# 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 $(s-1)\zeta(s) = O(e^{|s|})$) Apr 22, 2020 at 21:21
• @reuns maybe you can speak to my edit? May 9, 2020 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. May 9, 2020 at 19:14
• but I thought you said that there's no point in asking about the analytical continuation! May 9, 2020 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. May 9, 2020 at 19:50

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 this region further.

• I posted a remark in the parallel MO thread. Just thought you might be interested :-) Jan 7, 2021 at 14:29
• First thanks, I had my mind elsewhere so I'm just reading it a lot of days later so sorry about that, and also sorry because in those three expression you added you got me lost, not sure how the Stirling expression became that, then what happened to $|\zeta(1+ns)|$ and why $\tau$ must be the one satisfying that exactly, maybe I need to come back later if I ever get more math savvy. (I deleted what I asked previously as it seemed irrelevant now and was bloating the comments) May 29, 2021 at 6:19