sufficient conditions for integrability of a bivariate function

Question

Consider the bivariate function

$$f(x,y):=\frac{1}{x^a y^b |x-y|^{c}}$$

on $\left[0,\infty\right[\times [0,\infty[$.

I am trying to find sufficient conditions on the positive real exponents $a,b,c$ such that the integral $\;\int_0^\infty\int_0^\infty f(x,y) \,dx \,dy\;$ exists.

Necessarily we have the following set of conditions:

$$a<1\\b<1\\c<1\\a+c>1\\b+c>1$$

How figure these necessary conditions ? I split the double integral

$$\int_0^\infty\int_0^\infty= \int_0^1\int_0^1+\int_0^1\int_1^\infty+\int_1^\infty \int_0^1 + \int_1^\infty \int_1^\infty.$$

Are the conditions above also sufficient for $f$ to be in $L^1$?

Thanks a lot

Answer 1

Substituting $ (x,y) = (u \lambda, u(1-\lambda)) $ for $u > 0$ and $0 < \lambda < 1$, we have

$$ \left| \frac{\partial(x, y)}{\partial(u, \lambda)} \right| = u, $$

and so, we get

\begin{align*} \int_{(0,\infty)^2} f(x, y) \, \mathrm{d}x\mathrm{d}y &= \int_{0}^{1} \int_{0}^{\infty} u f(u\lambda, u(1-\lambda)) \, \mathrm{d}u\mathrm{d}\lambda \\ &= \int_{0}^{1} \int_{0}^{\infty} \frac{u}{u^{a+b+c} \lambda^{a}(1-\lambda)^{b}\left| 2\lambda - 1 \right|^{c}} \, \mathrm{d}u\mathrm{d}\lambda. \end{align*}

Since

$$ \int_{0}^{\infty} \frac{\mathrm{d}u}{u^{a+b+c-1}} = \infty $$

regardless of the values of $a, b, c$, it follows that the integral always diverges.