Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$

Question

I would like to evaluate the following integral: $$ \int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} $$ I tried various methods but without success.

Answer 1

Hint. One may write $$ \begin{align} \int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} &=\int_{-1}^0 \frac{dx}{(e^x+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} \\\\&=\int_0^1 \frac{dx}{(e^{-x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} \\\\&=\int_0^1 \frac{e^x\:dx}{(e^{x}+1)(x^2+1)} +\int_0^{1} \frac{dx}{(e^x+1)(x^2+1)} \\\\&=\int_0^1 \frac{(e^x+1)\:dx}{(e^{x}+1)(x^2+1)} \\\\&=\int_0^1 \frac{dx}{(x^2+1)} \end{align} $$ then it is easier.

Answer 2

Substitution $x \rightarrow -x$ gives $$\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^{-x}+1)(x^2+1)}=\int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}$$ Therefore $$ \int_{-1}^{1} \frac{dx}{(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} + \int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}=2\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} $$

Should be easy now.

Answer 3

HINT:

Use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

and $$2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$

$$2I=\int_{-1}^1\dfrac{dx}{x^2+1}$$

Answer 4

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

\begin{align} \color{#f00}{\int_{-1}^{1}{\dd x \over \pars{\expo{x} + 1}\pars{x^{2} + 1}}} & = \int_{-1}^{0}{\dd x \over x^{2} + 1}\ +\ \overbrace{\int_{-1}^{1}{\mathrm{sgn}\pars{x}\,\dd x \over \pars{\expo{\verts{x}} + 1}\pars{x^{2} + 1}}}^{\ds{=\ 0}}\ =\ \color{#f00}{{\pi \over 4}} \end{align}

where $\mathrm{sgn}$ is the sign function: $\ds{\,\mathrm{sgn}\pars{x} = \left\lbrace\begin{array}{rcrcl} \ds{-1} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\[2mm] \ds{1} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}\right.}$

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Why ?. Because $$ \mathrm{f}\pars{x} \equiv {1 \over \expo{x} + 1} = \left\lbrace\begin{array}{lcrcl} \ds{\Theta\pars{-x} + \mathrm{sgn}\pars{x}\mathrm{f}\pars{\verts{x}}} & \mbox{if} & \ds{x} & \ds{\not=} & \ds{0} \\[2mm] \ds{\half} & \mbox{if} & \ds{x} & \ds{=} & \ds{0} \end{array}\right. $$

$\ds{\Theta}$ is the Heaviside step function. $\ds{\Theta\pars{x} = \left\lbrace\begin{array}{rcrcl} \ds{0} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\[2mm] \ds{1} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}\right.}$

By the way, $\ds{\mathrm{f}\pars{x}}$ is quite usual in Statistical Physics.