Lebesgue integral in two dimensions over fraction

Let $X=[-1,1],Y=[0,1]$, and $$f(x,y)=\dfrac{xy}{(x^2+y^2)^2}$$ for $x\in X,y\in Y$. Let $\mu$ be the Lebesgue measure. Does the following integral exist:

$$\int_{X\times Y} f(x,y)d(\mu\times\mu)$$

It seems very hard to follow the definition of Lebesgue integrals using simple functions here. To do that we must find simple functions that are lower bounds for the function $f$. But the form of $f$ is quite complicated. What would be the way to compute this integral?

So, why not start by considering these separately: the positive part is precisely the part over $[0,1]\times[0,1]$, and the negative part is the part over $[-1,0]\times[0,1]$. If you can show that either does not exist, then you are done; otherwise, you will have shown that both exist, and are also done!
Think about $$\int\limits_{[0,1]\times[0,1]}\frac{xy}{(x^2+y^2)^2}\,d(\mu\times\mu).$$ In Riemann integration theory, we would consider this as an improper integral because of problems as we approach $(0,0)$; we would further find that the integral does not exist in that context, because it shoots of to $+\infty$ to quickly as we approach $(0,0)$. Can you use this sort of inuition to find some step functions that demonstrate that this integral will be $+\infty$?
Hint: Consider that the measure is $\sigma$ additive. Try to use this fact together with the improper Riemann integral to obtain your result.