real analysis - How to test this improper integral for convergence?

I'm supposed to test for convergence the following integral $$\int_1^{\infty}\frac{\ln x}{x\sqrt{x^2-1}}dx$$ I have tried using the comparison test with two different integrals but I've failed. I also tried using the Dirichlet test, however it doesn't work for this integral. I have thought about using the limit comparison test however I don't have any idea with what would I compare the expression I have.

Any hints?

Answer

Testing for convergence isn't so bad, simply note that for $x>\sqrt2$:

$$0<\frac1{x\sqrt{x^2-1}}<\frac1{x\sqrt{x^2-\frac12x^2}}=\frac{\sqrt2}{x^2}$$

Thus,

$$0<\int_{\sqrt2}^\infty\frac{\ln(x)}{x\sqrt{x^2-1}}~\mathrm dx<\sqrt 2\int_{\sqrt2}^\infty\frac{\ln(x)}{x^2}~\mathrm dx$$

Integration by parts,

$$\int_{\sqrt2}^\infty\frac{\ln(x)}{x^2}~\mathrm dx=\frac{\ln(2)}{2\sqrt2}+\int_{\sqrt2}^\infty\frac1{x^2}~\mathrm dx=\frac{\ln(2)}{2\sqrt2}+\frac1{\sqrt2}$$

For $1\le x\le\sqrt2$:

$$0\le\frac{\ln(x)}{x\sqrt{x^2-1}}\le1$$

$$0<\int_1^{\sqrt2}\frac{\ln(x)}{x\sqrt{x^2-1}}~\mathrm dx<\sqrt2-1$$

Thus, the integral converges and is bounded by $\displaystyle0