I'm studying the convergence of the following triple integral
$$I(\alpha) = \iiint \limits_{\Omega_{3}}\frac{dxdydz}{(x^{2}+y^{2}+z^{2})^\alpha}\tag{*}$$ Where $$\Omega_{3} = \{(x, y, z) \in \mathbb{R}^{3}: x^{2}+y^{2}+z^{2} \geq 1\}, \qquad \alpha \in \mathbb{R} $$
Now, to study the convergence of $I(\alpha)$ I am using the following theorem
Boundedness Criterion of Convergence:
The positive function $f: \Omega \rightarrow \mathbb{R}^{+}$ is improperly integrable if and only if there exits an increasing an exhausting sequence $(D_{n})_{n \in \mathbb{N}}$ of a measurable compact domain for which the sequence $\left( \int_{D_{n}} f d \mu \right)$ is bounded.
using this theorem I define my $D_{n}$ through $$D_{n} = \{(x,y,z) \in \Omega_{3}:1 \leq x^{2}+y^{2}+z^{2} \leq n^{2}\}, \quad n \in \mathbb{N^{*}}$$
Now, my question relates to the appropriateness of setting $D_{n}$ above and the method by which i evaluate the integral in $(*)$. For that evaluation; I use spherical polar coordinates $(x, y, z)=(r\sin \theta \cos \phi, r \sin \theta \sin \phi, r \cos \phi)$: Whence \begin{align} I(\alpha) &= \iiint \limits_{D_{n}}\frac{dxdydz}{(x^{2}+y^{2}+z^{2})^\alpha} \\ &= \int \limits_0^{2\pi} \int \limits_0^{\pi} \int \limits_{1}^{n}r^{2}\sin^{2} \phi r^{-2\alpha}dr d \theta d \phi \\ &=\int \limits_0^{2\pi} \int \limits_0^{\pi} \int \limits_{1}^{n}r^{2-2\alpha}\sin^{2} \phi dr d \theta d \phi \\ &=2\pi \int_1^nr^{2-2 \alpha}dr\int_0^{\pi}\left(\frac{1}{2}-\frac{1}{2}\cos 2 \phi \right)d \phi \\ &= 2\pi \int_1^nr^{2-2 \alpha}dr \left(\frac{\phi}{2}-\frac{1}{4}\sin 2 \phi\biggr \rvert_0^{\pi} \right) \\ &= \pi^{2} \int \limits_1^n r^{2-2 \alpha} dr \\ &= \frac{\pi^{2}}{3-2\alpha}\left(n^{3-2 \alpha}-1\right) \end{align} Now for the convergence part: clearly the integal is convergent for $\alpha > \frac{3}{2}$ and divergent for $\alpha \leq \frac{3}{2}$. The case $\alpha = \frac{3}{2}$ is clearly a sticking point, this must also be discussed.
After conversion to spherical coordinates (which you've done mostly correctly, though you should double-check the Jacobian), your integral boils down to (a constant multiple of) the improper integral $$ \int_{1}^{\infty} r^{2 - 2\alpha}\, dr = \lim_{n \to \infty} \int_{1}^{n} r^{2 - 2\alpha}\, dr. $$ If $\alpha = \frac{3}{2}$...