I have the following two iterated integrals: $\int_{-1}^1\int_{-1}^1 \frac{xy}{(x^2+y^2)^2} dx dy$ and $\int_{-1}^1\int_{-1}^1 \frac{xy}{(x^2+y^2)^2} dy dx$, where $f(x,y)$ is defined on $\mathbb{R}^2\backslash \{(0,0)\}$ and $f(0,0)=0$. I have to show that the integrals are well-defined and equal. I have split the inner integral at the singularity (0,0). However, I am not sure what the argument is that these integrals are well-defined. Is it that, for fixed y, the integrand in $\int_{-1}^1\int_{-1}^1 \frac{xy}{(x^2+y^2)^2} dx dy$ does not have a singularity at x=0? (and analogously for the other iterated integral) If so, I can consider the integrand continuous and bounded on a compact interval, and hence the integrals are Riemann integrable.
However, when it comes to the 2-dimensional Lebesgue integral, I don't know how to argue if it is well-defined or not. $\int_{[-1,1]^2}\frac{xy}{(x^2+y^2)^2}$d\beta^2(x,y)
Thank you in advance for your help.
Let's fix one of the arguments, then $f(x,y)$ is continuous function of the other, so $$\int\limits_{-1}^{1}\frac{xy}{(x^2+y^2)^2}dy$$ exists for all $x$ in $[-1,1]$. Thus, as integral of an odd function, integral $$\int\limits_{-1}^{1}dx\int\limits_{-1}^{1}\frac{xy}{(x^2+y^2)^2}dy=\int\limits_{-1}^{1}dy\int\limits_{-1}^{1}\frac{xy}{(x^2+y^2)^2}dx = 0$$ but $f(x,y)$ is not summable on given square. If it were summable, then it would be also summable on sub-square $[0,1]\times [0,1]$, then integral $$\int\limits_{0}^{1}dx\int\limits_{0}^{1}\frac{xy}{(x^2+y^2)^2}dy$$ must exists, but for $x\ne 0$ $$\int\limits_{0}^{1}\frac{xy}{(x^2+y^2)^2}dy=\frac{1}{2x}-\frac{x}{2(x^2+1)}$$ which is not summable on $(0,1]$.