# Does $\mathrm{Log}(\zeta)$ extend meromorphically past $\Re(s)=1$?

Let $$\zeta(s)=\sum_{n\ge 1}n^{-s}$$ be the Riemann zeta function. It is well-known that the infinite sum $$\mathrm{Log}(\zeta)=-\sum_{p\text{ prime}}\log(1-p^{-s})$$ converges to an analytic function on the right half-plane $$\Re(s)>1$$.

Question. Is it known whether this function also admits mermorphic contination to $$\Re(s)>1-\delta$$ for some $$\delta>0$$?

I assume the answer to this question should be well-documented, but I couldn't find it so far. Would very much appreciate a reference or a proof of existence or non-existence of meromorphic continuation.

Thank you!

The obstacles to defining a logarithm of a meromorphic function $$f$$ are the poles and zeros of $$f$$. That is, if $$U$$ is a simply-connected open region on which $$f$$ is analytic and has no zeros, it has an analytic logarithm in $$U$$. Conversely, if $$f$$ has zeros or poles in $$U$$, it can't have a meromorphic logarithm there (note that if $$g$$ has a pole at some point, $$\exp(g)$$ has an essential singularity there). In the case of $$\zeta$$, we know it has a simple pole at $$s = 1$$, so $$\log(\zeta(s))$$ can't be defined as a meromorphic function in a neighbourhood of $$1$$.