<s>[INST] Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.
Are there better methods?
 [/INST] Observe that for $x\geq 2$ we have $\frac{x}{2}\geq 1$ and $\frac{x^2}{2}\geq 2\geq 1$ . Hence,
\begin{align}
(x-1)\cdot \sqrt{x^2-1}&\geq (x-\frac{x}{2})\cdot \sqrt{x^2-\frac{x^2}{2}}\\
&=\frac{x}{2}\cdot \sqrt{\frac{x^2}{2}}\\
&=\frac{x^2}{2\sqrt{2}}
\end{align}
Hence for every $x\geq 2$
$$\frac{1}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{2\sqrt{2}}{x^2}$$
This shows that
\begin{align}
\int_{2}^{\infty}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx&\leq \int_{2}^\infty \frac{2\sqrt{2}\ln x}{x^2}\,dx\\
&=2\sqrt{2}\int_{2}^\infty \biggl(\frac{-1}{x}\biggr)'\cdot \ln x\,dx\\
&=2\sqrt{2}\biggl(-\frac{1}{x}\ln x \biggr|_{2}^{\infty}\biggr)+2\sqrt{2}\int_{2}^\infty\frac{1}{x^2}\,dx\\
&=\sqrt{2}\ln 2+2\sqrt{2}\biggl(-\frac{1}{x}\biggr|_{2}^\infty\biggr)\\
&=\sqrt{2}\ln 2+\sqrt{2}
\end{align}
Now, to treat the case where $x>1$ and $x$ being close to $1$, observe that
$$(x-1)\leq (x-1)\cdot \sqrt{x^2-1}\leq x\cdot (x-1)$$
for all $x>1$. Hence, for $x>1$
$$\frac{\ln x}{x\cdot (x-1)}\leq \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{\ln x}{x-1}$$
Using DLH to evaluate the limits of the right and left hand side of the above inequality we obtain
$$\lim_{x\to 1^+}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}=1$$
This shows also that
$$\int_{1}^2 \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx<\infty$$
</s>