A known generalization of Levi-Civita field is a field of formal power series of $\varepsilon$ of the form $\mathbb{R}[[\varepsilon^{\mathbb{Q}}]]$. Assuming $\varepsilon=1/\omega$, we can naturally embed a set of divergent integrals in our field as formal power series of $\omega$:
$F(\omega)=F(a)+\int_a^\infty F'(x)dx$
This way the ordering of divergent integrals will correspond to the ordering of their growth rates represented as powers series.
Assuming the Levi-Civita type of multiplication operation, we can obtain the multiplication rule for divergent integrals:
$\int_0^\infty f(x)dx\cdot\int_0^\infty g(x)dx=D^{-1}f(x)D^{-1}g(x)|_{x=0}+\int_0^\infty D_x[\int_0^x f(t)dt \int_0^x g(t)dt]dx-\operatorname{reg}\int_0^\infty D_x[\int_0^x f(t)dt \int_0^x g(t)dt]dx$
The $D^{-1}$ is assumed to be natural integral here.
The above formula can be coded in Mathematica system with the following code:
f[x_] := Exp[x]
g[x_] := Exp[x]
prod1[x_] :=
Evaluate[Refine[Integrate[f[x], x] Integrate[g[x], x], x > 0]]
prod2[x_] :=
Evaluate[Refine[
Integrate[f[x], {x, 0, x}] Integrate[g[x], {x, 0, x}], x > 0]]
Inactivate[
Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
g[x], {x, 0, Infinity}], Integrate] ==
FullSimplify[
prod1[0] +
Distribute[
Integrate[
ExpandAll[FullSimplify[D[prod2[x], x]]], {x,
0, \[Infinity]}]]] -
Limit[Sum[D[prod2[s x], x], {x, 1, Infinity},
Regularization -> "Borel"] // FullSimplify, s -> 0] //
ExpandAll // Quiet
Inactivate[
Reg[Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
g[x], {x, 0, Infinity}]], Integrate] == FullSimplify[prod1[0]]
For instance, as the program outputs, $\int _0^{\infty }e^xdx\cdot \int _0^{\infty }e^xdx=\int_0^{\infty } 2 e^{2 x} \, dx-\int_0^{\infty } 2 e^x \, dx$.
The code has two caveats. First, it assumes the indefinite integrals produced by Mathematica are natural integrals (which is the case for the basic elementary functions). Second, it uses a series regularization, in this case, Borel, but for other functions another method, like Dirichlet may be needed.
As follows from Levi-Civita multiplication, the regularized value of the product of two integrals is the product of the regularized values. Thus, knowing that $\operatorname{reg}\int_0^\infty e^x dx=-1$, we can conclude that this integral squared has the regularized value of $1$.
That said, I have the following questions.
Previously I already tried to define multiplication of divergent integrals in a different way. Now I see that Levi-Civita type of multiplication is the natural way (limit of the product should be equal to the product of limits, etc). For instance, the older approach would give
$\int_0^\infty (2x^3-3x^2+x-4) \, dx \cdot \int_0^\infty (2x^2-3x+1) \, dx= \int_0^\infty \left(\frac{7 x^6}{3}-\frac{17 x^5}{2} + 10 x^4-\frac{41 x^3}{3}+\frac{1007x^2}{60}-\frac{63 x}{10}-\frac{113}{120}\right) \, dx+\frac{127}{420}$
while Levi-Civita type of multiplication gives for the same integrals
$\int_0^\infty (2x^3-3x^2+x-4) \, dx \cdot \int_0^\infty (2x^2-3x+1) \, dx=\int_0^{\infty } \left(\frac{7 x^6}{3}-\frac{17 x^5}{2}+\frac{35 x^4}{3}-\frac{53 x^3}{3}+\frac{39 x^2}{2}-8 x\right) \, dx$
But the older approach was while more complicated, at the same time, more interesting because of connection with Bernoulli numbers and Zeta function.
As such, I wonder whether the older approach can be somehow embedded in the newer set by choosing a suitable basis of otherwise? Levi-Civita approach reminds me dual numbers while my old approach is similar to split-complexs numbers.
Can the Levi-Civita multiplication code and the formula be simplified?