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Prove this $$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$

My try: let $$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$

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    $\begingroup$ The square is causing me some confusion. Do you mean $\left(\ln y\right)^2$ or $\ln\left(y^2\right)$? $\endgroup$ – Fly by Night Oct 27 '13 at 18:10
  • $\begingroup$ @Katsu If $u(x) = \ln(1+x)$ then $u'(x) = \frac{1}{x+1}$. $\endgroup$ – Fly by Night Oct 27 '13 at 18:13
  • $\begingroup$ @Katsu Your corrected hint is no longer a hint, but an unrelated fact. The integral of $\frac{\ln(1\pm x)}{x}$ is non-trivial and requires the dilog function: en.wikipedia.org/wiki/Polylogarithm#Dilogarithm $\endgroup$ – Fly by Night Oct 27 '13 at 18:18
  • $\begingroup$ @Fly by Night: I guess square is inside log because he deducts that $ln(y²)=2.ln(y)$ $\endgroup$ – Katsu Oct 27 '13 at 18:19
  • $\begingroup$ @FlybyNight It's difficult to provide an answer when the OP refuses to answer your comment. So, we have to make a "guess" which is not a mathematical trick. $\endgroup$ – Felix Marin Aug 9 '14 at 22:19
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Here is an approach. Using the change of variables $\frac{1+x}{1-x}=y$ gives

$$ I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\left(\dfrac{1+x}{1-x}\right)^2\right)}dx=2\int_{-1}^{1}\frac{\ln(y^2)}{1-y^2}dy=8\int_{0}^{1}\frac{\ln y}{1-y^2}dy $$

$$ = 8\sum_{k=0}^{\infty} \int_{0}^{1} y^{2k} \ln y\, dy = 8\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = 8.\frac{\pi^2}{8} = \pi^2.$$

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    $\begingroup$ Mhenni Benghorbal well done +1 $\endgroup$ – mert Oct 27 '13 at 19:36
  • $\begingroup$ @mert: Thanks for the comment. I really appreciate it. $\endgroup$ – Mhenni Benghorbal Oct 27 '13 at 22:46
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    $\begingroup$ This is a very nice argument, but it changes one problem in to another. You can't just assume that $$\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}$$ as you have done here. Proving this summation is far from trivial. (The summation identity can be proven, for example, using Fourier Series, but you must justify the summation result.) $\endgroup$ – Fly by Night Oct 27 '13 at 22:47
  • $\begingroup$ @FlybyNight: Thanks for the comment. I really appreciate it. $\endgroup$ – Mhenni Benghorbal Oct 28 '13 at 2:32
  • $\begingroup$ @Downvoter: Can we know why the downvote? $\endgroup$ – Mhenni Benghorbal Apr 26 '14 at 2:15
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First perform the bilinear transformation: $$x={\frac {1-t}{t+1}}$$ to get: $$\int _{0}^{\infty }\!\ln \left( {\frac { \left( 1+x \right) ^{2}}{ \left( 1-x \right) ^{2}}} \right)\dfrac{1}{x} {dx}=4\,\int _{0}^{1}\!{ \frac {\ln \left( {t}^{2} \right) }{{t}^{2}-1}}{dt}=8\,\int _{0}^{1}\!{ \frac {\ln \left(t \right) }{{t}^{2}-1}}{dt} $$ then, from this answer, consider the integral: $$I(m)=\int _{0}^{1}\!{\frac { \ln\left( t \right) ^{m-1}}{ {t}^{2}-1}}{dt} \quad:\quad \mathfrak{R}(m)>1 $$ and the substitution $t=e^{-u}$: $$\begin{aligned} \int _{0}^{1}\!{\frac { \ln\left( t \right) ^{m-1}}{ {t}^{2}-1}}{dt}=& \left( -1 \right) ^{m-1}\int _{0}^{\infty }\!{\frac { {u}^{m-1}{{\rm e}^{-u}}}{-1+{{\rm e}^{-2\,u}}}}{du}\\ =&\left( -1 \right) ^{m-1} \int _{0}^{\infty }\!-{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}+{\frac {{u }^{m-1}}{-1+{{\rm e}^{2\,u}}}}{du}\\ =&\left( -1 \right) ^{m-1}\left( 1- \dfrac{1}{2^m} \right) \int _{0}^{\infty }\!{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}{du}\\ =&\left( -1 \right) ^{m}\left( 1- \dfrac{1}{2^m} \right) \Gamma \left( m \right) \zeta \left( m \right) \end{aligned}$$ 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).

One way to evaluate the Riemann zeta function at even integers is to use the Fourier series for Bernoulli polynomials. For example, calculating the Fourier series for the second Bernoulli polynomial tells you that: $$\displaystyle{x}^{2}-x+1/6=\frac{1}{\pi^2}\sum _{n=1}^{\infty }{\frac {\cos \left( nx \right) }{{n }^{2}}}\quad : \quad-\pi<x<\pi$$

and evaluating this series at $x=0$ then tells you that $\zeta(2)=\frac{\pi^2}{6}$ which together with $\Gamma(2)=1!=1$ leads to: $$I(2)=\frac{\pi^2}{8}$$ and the result follows.

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    $\begingroup$ This is a very nice argument, but you glibly use the fact that $\zeta(2) = \frac{\pi^2}{6}$. The OP needs to be able to both know , and also justify, why the following is true: $$\sum_{k=1}^{\infty}\frac{1}{k^2} = \frac{\pi^2}{6}$$ You need to justify this identity for this answer to be valid. $\endgroup$ – Fly by Night Oct 27 '13 at 23:45
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Allow me to present an approach that uses dilogarithms. Split the integral up into 2 and substitute $x \mapsto \dfrac{1}{x}$ for the second integral. This yields \begin{align} \int^\infty_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx &=\int^1_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx+\int^\infty_1\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx\\ &=4\int^1_0\frac{\ln(1+x)-\ln(1-x)}{x}dx\\ &=4\left(\operatorname{Li}_2(1)-\operatorname{Li}_2(-1)\right)\\ &=4\left(\frac{\pi^2}{6}+\frac{\pi^2}{12}\right)\\ &=\pi^2 \end{align}

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  • $\begingroup$ Nice solution. +1. BTW, instead of introducing dilogarithm function I think it would be better to use series expansion like this OP. $\endgroup$ – Tunk-Fey Aug 9 '14 at 4:53
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We can use an argument from Complex Analysis and the calculation of the integral is pretty simple and neat. Note that $$ \arctan z=\frac{1}{2i}\ln\frac{i-z}{i+z}. $$ So we have $$ \ln\frac{1+x}{1-x}=\ln\frac{i+ix}{i-ix}=-\ln\frac{i-ix}{i+ix}=-2i\arctan(ix) $$ and hence \begin{eqnarray} I&=&\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx\\ &=&2\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)}dx\\ &=&-4i\int_{0}^{\infty}\dfrac{1}{x}\arctan(ix)dx\\ &=&-4i\left(\arctan(ix)\ln x\bigg|_0^\infty-\int_0^\infty\frac{i\ln x}{-x^2+1}dx\right)\\ &=&4\int_0^\infty\frac{\ln x}{x^2-1}dx=-8\int_0^1\frac{\ln x}{1-x^2}dx\\ &=&-8\int_0^1\sum_{n=0}^\infty x^{2n}\ln xdx=-8\sum_{n=0}^\infty \frac{1}{(2n+1)^n}\\ &=&\pi^2. \end{eqnarray}

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  • $\begingroup$ +1) Introducing $\arctan$ is indeed a nice trick. You might want to justify the last integral to make this answer more complete though. $\endgroup$ – SuperAbound Aug 9 '14 at 2:16
  • $\begingroup$ @Superabound, thanks. $\endgroup$ – xpaul Aug 9 '14 at 21:54

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