My research has lead me to the following function that I'm trying to continue. 3 Months ago I posted this question to MSE, and have placed 3 bounties on the question, but haven't received an answer, so I've decided to ask here.

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

A natural question might be:

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

User @reuns noticed that $\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk).$

And an analytic continuation *is* indeed possible using the Cahen-Mellin integral to obtain the formula:

$$\varphi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns)$$

which is valid for $0<s<1.$

I noticed that:

$$e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$$

valid for $0<x<1$ and $\Re(z)>0$ if I'm not mistaken. Here $K_1$ is a modified Bessel function of the second kind.

Letting $x=e^{-n^{-s}}$ we obtain:

$$\varphi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$

I think the evaluation of this will give a new formula for $\varphi(s).$

Does anyone see how to accomplish this?

notreal analytic at any point $s>1$. I hope there is no mistake, but the argument is a bit involved. Maybe I'll post it as a proper answer later. The possibility still remains that you can go through the boundary of the half-plane $\Re z<1$ somewhere far from the real line but it is another story. $\endgroup$from the real lineis possible there. My proof promptly breaks down when $s<1$, metamorphy's series promptly diverges for $s>1$, so everything seems to fit together :-) $\endgroup$6more comments