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(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 $\zeta(s).$

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    $\begingroup$ "$\Phi$ and $\zeta$ are congruent structures. What I mean by that is that I think $\Phi$ also has a critical strip, nontrivial zeros, euler product, functional equation, etc" doesn't make any sense. It seems not crazy to ask if your function may be the analytic continuation of $\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk)$ wihch is analytic on $\Bbb{C}^* - \cup 1/k$ (because $(s-1)\zeta(s) = O(e^{|s|})$) $\endgroup$ – reuns Apr 22 '20 at 21:21
  • $\begingroup$ @reuns maybe you can speak to my edit? $\endgroup$ – geocalc33 May 9 '20 at 18:23
  • $\begingroup$ You should rename thisone into "analytic continuation of $\sum_{n\ge 1} e^{-n^s}$" and remove the remaining part. Did you try drawing $\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(-sk)$ and comparing ? As I said it seems obvious both functions are related. $\endgroup$ – reuns May 9 '20 at 19:14
  • $\begingroup$ but I thought you said that there's no point in asking about the analytical continuation! $\endgroup$ – geocalc33 May 9 '20 at 19:34
  • $\begingroup$ I never said that. I said your other question was a duplicate and I was upset because you didn't mention my result : that $\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk)$ is analytic away from the $1/k,k\ge 1$. And that asking for a functional equation doesn't make sense. $\endgroup$ – reuns May 9 '20 at 19:50
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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 $$\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).$$

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.

The remaining question is whether we can extend it further.

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    $\begingroup$ I posted a remark in the parallel MO thread. Just thought you might be interested :-) $\endgroup$ – fedja Jan 7 at 14:29
  • $\begingroup$ When you say $\zeta(s)$ is this the analytically continued zeta function? $\endgroup$ – geocalc33 Feb 13 at 22:46
  • $\begingroup$ and is it surprising that the zeta function appears? $\endgroup$ – geocalc33 Feb 13 at 23:09
  • $\begingroup$ @metamorphy okay thanks. do you think I could ask about possible roots of $\Phi(s)$? $\endgroup$ – geocalc33 Feb 25 at 18:35
  • $\begingroup$ @metamorphy What I said about "congruent structures" was an honest conjecture of mine based on what I knew months ago. I realize in hindsight that it was naive. Why do you wonder what's happening near $\Re(s)=1$? $\endgroup$ – geocalc33 Feb 28 at 19:33

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