I have worked out $\int_{[0,1]}(\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2} d\lambda(x)) d\lambda (y)=-\frac{\pi}{4}$ and $\int_{[0,1]}(\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2} d\lambda(y)) d\lambda (x)=\frac{\pi}{4}$
What should my exact reasoning be as to why the double integral does not exist according to $\lambda^2$?
Should I argue along the lines of $\int_{[0,1]\times[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda^2(x,y)$ not being well-defined as
$\int_{[0,1]\times[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda^2(x,y)=\int_{[0,1]}(\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2} d\lambda(x)) d\lambda (y)=-\frac{\pi}{4}$
and
$\int_{[0,1]\times[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda^2(x,y)=\int_{[0,1]}(\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2} d\lambda(y)) d\lambda (x)=\frac{\pi}{4} $
But I am not sure on this argument.
It's a direct application of Fubini: $f\notin \mathcal L^1(I\times I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=\left \{(x,y):) 0\le y\le \frac{1}{\sqrt 3}x\right \},$ we have $ |f(x,y)|=\frac{\cos 2t}{r^{2}}\ge \frac{1}{2r^{2}}$ so $\int|f|=\int_E |f|+\int_{E^c}|f|$ and since $\int_E |f|=\infty,\ f\notin L^1(I\times I)$.