For what $\alpha$ does $\iint_D\frac{1}{(x+y)^{\alpha}}\ dxdy$ converge? For what values of $\alpha$ does $$\iint_D\frac{1}{(x+y)^{\alpha}}\
 dxdy$$ converge? $D=\{0\leq y \leq 1-x, \quad 0 \leq \ x \leq 1\}.$


The double integral can be written as

$$\int_{0}^{1}\left(\int_{0}^{1-x} \frac{1}{(x+y)^{\alpha}} \ dy\right)dx.$$

How does one find a primitive to the inner integral?
In general, the answer might depend on whether this is considered to be the integral over a region in $\mathbb{R}^2$ with respect to the product measure,

$$\tag{1}I = \int_D (x+y)^{-\alpha}$$

or as iterated integrals

$$\tag{2} J= \int_0^1 \left(\int_0^{1-x} (x+y)^{-\alpha} \, dy \right) \, dx $$ 

In this case it doesn't matter, but I will answer with reference to (1).

The diagonal boundary of the region of integration falls in  between circular arcs with radii $1/2$ and $1$. 

Since the integrand is nonnegative,

$$\int_0^{\pi/2}\int_0^{1/2} \frac{r \, dr \, d\theta}{r^\alpha(\cos \theta + \sin \theta)^\alpha} \leqslant I \leqslant \int_0^{\pi/2}\int_0^{1} \frac{r \, dr \, d\theta}{r^\alpha(\cos \theta + \sin \theta)^\alpha}$$

Note that $\cos \theta + \sin \theta &gt; 0$ for $\theta \in [0,\pi/2]$ which implies that 

$$\int_0^{\pi/2} \frac{d \theta }{(\cos \theta + \sin \theta)^{\alpha}},$$

converges for all $\alpha$.

Thus, we have convergence of $I$ if $\alpha &lt; 2$ and divergence if $\alpha \geqslant 2$, determined  by the convergence/divergence of

$$\int_0^c \frac{dr}{r^{\alpha-1}}$$.
hint for the inner $\int $

For $\alpha\ne 1$

$$\int_0^{1-x}\frac {dy}{(x+y)^\alpha}=$$
$$\Bigl [\frac {(x+y)^{1-\alpha}}{1-\alpha}\Bigr]_0^{1-x} =$$

$$\frac {1}{1-\alpha}(1-x^{1-\alpha}) =$$

The double integral is
$$\frac {1}{2-\alpha} $$

If $\alpha=1$, we find $1$.

The double integral exists if $\alpha\ne 2$.