# "Values" of divergent integrals

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.

• Sounds like a duplicate of this older question: mathoverflow.net/questions/115743 Sep 19 '13 at 1:27
• Updated my answer Mar 24 '17 at 13:51
• My answer to this question is very outdated. In my current research, the quantities have alternative representations: $I_1=\int_0^{\infty } \, dx=\int_0^{\infty } \frac{1}{x^2} \, dx=\tau=\pi\delta(0)$ and $I_2=\int_0^{\infty} x \, dx=\int_0^\infty \frac2{x^3}dx=\frac{\tau ^2}{2}+\frac{1}{24}=\frac{\omega_+^3-\omega_-^3}6$. And, in general, $\int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)}$. I am not sure whether I should make a new answer. Jun 6 at 16:55

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).

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 $\mathbb{R}$.

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.

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.

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 $e^{-1/z}$.

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.

• Why the downvotes and the delete vote? Something is unclear? Oct 20 '16 at 3:15
• I think the downvoters object to metaphorical phrases like "the quantity of all even or all odd numbers". I hope the downvoters will phrase their objections in constructive fashion. Meanwhile, let me ask a concrete question: if the term in the RHS of the first equation has a plus-sign, why should its standard part have a minus sign? Oct 20 '16 at 14:45
• @James Propp powers of $\omega_-$ have standard part equal to Bernoulli numbers. First power has standard part -1/2, as you can see following the link. If we add 1/2, the standard part is zero. The standard part is what corresponds to the regularization of the series (as you can see en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%E2%8B%AF ). This integral is 1/2 greater than the quantity of positive integers, 1/2 less than quantity of non-negative integers and twice less than the quantity of all integers (integral over all real line would be equal to the sum of all integers). Oct 20 '16 at 16:37
• I think I now see why your posts got downvoted; the page you link to as a reference is unpublished and takes over-bold steps like "Now we define that to any divergent series there corresponds a non-standard number." This forum has a low level of tolerance for speculative, non-mainstream work, and I think your post fell afoul of that. If these rough ideas were developed more rigorously and published in a reputable peer-reviewed journal, they'd probably get more respect. Email me if you want to pursue this further. Oct 21 '16 at 16:37
• I just found out about your October updates to this question (I'm not sure why the Stack Exchange software didn't notify me sooner). You say "I wrote a e-mail to you" but I never received it. Please use my gmail address, and be sure to use the word "divergent" somewhere so I can search for it. Dec 4 '16 at 16:06