convergence and divergence of $\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$ I want to study the divergence or convergence of this integral : 
$$\iiint\limits_{x^2+y^2+z^2\geq 1}(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\,dx\,dy\,dz$$
I think that if $\alpha\geq-3$, then in the domain: 
$$(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\geq(x^2+y^2+z^2)^\alpha$$
so the integral on the domain over the R.H.S. function diverges. 
But know I am left with the situation when $\alpha\lt-3$. Can I say that $\alpha=-3-\epsilon$, so there exist some ball $B(0,r_\epsilon)$, such that in the domain $\mathbb{R}^3\setminus B(0,r_\epsilon)$:
$$\ln(x^2+y^2+z^2)\leq(x^2+y^2+z^2)^\frac{\epsilon}{2}$$
$$\iff(x^2+y^2+z^2)^\alpha \ln(x^2+y^2+z^2)\leq(x^2+y^2+z^2)^{-3-\frac{\epsilon}{2}}$$
so the integral on the domain over the R.H.S. function converges. Thus concluding that:
If $\alpha\geq-3$, our integral diverges, otherwise converges. Is it true?

The surface area of $x^2+y^2+z^2=\rho^2$ is given by $4\pi\rho^2$, hence the given integral equals
$$ \int_{1}^{+\infty} 4\pi\rho^2 \cdot \rho^{2\alpha}\cdot 2\log\rho\,d\rho $$
i.e. $\frac{8\pi}{(2\alpha+3)^2}$ as soon as $\alpha<-\frac{3}{2}$.
