How can $\log\zeta(s)$ be meromorphically continued to $\Re(s)>0$?

Question

I'm trying to understand how the function $P:\{s\in\mathbb{C}\ |\ \Re(s)>1\}\to\mathbb C,\ s\mapsto\sum_{p\text{ prime}}\frac{1}{p^s}$ can be continued meromorphically to the half-plane $\{s\in\mathbb{C}\ |\ \Re(s)>0\}$. Now I'm perfectly fine with the identity $\log\zeta(s)=\sum_{k=1}^{\infty}\frac{P(ks)}{k}$ for $\Re(s)>1$. I also understand that therefore $P(s)=\sum_{k=1}^{\infty}\frac{\mu(k)\log\zeta(ks)}{k}$ for $\Re(s)>1$. But now, every source I found jumps immediately to the conclusion that thus, $P$ can be continued meromorphically to $\Re(s)>0$, as the sum on the right converges uniformly on all compact subsets of its domain (I suppose) and the right hand sight is meromorphic. However, why can we meromorphically continue $\log\zeta(s)$ to $\Re(s)>0$? As $\zeta$ has many zeros in the critical strip, this seems not at all trivial to me. How can this be rigorously justified without presupposing a emromorphic continuation of $P$?

Answer 1

Neither $P(s)$ nor $\log \zeta(s)$ can be continued meromorphically to $\Re(s)>0$.

In each case, we are required to take infinitely many branch cuts (corresponding to the zeros and pole of $\zeta(s)$, in the case of its logarithm), which yields an analytic continuation to a large subset of $\Re(s)>0$; however, a meromorphic continuation would be a continuation to all of $\Re(s)>0$ other than isolated poles, and that is not possible here.

Answer 2

Usually we consider the branch of $\log \zeta(s)$ which is represented by $\sum_p \frac{p^{-sk}}{k}$ on $\Re(s) > 0$ and which is analytic on horizontal lines with no zeros as well as on $\rho+(0,\infty)$. Thus it is analytic on $\Bbb{C}-\{\rho+(-\infty,0)\}-(-2,1)-(-\infty,-4)$.

(any branch of) $\log(s-1)$ is locally analytic on $\Bbb{C}^*$, what it means is that starting with any branch, it can be analytically continued along any curve $\subset \Bbb{C}^*$.

$\log \zeta(s)$ is locally analytic on $\Bbb{C}-\{-2k\}-\{\rho\}-\{1\}$ and $P(s)$ is locally analytic on $\{\Re(s) > 0\} -\{\rho/k\}-\{1/k\}$.