Double Integral Question on unit square Hints on solving following double integral will be appreciated.

$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$
The integral does not converge. By various tricks, you can cause some kind of iterated calculation to give any value you like. 

You can cut up the domain into parts. For $x^2 + y^2 \geq 1$ there is no trouble. Furthermore, there is a $\pm$ symmetry across the $45^\circ$ line $y=x.$ So, we consider the part $$ x^2 + y^2 \leq 1, \; \; y \leq x. $$ On this part, the integral is, indeed, $$ \int_0^{\pi/4}  \int_0^{1} \frac{\cos 2 \theta}{r} dr d\theta = \frac{1}{2 }  \int_0^{1} \frac{1}{r} dr = \left.\frac{1}{2 } \log r \; \right\rvert_{r=0}^{r=1} $$ which does not work.

There is surely some good way to write the traditional vertical bar symbol denoting the evaluation step for a one-variable integral. I think it will work if I make the $r$ values into an array with a blank line or two in the middle. That's better, array, with matching \left|  and \right. which produces nothing as the right delimiter; if I had put \right| there would be a second vertical bar.

http://en.wikipedia.org/wiki/Multiple_integral#Multiple_improper_integral
0 by symmetry.Integrand changes sign across the diagonal