# 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 ?

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.