"Values" of divergent integrals

Question

Are there existing theories of integration in which $I_0 = \int_0^{\infty} dx$ and $I_1 = \int_0^{\infty} x \ dx$ are well-defined infinite elements in a non-archimedean extension of the reals? I can think of at least two different ways to set up such a theory; in one, $I_1 = I_0^2$, and in the other, $I_1 = I_0^2/2$. (Both are based on mollification. In the first framework, you introduce a mollified version of $dx$, namely $e^{-\lambda x} dx$, and you look at the behavior of the mollified integral for $\lambda$ near $0^+$; in the second case, you think of the integral as the area under the graph, and you mollify in the $y$-direction as well.) There are no problems as long as the integrand is sufficiently nice, e.g., a polynomial function. I'm wondering if any such theories have been written down and/or been found to be useful.

Answer 1

For an attempt of such a theory, see http://carlossicoli.free.fr/B/Burgin_M.-Hypernumbers_and_Extrafunctions__Extending_the_Classical_Calculus-Springer(2012).pdf (Hypernumbers and Extrafunctions: Extending the Classical Calculus, by Mark Burgin).

Link in amazon: http://www.amazon.com/Hypernumbers-Extrafunctions-Extending-SpringerBriefs-Mathematics/dp/1441998748

Answer 2

As I recall from QFT class the basic idea is to set the bound of integration to be $L$ and let $L \to \infty$.

$$ \int_{-L}^L dx = 2L \quad\text{ and }\quad \int_{-L}^L x \, dx = L^2 $$

The integrals in physics are badly divergent. The usual way is to set a parameter $\epsilon$ And there is a lot of talk about this method or that method being the canonical way of assigning a value.

Dual to these are badly oscillatory integrals. Revolving around the idea that $\sin n x \to 0$ is weakly convergent in $L^2[-\pi, \pi]$. And physicist might extract this limit exists generally over $\mathbb{R}$.

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I think there is too much attention drawn to zeta function regularization which is intimately involved with the eq $\zeta(-1) = - \frac{1}{12}$. As you note there are careful, it's possible to illustrate two plausible values for the same diveergent integrals and then you are in trouble.

• Henle & Kleinberg Infinitesimal Calculus
• John Bell A Primer of Infinitesimal

And the warning here is they use two different types of infinesimals. As I learned... one uses non-standard analysis, the other uses smooth infinitesimal analysis. In particular, there's difficulty with the mean value theorem.

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I have a feeling you know much more than this but I am not sure what to recommend. Lately there is the fascinating theory of resurgence. As one might know: $$ \log n! = \sum_{k=1}^n \log k \approx \int_1^n \log x \, dx = n \log n - n $$ However if you try to "correct" this series too much, the answer on the right hand side is horribly divegent. And this is tied to essential singularities such as $e^{-1/z}$.

Answer 3

Using [this approach][1] the values of these integrals are as follows:

$$\int_0^\infty 1\, dx = \omega_-+1/2=\tau$$

Its standard part (analog of regularization in case of series) is zero.

Here $$\omega_-=\sum_{k=1}^\infty 1$$, the quantity of natural numbers, while $\tau$ is the quantity of all even or all odd numbers, half of the quantity of all integers.

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

$$\int_0^\infty \cos x\,dx=0$$

The standard part of arbitrary improper integral of an analytic function can be calculated from this rule from divergent series:

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

The following Mathematica code does the trick:

Sum[f[s x],{x,1,Infinity},Regularization->"Borel"]//FullSimplify
Limit[s %,s→0]

Thus,

$$\operatorname{st} \int_0^\infty e^x\, dx=-1$$

which coincides with analytic continuation, by the way.