Prove that $\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x=\log 2$

Question

This integral popped up recently

$$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2$$

A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of $1/(1+e^{nx})$, however this did not quite make it.

\begin{align*} I & = 2\int_{0}^{\infty} \frac{e^{2x}-e^x}{x e^{3x}(1+e^{-2x})(1+e^{-x})}\mathrm{d}x \\ & = 2 \int_{0}^{\infty}\frac{e^{-2x} -e^{-x}}{x} \left(\sum_{n=0}^\infty (-1)^n e^{-n x}\right)\left(\sum_{m=0}^\infty (-1)^me^{-2mx}\right) \mathrm{d}x \end{align*} The first part reminds me of a Frullani integral (it evaluates to $\log 2$). However I am unsure if this is the correct path, any help would be appreciated. =)

Answer 1

I reckon you are on the right track with Frullani's Integral, although the series expansion of the exponential function in the latter part of your derivation complicates matters unnecessarily.

Using a partial fraction decomposition and symmetry,

$$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x =\int_{-\infty}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\=2\int_{0}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x$$ Setting $f(x)=\frac{1}{1+e^x}$, we are in a position to use Frullani's integral as follows $$I=2\int_0^{\infty}\frac{f(x)-f(2x)}{x}\mathrm{d}x=2[f(0)-f(\infty)]\log\left(\frac{2}{1}\right)=2\left(\frac{1}{2}\right)\log 2=\log 2$$

Answer 2

One can start from the identity $$ \frac{\mathrm e^{2x}-\mathrm e^x}{(1+\mathrm e^{2x})(1+\mathrm e^{x})}=\frac1{1+e^{x}}-\frac1{1+\mathrm e^{2x}}=-\int_1^2\frac{\mathrm d}{\mathrm du}\left(\frac1{1+\mathrm e^{ux}}\right)\cdot\mathrm du, $$ that is, $$ \frac{\mathrm e^{2x}-\mathrm e^x}{x(1+\mathrm e^{2x})(1+\mathrm e^{x})}=\int_1^2\frac{\mathrm e^{ux}}{(1+\mathrm e^{ux})^2}\mathrm du. $$ Note that, for every nonzero $u$, $$ \int_{-\infty}^\infty\frac{\mathrm e^{ux}}{(1+\mathrm e^{ux})^2}\mathrm dx=\left.\frac{-1}{u(1+\mathrm e^{ux})}\right|_{-\infty}^\infty=\frac1{|u|}, $$ hence Fubini theorem shows that the integral to be computed is $$ \int_{-\infty}^{\infty} \frac{\mathrm e^{2x}-\mathrm e^{x}}{x (1+\mathrm e^{2x})(1+\mathrm e^{x})}\mathrm{d}x = \int_1^2\frac{\mathrm du}u=\log2. $$ More generally, for every positive $a$ and $b$, $$ \int_{-\infty}^{\infty} \frac{\mathrm e^{ax}-\mathrm e^{bx}}{x (1+\mathrm e^{ax})(1+\mathrm e^{bx})}\mathrm{d}x = \log\left(\frac{a}b\right). $$ This is rediscovering the fact that, for every monotonous function $\varphi$ with limits at $\pm\infty$, $$ \int_\mathbb R\frac{\varphi(ax)-\varphi(bx)}x\mathrm dx=(\varphi(+\infty)-\varphi(-\infty))\cdot\log\left(\frac{b}a\right). $$ in the present case for the function $$ \varphi(x)=\frac1{1+\mathrm e^x}. $$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\expo{2x} -\expo{x} \over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x=\ln\pars{2}: \ {\large ?}}$

\begin{align}&\color{#c00000}{% \int_{-\infty}^{\infty}{\expo{2x} -\expo{x} \over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x} =2\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{1 \over x} \pars{{1 \over 1 + \expo{x}} - {1 \over 1 + \expo{2x}}}\,\dd x \\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{% \int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x} -\int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{2x}}\,{\dd x \over x}} \\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{% \int_{\epsilon}^{\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x} -\int_{2\epsilon}^{2\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}} \\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{% \int_{\epsilon}^{2\epsilon}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x} +\int_{2\epsilon}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x} -\int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x} -\int_{2\epsilon}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}} \\[3mm]&=2\lim_{\epsilon \to 0^{+} \atop \Lambda \to \infty}\bracks{% \half\int_{\epsilon}^{2\epsilon}{\dd x \over x} +\int_{\epsilon}^{2\epsilon}\pars{{1 \over 1 + \expo{x}} - \half}\,{\dd x \over x} -\int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}} \\[3mm]&=\ln\pars{2} -\half\ \overbrace{\lim_{\epsilon \to 0^{+}} \int_{\epsilon/2}^{\epsilon}{\tanh\pars{x} \over x}\,\dd x}^{\ds{=\ 0}}\ -\ \overbrace{\lim_{\Lambda \to \infty} \int_{\Lambda}^{2\Lambda}\!\!{1 \over 1 + \expo{x}}\,{\dd x \over x}} ^{\ds{=\ 0}}\,, \qquad\pars{~\large\tt\mbox{See below.}~} \end{align}

$$\color{#00f}{\large% \int_{-\infty}^{\infty}{\expo{2x} -\expo{x} \over x\pars{1 + \expo{2x}}\pars{1 + \expo{x}}}\,\dd x=\ln\pars{2}} $$

$$ {\tanh\pars{x} \over x} = \sech^{2}\pars{\xi} \leq 1\,,\qquad 0 < \xi < x $$ $$ \verts{\int_{\epsilon/2}^{\epsilon}{\tanh\pars{x} \over x}\,\dd x} <\half\,\epsilon\to 0\quad\mbox{when}\quad\epsilon\to 0^{+} $$

$$ \verts{\int_{\Lambda}^{2\Lambda}{1 \over 1 + \expo{x}}\,{\dd x \over x}} <{1 \over \Lambda}\int_{\Lambda}^{2\Lambda}{\expo{-x} \over 1 + \expo{-x}}\,\dd x ={1 \over \Lambda}\ln\pars{1 + \expo{-\Lambda} \over 1 + \expo{-2\Lambda}} \to 0\quad\mbox{when}\quad\Lambda\to\infty $$

Answer 4

$$\begin{align}\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x &=\int_{-\infty}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\&=2\int_{0}^{\infty} \frac{1}{x}\left[\frac{1}{1+e^x}-\frac{1}{1+e^{2x}}\right]\mathrm{d}x\\&=2\lim_{s\to0^+}\int_0^\infty\frac{x^{s-1}\ }{e^x+1}\mathrm dx-\int_0^\infty\frac{x^{s-1}}{e^{2x}+1}\ \mathrm dx\end{align}$$

By letting $x\mapsto\frac12x$ in the last integral, we get

$$\int_0^\infty\frac{x^{s-1}}{e^{2x}+1}\ \mathrm dx=2^{-s}\int_0^\infty\frac{x^{s-1}}{e^x+1}\ \mathrm dx$$

Likewise, using the Dirichlet eta function, we have

$$\int_0^\infty\frac{x^{s-1}\ }{e^x+1}\mathrm dx=\Gamma(s)\eta(s)$$

and so,

$$\begin{align}\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x&=2\lim_{s\to0^+}(1-2^{-s})\Gamma(s)\eta(s)\\&=2\lim_{s\to0^+}\color{red}{\frac{1-2^{-s}}s}\color{blue}{s\Gamma(s)}\eta(s)\\&=2\color{red}{\ln(2)}\color{blue}{(1)}\eta(0)\\&=\ln(2)\end{align}$$