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I am currently trying to prove

$$ N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\mathcal O(\log T) $$

in which $N(T)$ denotes the number of $\zeta$'s nontrivial zeros with imaginary part between $(0,T]$. Currently, using symmetric properties of $\xi(s)$, I am able to obtain

$$ N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\frac78+\frac1\pi\arg\zeta\left(\frac12+iT\right)+\mathcal O\left(\frac1T\right) $$

Apparently, the remaining job is to show that the argument of $\zeta$ on the critical line is of logarithmic growth, and I become stuck on interpreting the meaning of $\arg\zeta$. According to H. M. Edwards' Riemann's zeta function, this argument is bounded by the number of zeros of $\Re\zeta(s)$ on a certain curve (section 6.7 of his book), and I wonder if anybody could provide a more intuitive and clear explanation on that. Thank you!

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    $\begingroup$ I suggest to look up the survey paper of karatsuba and korolev, the argument of the zeta function (free pdf top google search) as it is best reference imho $\endgroup$ – Conrad Jan 17 at 17:24
  • $\begingroup$ @Conrad Although my problem is not solved yet, thank you for providing that source since it convinces me that approximating exponential sums are inevitable to learn :D $\endgroup$ – TravorLZH Feb 3 at 11:59
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Contour integral over three line segments

It can be shown that

$$ \frac1\pi\arg\zeta\left(\frac12+iT\right)={1\over2\pi i}\int_\mathcal L{\zeta'\over\zeta}(s)\mathrm ds $$

in which $\mathcal L$ represents line segments $1/2-iT\to2-iT\to2+iT\to1/2+iT$. Since $|\log\zeta(2+it)|\le\log\zeta(2)$ for all $t\in\mathbb R$, we see that the integral on the line segment $2-iT\to2+iT$ is bounded. Consequently, all we need is to estimate the upper bound for integrals on lines segments $1/2-iT\to2-iT$ and $2+iT\to1/2+iT$. Because the procedures are so similar, we here only present the proof for line segment $2+iT\to1/2+iT$.

Meromorphic expansion of $\zeta'/\zeta$

Since we know little information about the location of $\zeta$'s nontrivial zeros in the critical strip, we consider meromorphically expanding $\zeta'/\zeta$ near $2+iT$. Let $S=\{\rho\in\mathbb C:\zeta(\rho)=0\wedge|\rho-2-iT|\le2\}$, then define

$$ g(z)={\zeta(z+2+iT)\over\zeta(2+iT)}\prod_{\rho\in S}\left(1-{z\over\rho-2-iT}\right)^{-1} $$

Because $|1-z/(\rho-2-iT)|\ge1$ on $|z|=4$, it follows from maximum modulus principle that if we set

$$ M=\max_{|z|\le4}\left|\zeta(z+2+iT)\over\zeta(2+iT)\right| $$

then $\log|g(z)|\le\log M$. By Borel-Caretheodory lemma, we see that

$$ {g'\over g}(z)\ll\log M $$

for all $|z|\le3/2$. Now, all we need is to find an explicit estimate of $M$:

Growth of $\zeta$ and $\zeta'/\zeta$

According to The theory of Riemann zeta-function by E.C. Titchmarsh, we know that there exists a fixed positive constant such that $\zeta(\sigma+it)\ll|t|^A$ for large $|t|$ for all $\sigma\in\mathbb R$. This indicates

$$\log|\zeta(\sigma+it)|\ll\log|t|$$

Moreover because

$$ |\zeta^3(\sigma)\zeta^4(\sigma+it)\zeta(\sigma+2it)|\ge1 $$

we see that

$$ -\log|\zeta(\sigma+it)|\ll\log|t| $$

for all $\sigma>1$. Consequently we have $\log M\ll\log T$. This, combining with the definition of $g$, indicates that

$$ {\zeta'\over\zeta}=\sum_{\rho\in S}{1\over s-\rho}+\mathcal O(\log T) $$

when $s$ is on the line segment $[1/2+iT,2+iT]$.

Final shot: the integral is $\mathcal O(\log T)$

$$ \begin{aligned} \int_{2+iT}^{1/2+iT}{\zeta'\over\zeta}(s)\mathrm ds &=\sum_{\rho\in S}\int_{2+iT}^{1/2+iT}{\mathrm ds\over s-\rho}+\mathcal O(\log T) \\ &\ll\sum_{\rho\in S}1+\mathcal O(\log T) \end{aligned} $$

To estimate the sum, we apply Jensen's inequality:

$$ \sum_{\rho\in S}1\le{\log M\over4/2}\ll\log T $$

Therefore, the expansion of Riemann-von Mangoldt formula becomes

$$ N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\mathcal O(\log T) $$

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