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Let $\rho=\beta+i\gamma$ a non-trivial zeros of the Riemann zeta function and $s=\sigma+it$ a complex number. It is possible to prove that $$\frac{\zeta'}{\zeta}\left(s\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{s-\rho}+O\left(\log\left(t\right)\right) \tag{1}$$ uniformly for $-1\leq\sigma\leq2$ (see for example Titchmarsh, “The theory of the Riemann zeta function”, second ed., page $217$). So in particular if we take $\sigma=0$ holds $$\frac{\zeta'}{\zeta}\left(it\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}+O\left(\log\left(t\right)\right). $$ Question: is it possible to prove that $$ \sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}\ll\log\left(t\right)? $$ The problem is that it could be some zeros with real part very close to $0$ and so for $t=\gamma$ the sum is very "big". Thank you.

Addendum: My final goal it's to prove that $$\frac{\zeta'}{\zeta}\left(it\right)=O\left(\log\left(t\right)\right)$$ so also another proof of it (if exists) without the use of $(1)$ is welcome.

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    $\begingroup$ Note that $\beta\gg\frac{1}{\log t}$ by the classical zero-free region and the symmetry $s\leftrightarrow 1-s$ for the zeros, so each term in the sum is $\ll\log t$. In fact the whole sum is $\ll\log t$, as shown by my response below. $\endgroup$
    – GH from MO
    Mar 3, 2016 at 14:57

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I think your final goal follows by taking the logarithmic derivative of the functional equation: $$\frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right).$$ Applying this with $s=it$ and using the familiar asymptotic expansion of $\Gamma'/\Gamma$, we get $$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=O(\log t),\qquad t>2.$$ In fact we get $$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=-\log t+\log(2\pi)-\frac{i}{2t}+O(t^{-2}),\qquad t>2,$$ but observing $O(\log t)$ is sufficient on the right hand side. Indeed, the second term on the left hand side is $O(\log t)$, hence the same holds for the first term.

P.S. Of course this means that your sum is indeed $O(\log t)$. See also my comment below the original post.

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  • $\begingroup$ You say "Indeed, the second term on the left hand side is O(log t)". Sorry for my ignorance - Is this well known? $\endgroup$
    – Wolfgang
    Mar 5, 2016 at 12:45
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    $\begingroup$ @Wolfgang: Yes, this is a classical result. See, for example, Theorem 3.5 in Titchmarsh: The theory of the Riemann zeta-function. The improved bound $O((\log t)^{2/3})$ is also known, see Section 6.19 of the same book. $\endgroup$
    – GH from MO
    Mar 5, 2016 at 16:26
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    $\begingroup$ @Wolfgang: Let me correct my previous comment as I confused $\zeta'/\zeta$ with $\zeta$. Yes, this is a classical result. See, for example, Theorem 5.17 or Theorem 4.11 in Titchmarsh: The theory of the Riemann zeta-function (or Theorem 6.7 in Montgomery-Vaughan: Multiplicative number theory I). The improved bound $O((\log t)^{2/3}(\log\log t)^{1/3})$ is also known, see Section 6.19 of the same book. $\endgroup$
    – GH from MO
    Mar 5, 2016 at 16:50

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