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Question Number 41301 by math khazana by abdo last updated on 05/Aug/18

l e t f (x) = \int_{0}^{\infty} e^{- a x} l n (1 + e^{- b x}) d x w i t h a > 0 a n d

b > 0

1) c a l c u l a t e \frac{\partial f}{\partial a} (x)

2) c a l c u l a t e \frac{\partial f}{\partial b} (x)

3) f i n d t h e v a l u e o f \int_{0}^{\infty} e^{- 2 x} l n (1 + e^{- x}) d x a n d

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

Commented by prof Abdo imad last updated on 05/Aug/18

1) w e h a v e \frac{\partial f}{\partial a} (x) = - \int_{0}^{\infty} {x e}^{- a x} l n (1 + e^{- b x}) d x

b u t f o r ∣ u ∣< 1 l n (1 + u) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow

\frac{\partial f}{\partial a} (x) = - \int_{0}^{\infty} x^{} e^{- a x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- n b x}) d x

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{\infty} {x e}^{- (a + n b) x} d x b y p a r t s

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

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

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

= \frac{1}{{(a + n b)}^{2}} \Rightarrow \frac{\partial f}{\partial a} (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n {(a + n b)}^{2}} \dots

Commented by prof Abdo imad last updated on 05/Aug/18

2) w e h a v e \frac{\partial f}{\partial b} (x) = - \int_{0}^{\infty} e^{- a x} \frac{x}{1 + e^{- b x}} d x

= - \int_{0}^{\infty} x e^{- a x} (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- n b x}) d x

= \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} \int_{0}^{\infty} x e^{- (a + n b) x} d x b u t w e h a v e

p r o v e d t h a t \int_{0}^{\infty} x e^{- (a + n b) x} d x = \frac{1}{{(a + n b)}^{2}} \Rightarrow

\frac{\partial f}{\partial b} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(a + n b)}^{2}} \dots

Commented by prof Abdo imad last updated on 05/Aug/18

3) l e t I = \int_{0}^{\infty} e^{- 2 x} l n (1 + e^{- x)} d x

I = \int_{0}^{\infty} e^{- 2 x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- n x}) d x

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

\int_{0}^{\infty} e^{- (n + 2) x} d x = {[\frac{- 1}{n + 2} e^{- (n + 2) x}]}_{0}^{+ \infty} = \frac{1}{n + 2} \Rightarrow

I = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (n + 2)} = \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{n} - \frac{1}{n + 2}}

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n + 2} b u t

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2)

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n + 2} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 3}}{n}

= \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2) - {1 - \frac{1}{2}} = l n (2) - \frac{1}{2}

\Rightarrow I = \frac{1}{2} l n (2) - \frac{1}{2} {l n (2) - \frac{1}{2}} \Rightarrow I = \frac{1}{4}

Commented by prof Abdo imad last updated on 05/Aug/18

l e t J = \int_{0}^{\infty} e^{- x} l n (1 + e^{- 2 x}) d x w e h a v e

J = \int_{0}^{\infty} e^{- x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- 2 n x}) d x

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

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {[\frac{- 1}{2 n + 1} e^{- (2 n + 1) x}]}_{0}^{+ \infty}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (2 n + 1)} \Rightarrow

\frac{1}{2} J = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{2 n} - \frac{1}{2 n + 1}}

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}

= \frac{l n (2)}{2} + \frac{π}{4} - 1 \Rightarrow J = l n (2) + \frac{π}{2} - 2 .

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

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

a f t e r i n t r e g a t i o n t h e r e s u l t s w i l l b e i n t e r m s o f a a n d b

s o w e c a n w r i t e

d f = {(\frac{\partial f}{\partial a})}_{b} d a + {(\frac{\partial f}{\partial b})}_{a} d b

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

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

\frac{d f}{d a} = - a f

\frac{d f}{f} = - a d a

l n f = - \frac{a^{2}}{2} + c_{1}

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

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

\frac{d f}{d b} = \int_{0}^{\infty} e^{- a x} . \frac{- {b e}^{- b x}}{1 + e^{- b x}} d x

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

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

l e t t = 1 + e^{b x} d t = e^{b x} . b . d x

d x = \frac{d t}{b (t - 1)}

e^{b x} = t - 1 e^{x} = {(t - 1)}^{\frac{1}{b}} s o e^{- a x} = {(t - 1)}^{\frac{- a}{b}}

\frac{d f}{d b} = (- b) \int_{2}^{\infty} \frac{{(t - 1)}^{\frac{- a}{b}}}{t} \times \frac{d t}{b (t - 1)}

\frac{d f}{d b} = (- 1) \int_{2}^{\infty} \frac{{(t - 1)}^{\frac{- a}{b} - 1}}{t}

c o n t d \dots