The product
$$
F(s)=\prod\_{p}\frac1{(1-p^{-s})^p},
$$
converges for $\mathrm{Re}(s)>2$, when $p$ runs over all primes. Does it admit analytic continuation beyond the line $\mathrm{Re}(s)=2$? Any papers where it has been studied?



Does this product have analytic continuation?
$$P(s) = \sum\_p p^{-s}, \qquad \log F(s) = \sum\_{p^k} \frac{p^{1-sk}}{k} = \sum\_{k\ge 1} \frac{P(sk-1)}{k}$$


* $P(s) = \sum\_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P\_N(s) = \sum\_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ 
so that $$e^{N! P(s)} = e^{N! P\_N(s)}\prod\_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :


$P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$. 


Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.
* For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.


