# 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 $$\zeta(s).$$

• "$\Phi$ 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 $(s-1)\zeta(s) = O(e^{|s|})$) – reuns Apr 22 '20 at 21:21
• @reuns maybe you can speak to my edit? – geocalc33 May 9 '20 at 18:23
• 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. – reuns May 9 '20 at 19:14
• but I thought you said that there's no point in asking about the analytical continuation! – geocalc33 May 9 '20 at 19:34
• 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. – reuns May 9 '20 at 19:50

## 1 Answer

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 $$0, the integrand tends to $$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.

• I posted a remark in the parallel MO thread. Just thought you might be interested :-) – fedja Jan 7 at 14:29
• When you say $\zeta(s)$ is this the analytically continued zeta function? – geocalc33 Feb 13 at 22:46
• and is it surprising that the zeta function appears? – geocalc33 Feb 13 at 23:09
• @metamorphy okay thanks. do you think I could ask about possible roots of $\Phi(s)$? – geocalc33 Feb 25 at 18:35
• @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 $\Re(s)=1$? – geocalc33 Feb 28 at 19:33