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Question Number 7157 by Tawakalitu. last updated on 13/Aug/16

I f a a n d b a r e p o s i t i v e n u m b e r s

w h a t i s t h e v a l u e o f

\int_{0}^{\infty} \frac{e^{a x} - e^{b x}}{(1 + e^{a x}) (1 + e^{b x})} d x

Answered by Yozzia last updated on 14/Aug/16

\frac{1}{1 + e^{b x}} - \frac{1}{1 + e^{a x}} = \frac{e^{a x} - e^{b x}}{(1 + e^{a x}) (1 + e^{b x})} (a, b > 0)

I = \int_{0}^{\infty} \frac{e^{a x} - e^{b x}}{(1 + e^{a x}) (1 + e^{b x})} d x = \int_{0}^{\infty} [\frac{1}{1 + e^{b x}} - \frac{1}{1 + e^{a x}}] d x

I = \lim_{m \to \infty} [\int_{0}^{m} \frac{1}{1 + e^{b x}} d x - \int_{0}^{m} \frac{1}{1 + e^{a x}} d x]

u = e^{b x} \Rightarrow d u = {b e}^{b x} d x \Rightarrow d x = \frac{1}{b u} d u

\int \frac{1}{b u (1 + u)} d u = \frac{1}{b} \int (\frac{1}{u} - \frac{1}{1 + u}) d u

= \frac{1}{b} l n \frac{u}{1 + u} = \frac{1}{b} l n \frac{1}{1 + u^{- 1}} = \frac{1}{b} l n \frac{1}{1 + e^{- b x}}

∴ I = \lim_{m \to \infty} {[\frac{1}{b} l n (\frac{1}{1 + e^{- b x}}) - \frac{1}{a} l n (\frac{1}{1 + e^{- a x}})]}_{0}^{m}

I = \lim_{m \to \infty} [\frac{1}{b} l n \frac{1}{1 + e^{- b m}} - \frac{1}{a} l n \frac{1}{1 + e^{- a m}} - \frac{1}{b} l n \frac{1}{1 + 1} + \frac{1}{a} l n \frac{1}{1 + 1}]

S i n c e a, b > 0, I = \frac{1}{b} l n \frac{1}{1 + e^{- \infty}} - \frac{1}{a} l n \frac{1}{1 + e^{- \infty}} + (l n \frac{1}{2}) (\frac{1}{a} - \frac{1}{b})

I = \frac{1}{b} l n \frac{1}{1 + 0} - \frac{1}{a} l n \frac{1}{1 + 0} + \frac{(a - b) l n 2}{a b}

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

Commented by Tawakalitu. last updated on 14/Aug/16

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