Generalization of Levi-Civita type construction towards divergent integrals and corresponding questions

Question

A known generalization of Levi-Civita field is a field of Hahn 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 -> "Dirichlet"] // 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, Dirichlet, 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

$$\int_0^\infty (2x^3-3x^2+x-4) \, dx \cdot \int_0^\infty (2x^2-3x+1) \, dx$$ $$=\left(\frac{\omega ^4}{2}-\omega ^3+\frac{\omega ^2}{2}-4 \omega \right) \left(\frac{2 \omega ^3}{3}-\frac{3 \omega ^2}{2}+\omega \right)=$$ $$\frac{\omega ^7}{3}-\frac{17 \omega ^6}{12}+\frac{7 \omega ^5}{3}-\frac{53 \omega ^4}{12}+\frac{13 \omega ^3}{2}-4 \omega ^2=\int_0^{\infty } \left(\frac{7 x^6}{3}-\frac{17 x^5}{2}+\frac{35 x^4}{3}-\frac{53 x^3}{3}+\frac{39 x^2}{2}-8 x\right) \, dx$$

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-complex numbers.

• Can the Levi-Civita multiplication code and the formula be simplified?