Why the double integral does not exist? Consider $([-1,1]\times[-1,1],\mathcal{B}[-1,1]\times\mathcal{B}[-1,1],dx\,dy)$. Let $f(x,y)=\dfrac{xy}{(x^2+y^2)^2}$. Show the double integral does not exist.
My solution:
\begin{align}
\int_{[-1,1]\times[-1,1]}|f(x,y)|\,dx\,dy&\ge\int_{x^2+y^2\le1}\left|\frac{xy}{(x^2+y^2)^2}\right|\,dx\,dy\tag{1}\\
&=\int_0^{2\pi}d\theta\int_0^1\left|\frac{r\cos\theta\cdot r\sin\theta}{r^4}\right|\,r\,dr\tag{2}\\
&\ge\int_0^{\pi/2}\cos\theta\cdot\sin\theta \,d\theta\int_0^1\frac{1}{r}\,dr\tag{3}\\
&=\frac12\ln r|_0^1\tag{4}\\
&=+\infty.\tag{5}
\end{align}
Any comment is appreciated.
 A: You immediately changed the region over which you're integrating from a square to a circle, so that won't work.
I would construe the assertion that the double integral does not exist to mean that the integrals of the positive and negative parts are both infinite.  If one of them is infinite and the other is finite, I would say the integral exists and is either $+\infty$ or $-\infty$.  By Fubini's theorem, the positive and negative parts must both be infinite if the two iterated integrals are unequal, i.e.
$$
\int_{-1}^1\left(\int_{-1}^1 f(x,y)\,dx\right)\, dy \ne \int_{-1}^1\left(\int_{-1}^1 f(x,y)\,dy\right)\, dx.
$$
The symmetry of the function and of the region over which we're integrating should cut our work in half.
$$
\int_0^1 \frac{xy}{(x^2+y^2)^2}\,dx = \int_0^{\pi/4} \frac{y^2\tan u}{(y^2\tan^2 u+y^2)^2} \, \left(y\sec^2 u \, du\right)
$$
$$
= \int_0^{\pi/4} \frac{y^2\tan u}{(y^2\tan^2 u+y^2)^2} \, \left(y\sec^2 u \, du\right) = \int_0^{\pi/4} \frac{y^2\tan u}{y^4\sec^2 u} \, \left(y \sec^2 u\,du\right)
$$
$$
= \frac{1}{y}\int_0^{\pi/4} \tan u\,du = \left.\frac{-1}{y} \log(\cos u)\right|_0^{\pi/4} = \frac{\log 2}{2y}.
$$
The integral of this function of $y$ from $0$ to $1$ is $+\infty$.  So over that square we get $+\infty$, and by symmetry, over each of the second and fourth quadrants we get $-\infty$ (and over the third we get $+\infty$).
A: There's no need to attempt to evaluate the integral. Simply consider the neighbourhood around $(0,0)$.
As $(x,y) \rightarrow (0,0)$ from the $x$- or $y$-axis, $f(x,y) \rightarrow 0$.
As $(x,y) \rightarrow (0,0)$ along the line $y=x$, $f(x,y) \rightarrow \infty$.
This means that a partition around $(0,0)$ will have differing upper and lower sums for any arbitrarily small partition, so $f$ is not Riemann integrable on $[-1,1]\times[-1,1]$.
