How can we determine the closed form for $\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}\mathrm dx?$

Question

We know the closed form for $(1)$

$$\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}x\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)-{\zeta(3)\over 2^3}\tag1$$

Proposed:

How can we determine the closed form for $(2)?$

$$\int_{0}^{\infty}{\ln(e^x-1)\over e^x+1}\mathrm dx=I\tag2$$

My try:

$y=e^x$

$$\int_{1}^{\infty}{\ln(y-1)\over y+1}\cdot{\mathrm dy\over y}\tag3$$

Answer 1

On the path of Mark Viola,

$\displaystyle I=\int_0^\infty \frac{\log(e^x-1)}{e^x+1}\,dx \tag 1$

Perform the change of variable $y=\text{e}^{-x}$,

$\displaystyle I=\int_0^1 \frac{\log(1-x)-\log(x)}{1+x}\,dx\tag 2$

In $(1)$ perform the change of variable $y=\dfrac{1-\text{e}^{-x}}{1+\text{e}^{-x}}$,

$\begin{align} I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1-x}\right)}{1+x}dx\\ &=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1-x)}{1+x}dx\\ &=(\ln 2)^2+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1-x)}{1+x}dx \end{align}$

Using $(2)$,

$I=(\ln 2)^2-I$

Therefore,

$\boxed{I=\dfrac{(\ln 2)^2}{2}}$

Answer 2

Let $I$ be given by the integral

$$I=\int_0^\infty \frac{\log(e^x-1)}{e^x+1}\,dx \tag 1$$

Enforcing the substitution $x\to -\log(x)$ in $(1)$ we find

$$\begin{align} I&=\int_0^1 \frac{\log(1-x)-\log(x)}{1+x}\,dx\\\\ &=\int_0^1 \frac{\log(1-x)}{1+x}\,dx-\int_0^1 \frac{\log(x)}{1+x}\,dx\tag2 \end{align}$$

We enforce the substitution $x\to 2x-1$ in the first integral on the right-hand side of $(2)$ to reveal

$$\begin{align} \int_0^1 \frac{\log(1-x)}{1+x}\,dx&=\int_{1/2}^1 \frac{\log(2)+\log(1-x)}{x}\,dx\\\\ &=\log^2(2)-\text{Li}_2(1)+\text{Li}_2(1/2)\\\\ &=\log^2(2)-\frac{\pi^2}{6}+\frac{\pi^2}{12}-\frac12\log^2(2)\\\\ &=\frac12\log^2(2)-\frac{\pi^2}{12}\tag 3 \end{align}$$

We integrate by parts the second integral on the right-hand side of $(2)$ with $u=\log(x)$ and $v=\log(1+x)$ to reveal

$$\int_0^1 \frac{\log(x)}{1+x}\,dx=-\int_0^1 \frac{\log(1+x)}{x}\,dx\tag 4$$

Then letting $x\to -1$ in $(4)$, we find

$$\begin{align} \int_0^1 \frac{\log(x)}{1+x}\,dx&=-\int_0^{-1} \frac{\log(1-x)}{x}\,dx\\\\ &=\text{Li}_2(-1)\\\\ &=-\frac{\pi^2}{12}\tag 5 \end{align}$$

Finally, substituting $(3)$ and $(5)$ into $(2)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{I=\frac12\log^2(2)}$$

Answer 3

After the substitution $x=\ln(1+t)$ we have \begin{equation*} I = \int_{0}^{\infty}\dfrac{\ln(t)}{(t+1)(t+2)}\, dt \tag{1} \end{equation*} which can be evaluated by integrating $\dfrac{\log^{2}(z)}{(z+1)(z+2)}$ along a keyhole contour. As $\log(z)$ we choose the branch given by \begin{equation*} \log(z) = \ln|z| +i\arg(z), \quad -\pi<\arg(z) <\pi. \end{equation*} We get \begin{equation*} I = \dfrac{1}{2}\ln^{2}(2). \end{equation*}

Remark. Inspired by $@$FDP I realize that my answer can be simplified by the substitution $t = \frac{2}{s}$ in (1). Then \begin{gather*} I = \int_{0}^{\infty}\dfrac{\ln(2)- \ln(s)}{(s+1)(s+2)}\, ds = \ln(2 ) \int_{0}^{\infty}\left(\dfrac{1}{s+1}-\dfrac{1}{s+2}\right)\, ds -I = \\[2ex] \dfrac{1}{2}\ln(2)\left[\ln\left(\dfrac{s+1}{s+2}\right)\right]_{0}^{\infty} = \dfrac{1}{2}\ln^{2}(2). \end{gather*}