Lennie Davis

2022-01-03

How can I prove that?
${\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx=\frac{{\pi }^{2}}{8}$

Janet Young

${\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx=\frac{1}{2}\left({\int }_{0}^{1}\frac{\mathrm{ln}\left(u\right)}{{u}^{2}-1}du+{\int }_{0}^{1}\frac{\mathrm{ln}\left(v\right)}{{v}^{2}-1}dv\right)$
$=\frac{1}{2}\left({\int }_{0}^{1}\frac{\mathrm{ln}\left(u\right)}{{u}^{2}-1}du+{\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(v\right)}{{v}^{2}-1}dv\right)$
$=\frac{1}{2}\left({\int }_{0}^{1}\frac{\mathrm{ln}\left(u\right)}{{u}^{2}-1}du+{\int }_{1}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(u\right)}{{u}^{2}-1}du-{\int }_{0}^{1}\frac{\mathrm{ln}\left(u\right)}{{u}^{2}-1}du\right)$
$=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}\frac{1}{1-{u}^{2}}{\left[\mathrm{ln}\left(\frac{1+v}{1+{u}^{v}}\right)\right]}_{v=0}^{v=\mathrm{\infty }}du$
$=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}\frac{1}{1-{u}^{2}}\left({\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{1+v}-\frac{{u}^{2}}{1+{u}^{2}v}\right)dv\right)du$
$=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{1}{\left(1+v\right)\left(1+{u}^{2}v\right)}dvdu=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{1+v}{\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{u}^{2}v}du\right)dv$

Fasaniu

We have
${\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx={\int }_{0}^{1}\frac{\mathrm{ln}\left(1-x\right)}{{\left(1-x\right)}^{2}-1}dx={\int }_{0}^{1}\frac{\mathrm{ln}\left(1-x\right)}{x\left(x-2\right)}dx$
We will generalize by introducing parameter $\alpha$ such that
$I\left(\alpha \right)={\int }_{0}^{1}\frac{\mathrm{ln}\left(1-\alpha x\right\}\left\{x\left(x-2\right)\right\}dx}{}$
And we have $I\left(0\right)=0$ Then
${I}^{\prime }\left(\alpha \right)=-{\int }_{0}^{1}\frac{1}{\left(1-\alpha x\right)\left(x-2\right)}dx=\frac{1}{2\alpha -1}{\left[\mathrm{ln}\left(\frac{x-2}{1-\alpha x}\right)\right]}_{0}^{1}=\frac{\mathrm{ln}\left(2-2\alpha \right)}{1-2\alpha }$
And we have
${I}^{\prime }\left(\alpha \right)=\frac{\mathrm{ln}\left(2-2\alpha \right)}{1-2\alpha }$
$I\left(\alpha \right)=\frac{1}{2}L{i}_{2}\left(2\alpha -1\right)+c$
$I\left(0\right)=\frac{1}{2}L{i}_{2}\left(-1\right)+c=0⇒c=-\frac{1}{2}L{i}_{2}\left(-1\right)=\frac{{\pi }^{2}}{24}$
$I\left(\alpha \right)=\frac{12}{L}{i}_{2}\left(2\alpha -1\right)+\frac{{\pi }^{2}}{24}$
$I\left(1\right)=\frac{1}{2}L{i}_{2}\left(2\alpha -1\right)+\frac{{\pi }^{2}}{24}$
$=\frac{{\pi }^{2}}{12}+\frac{{\pi }^{2}}{24}$
$=\frac{{\pi }^{2}}{8}$
$I\left(1\right)=\frac{{\pi }^{2}}{8}$

karton

Write the integrand as a sum of fractions and use the polylogarithm function $L{i}_{2}$:
$f\left(x\right):=\int \frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx=\int \frac{1}{2}\left(\frac{\mathrm{ln}\left(x\right)}{x-1}-\frac{\mathrm{ln}\left(x\right)}{x+1}\right)dx$
$=\frac{1}{2}\int \frac{\mathrm{ln}\left(x\right)dx}{x-1}-\frac{1}{2}\int \frac{\mathrm{ln}\left(x\right)dx}{x+1}=-\frac{1}{2}L{i}_{2}\left(1-x\right)-\frac{1}{2}\left(L{i}_{2}\left(-x\right)+\mathrm{ln}\left(x\right)\mathrm{ln}\left(x+1\right)\right)$
Since $L{i}_{2}\left(0\right)=0$ and $\mathrm{ln}\left(x\right)\mathrm{ln}\left(x+1\right)$ vanishes at x=0 and x=1, we have
$f\left(0\right)=-\frac{1}{2}L{i}_{2}\left(1\right)=-\frac{{\pi }^{2}}{12}$
and
$f\left(1\right)=-\frac{1}{2}L{i}_{2}\left(-1\right)=\frac{{\pi }^{2}}{24}$
and the value of the integral is
${\int }_{0}^{1}\frac{\mathrm{ln}\left(x\right)}{{x}^{2}-1}dx=\frac{{\pi }^{2}}{24}+\frac{{\pi }^{2}}{12}=\frac{{\pi }^{2}}{8}$

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