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

Question

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!

Answer 1

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 $1$.