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The explicit formula for the logarithmic derivative $\frac{\zeta'(s)}{\zeta(s)}$ is illustrated in (1) below.


(1) $\quad\frac{\zeta'(s)}{\zeta(s)}=\frac{s}{1-s}+\log(2\,\pi)-\frac{1}{2}H_{\frac{s}{2}}+s\sum\limits_\rho\frac{1}{\rho\,\left(s-\rho\right)}$


Question (1): Is there a analogous explicit formula for $\log(\zeta(s))$?

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    $\begingroup$ It is $\sum_\rho \frac{1}{s-\rho}+\frac1\rho$ not $\sum_\rho \frac{1}{s-\rho}$ (a consequence of the density of zeros, itself a consequence of the functional equation). The series converges absolutely and locally uniformly so you can integrate it $\int_{s_0}^s$ without fear. Of course there is the problem of the branches. $\endgroup$ – reuns Dec 14 '18 at 2:09
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    $\begingroup$ I am not sure I understand the question but can't you use: $\zeta(s)=\prod_{p} (1-p^{-s})^{-1} \implies \ln(\zeta(s))=-\sum_{p}\ln(1-p^{-s}) $ $\endgroup$ – Mason Dec 14 '18 at 2:14
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    $\begingroup$ @Mason explicit formula means series over the zeros, not over the primes. Your sum (with the correct branch of $\log$) converges only for $\Re(s) > 1$ and assuming the RH can only be regularized for $\Re(s) > 1/2$. $\endgroup$ – reuns Dec 14 '18 at 2:22
  • $\begingroup$ @reuns I copied the wrong formula, so I replaced formula (1) above. I was investigating integration of the formula and having trouble with the branches which led to this question. $\endgroup$ – Steven Clark Dec 14 '18 at 14:05
  • $\begingroup$ If $\gamma$ is a curve $a \to b$ then $\int_\gamma \frac{1}{s-\rho} ds = \log(b-\rho)-\log(a-\rho)+2i\pi k$ and finding the correct $k=k_{\rho,\gamma}$ (equivalently the correct branch) is a problem treated in every text on complex analysis. The possible $k_{\rho,\gamma}$ is constrained by the fact the series must converge, so overall you have an uncertainty $\log \zeta(s)+2i\pi K$ with $K$ finite and depending on your path of integration. Usually we start with $\log \zeta(s)$ analytic for $\Re(s) > 1$ then we fix $K$ from the path of integration from $\Re(s) > 1$. $\endgroup$ – reuns Dec 14 '18 at 14:40

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