Integrability of $(x+y) ^{-3}$. I'm asked to determine for what positive values of $\alpha$ is $(x+y)^{-3}$ integrable in the region where $0<x<1$ and $0<y<x^\alpha$. I've found that the function is integrable when $\alpha \geq 3$, but I'm not sure about the case when $0<\alpha<3.$
 A: We have
$$\begin{align}
\int_0^1\int_0^{x^{\alpha}}\frac{1}{(x+y)^3}dy\,dx&=\frac12\int_0^1\left(\frac{1}{x^2}-\frac{1}{(x+x^{\alpha})^2}\right)dx\\\\
&=\frac12\int_0^1\frac{2x^{\alpha-1}+x^{2\alpha-2}}{(x+x^{\alpha})^2}dx\tag1\\\\
\end{align}$$
For $\alpha>1$, we can rewrite the right-hand side of $(1)$ as
$$\begin{align}
\frac12\int_0^1\frac{2x^{\alpha-1}+x^{2\alpha-2}}{(x+x^{\alpha})^2}dx=\frac12\int_0^1\frac{2x^{\alpha-3}+x^{2\alpha-4}}{(1+x^{\alpha-1})^2}dx \tag 2\\\\
\end{align}$$
In order for the integral to converge, we require both $\alpha-3>-1$ and $2\alpha -4>-1$.  Both of these conditions are met for $\alpha >2$.
If $\alpha \le 1$, we can rewrite the right-hand side of $(1)$ as
$$\begin{align}
\frac12\int_0^1\frac{2x^{\alpha-1}+x^{2\alpha-2}}{(x+x^{\alpha})^2}dx=\frac12\int_0^1\frac{2x^{-\alpha-1}+x^{-2}}{(1+x^{1-\alpha})^2}dx \tag 2\\\\
\end{align}$$
and we see that the integral diverges.
$$\bbox[5px,border:2px solid #C0A000]{\text{Thus, the integral converges for all}\,\,\alpha >2\,\,\text{and diverges otherwise.}}$$
