About the logarithmic derivative of the Riemann zeta function

Question

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.

Answer 1

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.