Let $$ \Phi(s)=\sum_{n=1}^\infty e^{-n^{s}} $$ converging iff $s>0$ is real.
So far the best method for an analytic continuation of $\Phi(s)$ is to use the Cahen-Mellin integral outlined below (see Analytic continuation of $\Phi(s)=\sum_{n \ge 1} e^{-n^s}$ for more context):
$$e^{-x}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)x^{-z}\,dz\qquad(x,c>0)$$ 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.
It occurred to me to try the aforementioned approach with a function other than $e^{-x}:$
$$e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$$
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:
$$\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$
It's not at all obvious to me how to evaluate this integral. Someone mentioned that the kernel doesn't converge and that it might be evaluated if the kernel converges in the distributional sense.
If $\Phi(s)$ is produced by these two very different looking integrals, are they equivalent in some sense?
Does anyone know how to evaluate the last integral? Can numerical methods shed light?
