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?
 A: Near $x=1$, since $\ln x=\ln(1+(x-1))\sim x-1$, and $\sqrt{x^2-1}\sim\sqrt{x-1}$, you can compare your integral with $\frac{1}{\sqrt{x-1}}$.
Near $x=\infty$ (let me express in this way), since $\ln x \ll x^\alpha$ for every $\alpha>0$, and $(x-1)\sqrt{x^2-1}\sim x^{3/2}$, you can compare your integral with $\frac{1}{x^{3/2-\alpha}}$ for any $\alpha>0$ such that $\frac{3}{2}-\alpha>1$ (take $\alpha=1/4$ for instance).
A: Hint:
Observe that $\;(x-1)\sqrt{x^2-1}=(x-1)^{3/2}(x+1)^{1/2}\;$ , and thus
$$\frac{\log x}{(x-1)^{3/2}\sqrt{x+1}}\le\frac{x^{1/4}}{(x-1)^{3/2}}\le\frac1{x^{9/8}}$$
A: 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
$$\int_{1}^2 \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx<\infty$$
A: First, note that
$$
\int_1^\infty \frac{\ln x\,dx}{(x-1)\sqrt{x^2-1}}=
\int_0^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}
=\int_0^1 \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}\,dt}+\int_1^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}
$$
Now, due to Mean Value Theorem, for $f(t)=\ln (1+t)$,
$$
\frac{\ln (1+t)}{t}=\frac{\ln(1+t)-\ln 1}{t-0}=(\ln (1+s))'=\frac{1}{1+s}<1, \quad s\in (0,t),
$$
and so
$$
0\le \int_0^1 \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}\le 
\int_0^1 \frac{dt}{\sqrt{t(t+2)}}\le \int_0^1 \frac{dt}{\sqrt{t}}=2.
$$
Meanwhile, for $t\ge 0$,
$$
\ln(1+t)=\int_0^t\frac{ds}{1+s}\le \int_0^t\frac{ds}{\sqrt{1+s}}=2\sqrt{1+t}-2<2\sqrt{1+t}
<2\sqrt{t+2},
$$
and hence
$$
0\le \int_1^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}\le 
\int_1^\infty \frac{2\sqrt{t+2}\,dt}{t\sqrt{t(t+2)}}=
\int_1^\infty \frac{2\,dt}{t\sqrt{t}}=\left.-t^{-1/2}\right|_1^\infty=1
$$
Altogether
$$
0\le \int_1^\infty \frac{\ln x\,dx}{(x-1)\sqrt{x^2-1}}\le 3
$$
A: Set $u:=\ln x$. The reason $\int_0^\infty\frac{ue^udu}{(e^u-1)\sqrt{e^{2u}-1}}$ converges is that its integrand is asymptotic to $\tfrac{1}{\sqrt{2}}u^{-1/2}$ for small $u>0$ and to $ue^{-u}$ for large $u$. Note $\int_0^\epsilon u^{-1/2}du=2\sqrt{\epsilon}$ for small $\epsilon>0$, and $\int_0^\infty ue^{-u}du=1$.
A: First of all $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\frac{\ln{x}}{\left(x-1\right)\sqrt{x+1}}\cdot\frac{1}{\sqrt{x-1}}\underset{x\to 1}{\sim}\frac{1}{\sqrt{2}\sqrt{x-1}} $, and we know that $ \int_{1}^{a}{\frac{\mathrm{d}x}{\sqrt{x-1}}} $ converges for any $ a>1 $.
We also have that $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{\sqrt{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\right)=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{1}{x\sqrt{x}}\right) $, and $ \int_{b}^{+\infty}{\frac{\mathrm{d}x}{x\sqrt{x}}} $ converges for any $ b>1 $.
Thus our integral $ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x} $ converges.
Let's use a substitution $ \left\lbrace\begin{matrix}x=\sec{y}\\ \sec{y}\,\mathrm{d}y=\frac{\mathrm{d}x}{\sqrt{x^{2}-1}}\ \ \ \ \ \ \ \end{matrix}\right. $, we get : \begin{aligned}\int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\int_{0}^{\frac{\pi}{2}}{\frac{\sec{y}\ln{\left(\sec{y}\right)}}{\sec{y}-1}\,\mathrm{d}y}\\ &=-\int_{0}^{\frac{\pi}{2}}{\frac{\ln{\left(\cos{y}\right)}}{1-\cos{y}}\,\mathrm{d}y}\\ &=\left[\left(\cot{\left(\frac{y}{2}\right)}-1\right)\ln{\left(\cos{y}\right)}\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{x}}{\sin{x}+\cos{x}}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{\left(\frac{\pi}{4}-x\right)}}{\cos{\left(\frac{\pi}{4}-x\right)}+\sin{\left(\frac{\pi}{4}-x\right)}}\,\mathrm{d}y}\\ &=2\int_{0}^{\frac{\pi}{4}}{\left(1+\tan{x}\right)\mathrm{d}x}\\ &=2\left[x-\ln{\left(\cos{x}\right)}\right]_{0}^{\frac{\pi}{4}}\\ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\frac{\pi}{2}+\ln{2}\end{aligned}
