# How to prove that the order of integration can be changed? For the function: $\frac{x^2-y^2}{(x^2+y^2)^{2a}}$

I want to prove that for the function: $$\frac{x^2-y^2}{(x^2+y^2)^{2a}}$$

The integrals can be exchanged when $$0, that is:

$$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{2a}})dxdy=\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^{2a}}dydx$$

Using Fubini's theorem, that is, I believe I would need to show that: $$\int_{(0,1)\times(0,1)}|\frac{x^2-y^2}{(x^2+y^2)^{2a}}|dP_1\otimes dP_2<\infty$$

How do I evaluate the integral?

Let $$I=\int_0^1\int_0^1\frac{x^2-y^2}{(x+2+y^2)^{2a}}dxdy$$.
For Fubini theorem proof. $$|x^2-y^2|\le x^2+y^2$$. Switch to polar coordinates and get $$|I|\le \int_0^{\frac{\pi}{2}}\int_0^{\sqrt{2}}\frac{r^3}{r^{4a}}drd\theta$$. Since $$4a-3\lt 1$$, the integral exists.
To evaluate the integral, interchange $$x$$ and $$y$$ and see that $$I=-I$$, so the integral $$=0$$.