for which $a$ is $f$ integratble for which $a$, with $a>0$ is this function integratble $f: [0,1] \times [0,1] \to [0,\infty]: \mathcal{f(x,y)} = \left\{ \begin{matrix} \frac{xy}{(x+y)^{a}} & (x,y-\neq(0,0) \\ \infty & (x,y)=(0,0) \end{matrix}\right.$
My intuition says that $a$ needs to be $< 1$. But I have no idea how I can prove it correctly.

*

*First I tried to solve the integral. Because $f$ is a positive continuous (so measurable) function, it's enough to see for which $a$ the integral is finite. But this didn't work.

*Second, I tried by saying $\frac{xy}{(x+y)^{a}}$<$\frac{y}{(x+y)^{a}}$. I tried with change of variables (x,y)=(u-y,y) an then we can say that $\frac{y}{(u)^{a}}$= $\theta(u^{a})$. I thought this meant that the function is only integrateable if $a< 1$.

EDIT: a good suggestion $\frac{xy}{(x+y)^{a}}$<$\frac{1}{(x+y)^{a}}$
 A: The function is integrable when $a<4$.
Due to the fact that $f(x,y)$ is always positive, Fubini's Theorem allows us to work with the iterated integrals. As the only possible problem occurs at $0$, we can also integrate on a smaller rectangle.
If $a\geq 4$, then
$$
\int_0^{1/2}\int_0^{1/2}\frac{xy}{(x+y)^a}\,dx\,dy
\geq\int_0^{1/2}\int_0^{1/2}\frac{xy}{(x+y)^4}\,dx\,dy=\int_0^{1/2}\frac{3y+1}{6y(y+1)^3}\,dy=\infty.
$$
And when $a<4$, I don't see a nice argument, but it can be done by brute force. We may assume that $a$ is not an integer, because on $[0,1/2]\times[0,1/2]$ we have $(x+y)^{-a}\leq (x+y)^{-a-\delta}$ for any $\delta>0$. We may also assume $a>0$, because the case $a\leq 0$ is trivial.
It turns out that the antiderivatives are not that hard to calculate, using parts and a bit of patience.
$$
\int_c^1\int_0^1 \frac{xy}{(x+y)^a}\,dx\,dy=b_0+\frac{b_1 c}{(c+1)^a}+\frac{b_2 c^{a+3}}{(c+1)^a}+b_3 \,c^{4-a}+\frac{b_4 c^{a+1}}{(c+1)^a}+\frac{b_5c^{a+2}}{(c+1)^a}
$$
for appropriate constants $b_0,\ldots,b_5$, where in particular
$$
b_3=\frac{1}{(a-4)(a-2)(a-1)},
$$
so nonzero. As we need to take the limit as $c\to0$, the limit exists as long as $a<4$ to guarantee that the $b_3$ term converges.
