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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 $\Re s<1$.

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    $\begingroup$ Erm... And what are the function values for $s<0$ that you would like to extend? The series, as written, certainly diverges there, so you surely meant something different from what you wrote :-) $\endgroup$ – fedja Feb 12 at 1:02
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    $\begingroup$ @M.G. In this interpretation it has already been established by metamorphy in the MSE thread (the function can be analytically extended to the half-plane $\Re s<1$ from $(0,1)$ with just one pole at $0$) $\endgroup$ – fedja Feb 12 at 10:01
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    $\begingroup$ @fedja maximal analytic continuation of $\varphi(s)$. Better? $\endgroup$ – geocalc33 Feb 12 at 22:43
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    $\begingroup$ @geocalc33 Yep. Basically at this point the main question is whether the line $\Re z=1$ is the natural boundary for metamorphy's analytic continuation. $\endgroup$ – fedja Feb 12 at 22:49
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    $\begingroup$ @GHfromMO The function is clearly analytic for $s>0$ as a real variable No, no, and once more no! When $s>1$, it is $C^\infty$ and even in a quasi-analytic class, but not real analytic. $\endgroup$ – fedja Mar 4 at 16:39

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