Is the function $f(x, y) :=$ $\sin(xy) \over x^2 + y^2$ Lebesgue-integrable? 
Given the function $f: \Bbb R^2 \rightarrow \Bbb R$,
$f(x, y) :=
\begin{cases}
\sin(xy) \over x^2 + y^2,  & \text{$(x, y) \in \Bbb R^2 \setminus ${0}$$} \\
0, & \text{otherwise}
\end{cases}$
decide whether it is Lebesgue-integrable or not. Hint: $\int_{(0, \infty)}$ $1 \over r$ $d\lambda(r) = \infty$.

Where to start about something like that? I know that a function is Lebesgue integrable if it is measurable and if $f(x, y)_+$ and $f(x, y)_-$ are integrable. I think it is measurable since it is continuous. So now, I would have to determine what $f(x, y)_+$ is and prove (or disprove) that 
$\int_{\Bbb R} f(x, y)_+ < \infty$? 
Edit:
Since I didn't receive further help in the comments, I am searching for an other approach. There is a similar problem to this:
Prove that function is not Lebesgue integrable
If you take a look at the answer with 8 upvotes, you'll find that the hint that I was given was applied there. So by switching to polar coordinates $(x, y) = (r \cos \phi, r \sin \phi)$, we would receive something like:
$\int_0^{2\pi} \int_0^{\infty}$ $\sin(r^2 \ \cos \phi \ \sin \phi) \over r$ $dr \ d\phi.$
If there was a way to "clear" the numerator here, it would be fairly easy to apply the hint here too, which would yield that the function is not Lebesgue-integrable directly. But is there a way to do it?
Edit 2:
On the other hand, isn't
$\int_0^{\infty}$ $\sin(r^2 \ \cos \phi \ \sin \phi) \over r$ $\le \int_0^{\infty}$ $1 \over r$?

The statement is equivalent with
$$
\iint \frac{|\sin(xy)|}{x^2+y^2} dxdy=\infty.
$$
Let us try substituting $t=xy$, $s=x^2+y^2$; then we will have
$$
dt\,ds =
\left|\det\begin{pmatrix}y&x\\2x&2y\end{pmatrix}\right| dx\,dy =
2|y^2-x^2| \,dx\,dy < 2s \,dx\,dy,
$$
so
$$
dx\,dy > \frac{dt\,ds}{2s}.
$$
If $s>2t>0$ then the system $x^2+y^2=s$, $xy=t$ has four solutions, so
$$
\iint \frac{|\sin(xy)|}{x^2+y^2} dx\,dy \ge
4\iint_{s>2t>0} \frac{|\sin t|}{s} \frac{dt\,ds}{2s} =\\=
2\int_{t=0}^\infty |\sin t| 
\left(\int_{s=2t}^\infty \frac{ds}{s^2}\right)  dt =
\int_{t=0}^\infty \frac{|\sin t|}{t} dt = \infty.
$$
