数理解析研究所講究録
第 1874 巻 2014 年 12-21

12The explicit upper bound of the multiple integral of $S(t)$ on the
Riemann Hypothesis
名古屋大学多元数理科学研究科若狭尊裕

Takahiro Wakasa
Graduate School of Mathematics, Nagoya University
Abstract

We prove explicit upper bounds of the function $S_{m}(T)$ , defined by the repeated integration of theargument of the Riemann zeta-function. The explicit upper bound of $S(T)$ and $S_{1}(T)$ have alreadybeen obtained by A. Fujii. Our result is a generalization of Fujii’s results.

1

Introduction, where $s=\sigma+ti$ is a complex variable, on
We consider the argument of the Riemann zeta function
the critical line
We shall give some explicit bounds on $S_{m}(T)$ defined below under the Riemann hypothesis.
( is not the ordinate of any zero of. When),We introduce the functions $S(t)$ andwe define$\zeta(s)$$\sigma=\frac{1}{2}.$$S_{1}(t)$$T\neq\gamma$$\zeta(s)$$\gamma$$S(T)= \frac{1}{\pi}\arg\zeta(\frac{1}{2}+Ti)$.

This is obtained by continuous variation along the straight lines connecting 2, $2+Ti$ , andwith the value zero. When $T=\gamma$ , we define$\frac{1}{2}+Ti$, starting$S(T)= \frac{1}{2}\{S(T+O)+S(T-O)\}.$

Next, we define $S_{1}(T)$ by$S_{1}(T)= \int_{0}^{T}S(t)dt+C,$$(C= \frac{1}{\pi}\int_{1}^{\infty}logz|\zeta(\sigma)|d\sigma$: $constant)$ .

It is a classical results of von Mangoldt (cf. chapter 9 of Titchmarsh [7]) that there exists a number
such that for $T>T_{0}$ we have$T_{0}>0$$S(T)=O(\log T) , S_{1}(T)=O(\log T)$ .

Further, it is a classical result of Littlewood [8] that under the Riemann Hypothesis we have
$S(T)=O( \frac{\log T}{\log\log T}) , S_{1}(T)=O(\frac{\log T}{(\log\log T)^{2}})$.

For explicit upper bounds of $|S(T)|$ and $|S_{1}(T)|$ , Karatsuba and Korolev (cf. Theorem 1 and Theorem
2 on [9] have shown that$)$$|S(T)|<8\log T, |S_{1}(T)|<1.2\log T$

13for $T>T_{0}$ . Also, under the Riemann Hypothesis, it was shown that$|S(T)| \leq 0.83\frac{\log T}{\log\log T}, |S_{1}(T)|\leq 0.51\frac{\log T}{(\log\log T)^{2}}$for $T>T_{0}$ by Fujii.
Next, we introduce the functions $S_{2}(T),$ $S_{3}(T),$. When, we put$\rho=\beta+\gamma i$$\cdots$. And the non-trivial zeros of$\zeta(s)$we denote by$T\neq\gamma$$\ovalbox{\tt\small REJECT}(T)=S(T) , S_{m}(T)=\int_{0}^{T}S_{m-1}(t)dt+C_{m}$for any integer $m\geq 1$ , where$C_{m}$’s are the constants which are defined by, for any integer $k\geq i,$$C_{2k-1}= \frac{1}{\pi}(-1)^{k-1}\underline{\int_{5}^{\infty}\int_{\sigma}^{\infty}\cdots\int_{\sigma}^{\infty}}\log|\zeta(\sigma)|(d\sigma)^{2k-1},$$(2k-1)$ -times

and$C_{2k}=(-1)^{k-1}(d \sigma)^{2k}=\frac{(-1)^{k-1}}{(2k)!2^{2k}}\frac{\int_{1}^{\infty}\int_{\sigma}^{\infty}\cdots\int_{\sigma}^{\infty}z}{2k-times}.$

When $T=\gamma$ , we put$S_{m}(T)= \frac{1}{2}\{S_{m}(T+0)+S_{m}(T-0)\}.$

Concerning $S_{m}(T)$ for $m\geq 2$ , Littlewood [8] have shown under the Riemann Hypothesis that
$S_{m}(T)=O( \frac{\log T}{(\log\log T)^{m+1}})$.

Theorem 1.
Under the Riemann Hypothesis for any integer $m\geq 1$ , if $m$ is odd,$|S_{m}(t)| \leq\frac{\log t}{(\log\log t)^{m+1}}\cdot\frac{1}{2\pi m!}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}})$$+ \frac{1}{m+1}\cdot 1-\frac{(1}{e}(1\frac{1}{e})\frac{1}{e}1+\frac{1}{+e})+\frac{1}{m(m+1)}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\}$$+O( \frac{\log t}{(\log\log t)^{m+2}})$

If

$m$

.

is even,$|S_{m}(t)| \leq\frac{\log t}{(\log\log t)^{m+1}}\cdot\frac{1}{2\pi m!}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}})$$+ \frac{1}{m+1}\cdot\frac{\frac{1}{e}(1+\frac{1}{e})}{1-\frac{1}{e}(1+\frac{1}{e})}+\frac{\pi}{2}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\}+O(\frac{\log t}{(\log\log t)^{m+2}})$.

This result is a generalization of the known explicit upper bounds for $S(T)$ and $S_{1}(T)$ . It is to be
stressed that the argument when the number of integration is odd is different from that when the numberof integration is even.

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The basic policy of the proof of this result is based on A. Fujii [1]. In the case when $m$ is odd, we candirectly generalize the proof of A. Fujii [1]. In the case when $m$ is even, it is an extension of the methodof A. Fujii [2].
To prove this result, we introduce some more notations. First, we define the function $I_{m}(T)$ as follows.
, we put for any integer $k\geq 1$
When$T\neq\gamma$$I_{2k-1}(T)= \frac{1}{\pi}(-1)^{k-1}\Re\{\underline{\int^{\infty}\int_{\sigma}^{\infty}\cdots\int_{\sigma}^{\infty}}\log\zeta(\sigma+Ti)(d\sigma)^{2k-1}\}$$(2k-1)$ -times

and$I_{2k}(T)= \frac{1}{\pi}(-1)^{k}\Im\{_{\frac{\int_{1}^{\infty}\int_{\sigma}^{\infty}\cdots\int_{\sigma}^{\infty}q}{2k-times}}\log\zeta(\sigma+Ti)(d\sigma)^{2k}\}.$

When $T=\gamma$ , we put for $m\geq 1$$I_{m}(T)= \frac{1}{2}\{I_{m}(T+0)+I_{m}(T-0)\}.$

Then, $I_{m}(T)$ can be expressed as a single integral of the following form (cf. Lemma 2 in Fujii [3]): for

any integer $m\geq 1$$I_{m}(T)=- \frac{1}{\pi}\Im\{\frac{i^{m}}{m!}\int_{ゴ}^{\infty}(\sigma-\frac{1}{2})^{m}\frac{\zeta’}{\zeta}(\sigma+Ti)d\sigma\}.$From this expression, it is known under the Riemann Hypothesis that $S_{m}(T)=I_{m}(T)$ by Lemma 2 inFujii [4].
Therefore, we should estimate $I_{m}(T)$ .

Some lemmas

2

Let $s=\sigma+ti$ . We suppose that
Also, we put$\sigma\geq\frac{1}{2}$$\sigma_{1}=\frac{1}{2}+\frac{1}{\log X},$

and $t\geq 2$ . Let $X$ be a positive number satisfying $4\leq X\leq t^{2}.$$\Lambda_{X}(n)=\{\begin{array}{ll}\Lambda(n) for 1\leq n\leq X,\Lambda(n)\frac{\log\frac{x^{2}}{n}}{\log X} for X\leq n\leq X^{2},\end{array}$with$\Lambda(n)=\{\begin{array}{ll}\log p if n=p^{k} with a prime p and an integer k\geq 1,0 otherwise.\end{array}$

Lemma 1.

Let $t\geq 2,$ $X>0$ such that $4\leq X\leq t^{2}$ . For$\sigma\geq\sigma_{1}=\frac{1}{2}+\frac{1}{\log X},$$\frac{\zeta’}{\zeta}(\sigma+ti)=-\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma+t1}}-\frac{(1+X^{1}z^{-\sigma})\omega X^{1}\pi^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})$$+ \frac{(1+X\xi-\sigma)\omega X^{1}l^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(X^{1}z^{-\sigma})$where$|\omega|\leq 1,$$-1\leq\omega’\leq 1.$,

15This has been proved in Fujii [1].

Lemma 2. (cf. 2.12.7 of Titchmarsh[7])$\frac{\zeta’}{\zeta}(s)=\log 2\pi-1-\frac{E}{2}-\frac{1}{s-1}-\frac{1}{2}\cdot\frac{\Gamma’}{\Gamma}(\frac{s}{2}+1)+\sum_{\rho}(\frac{1}{s-\rho}+\frac{1}{\rho})$$= \log 2\pi-1-\frac{E}{2}-\frac{1}{s-1}-\frac{1}{2}\log(\frac{s}{2}+1)+\sum_{\rho}(\frac{1}{s-\rho}+\frac{1}{\rho})+O(\frac{1}{|s|})$where $E$ is the Euler constant and$\rho$

runs through zeros of$\zeta(s)$.

Lemma 3. (Lemma 1 of Selberg [6])$ForX>1,$ $s\neq 1,$ $s\neq-2q(q=1,2,3, \cdots),$ $s\neq\rho)$$\frac{\zeta’}{\zeta}(s)=-\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{8}}+\frac{X^{2(1-s)}-X^{1-s}}{(1-s)^{2}\log X}+\frac{1}{\log X}\sum_{q=1}^{\infty}\frac{X^{-2q-s}-X^{-2(2q+\epsilon)}}{(2q+s)^{2}}+\frac{1}{\log X}\sum_{\rho}\frac{X^{\rho-s}-X^{2(\rho-e)}}{(s-\rho)^{2}}.$

By Lemma 2, we have

(1)$\Re\frac{\zeta’}{\zeta}(\sigma_{1}+ti)=-\frac{1}{2}\log t+\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}+O(1)$Since for$\sigma_{1}\leq\sigma$$\frac{1}{\log X}|\sum_{\rho}\frac{X^{\rho-\epsilon}-X^{2(\rho-\epsilon)}}{(s-\rho)^{2}}|\leq(1+X^{1}z^{-\sigma})X^{1}z^{-\sigma}\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}},$

we have$\frac{1}{\log X}\sum_{\rho}\frac{X^{\rho-\epsilon}-X^{2(\rho-\epsilon)}}{(s-\rho)^{2}}=(1+X^{1}z^{-\sigma})X^{1}z^{-\sigma}.\omega\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}},$

where$|\omega|\leq 1$. Since for$\sigma\geq\frac{1}{2}$and$X\leq t^{2}$$| \frac{X^{2(1-s)}-X^{1-s}}{(1-s)^{2}\log X}|\ll\frac{X^{2(l-\sigma)}}{t^{2}\log X}\leq\frac{X^{1}z^{-\sigma}}{\log X},$

we have for$\sigma_{1}\leq\sigma$$\frac{\zeta’}{\zeta}(\sigma+ti)=-\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma+ti}}+O(\frac{X^{1}\Sigma^{-\sigma}}{\log X})+(1+X^{\frac{1}{2}-\sigma})\omega X^{1}z^{-\sigma}\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}$by Lemma 3. Especially,,$\Re\frac{\zeta’}{\zeta}(\sigma_{1}+ti)=\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})+O(\frac{1}{\log X})+(1+\frac{1}{e})\frac{1}{e}\omega’\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}$(2)where-lHence by (1) and (2), we get$\leq\omega’\leq 1.$.

$\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}=\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)$

This relation will be used in the following proof of Theorem 1.

(3)

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3

Proof of Theorem 1 in the case when$m$is oddIf $m$ is odd, we have$I_{m}(t)$$=$$\frac{i^{m+1}}{\pi m!}\Im\{i\{$ $\int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\frac{\zeta’}{\zeta}(\sigma+ti)d\sigma+\frac{(\sigma_{1}-\frac{1}{2})^{m+1}}{m+1}\cdot\frac{\zeta’}{\zeta}(\sigma_{1}+ti)$$- \int_{1}^{\sigma_{1}}q(\sigma-\frac{1}{2})^{m}\{\frac{\zeta’}{\zeta}(\sigma_{1}+ti)-\frac{\zeta’}{\zeta}(\sigma+ti)\}d\sigma\}\}$$= \frac{i^{m+1}}{\pi m!}\Im\{i(J_{1}+J_{2}+J_{3})\},$say.
First, we estimate$J_{1}$. By Lemna 1,$J_{1}= \int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\{-\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma+ti}}-\frac{(1+X^{1}z^{-\sigma})\omega X^{1}z^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega}\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})$$+ \frac{(1+X^{1}z^{-\sigma})\omega X^{1}z^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega}\cdot\frac{1}{2}\log t+O(X^{1}z^{-\sigma})\}d\sigma$$=- \int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma+ti}}d\sigma+\eta_{1}(t)$,$=- \sum_{j=0}^{m}(\frac{m!}{(m-j)!}(\sigma_{1}-\frac{1}{2})^{m-j}\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}(\log n)^{j+1}})+\eta_{1}(t)$,say. And we have$| \eta_{1}(t)|=|\int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\frac{(1+X\int-\sigma)\omega Xz^{-\sigma}1}{1-\frac{1}{e}(1+\frac{1}{e})\omega}d\sigma|\cdot|-\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})+\frac{1}{2}\log t|$$+O \{\int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}X^{1}z^{-\sigma}d\sigma\}$$\leq\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\cdot\frac{1}{2}\log t\cdot\frac{1}{(\log X)^{m+1}}(\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}}))$$+O( \frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+t\dot{\iota}}}|)$$= \eta_{2}(t)+O(\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)$,

say, since by partial integration$\int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}(1+X^{1}z^{-\sigma})X^{S-\sigma}d\sigma=\frac{1}{(\log X)^{m+1}}(\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}}))$

Next, applying Lemma 1 to$J_{2}$, we get$J_{2}= \frac{l}{(m+1)(\log X)^{m+1}}\cdot\frac{(1+\frac{1}{e})\frac{1}{\epsilon}\omega}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O\{\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+t}1}|\}$$= \eta_{3}(t)+O\{\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|\},$

.

17
say.
Next, we estimate$J_{3}$. By Lemma 2, we have$\Im(iJ_{3})=-\sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}.$$\int_{z}^{\sigma_{1}}1(\sigma-\frac{1}{2})^{m}\frac{(\sigma_{1}-\sigma)\{(t-\gamma)^{2}-(\sigma_{1}-\frac{1}{2})(\sigma-\frac{1}{2})\}}{(\sigma-\frac{1}{2})^{2}+(t-\gamma)^{2}}d\sigma$$+O( \frac{1}{t(\log X)^{m+1}})$$=- \sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}\cdot K(\gamma)+O(\frac{1}{(\log X)^{m+1}})$say, where
If$\gamma$is the imaginary part of,$\rho=\beta+\gamma i.$$t=\gamma,$$K( \gamma)=-\frac{1}{m(m+1)}(\sigma_{1}-\frac{1}{2})^{m+2}$

If$t\neq\gamma$, by putting$\sigma-\frac{1}{2}=v,$$\sigma_{1}-\frac{1}{2}=\frac{1}{\log X}=\Delta$

and $|t-\gamma|=B$ , we get$K( \gamma)=\int_{0}^{\Delta}v^{m}\frac{(\triangle-v)(B^{2}-\triangle v)}{v^{2}+B^{2}}dv=\frac{\Delta^{m+2}}{m+1}-\frac{(B^{2}+\Delta^{2})\triangle^{m}}{m}+\int_{0}^{\Delta}\frac{(B^{2}+\Delta^{2})v^{m-1}}{(\frac{v}{B})^{2}+1}dv.$

Putting$\frac{v}{B}=u$, we have$K( \gamma)=\frac{\triangle^{m+2}}{m+1}-\frac{(B^{2}+\Delta^{2})\Delta^{m}}{m}+(B^{2}+\Delta^{2})\int_{0^{B}}^{\Delta}\frac{(uB)^{m-1}B}{1+u^{2}}$d鋭$= \Delta^{m+2}\{\frac{1}{m+1}-\frac{B^{2}}{m\triangle^{2}}-\frac{1}{m}+(\frac{B^{m+2}}{\Delta^{m+2}}+\frac{B^{m}}{\Delta^{m}})i^{m+1}\{\sum_{j=1}^{\underline{m}_{5}\underline{-1}}\frac{(-1)^{j-1}}{2j-1}(\frac{\triangle}{B})^{2j-1}-$

Putting$y= \frac{\Delta}{B}$arctan$( \frac{\Delta}{B})\}\}.$, we get$K( \gamma)=\Delta^{m+2}(g(y)-\frac{1}{m(m+1)})$

,

(4)

where
$g(y)= \{-i^{m+1}(\frac{1}{y^{m+2}}+\frac{1}{y^{m}})$Whenarctantends to $g(y)$ is convergent to$y>0$ , we get $g’(y)<0$ , so that
$y$

$0,$

$y- \frac{1}{my^{2}}+i^{m+1}(\frac{1}{y^{m+2}}+\frac{1}{y^{m}})\sum_{j=1}^{\underline{m}_{7}-\underline{1}}\frac{(-1)^{j-1}}{2j-1}y^{2j-1}\}.$$\frac{2}{m(m+2)}$. When$y$tends to infinity, $g(y)$ tends to . Hence for$- \frac{1}{m(m+1)}\leq g(y)-\frac{1}{m(m+1)}\leq\frac{1}{(m+1)(m+2)}$

$0$

.

(5)

Therefore by (4) and (5), we obtain$- \frac{1}{m(m+1)}(\sigma_{1}-\frac{1}{2})^{m+2}\leq K(\gamma)\leq\frac{1}{(m+1)(m+2)}(\sigma_{1}-\frac{1}{2})^{m+2}$

Hence$- \sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}K(\gamma)\leq\frac{(\sigma_{1}-\frac{1}{2})^{m+2}}{m(m+1)}\sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}$

(6)

18
and
$- \sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}K(\gamma)\geq-\frac{(\sigma_{1}-\frac{1}{2})^{m+2}}{(m+1)(m+2)}\sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}$.

(7)By (3), (6) and (7), we have$- \sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}K(\gamma)\leq\frac{(\sigma_{1}-\frac{1}{2})^{m+1}}{m(m+1)}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+0(|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)\}$

and$- \sum_{\gamma}\frac{1}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}K(\gamma)\geq-\frac{(\sigma_{1}-\frac{1}{2})^{m+1}}{(m+1)(m+2)}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)\}.$

Hence$|i^{m+1} \Im(iJ_{3})|\leq\frac{1}{m(m+1)}.$$\frac{1}{(\log X)^{m+1}}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)$$= \eta_{5}(t)+0(\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)$

Therefore, we obtain
漏$(t)= \frac{1}{\pi m!}\{-i^{m+1}\sum_{j=0}^{m}(\frac{m!}{(m-j)!}(\sigma_{1}-\frac{1}{2})^{m-j}\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}(\log n)^{j+1}})$,(8)$+O( \frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)\}+\frac{1}{\pi m!}\cdot\Xi(t)$where$\Xi(t)$satisfies the following inequalities.

$| \Xi(t)|\leq\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\cdot\frac{1}{2}\log t\cdot\frac{1}{(\log X)^{m+1}}(\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2e}))$$+ \frac{1}{m+1}\cdot\frac{(1+\frac{1}{e})\frac{1}{e}\omega}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t\cdot\frac{1}{(\log X)^{m+1}}$$+ \frac{1}{m(m+1)}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t\cdot\frac{1}{(\log X)^{m+1}}.$

In (8), we have$| \sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+t\dot{\iota}}}|\leq\sum_{n<X}\frac{\Lambda(n)}{n^{1}}+\sum_{x\leq n\leq X^{2}}\frac{\Lambda(n)\log\frac{X^{2}}{n}}{n^{l}r}\cdot\frac{1}{\log X}\ll\frac{X}{\log X}$

(9)

,

.

$| \sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+t}1(\log n)^{j+1}}|\leq\sum_{n<X}\frac{\Lambda(n)}{n^{1}z(\log n)^{j+1}}+\sum_{x\leq n\leq X^{2}}\frac{\Lambda(n)\log\frac{X^{2}}{n}}{n^{\iota}\tau(\log n)^{j+1}}\cdot\frac{1}{\log X}\ll\frac{X}{(\log X)^{j+2}}$

We estimate that the first term and the second term on the right-hand side of (8) is$\ll\frac{X}{(\log X)^{m+s}}.$

(10)

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Therefore, taking $X=\log t$ , we obtain$|I_{m}(t)|= \frac{1}{\pi m!}\Xi(t)+O(\frac{\log t}{(\log\log t)^{m+2}})$$= \frac{\log t}{(\log\log t)^{m+1}}\cdot\frac{1}{2\pi m!}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}})$$\frac{1}{e}(1+\frac{1}{e})$$+ \frac{1}{m+1}\cdot 1-\frac{1}{e}(1+\frac{1}{e})+\frac{1}{m(m+1)}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\}+O(\frac{\log t}{(\log\log t)^{m+2}})$.

This is the first part of the result.

4

Proof of Theorem 1 in the case when $m$ is even

If $m$ is even, we get similarly$I_{m}(t)$$=$$\frac{-i^{m}}{\pi m!}\Im\{\{$ $\int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\frac{\zeta’}{\zeta}(\sigma+ti)d\sigma+\frac{(\sigma_{1}-\frac{1}{2})^{m+1}}{m+1}\cdot\frac{\zeta’}{\zeta}(\sigma_{1}+ti)$$- \int_{1}^{\sigma_{1}}z(\sigma-\frac{1}{2})^{m}\{\frac{\zeta’}{\zeta}(\sigma_{1}+ti)-\frac{\zeta’}{\zeta}(\sigma+ti)\}d\sigma\}\}$$= \frac{-i^{m}}{\pi m!}\Im\{(J_{1}+J_{2}+J_{3})\},$

say. By Lemma 1 and (9), we have
$J_{1}=- \int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma+ti}}d\sigma+l_{1}^{\infty}(\sigma-\frac{1}{2})^{m}o(X^{1}z^{-\sigma})d\sigma$$+ \int_{\sigma_{1}}^{\infty}(\sigma-\frac{1}{2})^{m}\{-\frac{(1+X^{1}z^{-\sigma})\omega X^{1}a^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})+\frac{(1+X^{1}z^{-\sigma})\omega X^{1}z^{-\sigma}}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t\}d\sigma$$= \sum_{j=0}^{m}\frac{m!}{(m-j)!}(\sigma_{1}+\frac{1}{2})^{m-j}\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}(\log n)^{j+1}}+O(\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)+\eta_{1}’(t)$$\ll\frac{X}{(\log X)^{m+2}}+\eta_{1}’(t)$,say, and$J_{2}= \frac{l}{(m+1)(\log X)^{m+1}}\{\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}-\frac{(1+\frac{1}{e})\frac{1}{e}\omega}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\Re(\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}})$$+ \frac{(1+\frac{1}{e})\frac{1}{e}\omega}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(X^{1}z^{-\sigma_{1}})\}$$= \frac{l}{(m+1)(\log X)^{m+1}}\cdot\frac{(1+\frac{1}{e})\frac{1}{e}\omega}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O\{\frac{1}{(\log X)^{m+1}}|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|\}$$\ll\eta_{3}’(t)+\frac{X}{(\log X)^{m+2}},$say. As well as$\eta_{1}(t)$, we have$| \eta_{1}’(t)|\leq\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\cdot\frac{1}{2}\log t\cdot\frac{1}{(\log X)^{m+1}}(\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}}))$.

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Finally, we estimate$J_{3}$. By Stirling’s formula, we get$| \frac{\Gamma’}{\Gamma}(\frac{\sigma_{1}+ti}{2}+1)|=|\frac{i}{2}\log\frac{ti}{2}+(\frac{\sigma_{1}+ti+1}{2})\frac{1}{t}-\frac{i}{2}+O(\frac{1}{t})|\leq\frac{1}{2}\log t+O(\frac{1}{t})$

.

(11)

is estimated similarly.
Also
Hence by (11) and Lemma 2, we have$|_{T}^{\Gamma’}( \frac{\sigma+u}{2}+1)|$$| \Im\{\frac{\zeta’}{\zeta}(\sigma_{1}+ti)-\frac{\zeta’}{\zeta}(\sigma+ti)\}|\leq\sum_{\gamma}\frac{(t-\gamma)\{(\sigma-\frac{1}{2})^{2}-(\sigma_{1}-\frac{1}{2})^{2}\}}{\{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}\}\{(\sigma-\frac{1}{2})^{2}+(t-\gamma)^{2}\}}+O(\frac{1}{t})$.

Therefore,$| \Im(J_{3})|\leq|\int_{\pi}^{\sigma_{1}}(\sigma-\frac{1}{2})^{m}\sum_{\gamma}\frac{(t-\gamma)\{(\sigma-\frac{1}{2})^{2}-(\sigma_{1}-\frac{1}{2})^{2}\}}{\{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}\}\{(\sigma-\frac{1}{2})^{2}+(t-\gamma)^{2}\}}d\sigma|$$+ \int_{1}^{\sigma_{1}}z(\sigma-\frac{1}{2})^{m}\cdot 0(\frac{1}{t})d\sigma.$

If$t=\gamma$, the first term of the right-hand side of above inequality is . If
$0$$t\neq\gamma$, since$\sigma<\sigma_{1}$, we have$| \int_{1}^{\sigma_{1}}q(\sigma-\frac{1}{2})^{m}\{\sum_{\gamma}\frac{(t-\gamma)\{(\sigma-\frac{1}{2})^{2}-(\sigma_{1}-\frac{1}{2})^{2}\}}{\{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}\}\{(\sigma-\frac{1}{2})^{2}+(t-\gamma)^{2}\}}\}d\sigma|$$< \sum_{\gamma}\frac{(\sigma_{1}-\frac{1}{2})^{m+2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}\int^{\infty}\frac{|t-\gamma|}{(\sigma-\frac{1}{2})^{2}+(t-\gamma)^{2}}d\sigma\leq\frac{\pi}{2}(\sigma_{1}-\frac{1}{2})^{m+1}\sum_{\gamma}\frac{\sigma_{1}-\frac{1}{2}}{(\sigma_{1}-\frac{1}{2})^{2}+(t-\gamma)^{2}}.$

Applying (3) and (9), and taking $X=\log t$ lastly, the right-hand side of above inequality is$\leq\frac{\pi}{2}(\sigma_{1}-\frac{1}{2})^{m+1}\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})\omega’}\cdot\frac{1}{2}\log t+O(|\sum_{n<X^{2}}\frac{\Lambda_{X}(n)}{n^{\sigma_{1}+ti}}|)\}$$\leq\frac{\pi}{4}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\cdot\frac{\log t}{(\log\log t)^{m+1}}+O(\frac{\log t}{(\log\log t)^{m+2}})$

.

(12)

.

(13)

Also,
$\int_{1}^{\sigma_{1}}z(\sigma-\frac{1}{2})^{m}\cdot O(\frac{1}{t})d\sigma=0(\frac{1}{t(\log X)^{m+1}})$

By (12) and (13),$| \Im(J_{3})|\leq\frac{\pi}{4}\cdot\frac{1}{1-\frac{1}{\epsilon}(1+\frac{1}{e})}\cdot(\log\log t)^{m+1}\log t+O(\frac{l}{t(\log\log t)^{m+1}})+O(\frac{\log t}{(\log\log t)^{m+2}})$.

Therefore, we obtain$|S_{m}(t)| \leq\frac{1}{2\pi m!}\cdot(\log\log t)^{m+1}\log t\{\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\sum_{j=0}^{m}\frac{m!}{(m-j)!}(\frac{1}{e}+\frac{1}{2^{j+1}e^{2}})$$+ \frac{1}{m+1}\cdot 1-\frac{1}{e}(+\frac{1}{e})(1+\frac{1}{1e})\frac{1}{e}+\frac{\pi}{2}\cdot\frac{1}{1-\frac{1}{e}(1+\frac{1}{e})}\}+O(\frac{\log t}{(\log\log t)^{m+2}})$

.
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References[1] A. Fujii, A Note on the Distribution of the Argument of the Riemann Zeta Function, Comment. Math.
Univ. Sancti Pauli, 55, (2006), 135-147.
[2] A. Fujii, An explicit estimate in the theory of the distribution of the zeros of the Riemann zetafunction, Comment. Math. Univ. Sancti Pauli, 53, (2004), 85-114.
[3] A. Fujii, On the zeros of the Riemann zeta function, Comment. Math. Univ. Sancti Pauli 51, (2002),

1-17.

[4] A. Fujii, On the zeros of the Riemann zeta function(2003), 165-190.

$\Pi$, Comment. Math. Univ. Sancti Pauli 52,[5] A. Selberg, On the Remainder in the formula for $N(T)$ , the number of zeros of$0<t<T$ , Avh. Norske Vid. Akad. Oslo. I. No. 1, 1944.

$\zeta(s)$in the strip[6] A. Selberg, Collected Works, vol I, 1989, Springer.

[7] E. C. Titchmarsh, The Theory of the Riemann Zeta-function, Second Edition; Revised by D. R.
Heath-Brown. Clarendon Press Oxford, 1986.
[8] J. E. Littlewood, On the zeros of the Riemann zeta function, Proc. Camb. Phil. Soc, 22, (1924),
295-318.
[9] A. A. Karatsuba and M. A. Korolev, The argument of the Riemann zeta function, Russian Math.
Surveys, 60:3, (2005), 41-96.

[10] D. A. Goldston and S. Gonek,
note on $S(t)$ and the zeros of the Riemann zeta funcion, Bull.
London Math. Soc. 39, (2007), 482-486.
$A$