Separate double integrals with min/max

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Question: Is it possible to separate integrals of multiple functions which contain min/max functions?

In my case in particular I have an integral in the following form:

$$ \begin{equation} I = \iint_0^\infty f(x)g(y)\frac{min(x,y)^L}{max(x,y)^{L+1}}dxdy \end{equation} $$

Where L is some integer. I have a numerical solver which seems to give reasonable results, but I would like to have a closed-form solution, if possible. What I envision (if it is possible) is to have some sort of solution of the form:

$$ \begin{equation} I = I_{x>y} + I_{x<y} + I_{x=y} \end{equation} $$

(I argue that since $x=y$ only in some small infinitesimal case, it can be neglected)

What I have tried:

Attempt 1: Solve indefinite

I tried to solve the indefinite cases where $x>y$, $x<y$ and $x=y$, then add them thus:

$$ \begin{equation} I = I_{x>y} + I_{x<y}=\int_0^\infty f(x)x^{-L-1}dx\int_0^\infty g(y)y^Ldy +... \end{equation} $$

However this solution seems to deviate from the numerical one. For L=0, it seems to be almost exactly a factor of 10, at L=1 it seems to be a factor of 2, etc. (This solution is especially tempting as I have closed form solutions for all of the integrals).

Attempt 2: Change bounds of integration

In this attempt, I utilized the min/max to change the bounds of integration of each of the cases. So for example when $x>y$, I only integrate $y$ from 0 to $x$, something like:

$$ \begin{equation} I = I_{x>y} + I_{x<y}=\int_0^\infty f(x)x^{-L-1} \left(\int_0^x g(y)y^Ldy\right)dx + ... \end{equation} $$

This doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some clean factor like attempt 1.

Attempt 3: Add new bound of integration

In this attempt I again use the min/max to split the equation, but this time I add a new variable and integrate over all possible values of this new variable. So for $x>y$ I integrate $y$ from 0 to $z$, then I integrate $x$ from $z$ to $\infty$, then I integrate $z$ from 0 to $\infty$.

$$ \begin{equation} I = I_{x>y} + I_{x<y}=\int_0^\infty \left(\int_z^\infty f(x)x^{-L-1}dx\int_0^z g(y)y^Ldy\right)dz + ... \end{equation} $$

Again, this doesn't seem to correspond to the numerical solution, but also doesn't seem to be off by some nice clear factors like attempt 1.