Anyone with an interesting method to prove $\int_{0}^{\infty}e^{-2x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=1$? We wish to prove that

$$I=\int_{0}^{\infty}e^{-2x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\color{red}{1}\tag1$$

Apply substitution:
$u=e^{-x}\rightarrow du=-e^{-x}dx$
$x=\infty\rightarrow u=0$, $x=0\rightarrow u=1$
$$I=2\int_{0}^{1}u\tanh^{-1}udu\tag2$$
Recall
$$\int x\tanh^{-1}x dx={x\over 2}+{1\over 2}(x^2-1)\tanh^{-1}x\tag3$$
$$\int_{0}^{1}x\tanh^{-1}x dx={1\over 2}\tag4$$
Sub $(4)$ into $(2)$
Therefore $I=1$
Anyone with an interesting method to prove (1)?
 A: It's easy to show that
\begin{equation}
\ln\left(\!{1+e^{-x}\over 1-e^{-x}}\!\right)=2\sum_{n=0}^{\infty}{e^{-(2n+1)x}\over 2n+1}
\end{equation}
Using the generating function above, the considered integral becomes
\begin{align}
I&=\sum_{n=0}^{\infty}{2\over 2n+1}\int_0^\infty e^{-(2n+3)x}\ dx\\[10pt]
&=\sum_{n=0}^{\infty}{2\over {(2n+1)(2n+3)}}\\[10pt]
&=\sum_{n=0}^{\infty}\left[{1\over{2n+1}}-{1\over {2n+3}}\right]\\[10pt]
&=1
\end{align}
where the latter sum is a telescoping series.
A: For a more compact notation, rewrite $\dfrac{1+e^{-x}}{1-e^{-x}}=\coth\dfrac{x}{2}$. Denote a parameterized form of your integral by
$$\mathcal{I}_s=\int_0^\infty e^{-sx}\ln\left(\coth\frac{x}{2}\right)\,\mathrm{d}x$$
which you may observe to be the Laplace transform of $\ln\left(\coth\dfrac{x}{2}\right)$.
It so happens that the transform in this case has an interesting closed form in terms of the harmonic numbers $H_n$:
$$\mathcal{I}_s=\frac{H_{(s-1)/2}-H_{s/2}+2H_s+\ln4}{2s}$$
(computation courtesy of WolframAlpha
The value of your integral is then obtained when $s=2$, i.e.
$$\mathcal{I}_2=\frac{H_{1/2}-H_1+2H_2+\ln4}{4}=\frac{\left(2-2\ln2\right)-1+2\left(\frac{3}{2}\right)+\ln4}{4}=1$$
as desired.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{I} & =
\int_{0}^{\infty}\expo{-2x}\ln\pars{1 + \expo{-x} \over 1 - \expo{-x}}\,\dd x =
2\int_{0}^{\infty}\expo{-2x}\,\mathrm{arctanh}\pars{\expo{-x}}\,\dd x
\\[3mm] & =
2\int_{0}^{\infty}\expo{-2x}\,
\sum_{n = 0}^{\infty}{\expo{-\pars{2n + 1}x} \over 2n + 1}\,\dd x =
2\sum_{n = 0}^{\infty}{1 \over 2n + 1}
\int_{0}^{\infty}\expo{-\pars{2n + 3}x}\,\dd x
\\[3mm] & =
\half\sum_{n = 0}^{\infty}{1 \over \pars{n + 3/2}\pars{n + 1/2}} =
\half\sum_{n = 0}^{\infty}\pars{{1 \over n + 1/2} - {1 \over n + 3/2}} =
\half\,{1 \over 0 + 1/2} = \color{#f00}{1}
\end{align}
A: $\displaystyle \int_{0}^{\infty}e^{-2x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\left[\tfrac{1}{2}\ln\left(\tfrac{1+e^{-x}}{1-e^{-x}}\right)\left(1-e^{-2x}\right)-e^{-x}\right]_0^{\infty}=0-(-1)=\boxed{1}$
A: $$\begin{align*}
I &= \int_0^\infty e^{-2x} \log\left(\frac{1+e^{-x}}{1-e^{-x}}\right) \, dx \\[1ex]
&= \int_0^1 (1-y) \log\left(\frac{2-y}y\right) \, dy \tag{1} \\[1ex]
&= \int_0^1 dy \tag{2} \\[1ex]
&= \boxed{1}
\end{align*}$$


*

*$(1)$ : substitute $y=1-e^{-x}$

*$(2)$ : integrate by parts

A: It only comes down to basic high school math$$I=\int_0^1x\ln\left(\frac{1+x}{1-x}\right)dx=\int_0^1x\ln(1+x)dx-\int_0^1x\ln(1-x)$$For the first integral we substitute $u=1+x$ and for the second integral we substitute $v=1-x$ $$=\int_1^2(x-1)\ln{x}\space dx-\int_0^1(1-x)\ln x\space dx=\int_0^2(x-1)\ln x\space dx$$
The only thing left to do is integrate by parts
$$I=\left(\frac{x^2}{2}-x\right)\ln x\Bigg|_0^2-\int_0^2\left(\frac{x^2}{2}-x\right)\frac{1}{x}dx=\int_0^2\left(1-\frac{x}{2}\right)dx=\left(x-\frac{x^2}{4}\right)\Bigg|_0^2=1$$
