Maximal extension of domain of $\Psi(x)=\sum_{n=1}^\infty e^{\frac{\log(n)}{\log(x)}}$ using analytic continuation

Question

Given $$ \Psi(x)=\sum_{n=1}^\infty e^{\frac{\ln(n)}{\ln(x)}}= \prod_{p~ \mathrm{prime}}\frac{1}{1-e^{\frac{\ln(p)}{\ln(x)}}}$$

What is the maximal extension of $\Psi$?

I think that there is a barrier at $\Re(x)=0$ but don't know how to show that.

Answer 1

The function is simply a composition involving the Riemann zeta function: $$ \Psi(x) = \sum_{n=1}^\infty n^{1/\ln x} = \zeta\biggl( -\frac1{\ln x} \biggr). $$ The series originally converges when $-\frac1{\ln x}>1$, that is, for $x\in(\frac1e,1)$. But $\zeta(s)$ can be analytically continued to the entire complex plane except for a pole at $s=1$. Therefore, $\Psi(x)$ can be analytically continued to the entire complex plane except for a pole at $x=\frac1e$ and a branch cut (due to the $\log x$) from the origin out to infinity.