# 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)?

• Interesting as in how? I think your way is pretty interesting myself ... – John Jun 16 '16 at 20:25
• As in different style of approaching this problem. – gymbvghjkgkjkhgfkl Jun 16 '16 at 20:39
• I suppose something like substituting $u=1-e^{-x}$ is technically different... – user170231 Jun 16 '16 at 20:45

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\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}

• Q.E.D (+1) @Felix Marin – gymbvghjkgkjkhgfkl Jun 16 '16 at 21:39
• @Chinacat Thanks. You are welcome. – Felix Marin Jun 16 '16 at 21:49
• @Algebra Thanks. I'm sorry I didn't read your comment in June !!!. – Felix Marin Jan 18 '17 at 22:58

It's easy to show that

$$\ln\left(\!{1+e^{-x}\over 1-e^{-x}}\!\right)=2\sum_{n=0}^{\infty}{e^{-(2n+1)x}\over 2n+1}$$

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.

• Isn't this essentially the same as Felix's solution? – user170231 Jun 17 '16 at 15:10

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.

• This is a cool one! (+1) @user170231 – gymbvghjkgkjkhgfkl Jun 16 '16 at 21:35

$\displaystyle \int_{0}^{\infty}e^{-2x}\ln\left({1+e^{-x}\over 1-e^{-x}}\right)dx=\left[\tfrac{1}{2}\ln\left(\tfrac{1+e^{-x}}{1-e^{-x}}\right)\left(1-e^{-2x}\right)-e^{-x}\right]_0^{\infty}=0-(-1)=\boxed{1}$