Are there existing theories of integration in which $I_0 = \int_0^{\infty} dx$ and $I_1 = \int_0^{\infty} x \ dx$ are welldefined infinite elements in a nonarchimedean 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.

$\begingroup$ Sounds like a duplicate of this older question: mathoverflow.net/questions/115743 $\endgroup$– Igor KhavkineSep 19 '13 at 1:27

$\begingroup$ Updated my answer $\endgroup$– AnixxMar 24 '17 at 13:51

$\begingroup$ My answer to this question is very outdated.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. $\endgroup$– AnixxJun 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_CalculusSpringer(2012).pdf (Hypernumbers and Extrafunctions: Extending the Classical Calculus, by Mark Burgin).
Link in amazon: http://www.amazon.com/HypernumbersExtrafunctionsExtendingSpringerBriefsMathematics/dp/1441998748
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.
 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 nonstandard 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.

$\begingroup$ Why the downvotes and the delete vote? Something is unclear? $\endgroup$– AnixxOct 20 '16 at 3:15

2$\begingroup$ 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 plussign, why should its standard part have a minus sign? $\endgroup$ Oct 20 '16 at 14:45

$\begingroup$ @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 nonnegative integers and twice less than the quantity of all integers (integral over all real line would be equal to the sum of all integers). $\endgroup$– AnixxOct 20 '16 at 16:37

6$\begingroup$ I think I now see why your posts got downvoted; the page you link to as a reference is unpublished and takes overbold steps like "Now we define that to any divergent series there corresponds a nonstandard number." This forum has a low level of tolerance for speculative, nonmainstream work, and I think your post fell afoul of that. If these rough ideas were developed more rigorously and published in a reputable peerreviewed journal, they'd probably get more respect. Email me if you want to pursue this further. $\endgroup$ Oct 21 '16 at 16:37

2$\begingroup$ 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 email 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. $\endgroup$ Dec 4 '16 at 16:06