I want to show that
$$ \int\limits_{[0,1]\times[0,1]}\frac{xy}{(x^2+y^2)^2}\,d(\mu\times\mu). $$ equals $\infty$, where $\mu$ is the Lebesgue measure. I've tried to find simple functions that give lower bounds to $\dfrac{xy}{(x^2+y^2)^2}$ in the area close to $(0,0)$. I want to split up $[0,1]\times[0,1]$ into $R_1,R_2,\ldots$, where $R_i$ is the region such that $\dfrac{1}{i+1}\leq \max(x,y)\leq \dfrac{1}{i}$. The area of the region $R_i$ is $\dfrac{2i+1}{i^2(i+1)^2}$.
I want to find an lower bound for $\dfrac{xy}{(x^2+y^2)^2}$ in the region $R_i$, so that the eventual sum of the simple function diverges. But I can't seem to find a bound that works.
Consider the subset where $\frac12<\frac{y}{x}<2$. On this set, $4xy>x^2+y^2$. So your integrand is bounded below on this set by $\dfrac{1/4}{x^2+y^2}$. For simplicity, we can also consider just the part where $x\leq \frac12$ to get
$$ \begin{align*} \int\limits_{[0,1]\times[0,1]}\frac{xy}{(x^2+y^2)^2}\,d(\mu\times\mu) &\geq\dfrac{1}{4}\int_{0}^\frac{1}{2}\int_{\frac{1}{2}x}^{2x}\dfrac{1}{x^2+y^2}\,dy\,dx\\ &=\dfrac{1}{4}\int_0^{\frac12}\left(\dfrac{\arctan(2)}{x}-\dfrac{\arctan(\frac12)}{x}\right)\,dx\\ &=\dfrac{1}{4}\left(\arctan(2)-\arctan\left(\frac12\right)\right)\int_0^{\frac12}\dfrac{1}{x}\,dx. \end{align*}$$