Let $V=\{ (x, y ,z) \in \mathbb R^3 | z>0, \sqrt{x^2+y^2} \leq -z \log z\}$ For what value $a \in \mathbb R$ does this integral converge? $$ \int \int \int_V \frac{(x^2+y^2)^a}{z^2} dxdydz$$
My attempts: I replaced $x$ and $y$ as $x=r \cos θ$ and $y=r \sin θ$ and got $$V=\{(r, θ, z): r\leq -z\log z, z>0, 0\leq θ\leq 2π, 0\leq r\}$$ and the integral: $$2π \int \int_V \frac{r^{2a+1}}{z^2} drdz=2π \int_0^{\infty} z^{2a} (\log z)^{2a+1}dz$$
From here I completely have no idea what to do with this integral, I even don't know how to calculate that on $V$.
If $0\leq\sqrt{x^2+y^2}\leq -z\log z$, $z$ is at most $1$. It follows that the wanted integral is:
$$ \iiint_V \frac{(x^2+y^2)^\alpha}{z^2}\,dx\,dy\,dz = 2\pi \int_{0}^{1}\int_{0}^{-z\log z}\frac{\rho^{2\alpha+1}}{z^2}\,d\rho\,dz $$ or $$ \frac{2\pi}{2\alpha+2}\int_{0}^{1}z^{2\alpha}(\log z)^{2\alpha+2}\,dz \stackrel{z\mapsto e^{-t}}{=}\frac{\pi}{\alpha+1}\int_{0}^{+\infty}t^{2\alpha+2}e^{-(2\alpha+1)t}\,dt$$ that equals: $$ \frac{\pi (2\alpha+2)!}{(\alpha+1)(2\alpha+1)^{2\alpha+3}} $$ as soon as $2\alpha+1>0$, i.e. as soon as $\alpha>-\frac{1}{2}$.