# What is the Explicit Formula for $\log(\zeta(s))$?

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 $$\log(\zeta(s))$$?

• 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. – reuns Dec 14 '18 at 2:09
• I am not sure I understand the question but can't you use: $\zeta(s)=\prod_{p} (1-p^{-s})^{-1} \implies \ln(\zeta(s))=-\sum_{p}\ln(1-p^{-s})$ – Mason Dec 14 '18 at 2:14
• @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 $\Re(s) > 1/2$. – reuns Dec 14 '18 at 2:22
• @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. – Steven Clark Dec 14 '18 at 14:05
• 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 $\Re(s) > 1$. – reuns Dec 14 '18 at 14:40