Assigning values to a divergent integral?

Question

Question

If I can assign the series of the zeta function to: $$ \zeta(-1) \to 1+2+3+\dots$$ why can't we assign the integral $$ \int_{0}^{\infty} x dx \to 0$$ and it still have some physical significance?

Background

We know the zeta function at $1$ has a series representation (which is useful to physicists):

$$ \zeta(-1)=1 + 2 + 3 + \dots = \frac{-1}{12}$$

Multiplying both sided by $h^2$

$$ (h + 2h + 3h + \dots)h = \frac{-h^2}{12}$$

Taking limits both sides of $h \to 0$ and using the representation of an integral as a Riemann sum:

$$ \int_{0}^{\infty} x dx = 0 $$

P.S: I do not advocate or endorse $\int_{0}^{\infty} x dx = 0 $ ... Also I have asked a question like this before but was over-zealous with it and deleted it. Hopefully this a better asked version

Answer 1

Divergent summations arise in physics is when making shortcuts using formal manipulations when they are not not allowed. One can then recover finite and correct results using formal manipulations that are not valid either in that particular case, but this has the effect of correcting the initial problem leading to the divergence. It is easy to find simple examples involving diverging integrals. Suppose we need to evaluate:

$$\int_0^{\infty}\left(\frac{\sqrt{x}}{1+x} - \frac{1}{\sqrt{x}}\right)dx$$

This integral converges, but you can evaluate it using the divergent integral evaluations of:

$$\int_0^{\infty}\frac{\sqrt{x}}{1+x}dx = -\pi$$

and

$$\int_0^{\infty}\frac{1}{\sqrt{x}}dx = 0$$

This follows from:

$$\int_0^{\infty}\frac{x^{-p}}{1+x}dx = \frac{\pi}{\sin(\pi p)}$$

and

$$\int_0^1 x^{s}dx + \int_1^\infty x^{t}dx = \frac{1}{1+s} - \frac{1}{1+t}$$

whenever they converge, and you can then analytically continue the integrals when they don't converge. The analytic continuation makes the evaluation of the original converging integral using the divergent integrals rigorously valid, as you can put the parameters $p$, $s$, and $t$ in the original integral and argue that the result is an analytic function of these parameters.

Answer 2

As I've noted in several places on this site, a principal reason that (certain!) manipulations of non-convergent series and integrals nevertheless produce useful answers is the "identity principle" in complex analysis, namely, that if $f,g$ are two holomorphic functions on a connected open set $\Omega$ in $\mathbb C$, and if they agree on a set "with an accumulation point" (e.g., a non-empty interval, or a non-empty open subset), then they agree everywhere on $\Omega$. Since many idealized relationships in mathematics and physics and ... are analytic, if the participants admit analytic continuation in some parameter, and in part of the region various manipulations are legit, then the conclusion (and, to a possibly delicate degree, if properly interpreted, the manipulations themselves), are legitimate throughout the region.

But then, no, an integral or infinite sum might not be the literal integral or sum, but an extension with the same properties (if expressible in analytic terms).

But/and this means that instead of just "making sense of" some provocative (not-literally-convergent) expression, we should/must be explicit about what the integral/sum means, and how its result is required to interact with other mathematical objects. Not just "assign a number": the "assigment" should/must be consistent with many other things.