- #1

Anixx

- 81

- 12

Suppose we want to multiply ##\int_0^\infty e^x dx\cdot\int_0^\infty e^x dx##.

The partial sums of these improper integrals are ##\int_0^x e^x dx=e^x-1##.

Now we multiply the germs at infinity of these partial sums: ##(e^x-1)(e^x-1)=-2 e^x+e^{2 x}+1##.

Now we find the integral for which the germ at infinity of the partial sum is equal to this product by solving equation for ##u(x)##:

##\int_0^x u(x)dx=-2 e^x+e^{2 x}+1.##

We get ##u(x)=2 e^{2 x}-2 e^x##, so the answer is

##\int_0^\infty e^x dx\cdot\int_0^\infty e^x dx=\int_0^\infty (2 e^{2 x}-2 e^x)dx=2\int_0^\infty e^{2 x}dx-2\int_0^\infty e^x dx.##

This all looks very elementary to me and I have made several programs in Mathematica that can calculate products of divergent integrals this way. It can be even generalized beyond Hardy fields (so as to include periodic functions like sine for instance).

One can see using this method that

\begin{align*}

\int _0^{\infty }1dx\cdot \int _0^{\infty }1dx&{}=2\int_0^{\infty } x \, dx \\

\int _0^{\infty }xdx\cdot \int _0^{\infty }xdx&{}=\int_0^{\infty } x^3 \, dx \\

\int _0^{\infty }xdx\cdot \int _0^{\infty }e^xdx&{}=\int_0^{\infty } \frac{e^x x^2}{2} \, dx-\int_0^{\infty } x \, dx+\int_0^{\infty } e^x x \, dx\\

\int _0^{\infty }e^xdx\cdot \int _0^{\infty }1dx&{}=\int_0^{\infty } e^x \, dx+\int_0^{\infty } e^x x \, dx-\int_0^{\infty } 1 \, dx\\

\int _0^{\infty }\frac{dx}{x+1}\cdot \int _0^{\infty }\frac{dx}{x+1}&{}=\int_0^{\infty } \frac{2 \log (x+1)}{x+1} \, dx \\

\int _0^{\infty }\log x\,dx\cdot \int _0^{\infty }1dx&{}=2\int_0^{\infty } x \log x \, dx-\int_0^{\infty } x \, dx \\

\int _0^{\infty }\log x\,dx\cdot \int _0^{\infty }xdx&{}=\frac{3}{2}\int_0^{\infty } x^2 \log x \, dx-\int_0^{\infty } x^2 \, dx\\

\int _0^{\infty }\log x\,dx\cdot \int _0^{\infty }\log x\,dx&{}=2\int_0^{\infty } x \log ^2 x \, dx-2\int_0^{\infty } x \log x \, dx\\

\int _0^{\infty }x \log x\,dx\cdot \int _0^{\infty }\log x\,dx&{}=\frac{3}{2}\int_0^{\infty } x^2 \log ^2(x) \, dx-\frac{5}{4}\int_0^{\infty } x^2 \log x \, dx\\

\int _0^{\infty }e^xdx\cdot \int _0^{\infty }\log x\,dx&{}=\int_0^{\infty } e^x \log x \, dx+\int_0^{\infty } e^x x \log x \, dx-\int_0^{\infty } e^x x \, dx-\int_0^{\infty } \log x \, dx

\end{align*}

etc.

It seems, the regularized value of product of divergent integrals defined this way is the product of regularized values of the factors, for instance, in our case, ##\operatorname{reg}\left(\int _0^{\infty }e^xdx\cdot \int _0^{\infty }e^xdx\right)=(-1)(-1)=1##. This can be seen using the Laplace transform, for instance.

I wonder whether this has been already suggested and whether it makes sense?

Code for divergent integrals multiplicator for the Mathematica system:

Code:

```
f[x_] := Exp[x]
g[x_] := Exp[x]
u[x_] := D[
Integrate[f[t], {t, 0, x}] Integrate[g[t], {t, 0, x}] // Normal,
x] // Evaluate
reg := Limit[
s Sum[f[s x], {x, 1, Infinity}, Regularization -> "Dirichlet"] //
FullSimplify, s -> 0] Limit[
s Sum[g[s x], {x, 1, Infinity}, Regularization -> "Dirichlet"] //
FullSimplify, s -> 0] // Evaluate
regu := Limit[
s Sum[u[s x], {x, 1, Infinity}, Regularization -> "Dirichlet"] //
FullSimplify, s -> 0] // Evaluate
Inactivate[
Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
g[x], {x, 0, Infinity}], Integrate] ==
reg - regu +
Distribute[
Integrate[ExpandAll[FullSimplify[u[x]]], {x, 0, \[Infinity]}]] //
ExpandAll // Quiet
Inactivate[
Reg[Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
g[x], {x, 0, Infinity}]], Integrate] == reg
```