¿Cómo mostrar que $\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)?$

Question

Propuesto:

$$\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)\tag1$$

Mi intento:

$$u={e^x+1\over e^x-1}\implies -{2\over (u-1)(u+1)}\mathrm du=\mathrm dx$$

$$-\int_{1}^{\infty}{\ln(u)\over u+1}\ln\left({u+1\over u-1}\right){\mathrm du}\tag2$$

$$-\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u+1){\mathrm du}+\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u-1){\mathrm du}\tag3$$

$$\sum_{n=1}^{\infty}{(-1)^n\over n}\int_{1}^{\infty}u^n{\ln(u)\over u+1}{\mathrm du}+\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u-1){\mathrm du}\tag4$$

¿Cómo se puede demostrar $(1)?$

Answer 1

Bajo $t=e^{-x}$ como hizo Jack D'Aurizio, se tiene $$ I=\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\int_{0}^{1}\frac{-\ln(t)\ln\left(\frac{1+t}{1-t}\right)}{1-t}\mathrm dt.$$ Definir $$ I(a)=\int_{0}^{1}\frac{-\ln(t)\ln\left(\frac{1+at}{1-t}\right)}{1-t}\mathrm dt $$ y luego $I(-1)=0, I(1)=I$, y \begin{eqnarray} I'(a)&=&\int_{0}^{1}\frac{-t\ln(t)}{(1-t)(1+at)}\mathrm dt\\ &=&-\frac{1}{1+a}\int_0^1\bigg(\frac1{1-t}-\frac{1}{1+at}\bigg)\ln(t)\mathrm dt\\ &=&\frac{1}{1+a}\bigg(\frac{\pi^2}{6}+\frac1a\text{Li}_2(-a)\bigg). \end{eqnarray} Así que \begin{eqnarray} I(1) &=&\frac16\int_{-1}^1\frac{1}{a(1+a)}\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\mathrm da\\ &=&\frac{1}{6}\int_{-1}^1\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\mathrm d\ln(\frac{|a|}{1+a})\\ &=&\frac{1}{6}\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\ln(\frac{|a|}{1+a})\bigg|_{-1}^1\\ &&-\frac16\int_{-1}^1\ln(\frac{a}{1+a})\mathrm d\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\\ &=&-\frac{\pi^2}{12}\ln2-\frac16\int_{-1}^1\ln(\frac{|a|}{1+a})(\pi^2-\frac{6\ln(1+a)}{a})\mathrm da\\ &=&-\frac{\pi^2}{12}\ln2+\frac{\pi^2}{3}\ln 2\\ &=&\frac{\pi^2}{4}\ln2. \end{eqnarray}

Answer 2

Al establecer $x = \log t$ nos queda $$ \int_{1}^{+\infty}\frac{\log t}{t(t-1)}\log\left(\frac{t+1}{t-1}\right)\,dt = \int_{0}^{1}\frac{-\log(t)\log\left(\frac{1+t}{1-t}\right)}{1-t}\,dt$$ que se puede calcular fácilmente a través de las fórmulas de reflexión del dilogaritmo.

Answer 3

En el camino de Xpaul y Jack D'Aurizio,

La misma vieja canción,

$\displaystyle I=\int_{0}^{\infty}{x\over e^x-1}\ln\left({e^x+1\over e^x-1}\right)\mathrm dx$

Realiza el cambio de variable $y=\text{e}^{-x}$,

$\begin{align}I&=\displaystyle \int_{0}^{1}\frac{\mathrm{ln}\left( y\right) \mathrm{ln}\left( y+1\right) -\mathrm{ln}\left( 1-y\right) \mathrm{ln}\left( y\right) }{y-1}dy\\ &=\int_0^1 \dfrac{\ln(1-x)\ln x}{1-x}dx-\int_0^1 \dfrac{\ln(1+x)\ln x}{1-x}dx\\ &=A-B \end{align}$

Define, para $x\in [0;1],$

$\begin{align} R(x)&=\int_0^x \dfrac{\ln t}{1-t}dt\\ &=\int_0^1 \dfrac{x\ln(tx)}{1-tx}dt \end{align}$

Observa que,

$R(0)=0$ y $R(1)=-\dfrac{\pi^2}{6}$

$\begin{align} B&=\int_0^1 \dfrac{\ln(1+x)\ln x}{1-x}dx\\ &=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1 \dfrac{R(x)}{1+x}dx\\ &=-\dfrac{\pi^2}{6}\ln 2-\int_0^1 \int_0^1 \dfrac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\ &=-\dfrac{\pi^2}{6}\ln 2-\int_0^1 \int_0^1 \dfrac{x\ln(t)}{(1+x)(1-tx)}dtdx-\int_0^1 \int_0^1 \dfrac{x\ln(x)}{(1+x)(1-tx)}dtdx\\ &=-\dfrac{\pi^2}{6}\ln 2+\left(\int_0^1\int_0^1 \dfrac{\ln t}{(1+x)(1+t)}dtdx-\int_0^1\int_0^1 \dfrac{\ln t}{(1+t)(1-tx)}dtdx\right)-\\ &\int_0^1 \int_0^1 \dfrac{x\ln(x)}{(1+x)(1-tx)}dtdx\\ &=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \left[\dfrac{\ln(1-tx)\ln t}{t(t+1)}\right]_{x=0}^{x=1}dt+\int_0^1 \left[\dfrac{\ln(1-tx)\ln x}{1+x}\right]_{t=0}^{t=1}dx\\ &=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-t)\ln t}{t(t+1)}dt+\int_0^1 \dfrac{\ln(1-x)\ln x}{1+x}dx\\ &=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-t)\ln t}{t}dt \end{align}$

En la última integral haz el cambio de variable $x=1-t$,

$\displaystyle B=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-x)\ln x}{1-x}dx$

Es bien sabido que,

$\displaystyle \int_0^1 \dfrac{\ln x}{1+x}dx=-\dfrac{\pi^2}{12}$

(expansión de la serie de Taylor)

Por lo tanto,

$\begin{align}I&=A-B\\ &=\dfrac{\pi^2}{6}\ln 2+\dfrac{\pi^2}{12}\ln 2\\ &=\boxed{\dfrac{\pi^2}{4}\ln 2} \end{align}$