Test convergence, find $\alpha$ which makes integral converge I'm testing the convergence of this improper integral
$$\int_2^{\infty} x(\ln x)^{\alpha} dx$$
I used the limit comparison test with $\frac{1}{x}$ which is divergent, I found that this integral diverges for all values of $\alpha$. 
Am I correct ?

You are correct.
Observe that the integrand is positive and we have, for all real values of $\alpha$,
$$
\lim_{x \to +\infty}\left(x(\ln x)^{\alpha}\right)=+\infty 
$$ thus the initial integral is divergent.
