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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 ?

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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.

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