let-f-a-0-1-e-t-ln-1-e-at-dt-with-a-0-1-find-f-a-2-calculate-f-a-3-find-the-value-of-0-1-e-t-ln-1-e-3t-dt-

Question Number 36755 by prof Abdo imad last updated on 05/Jun/18

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

1) f i n d f (a)

2) c a l c u l a t e f^{^{'}} (a)

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

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

Commented by abdo.msup.com last updated on 05/Jun/18

f o r ∣ x ∣< 1 {l n}^{^{'}} (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

a n d l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1}

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

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

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

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \frac{1}{1 - n a} {e^{1 - n a} - 1}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} e^{1 - n a}}{n (1 - n a)}

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

l e t w (a) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (1 - n a)}

\frac{w (a)}{a} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n a (n a - 1)}

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

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

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n a - 1} + \frac{l n (2)}{a} = \dots . b e c o n t i n u e d \dots

Commented by abdo.msup.com last updated on 05/Jun/18

2) w e h a v e f^{^{'}} (a) = \int_{0}^{1} \frac{- {t e}^{- a t} e^{t}}{1 + e^{- a t}} d t

= - \int_{0}^{1} \frac{t e^{(1 - a) t}}{1 + e^{- a t}} d t

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

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{1} t e^{(1 - a - n a) t} d t = \dots .

Commented by abdo.msup.com last updated on 05/Jun/18

3) \int_{0}^{1} e^{t} l n (1 + e^{- 3 t}) d t

= \int_{0}^{1} e^{t} {\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- 3 n t})} d t

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} e^{(1 - 3 n) t} d t

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \frac{1}{1 - 3 n} {[e^{(1 - 3 n) t}]}_{0}^{1}

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

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

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

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

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

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

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

l e t f (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{x^{3 n - 1}}{3 n - 1}

f^{^{'}} (x) = \sum_{n = 1}^{\infty} {(- 1)}^{n} x^{3 n - 2}

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

f (x) = \int \frac{d x}{x^{2} (1 + x^{3})} + c \dots . b e c o n t i n u e d \dots

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

2) f (α) = \int_{0}^{1} e^{t} {l n (1 + e^{α t}) - {l n e}^{α t})} d t

= \int_{0}^{1} e^{t} {l n (1 + e^{α t}) - α t} d t

f^{'} (α) = \int_{0}^{1} \frac{\partial}{\partial α} e^{t} {l n (1 + e^{α t}) - α t} d t

= \int_{0}^{1} e^{t} {\frac{e^{α t} \times t}{1 + e^{α t}} - t} d t

= \int_{0}^{1} e^{t} {\frac{{t e}^{α t} - t - {t e}^{α t}}{1 + e^{α t}}} d t

= \int_{0}^{1} \frac{- {t e}^{t}}{1 + e^{α t}} d t \dots c o n t d