$\int_{[-1,1]}\int_{[-1,1]}\frac{xy}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)= \int_{[-1,1]}\int_{[-1,1]}\frac{xy}{(x^2+y^2)^2}d\lambda(y) d\lambda(x)$

Question

I have calculated

$\int_{[-1,1]}\int_{[-1,1]}\frac{xy}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)= \int_{[-1,1]}\int_{[-1,1]}\frac{xy}{(x^2+y^2)^2}d\lambda(y) d\lambda(x)=0$

Why would the double integral w.r.t. $\lambda ^2$ not exist?

Answer 1

The rectangle $[-1,1] \times [-1,1]$ contains the unit disk. Consider $\int _{{x^{2}+y^{2} \leq 1}} \frac {|xy|} {(x^{2}+y^{2})^{2}} dxdy$. If you use polar coordinates you will get a constant times $\int_0^{1} \frac 1 r \, dr$ which is $\infty$. [The constant is $\int_0^{2\pi} |\sin\, \theta \cos \, \theta|\, d\theta$].