How to prove the following

$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$

I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.

Any other ideas ?


After the change of variables $x=\tanh u$ (suggested by the square root) this integral reduces to $$\mathcal{I}=\int_0^{\infty}\frac{2u\,du}{\sinh u}.$$ Expanding $\displaystyle\frac{1}{\sinh u}=2\sum_{k=0}^{\infty}e^{-(2k+1)u}$ and exchanging summation and integration, we find that $$\mathcal{I}=4\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}.$$ Standard manipulations express the last sum in terms of $\zeta(2)=\frac{\pi^2}{6}$: $$\zeta(2)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}+\frac{\zeta(2)}{4}\quad \Longrightarrow \quad \displaystyle\mathcal{I}=3\zeta(2).$$


Let $-1\le a \le 1$ and: \begin{align*} I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx \tag 1\\ \frac{\partial}{\partial a}I(a) &= \int_{0}^{1} \, \frac{1}{(1+a\, x)\sqrt{1-x^2}} + \frac{1}{(1-a\, x)\sqrt{1-x^2}} \, dx\\ &= \frac{1}{\sqrt{1-a^2}}\, \left(\arcsin\left(\frac{x+a}{1+a\, x}\right)+\arcsin\left(\frac{x-a}{1-a\, x}\right) \right) \Big|_0^1\\ &= \frac{\pi}{\sqrt{1-a^2}}\\ \therefore I(a) &= \pi\, \arcsin{a} + C \tag 2\\ \end{align*} Putting $a=0$, in $(1)$ and $(2)$, we see that $C=0$

Hence, \begin{align*} I(a) &= \int_{0}^{1} \, \log\left(\frac{1+a\,x}{1-a\, x}\right)\frac{1}{x\sqrt{1-x^2}}\, dx = \pi\, \arcsin{a} \end{align*}

and for this problem $$I(1)=\frac{\pi^2}{2}$$

  • $\begingroup$ I had never seen the trick of derivating respect to an invented constant. Amazing! $\endgroup$ – chubakueno Jun 7 '14 at 3:18
  • $\begingroup$ Thanks, you will get used to it! $\endgroup$ – gar Jun 7 '14 at 7:04
  • $\begingroup$ Still, this is a very clever application of that technique, very nice, gar! $\endgroup$ – Travis Willse Dec 2 '14 at 2:02

The following approach uses contour integration.

First notice that

$$ \int_{0}^{1} \frac{\log \left(\frac{1+x}{1-x} \right)}{x\sqrt{1-x^{2}}} \ dx = \frac{1}{2} \int_{-1}^{1}\frac{\log \left(\frac{1+x}{1-x} \right)}{x\sqrt{1-x^{2}}} \ dx.$$

Now consider $$f(z) = \frac{\log (1+z) - \log(1-z)}{z\sqrt{1-z^{2}}} = \frac{\log(1+z) - \log(1-z)}{z \sqrt{|1+z|e^{i \arg(1+z)} |1-z|e^{i\arg(1-z)}}} $$ where $- \pi < \arg(1+z) \le \pi$ and $ 0 < \arg(1-z) \le 2 \pi$.

By omitting the line segment $[-1,1]$, $f(z)$ is well-defined on the complex plane.

Just above the branch cut, $f(z)$ has a simple pole at $z=0$.

So starting just above the branch cut and integrating clockwise around a dogbone contour,

$$ \begin{align} &\text{PV} \int_{-1}^{1} \frac{\log (1+x) - \log(1-x) - 2 \pi i}{x\sqrt{(1+x)e^{i0}(1-x)e^{2 \pi i}}} \ dx+ \int_{1}^{-1} \frac{\log (1+x) - \log(1-x)}{x\sqrt{(1+x)e^{i0}(1-x)e^{i0}}} \ dx \\ &= -2\int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right)}{x\sqrt{1-x^{2}}} \ dx+ 2 \pi i \ \text{PV} \int_{-1}^{1} \frac{1}{x\sqrt{1-x^{2}}} \ dx \\ &= \pi i \ \text{Res}[f(z),0] + 2 \pi i \ \text{Res}[f(z), \infty] \\ &= \pi i (2 \pi i) + 2 \pi i (0)\\ &= - 2 \pi^{2} . \end{align}$$

And equating the real parts on both sides of the equation,

$$\int_{0}^{1} \frac{\log \left(\frac{1+x}{1-x} \right)}{x\sqrt{1-x^{2}}} \ dx = \frac{\pi^{2}}{2}.$$

  • $\begingroup$ Wow, amazing. How did you 'guess' the right function to integrate? $\endgroup$ – user111187 Jun 6 '14 at 22:53
  • $\begingroup$ @user111187 Really all I'm doing is integrating $f(z) = \frac{\log \left(\frac{1+z}{1-z} \right)}{z\sqrt{1-z^{2}}}$ using those branches for $(1+z)$ and $(1-z)$. I just wrote $(1+z)$ and $(1-z)$ in polar form so it's easier to understand (for me at least) what's going on. $\endgroup$ – Random Variable Jun 6 '14 at 23:50
  • $\begingroup$ @user111187 , i know its a long time ;) , but if u are interested i posted a complex analysis apporach where neither need to 'guess' a function nor take care of branch cuts $\endgroup$ – tired Feb 11 '16 at 15:24

Consider the integral \begin{align} I = \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx \end{align} when the transformation $x = \tanh(t)$ is made. The resulting integral is given by \begin{align} I &= \int_{0}^{\infty} \ln\left( \frac{1+\tanh(t)}{1-\tanh(t)}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\ &= \int_{0}^{\infty} \ln\left( \frac{e^{t}}{e^{-t}}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\ &= 2 \int_{0}^{\infty} \frac{t \cosh^{2}(t)}{\sinh(t)} \ dt \\ &= 2 \left( \frac{\pi^{2}}{4} \right) \end{align} which yields \begin{align} \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx = \frac{\pi^{2}}{2}. \end{align}

  • $\begingroup$ How did you arrived at $\frac{\pi^2}{4}$? I know it might be something standard/obvious, but I don't know yet :) $\endgroup$ – chubakueno Jun 7 '14 at 3:12
  • $\begingroup$ @chubakueno I found the integral mentioned in a book I saved from Google Books. When I find the book again I will post it, but for now I seem to be at a loss of remembering which book. $\endgroup$ – Leucippus Jun 7 '14 at 4:03

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{1} {\ln\pars{1 + x \over 1 - x} \over x\root{1 - x^{2}}}\,\dd x ={\pi^{2} \over 2}:\ {\large ?}}$

With $\ds{x \equiv \cos\pars{\theta}}$: \begin{align}&\color{#c00000}{\int_{0}^{1}% \ln\pars{1 + x \over 1 - x}\,{\dd x \over x\root{1 - x^{2}}}} =\int_{\pi/2}^{0} \ln\pars{1 + \cos\pars{\theta} \over 1 - \cos\pars{\theta}}\, {-\,\dd\theta \over \cos\pars{\theta}} \\[3mm]&=-2\int_{0}^{\pi/2}{\ln\pars{\tan\pars{\theta/2}} \over \cos\pars{\theta}}\,\dd\theta \end{align}

Set $\ds{\tan\pars{\theta \over 2} \equiv t}$: \begin{align}&\color{#c00000}{\int_{0}^{1}% \ln\pars{1 + x \over 1 - x}\,{\dd x \over x\root{1 - x^{2}}}} =-4\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t =-4\int_{0}^{1}{\ln\pars{t^{1/2}} \over 1 - t}\,\half\,t^{-1/2}\,\dd t \\[3mm]&=-\int_{0}^{1}{t^{-1/2}\ln\pars{t} \over 1 - t}\,\dd t =\lim_{\mu \to -1/2}\partiald{}{\mu}\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd t \end{align}

With the identity ${\bf\mbox{6.3.22}}$ ( $\ds{\Psi\pars{z}}$ is the Digamma Function ${\bf\mbox{6.3.1}}$ and $\ds{\gamma}$ is the Euler-Mascheroni Constant ${\bf\mbox{6.1.3}}$ ) $$ \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t =\Psi\pars{z} + \gamma\tag{$\bf 6.3.22$} $$

\begin{align}&\color{#44f}{\large\int_{0}^{1}% \ln\pars{1 + x \over 1 - x}\,{\dd x \over x\root{1 - x^{2}}}} =\lim_{\mu \to -1/2}\partiald{\Psi\pars{\mu + 1}}{\mu} = \Psi'\pars{\half} =3\ \underbrace{\zeta\pars{2}}_{\ds{{\pi^{2} \over 6}}}= \color{#44f}{\Large{\pi^{2} \over 2}} \end{align}

See ${\bf\mbox{6.4.4}}$.


Denote the integral as $I$. Define $$ I(a)=\int_0^1 \frac{\log\left(\frac{1+ax}{1-x}\right)}{x\sqrt{1-x^2}} dx. $$ Then $I(-1)=0$, $I(1)=I$ and \begin{eqnarray} \frac{\partial I(a)}{\partial a}&=&\int_0^1 \frac{1}{(1+ax)\sqrt{1-x^2}}dx\\ &=&\int_0^{\frac{\pi}{2}} \frac{1}{1+a\sin t}dt \quad(x=\sin t). \end{eqnarray} From this post, we have $$ \frac{\partial I(a)}{\partial a}=\frac{2}{\sqrt{1-a^2}}\arctan \left(\sqrt{\frac{1-a}{1+a}}\right) $$ and hence \begin{eqnarray} I&=&\int_{-1}^1\frac{2}{\sqrt{1-a^2}}\arctan \left(\sqrt{\frac{1-a}{1+a}}\right)da\quad(a=\sin t)\\ &=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{1-\sin t}{1+\sin t }}\right)dt\\ &=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{1+\cos(\frac{\pi}{2}+t)}{1-\cos(\frac{\pi}{2}+ t)}}\right)dt\\ &=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \left(\sqrt{\frac{2\cos^2(\frac{\pi}{4}+\frac{t}{2})}{2\sin^2(\frac{\pi}{4}+\frac{t}{2})}}\right)dt\\ &=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\arctan \tan(\frac{\pi}{4}+\frac{t}{2})dt\\ &=&2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{\pi}{4}+\frac{t}{2})dt\\ &=&\frac{\pi^2}{2}. \end{eqnarray}


After this corps has rised from the dead anyway, let me give an additional solution which is based on complex analysis but avoids the use of branch cuts, so it should be viewed as complementary to @RandomVariables approach. First, perform an subsitution $x\rightarrow\sin(t)$ which brings our integral into the form

$$ I=\int_0^{\pi/2}\frac{\log(1+\sin(t))-\log(1-\sin(t))}{\sin(t)} dt $$

Now we introduce a parameter $a$ and differentiate w.r.t. it: $$ I'(a)=\int_0^{\pi/2}\frac{1}{(a+\sin(t))\sin(t)}-\frac{1}{(a-\sin(t))\sin(t)} dt $$

Please note the singularity at zero is removeable, so we don't need a Cauchy principle part. We can reduce the above to

$$ I'(a)=4\int_0^{\pi/2}\frac{1}{2a^2-1+\cos(2t)}dt=\int_0^{2\pi}\frac{1}{2a^2-1+\cos(t)}dt $$

we now can perform the ususal $e^{it}\rightarrow z$ substitution to map the problem onto a contour integral over the unit circle $C_+$ traveresed counterclockwise:

$$ I'(a)=\frac{2}{i}\int_{C_+}\frac{1}{2z(2a^2-1)+z^2+1}dt $$

the poles are given by $z_{\pm}=1-2a^2\pm a\sqrt{a^2-1}$. At this stage a little problem appears: For $a < 1$ the poles degenerate and become double poles on our contour which we don't like, so we want to choose $a>1$ and take the limit $a\rightarrow 1_+$ in the end. For this choice of $a$ only $z_+$ lies inside our contour of integration. The corresponding residue is

$$ \text{Res}(z=z_+)=\frac{1}{2 i a\sqrt{a^2-1} } $$

and therefore

$$ I'(a)=\frac{\pi}{a \sqrt{a^2-1}} $$

Now what is left is integrating back to $a$. The corresponding integral is elementary and yields

$$ I(a)=\pi\arctan\left(\sqrt{\frac{1}{a^2-1}}\right)+C $$

The constant of integration can be fixed by the observation that $I(\infty)=0$ yieldung $C=0$

Taking finally the limit $a\rightarrow 1_+$ yields

$$ I=I(1)=\pi \times \arctan(+\infty)=\pi \times \frac{\pi}{2}=\frac{\pi^2}{2} $$

in accordance with other answers

Remark: We see that we can avoid branch cuts completly but get instead new complications in choosing the correct limit of $a$ and finding the poles which are inside our contour. In summary both methods work and we just exchange some comperable hard problems at the end of the day, so it's just a matter of taste which one we choose

  • $\begingroup$ Thank you for your contribution! $\endgroup$ – user111187 Feb 12 '16 at 5:00

Here is a solution, which I discovered:

$$I=\int_{0}^{1} \frac{\ln\left(\frac{1-x}{1+x}\right)}{x(\sqrt{1-x^2})}dx=\int_{0}^{1} \frac{2\tanh^{-1}(x)}{x(\sqrt{1-x^2})}dx.$$ Observe $$I=\int_{0}^{1} \int_{0}^{1} \frac{2}{(1-x^2y^2)\sqrt{1-x^2}} dy dx$$ by using the fact that $$\int_{0}^{1}\frac{1}{1-x^2y^2}dy=\frac{\tanh^{-1}(x)}{x}.$$ Now to evaluate $I,$ make the change of variables $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(u)}$$ which has Jacobian $$\frac{\partial(x,y)}{\partial(u,v)}=1-x^2y^2.$$ The region of integration becomes the open triangle formed by the inequalities $0<u+v<\frac{\pi}{2}$ with $u,v>0.$ We now get $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2}{\sqrt{1-\left(\frac{\sin(u)}{\cos(v)}\right)^2}}dvdu=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2\cos(v)}{\sqrt{(\cos(v))^2-(\sin(u))^2}}dvdu.$$ We can rewrite $I$ as $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}-u}\frac{2\cos(v)}{\sqrt{1-(\sin(v))^2-(\sin(u))^2}}dvdu.$$ Letting $t=\sin(v),dt=\cos(v)dv$ we get $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos(u)}\frac{2}{\sqrt{1-t^2-(\sin(u))^2}}dtdu$$ Simplifying a bit more, we get $$I=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\cos(u)}\frac{2}{\sqrt{(\cos(u))^2-t^2}}dtdu.$$ Using the fact $$\int_{0}^{\cos(u)}\frac{1}{\sqrt{(\cos(u))^2-t^2}}dt=\lim_{t \rightarrow \cos(u)}\sin^{-1}\left(\frac{t}{\cos(u)}\right)=\frac{\pi}{2},$$ we see $$I= \int_{0}^{\frac{\pi}{2}} 2\left(\frac{\pi}{2}\right)du=\frac{\pi^2}{2}.$$


$\displaystyle J=\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$

Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,

$\displaystyle J=-4\int^1_0 \dfrac{\log(x)}{1-x^2}dx=-4\int_0^1 \left( \log x\times\sum_{n=0}^{+\infty}x^{2n}\right) dx=-4\sum_{n=0}^{+\infty}\left(\int_0^1 x^{2n}\log x dx\right)$

$\displaystyle J=4\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=4\left(\zeta(2)-\sum_{n=1}^{+\infty}\dfrac{1}{(2n)^2}\right)=4\times\left(1-\dfrac{1}{4}\right)\times\zeta(2)=4\times\dfrac{3}{4}\times\dfrac{\pi^2}{6}=\dfrac{\pi^2}{2}$

Note that:

0) For $s>1$, $\displaystyle \zeta(s)=\sum_{n=1}^{+\infty} \dfrac{1}{n^s}$

1) $\zeta(2)=\dfrac{\pi^2}{6}$

2) For $n\geq 0$, $\displaystyle \int_0^1 x^n\log x dx=\left[\dfrac{x^{n+1}}{n+1}\times\log x\right]_0^1-\int_0^1 \left(\dfrac{x^{n+1}}{n+1}\times\dfrac{1}{x}\right)dx=-\left[\dfrac{x^n}{n+1}\right]_0^1=-\dfrac{1}{(n+1)^2}$


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