The order of $\zeta'/\zeta$

Assume the Riemann Hypothesis. Define $\nu(\sigma)$ for $\sigma > 1/2$ to be the greatest lower bound of the numbers $\nu$ for which

\begin{displaymath}\frac{\zeta'}{\zeta}(\sigma+it) \ll_\epsilon (\log t)^{\nu +\epsilon}\end{displaymath}

holds as $t\to \infty$ for all $\epsilon >0$. It is a theorem that $\nu(\sigma)$ is a convex function of $\sigma$ which is continuous and decreasing for $\sigma > 1/2$ with $\nu(\sigma)=0$ for all $\sigma \ge 1.$ It can be shown that

\begin{displaymath}1-\sigma \le \nu(\sigma) \le 2 - 2\sigma \end{displaymath}

for $1/2<\sigma < 1$. There is an analogous function which can be defined for $\log\vert\zeta(\sigma+it))$ and it can be shown that this analogous function is, in fact, equal to $\nu(\sigma)$. See Titchmarsh for all of these facts. Which bound is corrct?

If the smaller bound is the correct one, then near the half-line we see that

\begin{displaymath}\frac{\zeta'}{\zeta}(1/2+a+it)\ll_\epsilon (\log t)^ {1/2-a+\epsilon}\end{displaymath}

for $a>0$. On the other hand,

\begin{displaymath}\Re \frac{\zeta'}{\zeta}(1/2+it)=\frac{1}{2}\frac{\chi'}{\chi}(1/2+it)=-\frac12\log t+O(1).\end{displaymath}

Thus, if the smaller bound holds, then there is a jump at 1/2. It could be that the smaller bound holds to the right of the critical-line but that there is radically different behavior on the critical line.

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