5
$\begingroup$

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)?

$\endgroup$
  • 4
    $\begingroup$ Interesting as in how? I think your way is pretty interesting myself ... $\endgroup$ – John Jun 16 '16 at 20:25
  • $\begingroup$ As in different style of approaching this problem. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 16 '16 at 20:39
  • $\begingroup$ I suppose something like substituting $u=1-e^{-x}$ is technically different... $\endgroup$ – user170231 Jun 16 '16 at 20:45
5
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\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}

$\endgroup$
  • $\begingroup$ Q.E.D (+1) @Felix Marin $\endgroup$ – gymbvghjkgkjkhgfkl Jun 16 '16 at 21:39
  • $\begingroup$ @Chinacat Thanks. You are welcome. $\endgroup$ – Felix Marin Jun 16 '16 at 21:49
  • $\begingroup$ @Algebra Thanks. I'm sorry I didn't read your comment in June !!!. $\endgroup$ – Felix Marin Jan 18 '17 at 22:58
6
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Isn't this essentially the same as Felix's solution? $\endgroup$ – user170231 Jun 17 '16 at 15:10
5
$\begingroup$

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.

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.