Convergence of improper double integral. Please help me to determine $\alpha$ and $p$, such that the integral
$$
I = \iint_G \frac{1}{(x^{\alpha}+y^3)^p} \ dx dy
$$
converges, where $G = {x>0, y >0, x+y <1}$ and $\alpha >0, p>0$. 
I am comfortable with proper double integrals. I am also comfortable with improper double integrals when $f$ is continuous in $G$ everywhere except the one point. However, I do not understand how to handle this case - it looks like $f$ is not continuous at 2 segments. I am also not sure why and when I can change the order of integration in this case. My textbook shows example when double integral diverges, while both iterated integral converges. Unfortunately I do no know where to start. I guess that I should compute the integral itself in terms of $\alpha$ and $p$ first, but again, I am not sure.
Thanks a lot for your help!
 A: We can write the integral of interest $I$ as
$$\begin{align}
I&=\int_0^1 \int_0^{1-y}\frac{1}{(x^\alpha+y^3)^p}\,dx\,dy\\\\
\end{align}$$
Next we enforce the substitution $x=u^{2/\alpha}\cos^{2/\alpha}(v)$ and $y=u^{2/3}\sin^{2/3}(v)$.  The determinant of the Jacobian, $J$, is given by 
$$
\begin{align}
|J|&=\begin{vmatrix}\frac{\partial x}{\partial u}&&\frac{\partial x}{\partial v}\\\\\frac{\partial y}{\partial u}&&\frac{\partial y}{\partial v}\end{vmatrix}\\\\
&=\frac{4}{3\alpha}u^{2/\alpha+2/3-1}\cos^{2/\alpha-1}(v)\sin^{-1/3}(v)
\end{align}$$

Then, $I$ can be written
$$I=\frac{4}{3\alpha}\int_{D_{u,v}}u^{2/\alpha+2/3-1-2p}\cos^{2/\alpha-1}(v)\sin^{-1/3}(v)\,du\,dv$$
where the domain $D_{u,v}$ contains $(u,v)=(0,0)$.  The integral over $v$ is well-behaved for all $\alpha$.  
The integral over $u$ converges if and only if $\frac2\alpha+\frac23-1-2p>-1$ or $p<\frac1\alpha+\frac13$.

$I$ converges if and only if $p<\frac1\alpha+\frac13$.

A: Here's a different approach: For all $a,p >0,$ the integrand is bounded on the upper right triangle of the square $[0,1]^2,$ so it's enough to consider the integral over this square. This makes life a little easier.
Now let $x =y^{3/a}t.$ Then we have
$$\tag 1 \int_0^1\int_0^1 \frac{1}{(x^a+y^3)^p}\,dx\,dy = \int_0^1y^{3/a-3p}\int_0^{y^{-3/a}} \frac{1}{(t^a+1)^p}\,dt\,dy.$$
The inner integral on the right of $(1),$ as a function of $y,$ is bounded below by $\int_0^{1} 1/(t^a+1)^p\,dt.$ Thus a necessary condition for convergence of $(1)$ is $3/a-3p > -1,$ or $p<1/a +1/3.$ To see this is also a sufficient condition requires a little work in estimating this inner integral (as a function of $y$). I'll leave that to the reader for now.
