I am trying to determine whether or not $f(x,y) = \dfrac{\sin(x)\sin(y)}{x^2+y^2}$ is integrable on $E = \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \times \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
I'm thinking this is a bit of a trick question:
$f$ is continuous a.e. so it is measurable on $E$ and for any $y \neq 0$, $x \mapsto \dfrac{\sin(x)\sin(y)}{x^2+y^2}$ is continous on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , hence integrable. Also, since $f$ is an odd function:
$\displaystyle\int_{-\pi/2}^{\pi/2} f(x,y) \ dx = \underbrace{\int_{-\pi/2}^{\pi/2} f(x,y) \ dy}_\text{by symmetry} = 0$
and therefore
$\displaystyle\int_{-\pi/2}^{\pi/2} \int_{-\pi/2}^{\pi/2} f(x,y) \ dx \ dy = \int_{-\pi/2}^{\pi/2} \int_{-\pi/2}^{\pi/2} f(x,y) \ dy \ dx = 0$
However, since $f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable the above result doesn't imply that $f$ is integrable on $E$ - and I have a feeling that
$\displaystyle\int_{-\pi/2}^{\pi/2} \int_{-\pi/2}^{\pi/2} |f(x,y)| \ dx \ dy \ $ is not finite, but I'm having trouble showing this. I've been trying to evaluate:
$\displaystyle\int_{-\pi/2}^{\pi/2} |f(x,y)| \ dx $ to argue that the double integral doesn't exist, but I haven't been able to get anywhere in evaluating the integral. Any help would be appreciated!
Hint: Note that $|\sin x|\le |x|$ and $x^2+y^2\ge 2|xy|$.