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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 $0<s<1.$

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 $\varphi(s).$ Potentially we could use the distributional version of the kernel to evaluate the integral if one exists.

Does anyone see how to accomplish this?

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    $\begingroup$ To lay the groundwork for this question, in what region of the complex plane does the given series already converge to an analytic function? $\endgroup$
    – Gerry Myerson
    Commented Aug 8, 2020 at 23:54
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    $\begingroup$ @GerryMyerson Re s > 0 or am I missing something obvious? $\endgroup$
    – John Jiang
    Commented Aug 9, 2020 at 5:00
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    $\begingroup$ @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
    Commented Aug 13, 2020 at 7:07
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    $\begingroup$ @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
    Commented Jan 6, 2021 at 21:37
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    $\begingroup$ Sure. If you need to know more, just state clearly what you are now interested in and link to what has been done already to give everybody a clear picture of where the things stand. Follow-ups are pretty common on this site. $\endgroup$
    – fedja
    Commented Feb 12, 2021 at 0:21

1 Answer 1

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This is nowhere near a formal answer, but it might contain some useful starting points for doing computation.

The problematic aspect of the function $$ \varphi(s) = \Gamma\left(1+\frac{1}{s}\right) + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \zeta(-ns)$$ is that $\zeta(-ns)$ grows faster than the factorial function when $s>1$, so the series cannot converge. However, the series is alternating, so from a certain viewpoint, it should 'morally' cancel out.

From a divergent series regularization point of the view, the simplest way to obtain a finite value from the function $\varphi(x)$ is truncate the series early to approximate its true value. Thus, defining $$\varphi(s) \approx \Gamma\left(1+\frac{1}{s}\right)+ \sum_{n=0}^{N} \frac{(-1)^n}{n!} \zeta(-ns) $$ gives a good approximation to the true value of the function near $s=1$. The optimal place to truncate the series is generally at the point where the the size of the term is smallest. Here is a graph of this approximation with $N=10$ on the real line, with the infinite series shown in orange and the finite series in black

For some values of $\mathfrak{R}(s)>1$, the cancellation automatically happens on its own if we look at the integral representation rather than the sum of the residues. Thus, the integral $$ \varphi(s) = \frac{1}{2 \pi i s} \int_{c - i N}^{c + i N} \Gamma\left(\frac{t}{s}\right) \zeta(s)dt $$ provides another way to approximate the values of $\varphi(x)$ outside of its usual realm of convergence. The two methods unsuprisingly agree with each other, but they tend to converge well in different areas.

If we want to obtain the value of $\varphi(s)$ somewhere far from $s=1$, or we want to get an arbitrarily good approximation, we can rewrite zeta using its functional equation to obtain

$$\varphi(s) = \Gamma(1+\frac{1}{s}) - \frac{1}{2} + \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n!}\left(\frac{\left(2\pi\right)^{-ns}}{\pi}\sin\left(-\frac{\pi ns}{2}\right)\left(ns\right)!\zeta\left(1+ns\right)\right)$$ The part that causes it to diverge is the factorial, so we can replace it by its integral representation and simplify to obtain $$\varphi(s) = \Gamma(1+\frac{1}{s}) - \frac{1}{2} - \frac{1}{\pi}\int_{0}^{N}e^{-t_{2}}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n!}\left(\frac{t_{2}}{2\pi}\right)^{ns}\sin\left(\frac{\pi ns}{2}\right)\zeta\left(1+ns\right)dt_{2}$$

This integral agrees with the other two methods in areas where they converge. There are likely ways to simplify this last integral, though I'm not sure if such simplifications will actually make the function easier to compute.

Update: Here is a different integral that is only valid for $\mathfrak{R}(s) >1$

$$\varphi(s) = \Gamma\left(1+\frac{1}{s}\right)-\frac{1}{2} - \\ \frac{1}{\pi} \int_0^\infty \frac{e^{-\left(\frac{t}{2\pi}\right)^{s}\cos\left(-\frac{\pi s}{2}\right)}}{e^{t}-1}\left(\sin\left(\ \left(\frac{t}{2\pi}\right)^{s}\sin\left(-\frac{\pi s}{2}\right)\right)\right) dt$$

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  • $\begingroup$ thanks for the useful starting points. Just wanted to ask your thoughts on the integral $\varphi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$ providing a new formula $\endgroup$
    – geocalc33
    Commented Nov 2, 2022 at 15:45
  • $\begingroup$ and also what do you think about continuing the function for real $s>1$ using quasi analytic continuation, since the function is smooth and in a quasi analytic class in this domain $\endgroup$
    – geocalc33
    Commented Nov 2, 2022 at 15:46
  • $\begingroup$ @geocalc33 Doing some basic numerical calculations, your integral appears to agree with the ones I have presented here, which would suggest there is some sort of consistency in the value to assign the function when $\mathfrak{R}(s)>1$. I'll look more into the integral and update my answer if I discover anything $\endgroup$
    – Caleb Briggs
    Commented Nov 2, 2022 at 20:46
  • $\begingroup$ My integral agrees with $\varphi(s) = \Gamma\left(1+\frac{1}{s}\right) + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \zeta(-ns)$ on $0<s<1$? $\endgroup$
    – geocalc33
    Commented Nov 9, 2022 at 15:35

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