j_1\ge 1+\theta }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}{\int }_1^{j_2/(1+\theta )}{e}^{\lambda }{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{\int }_{1+\theta }^{j_2}{e}^{\lambda }{v}^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{T}^{-8}{\lambda }^{-1/2}\ll T^{-8}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{12,2}& \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}\sum _{j_2/j_1\le 1-\theta }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}{\int }_{j_2/(1-\theta )}^{\infty }{e}^{\lambda }{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{\int }_0^{1-\theta }{e}^{\lambda }{v}^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{T}^{-10}\ll T^{-9}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{12,3}& \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2}{\displaystyle \frac{log{j}_2}{j_2^{2\beta }}}\sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{e}^{\lambda }{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{log{j}_2}{j_2^{2\beta }}}\left({\displaystyle \frac{j_2}{1-\theta }}-{\displaystyle \frac{j_2}{1+\theta }}\right)T^{1/10}\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{2\theta log{j}_2}{j_2^{2\beta -1}(1-{\theta }^2)}}{T}^{1/10}\hfill \\ \hfill & \ll (U+2\Delta ){\theta }^{2\beta -1}{T}^{1/10}\ll T^{-10}.\hfill \end{array}$$

For $I_{21}$, by Lemma 2.2,

$$\begin{array}{cc}\hfill I_{21}=& {\displaystyle \frac{1}{L}}{e}^{\lambda }{\int }_{T_1-\Delta }^{T_2+\Delta }\sum _{j_1,j_2}{\displaystyle \frac{-log{j}_1}{j_2^{2\beta }}}{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill =& I_{21,0}+I_{21,1}+I_{21,2}+I_{21,3},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill I_{21,0}={\displaystyle \frac{1}{L}}& \sum _{j_1=j_2}{\displaystyle \frac{-e^{\lambda }log{j}_1}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du,\hfill \\ \hfill I_{21,1}={\displaystyle \frac{1}{L}}& \sum _{\rho \ge 1+\theta }{\displaystyle \frac{-e^{\lambda }log{j}_1}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{21,2}={\displaystyle \frac{1}{L}}& \sum _{\rho \le 1-\theta }{\displaystyle \frac{-e^{\lambda }log{j}_1}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{21,3}={\displaystyle \frac{1}{L}}& \sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{\displaystyle \frac{-e^{\lambda }log{j}_1}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du.\hfill \end{array}$$

Clearly,

$$I_{21,0}=(U+2\Delta ){\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{-log{j}_2}{j_2^{2\beta }}}=c_{21}(U+2\Delta )$$

By Lemma 2.3,

$$\begin{array}{cc}\hfill I_{21,1}& \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}\sum _{j_2/j_1\ge 1+\theta }{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}{\int }_1^{j_2/(1+\theta )}{e}^{\lambda }logx{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{\int }_{1+\theta }^{j_2}{e}^{\lambda }log(j_2/v)v^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{T}^{-8}{\lambda }^{-1/2}\ll T^{-8}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{21,2}& \ll \sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}\sum _{j_2/j_1\le 1-\theta }{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta }}}{\int }_{j_2/(1-\theta )}^{\infty }{e}^{\lambda }logx{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}}{j_2^{2\beta -1}}}{\int }_0^{1-\theta }{e}^{\lambda }log(j_2/v)v^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}(\theta +log{j}_2)}{j_2^{2\beta -1}}}{T}^{-10}\ll T^{-9}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{21,3}& \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2}{\displaystyle \frac{1}{j_2^{2\beta }}}\sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{log{j}_2}{j_2^{2\beta }}}\left({\displaystyle \frac{j_2}{1-\theta }}-{\displaystyle \frac{j_2}{1+\theta }}\right)T^{1/10}\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{2\theta log{j}_2}{j_2^{2\beta -1}(1-{\theta }^2)}}{T}^{1/10}\hfill \\ \hfill & \ll (U+2\Delta ){\theta }^{2\beta -1}{T}^{1/10}\ll T^{-10}.\hfill \end{array}$$

For $I_{22}$, by Lemma 2.2,

$$\begin{array}{cc}\hfill I_{22}={\displaystyle \frac{1}{L^2}}& e^{\lambda }{\int }_{T_1-\Delta }^{T_2+\Delta }\sum _{j_1,j_2}{\displaystyle \frac{log{j}_{1}log{j}_2}{j_2^{2\beta }}}{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill =& I_{22,0}+I_{22,1}+I_{22,2}+I_{22,3},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill I_{22,0}={\displaystyle \frac{1}{L^2}}& \sum _{j_1=j_2}{\displaystyle \frac{e^{\lambda }log{j}_{1}log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du,\hfill \\ \hfill I_{22,1}={\displaystyle \frac{1}{L^2}}& \sum _{\rho \ge 1+\theta }{\displaystyle \frac{e^{\lambda }log{j}_{1}log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{22,2}={\displaystyle \frac{1}{L^2}}& \sum _{\rho \le 1-\theta }{\displaystyle \frac{e^{\lambda }log{j}_{1}log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du\hfill \\ \hfill I_{22,3}={\displaystyle \frac{1}{L^2}}& \sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{\displaystyle \frac{e^{\lambda }log{j}_{1}log{j}_2}{j_2^{2\beta }}}{\int }_{T_1-\Delta }^{T_2+\Delta }{\rho }^{s_1}exp(-\lambda \rho )du.\hfill \end{array}$$

Clearly,

$$I_{22,0}=(U+2\Delta ){\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{log}^2{j}_2}{j_2^{2\beta }}}=c_{22}(U+2\Delta ).$$

By Lemma 2.3,

$$\begin{array}{cc}\hfill I_{22,1}& \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}\sum _{j_2/j_1\ge 1+\theta }{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}{\int }_1^{j_2/(1+\theta )}{e}^{\lambda }logx{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{\int }_{1+\theta }^{j_2}{e}^{\lambda }log(j_2/v)v^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}{log}^2{j}_2}{j_2^{2\beta -1}}}{T}^{-8}{\lambda }^{-1/2}\ll T^{-8}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{22,2}& \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}\sum _{j_2/j_1\le 1-\theta }{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta }}}{\int }_{j_2/(1-\theta )}^{\infty }{e}^{\lambda }logx{\left({\displaystyle \frac{j_2}{x}}\right)}^{\lambda +c}exp(-\lambda j_2/x)dx\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}log{j}_2}{j_2^{2\beta -1}}}{\int }_0^{1-\theta }{e}^{\lambda }log(j_2/v)v^{\lambda +c-2}exp(-\lambda v)dv\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}\sum _{j_2}{\displaystyle \frac{{\theta }^{-1}(log{j}_2+\theta )log{j}_2}{j_2^{2\beta -1}}}{T}^{-10}\ll T^{-9}.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_{22,3}& \ll {\displaystyle \frac{1}{L^2}}(U+2\Delta )\sum _{j_2}{\displaystyle \frac{log{j}_2}{j_2^{2\beta }}}\sum _{\begin{array}{c}1-\theta \le \rho \le 1+\theta \\ j_1\ne j_2\end{array}}{e}^{\lambda }log{j}_1{\left({\displaystyle \frac{j_2}{j_1}}\right)}^{\lambda +c}exp(-\lambda j_2/j_1)\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{{log}^2{j}_2}{j_2^{2\beta }}}\left({\displaystyle \frac{j_2}{1-\theta }}-{\displaystyle \frac{j_2}{1+\theta }}\right)T^{1/10}\hfill \\ \hfill & \ll {\displaystyle \frac{1}{L^2}}(U+2\Delta )\sum _{j_2\ge 1/\theta }{\displaystyle \frac{2\theta {log}^2{j}_2}{j_2^{2\beta -1}(1-{\theta }^2)}}{T}^{1/10}\hfill \\ \hfill & \ll (U+2\Delta ){\theta }^{2\beta -1}{T}^{1/10}\ll T^{-10}.\hfill \end{array}$$

Combining all the evaluations above, and recall (3.22), it has

$${\int }_T^{T+U}{\left|\mathcal{G}(a+it)\right|}^{2}dt=c_0(U+2\Delta )(1+O(L^3/\lambda ))+ϵ.$$

where

$$\begin{array}{cc}\hfill c_0=& c_{11}+c_{12}+c_{21}+c_{22}\hfill \\ \hfill =& 1+\sum _{j=2}^{\infty }{\displaystyle \frac{{(1-logj/L)}^2}{j^{2\beta }}}\hfill \\ \hfill =& 1+O\left(T^{-10}\right).(\beta \ge 10logT)\hfill \end{array}$$

Let $U=T$, by (3.15), it follows

$$\begin{array}{cc}\hfill & {\int }_T^{T+U}log\left|\mathcal{G}(a+it)\right|dt\le {\displaystyle \frac{T}{2}}log\left(1+{\displaystyle \frac{2\Delta }{T}}\right)+{\displaystyle \frac{T}{2}}log(1+O(L^3/\lambda ))+ϵ\hfill \\ \hfill & \le \Delta +O(T{L}^3/\lambda ).\hfill \end{array}$$

With (3.12), (3.14), (3.19),(3.20) and (3.23), and recall that $a=1/2-1/L$, it follows

$$2\pi N_G\left(D\right)\le {\displaystyle \frac{\Delta +O(T{L}^3/\lambda )+O(c{\Delta }^3/{\lambda }^2)+O\left(L\right)}{1/2-a}}\ll T^{1/3}{L}^{7/3}.$$

i.e.

$${\Delta }_{2T}^{T}argG(1/2+it)\le O\left(T^{1/3}{L}^{7/3}\right).$$

and

$$(N\left(2T\right)-N\left(T\right))-(N_0\left(2T\right)-N_0\left(T\right))\le O\left(T^{1/3}{L}^{7/3}\right).$$

Then let T be $T/2^k,1\le k\le {log}_2\left(T\right)$, and summing. This proves Theorem 1.1 in the case that there are no zeros of $G\left(s\right)$ on the boundary of D.

For the rest case, let $N_1$ and $N_2$ be the numbers of zeros of $G\left(s\right)$ on the left side of D, $\sigma =1/2$, and in D with $\sigma >1/2$, respectively. Indent the left side of D with small semicircles with centers at the zeros and lying in $\sigma \ge 1/2$. Let $N_1^{\prime }$ be the number of distinct zeros in the $N_1$ zeros. Let $V_j$ be the variation in $argG$ in the jth interval between the successive semicircles. Then by the principle of argument, it has

$$\sum _j{V}_j-\pi N_1=2\pi N_2+O\left(L\right),$$

Let $W_j$ be the variation of argument of

$$h\left(s\right)(f^{\prime }\left(s\right)+f^{\prime }(1-s))G\left(s\right)$$

in the jth interval, where $W_j$ is taken for increasing t, while $V_j$ is taken for decreasing t. With (3.2) and (3.24), it has

$$\begin{array}{cc}\hfill \sum _j{W}_j& =\mathrm{Im}\left(f\right){|}_T^{T+U}-\sum _j{V}_j\hfill \\ \hfill & =\mathrm{Im}\left(f\right){|}_T^{T+U}-(2\pi N_2+\pi N_1)+O\left(L\right)\hfill \end{array}$$

By (3.8), in the jth open interval, the number of zeros of $\zeta (1/2+it)$ is at least

$$(W_j/\pi )-1.$$

and in all the open intervals, the number of zeros is at least

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{\pi }}\sum _j{W}_j-N_1^{\prime }-1& ={\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-(2N_2+N_1)-N_1^{\prime }-1+O\left(L\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-2N_G\left(D\right)+N_1-N_1^{\prime }+O\left(L\right)\hfill \end{array}$$

Moreover, by (3.7), we can know that on the side $\sigma =1/2$, a zero of $G\left(s\right)$ is also a zero of ${\zeta }^{\prime }\left(s\right)$, and so a zero of $\zeta \left(s\right)$, with multiplicity one greater, so there are $N_1+N_1^{\prime }$ such zeros of $\zeta (1/2+it)$, adding to (3.26), in total, it has

$$N_0(T+U)-N_0\left(T\right)\ge {\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}-2N_G\left(D\right)+2N_1+O\left(L\right).$$

By (3.11), we can know

$${\displaystyle \frac{1}{\pi }}\mathrm{Im}\left(f\right){|}_T^{T+U}=N(T+U)-N\left(T\right)+O\left(L\right).$$

i.e.

$$(N(T+U)-N\left(T\right))-(N_0(T+U)-N_0\left(T\right))\le O\left(T^{1/3}{L}^{7/3}\right).$$

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Besides, we know that on the critical line a zero of $G\left(s\right)$ is also a zero of ${\zeta }^{\prime }\left(s\right)$, and so a zero of $\zeta \left(s\right)$, with multiplicity one greater. Hence

$$\sum (m-1)\le N_G\left(D\right).$$

where sum is over the distinct zeros of $\zeta \left(s\right)$ on the left side of D, m is the multiplicity of a zero.

And so,

$$\sum _{m\ge 2}m\le 2N_G\left(D\right)\le O\left(T^{1/3}{L}^{7/3}\right).$$

This means that the non-trivial zeros of $\zeta \left(s\right)$ are all on the critical line, and all are simple, with at most $O\left(T^{1/3}{(logT)}^{7/3}\right)$ ones excepted.