How do you "regularize" infinite integrals?

Question

This question was inspired by the post: " Is there a solid reason why some people assume the fundamental theorem of calculus should still hold for divergent integrals with improper bounds? " (and the follow-up discussion). Long story short, the OP was using some technique to "regularize" infinite integrals (and claims, for example, that $\int_2^{\infty}\frac{dx}{x}$ has a "regularized value" of $-\log 2$) - and then complaining that whatever regularization they were using did not play well with the rules for change of variable in an integral.

So... What is the "regularization" of an infinite integral? Are there multiple ways to do it, or do all of them boil down to the same thing whenever they "work"? Is there a preferred way of doing it? What generalizations of this notion have been used/studied/are practical? Is it a topic of active research? What are known unsolved problems related to it?

(Of course, I don't expect all those questions to be answered here, I am more after some sort of reference that would possibly be understandable with basic knowledge of real and complex analysis and Lebesgue integration, or perhaps slightly more than that. Ideally an article of the type "Everything you've always wanted to know about regularization but were afraid to ask...")

Answer 1

Basic answer :There is no such thing as regularization of an integral. Either it is convergent, or it makes no sense. The examples in the cited post are just another example of manipulations involving infinity that lead to absurd result (hence this should not be done).

Longer answer : -In courses on complex analysis you have integrals with (complex) parameters that define holomorphic functions and can be extended on a larger set than the set where the integral is convergent. See for example the holomorphic extension of the Gamma function, this is in any book on complex analysis.

The next answers are well beyond undergrad level.

-Physicists often use the notion of renormalization, which consists in removing the divergent part of the integral of some physical quantity, unfortunately I never really understood what part of it was rigorous, and what part was just cooking.

-To some extent, the work of Martin Hairer (Fields medalist 2014) is also centered on renormalization issues for stochastic partial differential equations. I do not know much about it, but I can tell that it requires graduate level knowledge in probabilities to understand the issues. Survey paper https://arxiv.org/pdf/1803.03044.pdf

Answer 2

Well.

First, you can regularize some integrals with the same techniques as you regularize divergent series. For instance, you can regularize some integrals using Cesaro summation. Basically it means finding the average of the integral as the integration limit goes to infinity.

Using these and related techniques one can find that

$$\int_0^\infty \sin x\, dx=1$$

and even

$$\int_0^\infty \tan x\,dx=\ln 2$$ (see here)

But these integrals go to some limit at least in a sense of mean value. There are other integrals that go to infinity, and they allow regularization as well.

For instance, we know that the harmonic series is regularized to Euler-Mascheroni constant (by Zeta regularization or Ramanujan):

$$\operatorname{reg}\sum_{k=1}^\infty \frac1k=\gamma$$

At the same time, the difference between the partial sum and integral goes to $\gamma$ as well (the blue area):

$$\gamma = \lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\int_1^n\frac1t dt\right)$$

This allows us to conclude that $$\operatorname{reg}\int_1^\infty\frac1t dt=\operatorname{reg}\sum_{k=1}^\infty \frac1{k}-\gamma=0$$ as regularization is always a linear operator.

In general, one can reduce regularizing an integral to regularizing the corresponding series:

$$\operatorname{reg}\int_0^\infty f(x)\,dx=\lim_{s\to0}\left( s \operatorname{reg} \sum_{k=1}^\infty f(sk)\right)$$

Answer 3

This code in Mathematica works just well for functions with no singularities:

If the function under integral has a singularity at zero, one can transform it using this operator:

The integral of the resulting function from zero to infinity will be the same as of the original function.