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?
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 $\displaystyle0<I<\frac{\ln(2)}2+\sqrt2$.
After substitution $x=\frac{1}{\sin{t}}$ use the following Euler: $\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d t$.
The answer is $\frac{\pi}{2}\ln2$.
The Euler's work: $$\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}=2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{2t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}+2\int\limits_0^{\frac{\pi}{4}}\ln{\cos{t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}+2\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}=$$ $$=\frac{\pi}{2}\ln2+2\int\limits_0^{\frac{\pi}{2}}\ln{\sin{t}}\,\mathrm d{t}.$$ Thus, $$\int\limits_0^{\frac{\pi}{4}}\ln{\sin{t}}\,\mathrm d{t}=-\frac{\pi}{2}\ln2,$$ which says that your integral converges.
For $x$ close to $1:$ Since $\ln x \sim x-1$ as $x\to 1^+,$ the integrand has limit $0$ from the right at $1.$ Hence there is no problem at $1.$
For large $x,$ the integrand looks like
$$\tag 1 \frac{\ln x }{x^2}.$$
Now $\ln x \to \infty$ as $x\to \infty,$ but it does so at a laughably small rate. So $\ln x < x^{1/2}$ for large $x.$ Thus $(1)$ is bounded above by $1/x^{3/2}$ for large $x.$ Since $\int_1^\infty (1/x^{3/2})\, dx <\infty,$ we have convergence of the original integral by the comparsion test.
I separated the integral: $$(1)\int_1^{3}\frac{\ln x}{x\sqrt{x^2-1}}dx, (2)\int_3^{\infty}\frac{\ln x}{x\sqrt{x^2-1}}dx$$
$$(1)\backsim \int_1^{3}\dfrac{1}{\sqrt{x^2-1}}dx$$ which converges because it is the derivative of $\sin^{-1}(x)$
$$(2)\backsim \int_3^{\infty}\frac{1}{x^2\log^{-1}(x)}dx $$ which converges because $2$ (the exponent)$>1$
So the entire integral converges.
