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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

$\int_0^{\infty}$ $\sin(r^2 \ \cos \phi \ \sin \phi) \over r$ $\le \int_0^{\infty}$ $1 \over r$?

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    If $f$ is integrable, the change of variables $(x,y)=(r\cos\theta,r\sin\theta)$ yields that$$\int_0^{2\pi}I(\cos\theta\sin\theta)d\theta$$ is finite, where, for every $a$, $$I(a)=\int_0^\infty|\sin(ar^2)|\,\frac{dr}r$$ Now, $I(0)=0$ and for every $a\ne0$, $I(a)=I(1)$ (use the change of variable $r\to r/\sqrt{a}$) hence $f$ is integrable if and only if $I(1)$ is finite. The change of variable $r=\sqrt{s}$ shows that $$2I(1)=\int_0^\infty|\sin(s)|\,\frac{ds}s$$ Now, for every integer $n\geqslant0$, $|\sin(s)|\geqslant\frac12$ for every $n\pi+\frac{\pi}6\leqslant s\leqslant n\pi+\frac{5\pi}6$ ... – 2017-01-25
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    ... hence $$\int_{n\pi+\pi/6}^{n\pi+5\pi/6}|\sin(s)|\,\frac{ds}s\geqslant\int_{n\pi+\pi/6}^{n\pi+5\pi/6}\frac{ds}{2s}\geqslant\int_{n\pi+\pi/6}^{n\pi+5\pi/6}\frac{ds}{2(n+1)\pi}=\frac1{3(n+1)}$$ hence, by summation over $n\geqslant0$, $I(1)$ diverges, which proves that $f$ is not integrable on $\mathbb R^2$. – 2017-01-25
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    Thanks for your solution, but I don't quite understand it. What is your $I$ exactly? An integral? How did you come up with it? – 2017-01-26
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    And where did you use the hint that was given? – 2017-01-26
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    ?? "What is your I exactly?" See comment: I is a function of a which happens to be constant. "An integral?" See comment (second displayed equation). I am wondering: do you know what a change of variables from Cartesian coordinates to polar ones, is? Have you ever used one? – 2017-01-26
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    We talked about it, but I never used it in such a scenario. – 2017-01-26
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    Is there another solution for it that doesn't use polar coordinates? – 2017-01-26
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    What for? What are the tools at your disposal if not this one? Note that the hint given at the end of your exercise clearly points at polar coordinates so why are you even trying to solve this problem if polar coordinates are a mystery to you? – 2017-01-26
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    It's an exam question from the past. – 2017-01-26
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    I'm sorry, I took a look in my notes and realized that you applied $\int_{\Bbb R^2} f(x, y) dx dy = \int_0^{2\pi} \int_0^\infty f(r cos \phi, r sin \phi) r dr d\phi$. – 2017-01-26
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    Could you then elaborate how you derived your representation of $I(a)$? I understand the part with $|sin(ar^2)|$, but what happens in the denominator? For $x^2 + y^2$, switching the variables leads to $(r cos \phi)^2 + (r sin \phi)^2$. Since $cos^2 \phi + sin^2 \phi = 1$, we have $2r^2$ in the denominator. We can simplify this to $2/r$, but it seems like you simplified to $1/r$. Where is your $2$? – 2017-01-26
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    Look better: $x^2+y^2=r^2$, not $2r^2$. – 2017-01-26
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    Ah, you're right, now I see it. :-) Thanks! May I ask a last question? What do you mean by "use the change of variable $r \rightarrow r/\sqrt{a}$"? – 2017-01-26

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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 $$ \iint \frac{|\sin(xy)|}{x^2+y^2} dx\,dy \ge 4\iint_{s>2t>0} \frac{|\sin t|}{s} \frac{dt\,ds}{2s} =\\= 2\int_{t=0}^\infty |\sin t| \left(\int_{s=2t}^\infty \frac{ds}{s^2}\right) dt = \int_{t=0}^\infty \frac{|\sin t|}{t} dt = \infty. $$