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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 ?

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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.

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    $\begingroup$ Wow, that's really ingenious of you ! Thanks ! $\endgroup$ – ConsiderTheMathOverFlaw Dec 15 '18 at 16:56
  • $\begingroup$ @GHfroMO, done. $\endgroup$ – ConsiderTheMathOverFlaw Dec 15 '18 at 18:42
  • $\begingroup$ @ConsiderTheMathOverFlaw: At mathoverflow.net/questions/316668 I claimed that $\int_T^{2T}\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma})$ implies $\int_0^T\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma})$. This is can be seen as follows. Let $T\geq 2$ be fixed, and let $K$ be the largest integer for which $2^K\leq T$. Then, for any integer $1\leq k\leq K$, we have $\int_{T/2^k}^{2T/2^k}\zeta(\sigma+it)\,dt = T/2^k+O_\sigma(T^{1-\sigma}/2^{k(1-\sigma)})$. The sum of these yields $\int_{T_0}^T\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma})$, where $T_0:=T/2^K$ lies in $[1,2]$. Done. $\endgroup$ – GH from MO Dec 15 '18 at 18:43
  • $\begingroup$ Thanks, but where did $$\int_{T/2^k}^{2T/2^{k}} \zeta(\sigma+ it) \mathrm{d}t = T/2^{k} + O_{\sigma}(T^{1-\sigma}/2^{k(1-\sigma)})$$ come from ? $\endgroup$ – ConsiderTheMathOverFlaw Dec 15 '18 at 18:51
  • $\begingroup$ @ConsiderTheMathOverFlaw: It came from $\int_T^{2T}\zeta(\sigma+it)\,dt = T+O_\sigma(T^{1-\sigma})$ that I proved under that link. $T$ is arbitrary, so one can replace it by $T/2^k$. $\endgroup$ – GH from MO Dec 15 '18 at 20:47

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