Double integral of $f(x,y)= xy/\left((1+|x|)^2+(1+|y|)^2\right)$ from $(-1,-1)$ to $(1,1)$

Question

Given

$$f(x,y) = \frac{xy}{(1+|x|)^2+(1+|y|)^2},$$

show that

$$\int_{-1}^1\int_{-1}^1f(x,y)\,dx\,dy = 0.$$

I have $$\int_{-1}^1\left(\int_{-1}^0f(x,y)\,dx+\int_0^1f(x,y)\,dx\right)dy,$$ and I think those inner integrals sum to $0$, but I don't know how to show that.

I then need to show that $f$ is not (Lebesgue) integrable on $\mathbb{R}$ but I should get the first part before moving on to that.

Answer 1

For every $ab$ in $[0,1] \times[0,1]$ there is $cd$ in $[-1,0] \times [0,1]$ such that $ab=-cd$ and $ab$ and $cd$ are symmetric with respect to $y$-axis. For every $ef$ in $[-1,0] \times [-1,0]$ there is $gh$ in $[0,1] \times [-1,0]$ such that $ef=-gh$ and $ef$ and $gh$ are symmetric with respect to $y$-axis.

I think that this shows that the integral is indeed zero.