Show that $\int_{1/e}^e \left|\frac{\ln x}{1+x}\right|\,\mathrm{d}x=\frac{1}{2}$

Question

How can one show that

$$I=\int_{1/e}^e \left|\frac{\ln x}{1+x}\right|\,\mathrm{d}x=\frac{1}{2}$$

preferably without resorting to the polylogarithm function?

Mathematica returns

1 + Pi^2/6 + PolyLog[2, -1/E] + PolyLog[2, -E]

which is numerically equivalent to $0.5$.

The first thing that comes to mind is to remove the absolute value and consider the two integrals

$$I=\color{red}{I_1}-\color{blue}{I_2}=\color{red}{\int_1^e\frac{\ln x}{1+x}\,\mathrm{d}x}-\color{blue}{\int_{1/e}^1\frac{\ln x}{1+x}\,\mathrm{d}x}$$

Is there a substitution that would reduce this to $\frac{1}{2}$?

Another method I've considered is to integrate by parts:

$$I=1-\int_1^e\frac{\ln(1+x)}{x}\,\mathrm{d}x+\int_{1/e}^1\frac{\ln(1+x)}{x}\,\mathrm{d}x$$

but this only seems to force me down a path of using the Mercator series (at least for the second integral, considering the interval of convergence of the series) and polylogarithm.

Answer 1

By setting $x=e^t$ we are left with

$$ I = \int_{-1}^{1}e^t\left|\frac{t}{1+e^t}\right|\,dt = \int_{0}^{1}\frac{t}{1+e^{-t}}\,dt+\int_{0}^{1}\frac{t}{1+e^t}\,dt=\int_{0}^{1}t\,dt=\color{red}{\frac{1}{2}},$$ sic et simpliciter.

Answer 2

Jack D'Aurizio posted a simpler answer above, but since I was already in the process of typing:

Make the substitution $x \mapsto 1/x$ to see $$\int_{1/e}^1 \frac{\log x}{1+x} dx= \int^1_{e} \frac{\log(1/x)}{1+1/x} \left( -\frac 1 {x^2} \right)dx = -\int^e_1 \frac{\log(x)}{x^2+x}dx.$$ Thus $$I = \int^e_1 \frac{\log x}{x+1} + \frac{\log x}{x^2+x} dx = \int^e_1 \frac{(x+1)\log(x)}{x^2+x}dx= \int^e_1 \frac{\log x}{x}dx = \left.\frac{1}{2}\log(x)^2\right|^e_1 = \frac 1 2.$$

Answer 3

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & = \color{#f00}{\int_{1/\expo{}}^{\expo{}}\verts{\ln\pars{x} \over 1 + x}\,\dd x} \,\,\,\stackrel{x\ \equiv\ \expo{-t}}{=}\,\,\, \int_{1}^{-1}\verts{-t \over \expo{-t} + 1}\pars{-\expo{-t}}\,\dd x = \int_{-1}^{1}{\verts{t} \over \expo{t} + 1}\,\dd x \\[5mm] & = \int_{-1}^{1}\verts{t}\bracks{\Theta\pars{-t} + {\mrm{sgn}\pars{t} \over \expo{\verts{t}} + 1}}\,\dd x\qquad\qquad \pars{ \begin{array}{rll} \ds{\Theta} & \ds{:} & Heaviside\ Step\ Function \\ \ds{\mrm{sgn}} & \ds{:} & Sign\ Function \end{array}} \\[5mm] & = \int_{-1}^{0}\pars{-t}\,\dd t = \color{#f00}{1 \over 2} \end{align}

Note that $\ds{\,\mrm{f}\pars{x} \equiv {1 \over \expo{x} + 1} = 1 - {1 \over \expo{-x} + 1} = 1 - \mrm{f}\pars{-x}}$ which satisfies: \begin{align} \mrm{f}\pars{x} & = \left\{\begin{array}{lcrcl} \ds{1 - {1 \over \expo{\verts{x}} + 1}} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\ \ds{1 \over 2} & \mbox{if} & \ds{x} & \ds{=} & \ds{0} \\ \ds{1 \over \expo{\verts{x}} + 1} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}\right. \\[5mm] \mbox{It's equivalent to}\ \mrm{f}\pars{x} & = \left\{\begin{array}{lcrcl} \ds{\Theta\pars{-x} + {\mrm{sgn}\pars{x} \over \expo{\verts{x}} + 1}} & \mbox{if} & \ds{x} & \ds{\not=} & \ds{0} \\ \ds{1 \over 2} & \mbox{if} & \ds{x} & \ds{=} & \ds{0} \end{array}\right. \end{align}

Answer 4

$$ \begin{align} \int_{1/e}^e\left|\frac{\log(x)}{1+x}\right|\,\mathrm{d}x &=\int_1^e\frac{\log(x)}{1+x}\,\mathrm{d}x-\int_{1/e}^1\frac{\log(x)}{1+x}\,\mathrm{d}x\\ &=\int_1^e\frac{\log(x)}{1+x}\,\mathrm{d}x+\int_1^e\frac{\log(x)}{x(1+x)}\,\mathrm{d}x\\ &=\int_1^e\frac{\log(x)}{x}\,\mathrm{d}x\\ &=\left[\frac12\log(x)^2\right]_1^e\\[3pt] &=\frac12 \end{align} $$