Lebesgue integral on $[0,1]^{2}$

Question

Let $f:[0,1]^2->\mathbb{R}$ with

$f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y)\neq(0,0)$

$f(x,y)=0$ else

I want to calculate

$\lim_{a\to 0}\int_{[a,1]^2}f(x,y)d\mathscr{L}^2$

But how do Ido this? I already calculated

$\int_{0}^1\int_{0}^1f(x,y)dxdy=-\pi/4$

$\int_{0}^1\int_{0}^1f(x,y)dydx=\pi/4$

Doesn't that mean that I cannot use Fubini-Tonelli? But what other choice do I have?

Answer 1

All right, the function isn't absolutely integrable on the whole square. But that's not the question; we're looking at an improper integral, approaching the full square in a specific way akin to a principal value.

For any $a>0$, $$I(a) = \int_a^1\int_a^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx = \int_a^1\int_a^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy$$ $$\int_a^1\int_a^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy = \int_a^1\int_a^1 \frac{y^2-x^2}{(x^2+y^2)^2}\,dy\,dx = -I(a)$$ In the first line, we use Fubini's theorem to switch the order (it's a bounded continuous function). In the second line, we just swap the names of the two variables - that isn't really a change of order. Combining the two, we get $I(a)=-I(a)$, so $I(a)=0$. Take the limit as $a\to 0$, and we get zero.

Answer 2

Hint: the integral is $not$ over $[0,1]\times [0,1].$ For fixed $a: 0<a\le x,y\le 1,$ show that $f\in L^1([a,1]\times [a,1]).$ Then, apply Fubini and finally, take the limit as $a\to 0.$