# Equivalence of regularization schemes of divergent series

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?

• 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. – level1807 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. – level1807 Oct 12 '16 at 23:06