# What is the analytic continuation of $\varphi(s)=\sum_{n \ge 1} e^{-n^s}?$

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.

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

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

User @reuns had an insightful point that maybe, $$\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk).$$

Edit 09/20/2020: There is now a partial answer found via the link above.

• To lay the groundwork for this question, in what region of the complex plane does the given series already converge to an analytic function? – Gerry Myerson Aug 8 '20 at 23:54
• @GerryMyerson Re s > 0 or am I missing something obvious? – John Jiang Aug 9 '20 at 5:00
• @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. – fedja Aug 13 '20 at 7:07
• @JackZimmerman 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. – fedja Jan 6 at 21:37
• 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 :-) – fedja Jan 9 at 3:13