# 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.

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

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:

$$\varphi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns)$$

which is valid for $$0

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 $$0 and $$\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?

• To lay the groundwork for this question, in what region of the complex plane does the given series already converge to an analytic function? Aug 8, 2020 at 23:54
• @GerryMyerson Re s > 0 or am I missing something obvious? Aug 9, 2020 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. Aug 13, 2020 at 7:07
• @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. Jan 6, 2021 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 :-) Jan 9, 2021 at 3:13