Lebesgue-integrability on a compact set

Question

Let $f: [-1, 1]^2 \to \mathbb{R}$, with $$ f(x, y) = \begin{cases} \frac{x^2 - y^2}{(x^2 + y^2)^2}, & (x, y) \ne (0, 0) \\ 0, & (x, y) = (0, 0) \end{cases} $$

I want to verify if $f$ is Lebesgue-integrable over $[-1, 1]^2$.

I have yet determined that the $\int f = \pi$, so I assume f to be $\lambda_2$ integrable.

Now I am trying to find a theorem to prove my assumption.

Tonelli requires $f$ to be non-negative, but $f(x, y) < 0$ for $x < y$.

Should I better use the fact that $f$ is improper Lebesgue-integrable only if $|f|$ is an improper Riemann-integrable function, and show the improper Riemann-integrability from $|f|$ first? Perhaps there may be a better approach to solve that problem.

Kindly regards

Answer 1

The function is not Lebesgue integrable.

Denote $A$ the integral of the absolute value of the function.Then $$ A=\int_{[-1,1]}\int_{[-1,1]}\frac{|x^2-y^2|}{(x^2+y^2)^2}dxdy$$ $$\geq \int_{B((0,1)}\frac{|x^2-y^2|}{(x^2+y^2)^2}dxdy$$

Now use polar coordinates and the fact that $\int_0^{2\pi}|\cos^2{x}-\sin^2{x}|dx=4$

Answer 2

Let $T$ be the triangle with vertices $(0,0),(1,0),(1,1).$ Then

$$\int_{[-1,1]^2} |f|\,dx\,dy \ge \int_T |f|\,dx\,dy=\int_0^1\int_0^x \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy.$$

In the last integral, let $y=xt$ in the inner integral. The integral becomes

$$\tag 1\int_0^1 \frac{1}{x}\int_0^1\frac{1-t^2}{1+t^2}\,dt\,dx.$$

The inner integral in $(1)$ is some positive number, and $\int_0^1 \frac{1}{x}\,dx=\infty.$ Thus $(1)=\infty,$ hence $\int_{[-1,1]^2} |f|\,dx\,dy=\infty$