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:


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?

  • 2
    $\begingroup$ To lay the groundwork for this question, in what region of the complex plane does the given series already converge to an analytic function? $\endgroup$ Aug 8, 2020 at 23:54
  • 2
    $\begingroup$ @GerryMyerson Re s > 0 or am I missing something obvious? $\endgroup$
    – John Jiang
    Aug 9, 2020 at 5:00
  • 14
    $\begingroup$ @JohnJiang You are missing the fact that the exponents $n^s$ not only grow in size but also rotate for complex $s$, so the minus sign becomes quite useless and you get arbitrarily huge terms. I suspect that there is no continuation from the real line anywhere though I cannot offer a proof off hand. $\endgroup$
    – fedja
    Aug 13, 2020 at 7:07
  • 3
    $\begingroup$ @geocalc33 I discussed the problem with Misha Sodin and you can find the result of this discussion in the set of handwritten notes at drive.google.com/file/d/191PhSQzr5Q-MbfuMzmuiogJZrk2bh9Ko/… .It is supposed to show that the sum of the series is not real 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. $\endgroup$
    – fedja
    Jan 6, 2021 at 21:37
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    $\begingroup$ I see nothing wrong with it (though I haven't checked all the statements about fast decay carefully, I'm pretty sure that @metamorphy will supply all details if you request clarification). At least, it is in good agreement with the notes I linked to: the sum of the series is real analytic for $0<s<1$ and extends to the half-plane $\Re z<1$ but it loses the real analyticity property on the line for $s>1$, so no extension from the real line is possible there. My proof promptly breaks down when $s<1$, metamorphy's series promptly diverges for $s>1$, so everything seems to fit together :-) $\endgroup$
    – fedja
    Jan 9, 2021 at 3:13


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