On the asymptotic bound for $\arg\zeta(s)$ on the critical line 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!
 A: 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
$$
N(T)={T\over2\pi}\log{T\over2\pi}-{T\over2\pi}+\mathcal O(\log T)
$$
