This is a very simple question; I apologize if it has already been asked here. Define the following function (superficially similar to a theta function):

$$\varsigma(x)=\sum_{n=1}^\infty e^{-xn^3}$$

I am interested in knowing the Laurent series about $x=0$ of this series if it exists, i.e. I would like to know if there exist $\{a_n\}$ such that:

$$\varsigma(x)=\sum_{n=-\infty}^\infty a_nx^n\tag{1}$$

for small enough $x>0$. I'm pretty sure that $\varsigma(x)$ diverges to infinity at $x=0$ so I assume that some of the $a_n$ for $n<0$ will be nonzero. Ideally I would love a closed form for the $a_n$ but I am especially interested for the moment in $a_1$. I have no idea how to find these terms, since this is not a Taylor series and since I do not know the complex behaviour of $\varsigma(z)$. Wolfram Alpha doesn't help me either. I am aware of the formula for Laurent series coefficients, and that for instance we will have:

$$a_1=\frac{1}{2\pi i}\oint_C \frac{\varsigma(z)}{z^2}\;dz$$

where $C$ is a closed contour around $z=0$, but I am not sure how I should go about evaluating this; formally interchanging integral and sum only gives me a divergent sum: for instance making use of the fact that $e^{az}=\sum\limits_{n=0}^\infty \frac{a^nz^n}{n!}$ I formally get $a_1=-\sum\limits_{n=1}^\infty n^3$ but I don't know whether I can make this argument rigorous to get $a_1=-\zeta(-3)$.

Background: This question arose from some recreational thoughts of mine on summing divergent series; this answer used an $ne^{-\epsilon n}$ regularization rather than the usual $n^s$ to 'evaluate' $\sum\limits_{n=1}^\infty n$ and curiously obtained a constant term of $-\frac{1}{12}$ in the Laurent series in $\epsilon$; this interested me and made me wonder what an $n^3e^{-\epsilon n^3}$ regularization of $\sum\limits_{n=1}^\infty n^3$ would give. We have $\sum\limits_{n=1}^\infty n^3e^{-\epsilon n^3}=-\varsigma'(\epsilon)$ so the constant term in the Laurent series expansion of this function will be $-a_1$; thus I would conjecture that $a_1=-\frac{1}{120}$ (i.e. $-\zeta(-3)$; see here). The above calculation supports this, but I don't know whether it can be made rigorous.

Thus I have the following questions: Do there exist $a_n$ such that $(1)$ holds for small $x>0$? If so, does $a_1=-\frac{1}{120}$? Is it possible to write a closed form for the $a_n$?


In order for $\varsigma(x)$ to have a Laurent series in a deleted neighbourhood of $0$, $f$ must be analytic in that deleted neighbourhood. But I'm pretty sure the function $\sum_n z^{n^3} = \varsigma(-\log(z))$ has a natural boundary on $|z|=1$. Therefore no such Laurent series is possible.

EDIT: The Fabry gap theorem says that if $\sum_n \alpha_n z^{p_n}$ has radius of convergence $1$, where $p_n$ is an increasing sequence of integers with $p_n/n \to \infty$, then the unit circle is a natural boundary for this series.

  • $\begingroup$ I see. How do we show that it has a natural boundary? $\endgroup$ – Anon Feb 1 '17 at 7:25
  • $\begingroup$ Yes, I see. +1. Thank you for your answer. I can see now that this will be impossible. It's still interesting that formally applying the contour integration for the term gave $-\sum\limits_{n=1}^\infty n^3$ although maybe this was to be expected. $\endgroup$ – Anon Feb 1 '17 at 22:58

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