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Define $$\Psi(s)=\prod_{n=1}^\infty e^{n^{-s}}$$

Noting that

$$ \log\big(\Psi(s)\big)=\zeta(s) $$

where $\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s},$ which converges for $\Re(s)>1.$

How do you analytically continue $\Psi?$ Is it pretty much the same as analytically continuing $\zeta?$

$\Psi$ converges for $\Re(s)>1$ and so does $\zeta,$ but I still don't understand completely. Do I have to do a composition of power series? I know that $\zeta$ can be analytically continued. I still don't fully understand what happens when you do $e^{\zeta(s)}$ and try to continue that. Thanks so much.

Arctic Char
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John Zimmerman
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1 Answers1

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Suppose $\Phi$ is some analytic continuation of $\Psi$; that is, $\Phi:U\to\Bbb C$ is holomorphic on some open $U$ (which has nonempty intersection with $\{\Re(s)>1\}$) and $\Phi(s)=\Psi(s)$ for $s\in U$ for $\Re(s)>1$ (there is an open set of such $s$ to make sure this really is an analytic continuation). Then, $\Phi(s)=e^{\zeta(s)}$ for all $s\neq1$ in $U$, where I've liberally analytically extended $\zeta$ to $\Bbb C\setminus\{1\}$. Indeed, this is because $e^{\zeta(s)}$ is the composition of holomorphic functions in $s$ and is thus holomorphic (this is by virtue of the chain rule, or an explicit argument is given here), so knowing $\Phi(s)=\Psi(s)=e^{\zeta(s)}$ for $s\in U$ with $\Re(s)>1$ implies that $\Phi(s)=e^{\zeta(s)}$ has to hold wherever both are defined.

In particular, since $\zeta(s)$ can be analytically continued to all $s\neq1$, so can $\Psi$ and the analytic continuation of $\Psi$ onto $s\neq1$ will coincide with $e^{\zeta(s)}$.

Zach Goldthorpe
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