How to prove that $\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$ converges I have been trying really hard to prove that this integral $$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$$
converges. However, I did not succeed. For now, I proved that
$$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx =  2  \int_1^\infty \frac{\ln(x)}{x^2-1}\, dx,$$
and that $f(x) = \frac{\ln(x)}{x^2-1}$ is continuous on $[1,+\infty)$.
If you can give me a hint, that would be nice.
 A: As you noticed, we have that
$$\int_0^\infty \frac{\ln x}{x^2-1} dx=\int_0^1\frac{\ln x}{x^2-1} dx+\int_1^\infty \frac{\ln x}{x^2-1} dx$$
and by $x=\frac1y$
$$\int_0^1\frac{\ln x}{x^2-1} dx=\int_{\infty}^1\frac{\ln\left(\frac1y\right)}{\frac1{y^2}-1} \left(-\frac1{y^2}\right)dy=\int_1^\infty \frac{\ln y}{y^2-1} dy$$
then
$$\int_1^\infty \frac{\ln x}{x^2-1} dx=\int_1^2 \frac{\ln x}{x^2-1} dx+\int_2^\infty \frac{\ln x}{x^2-1} dx$$
and the second integral converges bt limit comparison test with $\int \frac{dx}{x^{1.5}}$ while for the first integral by $t=x-1$ we have
$$\int_1^2 \frac{\ln x}{x^2-1} dx=\int_0^1 \frac{\ln (1+t)}{t(t+2)} dt$$
and as $t \to 0^+$
$$\frac{\ln (1+t)}{t(t+2)} \sim \frac{1}{t+2}$$
A: Let
$I=\int_0^\infty \frac{\ln(x)}{(x^2-1)} dx$.
We can rewrite this as
$-I=\int_0^\infty \frac{\ln(x)}{(1-x^2)} dx$.
Now the problem arises because a singularity point exists in this. You can try to resolve that in any one of the two ways. You can take the function as complex logarithm and integrate by finding the correct contour. Another way is to repeat your steps and get the equality
$\int_0^\infty \frac{\ln(x)}{(x^2-1)} dx=2\int_0^1 \frac{\ln(x)}{(x^2-1)} dx$.
Now in $(0, 1)$ we have taylor series
$\frac{1}{(1-x^2)}=\Sigma_{k=0}^\infty x^{2k}$
Now using dominated convergence theorem you can convert above integration into a summation of integrals and get the value for original integral.
A: Combine
\begin{align}
\int_0^1 \frac{\ln x}{1-x^2}dx
= \int_0^1 \frac{\ln x}{1-x}dx - \int_0^1 \overset{x^2\to x}{\frac{x\ln x}{1-x^2} }dx = \frac34\int_0^1\frac{\ln x}{1-x}dx
\end{align}
$$2\int_0^1  \frac{\ln x }{1-x^2}dx 
= \int_0^1\frac{\ln x}{1-x}dx+ \int_0^1\frac{\ln x}{1+x}dx
$$
to establish
$$ \int_0^1  \frac{\ln x }{x^2-1}dx=- \frac32 \int_0^1\frac{\ln x}{1+x}dx
$$
Then
$$ \int_0^\infty \frac{\ln x }{x^2-1}dx = 2\int_0^1  \frac{\ln x }{x^2-1}dx=- 3\int_0^1\frac{\ln x}{1+x}dx <-3\int_0^1{\ln x}dx=3
$$
hence, converging.
A: Near $1$,
$$\lim_{x \rightarrow 1 \\ x >1}\frac{\ln(x)}{x^2-1}  = \frac{1}{2}$$ so the function  is integrable near $1$, and near $+\infty$,
$$\frac{\ln(x)}{x^2-1} = o \left( \frac{1}{x^{3/2}}\right)$$ which is integrable near $+\infty$. Because the function is continuous over $(1,+\infty)$, then it is integrable over $(1,+\infty)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x} \leq
\int_{0}^{\infty}{2\pars{\root{x} - 1} \over x^{2} - 1}\,\dd x
\end{align}
See $\ds{\color{black}{\bf 4.1.37}\,\,}$ in A & S Table.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x} \leq
2\int_{0}^{\infty}{\dd x \over \pars{\root{x} + 1}
\pars{x + 1}}
\end{align}
The RHS is convergent since its integrand is
$\ds{\sim x^{-3/2}}$ as $\ds{x \to \infty}$.
A: It's trivial that $x\geq\ln(x)$ for all $x>0$. Use this to show that $$\int_{1}^{\infty}\frac{x}{x^2-1}dx\geq\int_{1}^{\infty}\frac{\ln(x)}{x^2-1}dx$$ and use the equation you proved to show that this integral converges by the comparison test.
