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?
On
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 $\alpha\ne 2$.
On
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 $\cos \theta + \sin \theta > 0$ 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 $\alpha < 2$ and divergence if $\alpha \geqslant 2$, determined by the convergence/divergence of
$$\int_0^c \frac{dr}{r^{\alpha-1}}$$.
On
How does one find a primitive to the inner integral?
Separete $\alpha$ in two cases and substitute $u = x + y$.
If $\alpha = 1$ then $$ \int_{0}^{1-x} \frac{1}{x+y}\ dy = \int_{x}^{1} \frac{1}{u}\ du = \log(1) - \log(x) = -\log(x)$$
Else $$ \int \frac{1}{(x+y)^\alpha}\ dy = \int_{x}^{1} u^{-\alpha}\ du = \left.\frac{u^{1-\alpha}}{1-\alpha} \right|_{u=x}^{u=1} = \frac{1 - x^{1-\alpha}}{1-\alpha} $$
From here I believe it already can be seen that the second integral will converge only when $\alpha < 2$.
$$ \begin{align} \int_0^1\int_0^{1-x}\frac1{(x+y)^\alpha}\,\mathrm{d}y\,\mathrm{d}x &=\int_0^1\int_x^1\frac1{t^\alpha}\,\mathrm{d}t\,\mathrm{d}x\tag1\\ &=\int_0^1\int_0^t\frac1{t^\alpha}\,\mathrm{d}x\,\mathrm{d}t\tag2\\ &=\int_0^1t^{1-\alpha}\,\mathrm{d}t\tag3\\ &=\frac1{2-\alpha}\tag4 \end{align} $$ Explanation:
$(1)$: substitute $y=t-x$
$(2)$: change order of integration
$(3)$: integrate in $x$
$(4)$: integrate in $t$, integral converges iff $1-a\gt-1$
Thus, the integral converges iff $\alpha\lt2$.
Primitive of the Inner Integral $$ \int\frac1{(x+y)^\alpha}\,\mathrm{d}y=\left\{\begin{array}{} \frac1{1-\alpha}(x+y)^{1-\alpha}+C&\text{if $\alpha\ne-1$}\\ \log|x+y|+C&\text{if $\alpha=-1$} \end{array}\right. $$