Maximal analytic continuation of $\varphi(s)=\sum_{n\ge1} e^{-n^s}$ About 6 months ago I asked for an analytic continuation of $\varphi(s)=\sum_{n\ge1} e^{-n^s}.$

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

Doing this will help me better understand how the function behaves.
As is stated in the comments, the main question is whether the line $\Re z=1$ is the natural boundary for the analytic continuation:
$$ \varphi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).$$
As noted by metamorphy, this series converges for complex $s\ne0$ with $\Re s<1$.
 A: I think the issue with trying to analytically continue this function is that it really defines two different, incompatible, functions. This is suggested, for instance, since
$$\sum_{n=1}^\infty (e^{-n^s}-1) = \sum_{k=1}^\infty \frac{(-1)^k}{k!} \zeta(-sk)$$
When we look at the left half-plane. However, when we construct an expansion on the right half-plane we instead obtain
$$\sum_{n=1}^\infty e^{-n^s} = \sum_{k=0}^\infty \frac{ (-1)^k}{k!} \zeta(-sk) + \mathbf{\Gamma\left(1+\frac{1}{s}\right)}$$
If we were dealing with a single function, we would expect that
$$\sum_{n=1}^\infty e^{-n^s} - \sum_{n=1}^\infty (e^{-n^s}-1) = \sum_{n=1}^\infty 1 = \zeta(0)$$
And without the bolded Gamma function term, we do find this. Thus, our sum is probably best represented as
$$\sum_{n=1}^\infty e^{-n^s} = \sum_{k=0}^\infty \frac{ (-1)^k}{k!} \zeta(-sk)+ \begin{cases} 
\Gamma\left(1+\frac{1}{s}\right) & \mathfrak{Re}(s)<0  \\
0 & \mathfrak{Re}(s)>0 
\end{cases}$$
To see this, let us add in an extra term to our sum so that converges on the whole real line. Let us consider $g(s,k) = \sum_{n=1}^\infty 
\frac{e^{-n^s}}{n^w}$ with $k>1$ so that it absolutely converges. Then we have
$$\sum_{n=1}^\infty \frac{e^{-n^s}}{n^w} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \sum_{n=1}^\infty n^{sk-w} = \sum_{k=0}^\infty \frac{(-1)^k}{k!} \zeta(w-sk)$$
Notice that in this step, if $\mathfrak{Re}(s)>0$ the inner series is diverging, and so it is only equal to the $\zeta$ function by analytical continuation. When we do a step like this, we have to pick up the residues created by the function we analytically continued (in this case the $\zeta$ function). (Note: to see a bit more about why this step is necessary, see this question of my mine). Thus, we pick up the residue $$ \frac{\csc(\pi z) \zeta(w-zs)}{\Gamma(z+1)2i} \bigg|_{z=\frac{w-1}{s}} = \frac{\Gamma\left(1-\frac{-1+s+w}{s}\right)}{s}$$
However, when $\mathfrak{Re}(s)<0$ the sums all converge, so we don't pick up that residue.
Thus, we obtain that on the real line when $\mathfrak{Re}(w)>1$
$$\sum_{n=1}^\infty \frac{e^{-n^s}}{n^w} = \sum_{k=0}^\infty \frac{ (-1)^k}{k!} \zeta(w-sk)+ \begin{cases} 
\frac{\Gamma\left(1-\frac{-1+s+w}{s}\right)}{s} & \mathfrak{Re}(s)<0  \\
0 & \mathfrak{Re}(s)>0 
\end{cases}$$
One short note is in order, which is that $\sum_{k=0}^\infty \frac{ (-1)^k}{k!} \zeta(w-sk)$ does not converge for for some values. This is not a problem, we just replace the sum using the residue theorem with the integral $\int_{C}  \frac{\csc(\pi z) \zeta(w-sz)}{z!2i} dz$ and making sure we choose $C$ so that we only pick up the residues at the natural numbers.
