A known generalization of [Levi-Civita field][1] 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][2] 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, 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][3] 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-complexs numbers. * Can the Levi-Civita multiplication code and the formula be simplified? [1]: https://en.wikipedia.org/wiki/Levi-Civita_field [2]: https://physics.stackexchange.com/questions/552317/what-are-some-areas-of-physics-where-the-concept-of-natural-integral-may-aris [3]: https://mathoverflow.net/a/403727/10059