Is the function $f(x,y) = \frac{x-y}{(x + y)^3}$ lebesgue integrable.

I'm trying to show whether the function $f(x,y) = \frac{x-y}{(x + y)^3}$ is Lebesgue integrable on $[0,1]\times[0,1]$.

I've split the region into two parts $x>y$ and $x<y$ (by the symmetry of the function $f(x,y) = -f(y,x)$).

I'm trying to show that it's either integrable or not integrable in one of these regions. In $x<y$ the function is positive, so by Fubini's theorem I want to show that either one of the repeated integrals is finite. However I'm having trouble finding these integrals explicitly, or finding a control function to show that they are integrable.

Any help is much appreciated,

Thanks

Solution 1:

If $f$ were integrable, then by Fubini's theorem the functions $f_y:x\in[0,1]\mapsto f(x,y)\in\mathbb{R}$ would be integrable for a.e. $y\in [0,1]$. But notice that for every $y$, $$\int_0^1|f_y|(x)dx=\int_0^1\left|\frac{x-y}{(x+y)^3}\right|dx=\int_0^1\frac{|z-2y|}{z^3}dz=\infty,$$ so $f_y$ is never Lebesgue-integrable. Hence, $f$ is not Lebesgue-integrable.