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

The integrals can be exchanged when $0<a<1$, 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?

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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 $=0$.

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