# Multiplying divergent integrals using Hardy fields approach

• A
• Anixx
In summary, the conversation discusses the idea of multiplying two improper integrals, specifically ##\int_0^\infty e^x dx\cdot\int_0^\infty e^x dx##, by looking at the partial sums of the integrals and finding the product of their germs at infinity. The result is an unbounded and positive product, as shown by taking the limit of the partial sums. The conversation also explores other examples of products of divergent integrals and suggests a method for calculating them using the Laplace transform.
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:

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

Anixx said:
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.
Anixx said:
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.
Anixx said:
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?
Anixx said:
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?

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

Office_Shredder said:
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:
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] // 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.

Last edited:

## 1. What is the Hardy fields approach?

The Hardy fields approach is a mathematical method used to solve divergent integrals, or integrals that do not have a definite value. It involves using a sequence of functions to approximate the divergent integral and then taking the limit as the sequence approaches infinity.

## 2. Why is the Hardy fields approach useful for multiplying divergent integrals?

The Hardy fields approach allows for the multiplication of divergent integrals by using a sequence of functions to approximate each integral separately and then taking the product of the limits of these sequences. This method is useful because it allows for the manipulation of divergent integrals, which would otherwise be impossible to solve.

## 3. What are some applications of the Hardy fields approach?

The Hardy fields approach has applications in various areas of mathematics, including differential equations, number theory, and analysis. It is also used in physics and engineering to solve problems involving divergent integrals.

## 4. Is the Hardy fields approach the only method for solving divergent integrals?

No, there are other methods for solving divergent integrals, such as the Borel summation method and the zeta function regularization method. However, the Hardy fields approach is a commonly used and effective method for solving these types of integrals.

## 5. Are there any limitations to the Hardy fields approach?

Yes, the Hardy fields approach may not work for all types of divergent integrals. It is also a complex and time-consuming method, so it may not be practical for every situation. Additionally, the convergence of the sequence of functions used in this approach must be carefully analyzed to ensure accurate results.

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