Trying to learn the path integral in Quantum Field Theory I came across some infinite products in Weinberg's book "The Quantum Theory of Fields".
Heuristically, the author pretends that what can be done with finitely many degrees of freedom also holds for infinite degrees of freedom to define the path integral.
The point is that searching more on the matter I've found out that to make any of this make sense, one needs regularization and more specially, zeta function regularization.
Searching on Wikipedia, we have the page Zeta function regularization which says:
In mathematics and theoretical physics, zeta function regularization is a type of regularization or summability method that assigns finite values to divergent sums or products, and in particular can be used to define determinants and traces of some self-adjoint operators. The technique is now commonly applied to problems in physics, but has its origins in attempts to give precise meanings to ill-conditioned sums appearing in number theory.
One method is to define its zeta regularized sum to be $\zeta_A(−1)$ if this is defined, where the zeta function is defined for $\Re(s)$ large by
$${\displaystyle \zeta _{A}(s)={\frac {1}{a_{1}^{s}}}+{\frac {1}{a_{2}^{s}}}+\cdots }$$
if this sum converges, and by analytic continuation elsewhere.
Now, this seems to be a technique used in Physics but quite common in math. I want to understand the idea behind it.
Why does it make sense to take one infinite sum or one infinite product that is divergent and define its sum to be $\zeta_A(-1)$?
It is clear that
$$\zeta_A(-1)=a_1+a_2+\cdots = \sum_{n=0}^\infty a_n. $$
It is also clear that if we have an infinite product then formally
$$\prod_{n=0}^\infty a_n=\exp \sum_{n=0}^\infty a_n=\exp\zeta_A(-1)$$
So things start to make sense, but the method is still awkward. I mean, one concludes that
$$\sum n = -1/12$$
and this makes no sense at all. Why would one consider this to be valid?
So what is the idea behind zeta-function regularization, why would it be meaningful to associate the sum to $\zeta_A(-1)$ if already on the simplest case it gives one crazy result like $\sum n = -1/12$?
Best Answer
Suppose some minimal $s_0\in\mathbb{R}$ exists for which $\sigma_n a_n^{-s}$ is finite for all $s\in\mathbb{C}$ with $\Re s>s_0$. Define $\zeta_A (s)$ as the analytic continuation, if it exists, of this sum; "the" is appropriate here, since the continuation if existent will be unique. For example, if $a_n=n$ the result is the Riemann zeta function, which most emphatically is not $\sum_{n\ge 1}n^{-s}$ in general, and it's not supposed to be. In the result $\zeta(-1)=-\frac{1}{12}$, the left-hand side should not be interpreted as $\lim_{N\to\infty}\sum_{n=1}^N n$, because that's not what it is.
But physical uses of zeta function regularisation should be interpreted as expressing a result as one of these continuation-defined functions, evaluated at an appropriate value of $s$. For example, a naïve treatment of the Casimir effect in $3$-dimensional space obtains a potential proportional to $\sum_{n\ge 1}n^3$ to include all modes, but of course this is just $\zeta_{\mathbb{N}}(-3)$. Why are we allowed to go looking for "reinterpretations" of that sum? Because there's no a priori reason Nature should "add" infinitely many terms in the first definition of addition mathematicians thought of. Instead, we're using Ramanujan summation.