Multivariable function Integrable for what values? For what values of $\alpha,\beta \in \mathbb{R}$ will the function
$$f:(0,1]\times(0,1] \to \mathbb{R}: (x,y) \mapsto x^\alpha y^\alpha (x+y)^\beta$$
be integrable?
Normally, I don't have problems solving these integrals using Fubini's theorem and by a change of variables. However, I don't seem to find the right change of variables? Any hints? Thank you!
 A: Let $C = (0, 1]^2$. I assume by $f$ being integrable you mean that $\int_C f < \infty$ (here and throughout I use $\int_A f$ to denote $\iint_A f \,dxdy$). 
First we'll show that $\int_C f < \infty$ implies $\alpha > -1$. Let $D = (0, 1] \times [1/2, 1]$, so since $D \subset C$ we have $\int_C f \geq \int_D f 
 $. But for $(x, y) \in D$ we have $1/2 \leq y \leq 1$ and $1/2 \leq x+y \leq 2$, hence $y^{\alpha} \geq 2^{-|\alpha|}$ and $(x+y)^{\beta} \geq 2^{-|\beta|}$, giving
$$\int_D f = \int_0^1 \int_{1/2}^1 x^\alpha y^\alpha (x+y)^\beta \,dy \,dx \geq 2^{-|\alpha|-|\beta|-1}\int_0^1 x^{\alpha} \,dx$$
so if $\int_C f < \infty$ we must have $\int_0^1 x^\alpha \,dx < \infty$, hence $\alpha > -1$.
Now assume $\alpha > -1$. Define $E = \{(x, y) \mid x, y > 0, x+y \leq 1\}$. Then clearly $E \subset C \subset 2E$, where $2E = \{(x, y) \mid x, y > 0, x+y \leq 2\}$, hence $\int_E f \leq \int_C f \leq \int_{2E} f$. But it's easy to check that $\int_{2E} f = 2^{2\alpha + \beta+2} \int_E f$ since $f$ is homogeneous of degree $2\alpha + \beta$, which gives $\int_E f \leq \int_C f \leq 2^{2\alpha + \beta + 2} \int_E f$, so it follows that $\int_C f < \infty$ if and only if $\int_E f < \infty$.
But we can evaluate $\int_E f$ using the change of variables $s = x + y$: 
$$\int_E f = \int_0^1 \int_0^s x^\alpha(s - x)^{\alpha} s^\beta \,dx \,ds = B(\alpha + 1, \alpha + 1)\int_0^1 s^{2\alpha + \beta + 1} \,ds$$
where $B(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} = \int_0^1 x^{a-1}(1-x)^{b-1} \,dx$ is the Beta function, defined for $a, b > 0$. Since $\alpha + 1 > 0$, this means $\int_E f < \infty$ iff $\int_0^1 s^{2\alpha + \beta + 1} \,ds < \infty$, which is true iff $2\alpha + \beta + 1 > -1$. 
We conclude that $\int_C f < \infty$ iff $\alpha > -1$ and $\beta > -2(\alpha + 1)$.
A: In the meantime, I did find some change of variables which might allow a simpler way to verify the integrability. If we perform a change of variables $y\mapsto xy$, this gives the integral
$$\int_0^1\int_0^{1/x}x^{2\alpha +\beta+1} y^\alpha(1+y)^\beta dy\, dx$$
where the integrability can be easily verified.
