Why does this integral diverge when $\alpha \gt 2$? Problem text:
For which values of $\alpha$ does the integral below converge?
$$\int\int_{D}\frac{1}{(x+y)^{\alpha}}dxdy$$
Where $D: 0 \le y \le 1 - x, 0 \le x \le 1$.
Answer:
$$\int\int_{D}\frac{1}{(x+y)^{\alpha}}dxdy=\int _0^1\left(\:\int _0^{1-x}\:\frac{1}{\left(x+y\right)^{\alpha \:}}dy\right)dx=\frac{1}{-α+2} \implies \alpha \lt 2$$
It's clear to me that $a \ne 2$, but I'm not convinced that $\alpha$ can't be greater than 2.
For example, if $\alpha = 3$, then $\frac{1}{-\alpha + 2}=-1$.
If I try to calculate the integral with $\alpha = 3$, I quickly realize that the integral does in fact diverge, but I'm not sure why, since it seems perfectly fine for $\alpha$ to be any value other than $2$ according to $\frac{1}{-\alpha + 2}$.
 A: The reason $\alpha$ can't be greater or equal than $2$ is analogous to the reason $\beta$ can't be greater or equal than $1$ in this other one-dimensional integral:
$$\int_0^1 \frac{1}{r^\beta}dr = \frac{1^{1-\beta}}{1-\beta} - \lim_{r\to 0^+} \frac{r^{1-\beta}}{1-\beta}. $$
The limit only exists if $\beta < 1$ because $\lim_{r\to0^+}r^c = \infty$ when $c<0$.
We need to evaluate the limit of the integral through a limit because the function is not continuous at $r=0$. In such cases, we call the integral an improper integral.
In the original problem, we have that
$$
\begin{align*}
\iint_{D}\frac{1}{(x+y)^{\alpha}}dxdy&=\int _0^1\left(\:\int _0^{1-x}\:\frac{1}{\left(x+y\right)^{\alpha \:}}dy\right)dx\\
&= \int_0^1 \frac{1-x^{1-\alpha}}{1-\alpha} dx\\
&= \frac{1}{1-\alpha} - \left(\frac{1^{2-\alpha}}{(1-\alpha)(2-\alpha)} - \lim_{x\to 0^+}\frac{x^{2-\alpha}}{(1-\alpha)(2-\alpha)}\right).
\end{align*}
$$
Again, the limit exists only if $\alpha<2$.
A: Notice that for $\alpha\not=1,2$,
$$\int _0^{1-x}\frac{1}{\left(x+y\right)^{\alpha}}dy=\left[\frac{(x+y)^{1-\alpha}}{1-\alpha}\right]_{0^+}^{1-x}
=\frac{1-x^{1-\alpha}}{1-\alpha},$$
and therefore (notice that the integral is improper for $\alpha>0$)
$$\iint_{D}\frac{1}{(x+y)^{\alpha}}dxdy=\int_0^1 \frac{1-x^{1-\alpha}}{1-\alpha}\, dx=\frac{1}{1-\alpha}\left[x-\frac{x^{2-\alpha}}{2-\alpha}\right]_{0^+}^1\\
=\frac{1}{2-\alpha}+\frac{\lim_{x\to 0^+}x^{2-\alpha}}{(2-\alpha)(1-\alpha)}.$$
The limit on the right-hand side gives the final result: for
$\alpha>2$, $\lim_{x\to 0^+}x^{2-\alpha}=+\infty$ and the integral is divergent, otherwise, for
$\alpha<2$ and $\alpha\not=1$, the limit is zero and the integral is equal to $\frac{1}{2-\alpha}>0$.
The special cases $\alpha=1,2$ are easy to handle: the given integral is equal to $1$ for $\alpha=1$ and it is divergent for $\alpha=2$.
