Multiplying divergent integrals using Hardy fields approach

[Anixx]

I wonder if the following makes sense.

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:

    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

 

[Mark44]

I wonder if the following makes sense.
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 $\int_0^x e^x dx=e^x-1$.

Yes, but to go from this proper definite integral to the improper with an infinite limit requires taking the limit as x increases without bound. Doing so shows that each integral is unbounded and positive, so the product is also unbounded and positive.

A small quibble is that you have used x both as a dummy variable in the integral as well as one of the integration limits. The usual practice is to use a different letter for one of these.

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

Germs? I'm not familiar with that term. Is it a synonym for antiderivative? In any case, once you take the limit, the expression on the right has a "limit" of infinity.

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.

Every one of the integrals above diverges, so how is this useful?

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?

 

[Office_Shredder]

Can you be very specific aboutv what regularization technique you are using?

 

[Anixx]

Can you be very specific aboutv what regularization technique you are using?

Well, this post is not about regularization really, but in the code I use Dirichlet regularization, provided by Mathematica system. This is not necessary though, it is a workaround for some strange glitch or a bug that I encountered.

Below is a more simple code that does the same:

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] // FullSimplify // Evaluate
Inactivate[
    Integrate[f[x], {x, 0, Infinity}]\[CenterDot]Integrate[
      g[x], {x, 0, Infinity}], Integrate] ==
   Distribute[
    Integrate[ExpandAll[FullSimplify[u[x]]], {x, 0, Infinity}]] //
  ExpandAll // Quiet

The only problem with it is that it gives a wrong answer when f[x]=Exp[x] and g[x]=Exp[-x] (should be greater by 1), due to the discontinuity of some intermediate expression of 0/0 type. In all other cases the result is the same as in the longer code.