# Algebra of divergent integrals

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Mark44
Mentor
Divergent integral is such integral whose partial Riemannian sum goes to infinity, of course.
Not necessarily. This would also be a divergent integral, wouldn't it?
##\int_0^\infty \sin(x) dx##

In any case, I don't see the point in an algebra of integrals that diverge.

That integral diverges, but it is Cesaro-summable, so it is equal to a real number (1).

Office_Shredder
Staff Emeritus
Gold Member
Not necessarily. This would also be a divergent integral, wouldn't it?
##\int_0^\infty \sin(x) dx##

In any case, I don't see the point in an algebra of integrals that diverge.
There's a huge field of physics that involves taking divergent integrals and getting numbers

https://en.m.wikipedia.org/wiki/Renormalization

Renormalization was first developed in quantum electrodynamics (QED) to make sense of infinite integrals in perturbation theory. Initially viewed as a suspect provisional procedure even by some of its originators, renormalization eventually was embraced as an important and self-consistent actual mechanism of scale physics in several fields of physics and mathematics.
there are also other areas where performing algebra on divergent values can be useful. As one example, you can count the number of integer points in a polytope by writing down a polynomial and evaluating it at the point 1. It turns out the easiest way to do this (in some sense) is to write down infinite sums that don't converge at 1, but can be realized as simple functions like 1/(1-x), and then you can cancel out the singularities carefully and end up getting the right number.

Stephen Tashi
There's a huge field of physics that involves taking divergent integrals and getting numbers
Whatever the practical uses of divergent integrals, the mathematical task of defining an algebra of such integrals (to currrent standards of rigor used in abstract algebra) requires formulating a precise definition of "divergent integral".

To a student of calculus, it is completely obvious what a divergent integral is. However, to say what it is with sufficient precision to use in doing abstract algebra is not simple. The definition of "definite integral" in calculus is precise enough to define a mathematical object only when the definite integral exists .

To make an analogy, suppose a person wishes to develop an algebra of arithmetic expressions that includes expressions that do not define numbers ( e.g "##\frac{2}{1-1} + \frac{0}{4-(3+1)}##"). The elements of such an algebra would be defined by saying ##X \in \mathbb{A}## if and only if ##X## is a pattern of symbols that obey a certain syntax.

Applying this to the case of divergent integrals, we could define an integration-template as a triple ##(a,b,f)## where each of ##a,b## is a number or one of the symbols ##{-\infty, \infty}## and ##f## is a function.

A divergent integral ##(a,b,f)## could be defined as integration-template such that (using the ordinary notation of Riemann integration) ##\int_a^b f(x) dx## does not exist.

I doubt that a useful algebra for the set of all integration-templates can be created. I think getting something useful requires assuming additional properties that restrict the integration-templates under consideration to a subset that has nice properties.

The original poster is assuming the intuitive idea of "divergent integral" is sufficient to define an element of an abstract algebra. His viewpoint seems to be that an "integral" is an attempted computation - i.e. it is not a mathematical object representing a number, but rather a mathematical object representing a process (i.e. an algorithm). His classification of "integrals" seems to be based on ways in which this process can produce a number, or produce a number in some alternative way to Riemann integration, or fail to produce a number by any of the allowed methods.

It may be possible define an abstract algebra on a set of algorithms. However, it requires going far beyond the calculus-level concept of integrals.