Fubini's Theorem contradiction Why does the fact that
$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dy \,dx = \frac{\pi}{4}$ and $ \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dx \,dy = -\frac{\pi}{4}$ doesn't contradict Fubini's Theorem? Does the tangent function have something to do with that matter?

Fubini's theorem says that if $\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) < \infty$ then $$\int_X \left( \int_Y f(x,y) dy \right) dx = \int_Y \left( \int_X f(x,y) dx \right) dy = \int_{X \times Y} f(x,y) d(x,y)$$
But here, $\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) = \infty$ so we should not expect the integrals to be equal.
Specifically, 
$$\begin{align*} \int_{I \times I} \left\vert \frac{x^2-y^2}{(x^2+y^2)^2} \right\vert d(x,y) &\ge \int_0^1 \left(\int_0^ x\frac{x^2-y^2}{(x^2+y^2)^2} dy \right) dx\\ &=\int_0^1 \left[ \frac{y}{x^2+y^2} \right]_{y=0}^x dx\\ &= \int_0^1 \frac{1}{2x} dx = + \infty. \end{align*}$$
