Prove that $\iint\limits_ {[0,1] \times [0,1]} \frac{x^2-y^2}{(x^2 + y^2)^2}\,\mathrm dx\,\mathrm dy$ is not integrable I have to prove that the following integral does not exist:
$$\iint \limits _{[0,1] \times [0,1]} \frac{x^2-y^2}{(x^2 + y^2)^2}\,\mathrm dx\,\mathrm dy .$$
I think I can use Fubini's Theorem, ie. if I show that iterated integrals are not equal to each other, then the corresponding double integral can't be integrable, because otherwise it would be "counter example" to Fubini's Theorem. Is that a good idea?
 A: The region $\{(x,y): x\ge y\ge 0,\;x^2+y^2\le 1\}$ is contaied in $[0,1]\times[0,1]$, and the integrand is nonnegative there. Furthermore, we can easily parametrize it in polar coordinates:
$$\int_0^{\pi/4}\int_0^1r\frac{r^2\cos2\theta}{r^4}drd\theta$$
But $1/r$ is not integrable at $r\in(0,1]$.
A: What you wrote is not exactly right. The function $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ is not integrable with respect to the Lebesgue measure on the unit square $(0,1)\times(0,1)$ (so it does not fulfill the hypothesis of Fubini's theorem), but the integral
$$ I=\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx$$
does exist, since:
$$ \int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy = \frac{1}{1+x^2} $$
hence:
$$ I = \int_{0}^{1}\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}.$$
However, it is straightforward to check that:
$$ \frac{\pi}{4}=\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx\neq \int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy = -\frac{\pi}{4}. $$
The integral that is not defined is:
$$ \iint_{(0,1)^2} f(x,y) \,d\mu $$
where $\mu$ is the Lebesgue measure on the unit square.
