Analytic continuation of $\Phi(s)=\sum_{n \ge 1} e^{-n^s}$

(After 3 bounties I've also posted on mathoverflow).

While discussing theta functions, I thought:

$\zeta(s)=\sum n^{-s}=1+2^{-s}+3^{-s}+ \cdot\cdot\cdot$

and

$\Phi(s)=\sum e^{-n^s}=e^{-1}+e^{-2^s}+e^{-3^s}+\cdot\cdot\cdot $

What is the analytic continuation of $\Phi(s)?$

User @reuns had an insightful point that maybe, $\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk).$

If the sum were instead a product, then the analytic continuation would coincide with the analytic continuation of $\zeta(s).$

Solution 1:

This is currently a partial answer, refining the idea given by @reuns.

The series $\Phi(s)=\sum_{n=1}^\infty\ e^{-n^s}$ converges iff $s>0$ is real. Using the Cahen–Mellin integral $$e^{-x}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)x^{-z}\,dz\qquad(x,c>0)$$ with $x=n^s$ and $c>1/s$, we get $$\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)\zeta(sz)\,dz.$$

For $0<s<1$, the integrand tends to $0$ rapidly enough when $z\to\infty$ in the half-plane $\Re z\leqslant c$ and out of a neighborhood of the line $L=\{z : \Im z=0\wedge\Re z\leqslant 1/s\}$. This allows us to deform the path of integration, making it encircle $L$, and we see that $\Phi(s)$ is equal to the (infinite) sum of residues of the integrand at its poles (which are $z=1/s$ and $z=-n$ for nonnegative integers $n$). Computing these, we get $$\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).\tag{*}\label{theseries}$$

This series converges for complex $s\neq 0$ with $\Re s<1$ (at least; the singularities at $s=-1/n$ for $n\in\mathbb{Z}_{>0}$ are removable), and gives the analytic continuation of $\Phi(s)$ in this region.

Update. In fact the series \eqref{theseries} converges also for some $s$ with $\Re s=1$. This (as well as the convergence for $\Re s<1$) is seen using the functional equation for $\zeta$: $$\zeta(-ns)=-(2\pi)^{-ns}\frac1\pi\sin\frac{n\pi s}{2}\Gamma(1+ns)\zeta(1+ns)$$ and Stirling's asymptotics for $\Gamma$. At $s=1+it$, as $n\to\infty$, the latter gives $$\frac1{n!}\Big|\Gamma(1+ns)\Big|\asymp(1+t^2)^{(2n+1)/4}e^{-nt\arctan t},$$ so that $$\frac1{n!}\Big|\zeta(-ns)\Big|\asymp(2\pi)^{-n-1}(1+t^2)^{(2n+1)/4}e^{nt\arctan 1/t}.$$

Thus, \eqref{theseries} converges at $s=1+it$ with $|t|<\tau$, where $\tau\approx 2.24$ satisfies $$(1+\tau^2)e^{2\tau\arctan 1/\tau}=4\pi^2.$$

The remaining question is whether we can extend this region further.