# How to integrate ? int_0^oo 1/(1+e^x)*dx

May 6, 2017

$\ln \left(2\right)$

#### Explanation:

First working with the indefinite integral, divide the numerator and denominator by ${e}^{x}$:

$\int \frac{1}{1 + {e}^{x}} \mathrm{dx} = \int {e}^{-} \frac{x}{{e}^{-} x + 1} \mathrm{dx}$

Let $u = {e}^{-} x + 1$, implying that $\mathrm{du} = - {e}^{-} x \mathrm{dx}$:

$= - \int \frac{- {e}^{-} x}{{e}^{-} x + 1} \mathrm{dx} = - \int \frac{1}{u} \mathrm{du} = - \ln \left\mid u \right\mid$

$= - \ln \left({e}^{-} x + 1\right) + C$

So:

${\int}_{0}^{\infty} \frac{1}{1 + {e}^{x}} \mathrm{dx}$

$= \left({\lim}_{x \rightarrow \infty} \left(- \ln \left({e}^{-} x + 1\right)\right)\right) - \left(- \ln \left({e}^{0} + 1\right)\right)$

Note that ${\lim}_{x \rightarrow \infty} {e}^{-} x = 0$:

$= - \ln \left(1\right) + \ln \left(1 + 1\right)$

$= \ln \left(2\right)$