Let $\zeta$ be the zeta function of Riemann. Is the bound for 


$$I\_{T}=\int\_{0}^{T} \Big|\log|\zeta(1/2 + it)| \Big| \mathrm{d}t$$
known ? 


It seems to me that $I\_{T} \ll T\log T$ since $\log|\zeta(1/2+it)|\ll \log t$ if $t$ is not an ordinate of a zero. But certainly this is too naive. Is there a better bound ?



Inquiry on the bound for $\int_{0}^{T} \Big|\log|\zeta(1/2 + it)| \Big| \mathrm{d}t$
We have $I\_{T} \ll T\log T$ for $T\geq 2$. For this it suffices to verify that
$$ \int\_{T}^{T+1} \Big|\log|\zeta(1/2 + it)|\Big|\ dt\ll\log T,\qquad T\geq 2.\tag{$\ast$}$$
We can deduce this local bound from Theorem 9.6 (B) and surrounding material in Titchmarsh: The theory of the Riemann zeta-function. Indeed, this theorem implies (after taking the real part and applying the triangle inequality) that
$$\Big|\log|\zeta(1/2 + it)|\Big|\leq\sum\_{|t-\gamma|\leq 1}
\Big|\log|1/2 + it-\rho)|\Big|+O(\log t),\qquad t\geq 2,$$
where $\rho=\beta+i\gamma$ runs through the zeros of $\zeta(s)$. It follows that
$$\Big|\log|\zeta(1/2 + it)|\Big|\leq\sum\_{|t-\gamma|\leq 1}
\Big(1-\log|t-\gamma|\Big)+O(\log t),\qquad t\geq 2,$$
since $\log|t-\gamma|$ is nonpositive throughout the sum. The contribution of the term $1$ in the sum is $O(\log t)$ by Theorem 9.2, hence in fact
$$\Big|\log|\zeta(1/2 + it)|\Big|\leq-\sum\_{|t-\gamma|\leq 1}
\log|t-\gamma|+O(\log t),\qquad t\geq 2.$$
On the other hand, from the proof of Theorem 9.7 we learn that there is a constant $A>0$ such that
$$\int\_T^{T+1}\sum\_{|t-\gamma|\leq 1}\log|t-\gamma|\ dt\geq -A\log T,\qquad T\geq 2.$$
Combining the previous two inequalities, the bound $(\ast)$ follows.


