# How do you find the integral of int dx/(e^x +e^-x) from negative infinity to infinity?

Sep 23, 2015

$\frac{\pi}{2}$

#### Explanation:

$f \left(x\right) = \frac{1}{{e}^{x} + {e}^{- x}}$
$f \left(- x\right) = \frac{1}{{e}^{- x} + {e}^{x}}$
=>f(x)=f(-x)
=> the given function is even function
son${\int}_{-} {\infty}^{\infty} \frac{1}{{e}^{x} + {e}^{- x}} \mathrm{dx} = 2 {\int}_{0}^{\infty} \frac{1}{{e}^{x} + {e}^{- x}} \mathrm{dx} = 2 {\int}_{0}^{\infty} {e}^{x} / \left({e}^{2 x} + 1\right) \mathrm{dx} = 2 {\int}_{0}^{\infty} {e}^{x} / \left({\left({e}^{x}\right)}^{2} + 1\right) \mathrm{dx}$
if $y = {\tan}^{-} 1 x$
$\mathrm{dy} = \frac{1}{{x}^{2} + 1} \mathrm{dx}$
=>$y = \int \frac{1}{{x}^{2} + 1} \mathrm{dx} = {\tan}^{-} 1 x$
if you substitute $x = {e}^{x}$ then $\mathrm{dt} a {n}^{-} 1 \left({e}^{x}\right) = \frac{1}{{\left({e}^{x}\right)}^{2} + 1} {e}^{x} \mathrm{dx}$
=>$2 {\int}_{0}^{\infty} {e}^{x} / \left({\left({e}^{x}\right)}^{2} + 1\right) \mathrm{dx} = 2 {\int}_{0}^{\infty} d \left({\tan}^{-} 1 {e}^{x}\right) \mathrm{dx} = 2 {\left[{\tan}^{-} 1 {e}^{x}\right]}_{0}^{\infty} = 2 \left[{\tan}^{-} 1 \infty - {\tan}^{-} 1 \left(1\right)\right] = 2 \left[\frac{\pi}{2} - \frac{\pi}{4}\right] = \frac{\pi}{2}$