0-1-1-e-x-dx-

Question Number 21398 by math1967 last updated on 23/Sep/17

\underset{0}{\int^{\infty}} \frac{1}{1 + e^{x}} d x =

Answered by sma3l2996 last updated on 23/Sep/17

e^{x} = t \Rightarrow d x = \frac{d t}{t}

\int_{0}^{\infty} \frac{d x}{1 + e^{x}} = \int_{1}^{\infty} \frac{d t}{t (t + 1)} = \int_{1}^{\infty} \frac{d t}{t} - \int_{1}^{\infty} \frac{d x}{1 + t} = {[l n (\frac{t}{1 + t})]}_{1}^{\infty}

\int_{0}^{\infty} \frac{d x}{1 + e^{x}} = \underset{x \to \infty}{l i m l n} (\frac{t}{1 + t}) - l n (\frac{1}{2}) = l n (\frac{1}{\frac{1}{t} + 1}) + l n 2

\int_{0}^{\infty} \frac{d x}{1 + e^{x}} = l n (1) + l n 2 = l n (2)