what a and b make the integral convergence? Consider the following integral
$$\iint_Ax^\alpha y^\beta \space dA$$
where $A=\{(x,y)\space|\space0\leq y\leq1-x,x\geq0\}$. Find all possible values of $\alpha$ and $\beta$, for which this integral converges. Justify your answer.
I have got that when $b > -1$, the first part of the integral is convergent, then it becomes 
$$\int_0^\infty\frac{x^\alpha(1-x)^{b+1}}{b+1}dx$$
but I have no idea how to find a. Does anyone could help me, thanks!

The first inequality defining the region $A$ states $0\leq y \leq 1-x$. Well, that's actually two inequalities stringed together, and by transitivity they imply $0\leq 1-x$, or equivalently $x\leq 1$. Combining this implication with the second inequality defining $A$, $0\leq x$, we see that the $x$ coordinates of points in $A$ actually have finite bounds, despite first appearances:
$$0\leq x \leq 1$$.
So the integral you should be evaluating is actually,
$$\int_0^1\frac{x^\alpha(1-x)^{b+1}}{b+1}dx,$$
which is essentially the Beta function.
