How can I prove that?


I know that $$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$ but I want another method.


$$\begin{align}\int_0^1\frac{\ln(x)}{x^2-1}dx&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)\\ &=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)\\ &=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)\\ &=\frac{1}{2}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{1}{2}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du =\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\ln\left(\frac{1}{u^2}\right)du\\ &=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du\\ &=\frac{1}{4}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du\\ &=\frac{1}{4}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\right)dv\\ &=\frac{1}{4}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\tan^{-1}(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{4}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv\\ &=\frac{\pi}{8}\int_0^{\infty}\frac{2w}{w(1+w^2)}dw=\frac{\pi}{8}\times2\left[\tan^{-1}(w)\right]_0^{\infty}=\frac{\pi}{8}\times2\times\frac{\pi}{2}=\frac{\pi^2}{8}.\end{align}$$


Consider the integral: $$I(m)=\int _{0}^{1}\!{\frac { \ln\left( x \right) ^{m-1}}{ {x}^{2}-1}}{dx} \quad:\quad \mathfrak{R}(m)>1 $$ and the substitution $x=e^{-u}$: $$\begin{aligned} \int _{0}^{1}\!{\frac { \ln\left( x \right) ^{m-1}}{ {x}^{2}-1}}{dx}=& \left( -1 \right) ^{m-1}\int _{0}^{\infty }\!{\frac { {u}^{m-1}{{\rm e}^{-u}}}{-1+{{\rm e}^{-2\,u}}}}{du}\\ =&\left( -1 \right) ^{m-1} \int _{0}^{\infty }\!-{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}+{\frac {{u }^{m-1}}{-1+{{\rm e}^{2\,u}}}}{du}\\ =&\left( -1 \right) ^{m-1}\left( 1- \dfrac{1}{2^m} \right) \int _{0}^{\infty }\!{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}{du}\\ =&\left( -1 \right) ^{m}\left( 1- \dfrac{1}{2^m} \right) \Gamma \left( m \right) \zeta \left( m \right) \end{aligned}$$ where we have used Riemann's integral representation of the zeta function and we also made the substitution $u\rightarrow\frac{u}{2}$ in the second term of the second line to pass to line three (having noted that convergence of both terms individually is assured by comparison with Riemanns integral). It follows from $\Gamma(2)=1, \zeta(2)=\frac{\pi^2}{6}$ that: $$I(2)=\frac{\pi^2}{8}$$


We have$$\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{(1-x)^2-1}\,\mathrm dx=\int_0^1\frac{\ln(1-x)}{x(x-2)}\,\mathrm dx$$ We will generalize by introducing parameter $\alpha$ such that $$I(\alpha )=\int_0^1\frac{\ln(1-\alpha x )}{x(x-2)}\,\mathrm dx$$ And we have $I(0)=0$ Then $$I'(\alpha )=-\int_0^1\frac{1}{(1-\alpha x)(x-2)}\,\mathrm dx=\frac{1}{2\alpha -1}\left[\ln\left(\frac{x-2}{1-\alpha x}\right)\right]_0^1=\frac{\ln(2-2\alpha )}{1-2\alpha }$$ And we have $$I'(\alpha )=\frac{\ln(2-2\alpha )}{1-2\alpha }$$

$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+c$$ $$I(0)=\frac{1}{2}\text{Li}_2(-1)+c=0\implies c=-\frac{1}{2}\text{Li}_2(-1)=\frac{\pi^2}{24}$$

$$I(\alpha )=\frac{1}{2}\text{Li}_2(2\alpha -1)+\frac{\pi^2}{24}$$ $$\begin{align}I(1)&=\frac{1}{2}\text{Li}_2(1)+\frac{\pi^2}{24}\\ &=\frac{\pi^2}{12}+\frac{\pi^2}{24}\\ &=\frac{\pi^2}{8}\\ \end{align}$$

$$I(1)=\int_0^1\frac{\ln(x)}{x^2-1}\,\mathrm dx=\frac{\pi^2}{8}$$

  • 1
    $\begingroup$ Lovely way! +1 ${}$ $\endgroup$ – Venus Nov 25 '14 at 11:24

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

$$ {\cal I} \equiv \int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\,\dd x = \int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x \quad\imp\quad {\cal I} \equiv {1 \over 2}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x $$

Let's consider the integral $$ {\cal W} \equiv \int_{C}{\ln^{2}\pars{z} \over z^{2} - 1}\,\dd z = 0 $$ enter image description here Then $\pars{~\mbox{with}\ z_{\pm} = x \pm \ic 0^{+}\ \mbox{and}\ z_{\pm}^{2} = x^{2} \pm \ic\sgn\pars{x}0^{+}~}$ \begin{align} &\int_{-\infty}^{0}{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over z_{+}^{2} - 1}\,\dd x + \int_{0}^{-\infty}{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over z_{-}^{2} - 1}\,\dd x \\[3mm]&= \sum_{\sigma = \pm 1}\sigma\int_{-\infty}^{0} {\ln^{2}\pars{-x} + 2\ic\pi\sigma\ln\pars{-x} - \pi^{2} \over x^{2} - 1 - \sigma\,\ic 0^{+}}\,\dd x \\[3mm]&= \sum_{\sigma = \pm 1}\sigma\,\braces{{\cal P}\int_{0}^{\infty} {\ln^{2}\pars{x} + 2\ic\pi\sigma\ln\pars{x} - \pi^{2} \over x^{2} - 1 }\,\dd x + \int_{0}^{\infty} {\bracks{\ln^{2}\pars{x} + 2\ic\pi\sigma\ln\pars{x} - \pi^{2}} \bracks{\ic\pi\sigma\delta\pars{x^{2} - 1}} }\,\dd x} \\[3mm]&= 4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x + \pars{-\pi^{2}}2\pars{\ic\pi \over 2} = 0 \quad\imp\quad \int_{0}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x = {\pi^{2} \over 4} \end{align} $${\large% {\cal I} = \int_{0}^{1}{\ln\pars{x} \over x^{2} - 1}\,\dd x = {\pi^{2} \over 8}} $$

  • $\begingroup$ (+1) for the similar approach to mine, although there is a subtle difference. Answer's the same of course. $\endgroup$ – Ron Gordon Oct 24 '13 at 13:41
  • $\begingroup$ @RonGordon I like those expressions $\large\left(x \pm {\rm i}0^{+}\right)^{-1} = {\cal P}\left(1/x\right) \mp \ic\delta\left(x\right)$ which takes care of the $\large\left(x^{2} - 1\right)^{-1}$ pole on the logarithm branch cut. They are analogous to your $\large\epsilon$ limit procedure. The use of the $\large\sigma$ sum let's see very easily which term vanishes out. Thanks. $\endgroup$ – Felix Marin Oct 24 '13 at 17:56

Here is another way to solve. Using the substitute $u=x^2$ gives \begin{eqnarray} \int_0^1\frac{\ln x}{x^2-1}dx&=&-\frac{1}{4}\int_0^1\frac{\ln u}{(1-u)\sqrt u}dx\\ &=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\partial}{\partial \mu}\int_0^1(1-u)^{\varepsilon-1}u^{\mu-1}du\\ &=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\partial}{\partial \mu}B(\varepsilon,\mu)\\ &=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}B(\varepsilon,\mu)(\psi(\mu)-\psi(\mu+\varepsilon))\\ &=&-\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\Gamma(\varepsilon+\mu)}{\Gamma(\varepsilon)\Gamma(\mu)}(\psi(\mu)-\psi(\mu+\varepsilon))\\ &=&\frac14\lim_{\varepsilon\to0,\mu\to\frac12}\frac{\psi(\mu+\varepsilon)-\psi(\mu)}{\varepsilon}\frac{\varepsilon}{\Gamma(\varepsilon)}\frac{\Gamma(\varepsilon+\mu)}{\Gamma(\mu)}\\ &=&\frac{1}4\psi'(\frac{1}{2})\\ &=&\frac{\pi^2}{8}. \end{eqnarray} Here we used $$ \Gamma(\varepsilon)\approx\varepsilon \text{ for small }\varepsilon>0. $$


Write the integrand as a sum of fractions and use the polylogarithm function $\mathrm{Li}_2:$ $$f(x):=\int\frac{\ln(x)}{x^2-1}dx=\int\frac{1}{2}\left(\frac{\ln(x)}{x-1}- \frac{\ln(x)}{x+1}\right) dx \\ =\frac{1}{2} \int \frac{\ln(x) dx}{x-1}- \frac{1}{2} \int \frac{\ln(x)dx }{x+1} =-\frac{1}{2}\mathrm{Li}_2(1-x) -\frac{1}{2} \left(\mathrm{Li}_2(-x) + \ln(x)\ln(x+1)\right)$$ Since $\mathrm{Li}_2(0)=0$ and $\ln(x)\ln(x+1)$ vanishes at $x=0$ and $x=1$, we have $$f(0) = -\frac{1}{2}\mathrm{Li}_2(1)= -\frac{\pi^2}{12}$$ and $$f(1) = -\frac{1}{2}\mathrm{Li}_2(-1)= \frac{\pi^2}{24}$$ and the value of the integral is $$ \int_0^1\frac{\ln(x)}{x^2-1}dx=\frac{\pi^2}{24}+\frac{\pi^2}{12} = \frac{\pi^2}{8} $$

  • $\begingroup$ Why $\lim\limits_{x\rightarrow0}f(x)=-\frac{\pi^2}{12}$ and $f(1)=\frac{\pi^2}{24}$? Can you more explain about the $Li_2$? $\endgroup$ – user97601 Oct 24 '13 at 8:47
  • $\begingroup$ Since $f(0)$ is undefined you have to take the limit, the series at $x=0$ starts with $$ f(x) = -\frac{\pi^2}{12} + (1-\ln(x))x + \left(\frac{1}{9}-\frac{1}{2}\ln(x)\right)x^3 + O(x^5).$$ $\mathrm{Li_2}(x)$ is the dilogarithm $$\mathrm{Li_2}(x) = \sum\limits_{k=1}^{\infty}\frac{x^k}{k^2}$$ see e.g. mathworld.wolfram.com/Dilogarithm.html, where you can find among others e.g. $\mathrm{Li_2}(-1)=-\pi^2/12$ which gives $f(1)$ because $\mathrm{Li_2}(0)=0$ $\endgroup$ – gammatester Oct 24 '13 at 9:13

This comes from my answer here. Let

$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$

Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value. We evaluate this integral by once again appealing to the residue theorem, but this time, we consider

$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1}$$

where $C'$ is a keyhole contour with respect to the positive real axis. By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C'$ increases without bound, we get

$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$

This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C'$. As the only pole is at $z=-1$, we see that

$$\begin{align}\oint_{C'} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi \frac{\pi^2}{2}\end{align}$$

Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$. The Cauchy principal value is zero:

$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon} \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$

The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:

$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$

so that

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$


$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{4}$$

But the sought-after integral is

$$\int_0^{1} dx \frac{\log{x}}{x^2-1} = \frac12 I = \frac{\pi^2}{8}$$


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