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I'm learning about different methods of regularization of divergent series like $$\sum_{n=0}^\infty \epsilon_n$$ used in theoretical physics, for instance:

  • Heat kernel regularization $$\sum_{n=0}^\infty \epsilon_n e^{-\epsilon_n t}$$
  • Zeta-function regularization $$\sum_{n=0}^\infty \epsilon_n^{1-t}$$

The goal is to show that the constant terms in the asymptotic expansions of these sums at $t\to 0$ coincide. I came up with the following proof, which however involves a few assumptions, any justification for which I couldn't find in my complex analysis books.

If we assume that the "heat kernel" $$K(t)=\sum_{n=0}^\infty e^{-\epsilon_n t}$$ can be analytically continued into a punctured neighborhood of zero, where it has a Laurent series $$K(t)=\sum_{n\in\mathbb{Z}}a_n t^n,$$ then the regularized sum we are looking for is clearly just $-a_1$.

Moreover, we can relate the zeta function $$\zeta(s)=\sum_{n=0}^\infty \epsilon_n^{-s}$$ to the heat kernel via the Mellin transform: $$\zeta(t-1)=\frac{1}{\Gamma(t-1)}\int_0^\infty x^{t-2}K(x)dx=\frac{1}{\Gamma(t-1)}\frac{1}{e^{2\pi it}-1}\int_\gamma x^{t-2}K(x)dx,$$ where $\gamma$ is a contour going around the positive half-line in the positive direction. Here we again have to assume that $K(x)$ decays at large $x$ sufficiently quickly and doesn't have any poles at positive $x$ (the former is obviously true assuming all $\epsilon_n$ are positive, increasing in $n$ and tend to infinity so that the sum for $K(t)$ converges in a half-plane $\Re t>t_0$). The last contour integral manifestly defines an analytic function which can now be evaluated at $t=0$. Taking the limit and using the above Laurent expansion of $K(x)$, we find $$\zeta(-1)=-\frac{1}{2\pi i}\int_\gamma x^{-2}K(x)dx=-a_1,$$ which is the desired result.

My questions are: what are the most general results about the analytic continuation of a series of the type $\sum_n e^{-\epsilon_n t}$ (or equivalently $\sum_n a_n z^{\epsilon_n}$)? Do we know that the heat kernel $K(t)$ doesn't have poles at positive $t$ and is meromorphic around zero?

Alex Bogatskiy
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  • If you assume that $\epsilon_n$ is increasing faster than $n^a$ for some $c> 0$, then $K(x)= o(e^{-x^{c/2}})$ and $\zeta(s) \Gamma(s) = \int_0^\infty x^{s-2} K(x)dx$ where $\int_c^\infty x^{s-2} K(x)dx$ is entire for any $c > 0$. And your Laurent series expansion tell you the poles of $\int_0^c x^{s-2} K(x)dx$ are at the integers, where $\zeta(s)\Gamma(s) \sim a_n \frac1{s-1+n}$. So this is indeed very rare that such a Laurent series expansion of $K(x)$ at $x=0$ exists. – reuns Oct 12 '16 at 22:33
  • And (assuming $K(x)$ has a Laurent series at $x=0$) your series for $\zeta(s)$ converges for $Re(s)$ large enough only if $a_n = 0$ for $n < M$. – reuns Oct 12 '16 at 22:39
  • I don't think you can say anything about the structure of the poles by looking at the integral from 0 to c because it diverges for interesting (or possibly all) s. One has to use the countour integral representation to analytically continue. Also the poles that you found are at negative integers, just like for the Gamma, and I don't see a problem with that. Assuming power-law behavior of the sequence seems to make sense though, based on the convergence of zeta. – Alex Bogatskiy Oct 12 '16 at 22:48
  • If it diverges for every $s$, then your series $\zeta(s)$ never converges. And $n \in \mathbb{Z}$ as in your Laurent series expansion. – reuns Oct 12 '16 at 22:56
  • @user1952009 what you are saying about the poles is not true even for $\epsilon_n=n$ (there both sums have only one pole at zero). Plus, the wide equivalence of these methods is an empirically verified fact. Unfortunately I don't know much about the rigorous results in this area, thus the post. – Alex Bogatskiy Oct 12 '16 at 23:06
  • yes, and, considering that both K and zeta in this case have only one pole on the real line, doesn't this go against what you said before about K not having a Laurent expansion and zeta having poles at positive or all integers? Even if the zeta did have poles all over the place, it wouldn't be an issue because I'm only worried about the positive poles of K – Alex Bogatskiy Oct 12 '16 at 23:29
  • You assumed that $K(x)$ has a Laurent expansion $\sum_k a_k t^k$, I told you it means the poles of $\zeta(s-1)\Gamma(s)$ are at $s = 1-k$ of residue $a_k$, and hence the series for $\zeta(s-1)$ never converges if you don't have $a_k = 0$ when $k < M$ for some $M\in \mathbb{Z}$ (at least when $|\epsilon_n| \to \infty$ so that $\zeta(s-1)$ is a generalized Dirichlet series, with an abscissa of convergence) – reuns Oct 12 '16 at 23:35
  • I don't see anything restrictive, since in applications it does have a Laurent series (and is even meromorphic). Yes, if K has an essential singularity, then zeta has poles at all positive integers and the series can't converge. There's still a lot of room for series that give meromorphic K. – Alex Bogatskiy Oct 12 '16 at 23:58
  • How did you compute this K? – Alex Bogatskiy Oct 13 '16 at 00:07
  • I made a mistake, but the pole of $\sum_{n \ge 1} (n^c)^{1-s}$ is at $s = 1+1/c$ so $K(x) = \sum_{n \ge 1} e^{-n^c x} \sim A x^{-1-1/c}$ as $x \to 0$ – reuns Oct 13 '16 at 00:12
  • Thats a good point. However, I just checked that $\sum_{n=1}^\infty \sqrt{n} e^{-t \sqrt{n}}$ has Laurent series $4t^{-3}+\zeta(-1/2)+...$, which agrees with the zeta function regularization... this really does seem to be a very general procedure. – Alex Bogatskiy Oct 13 '16 at 00:26
  • I wrote all this because I use it frequently, see this, but only when everything works well (your two series well-defined for $x$ and $Re(s)$ large enough) – reuns Oct 13 '16 at 00:35
  • That's quite interesting. Do you know if the regularizations would give consistent results for irrational $c$? – Alex Bogatskiy Oct 13 '16 at 00:37
  • I don't know. It could be easier to find a possible counter-example by generalizing to $k(s) = \sum_{n=1}^\infty a_n (b_n)^{-s}, K(x) =\sum_{n=1}^\infty a_n e^{-b_n x}$ – reuns Oct 13 '16 at 01:21

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