let-f-x-0-1-ln-1-xe-t-dt-1-find-a-explicit-form-of-f-x-2-calculate-0-1-ln-1-2e-t-dt-3-developp-f-x-at-integr-serie-if-x-lt-1-

Question Number 59275 by Mr X pcx last updated on 07/May/19

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

1) f i n d a e x p l i c i t f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{1} l n (1 + 2 e^{- t}) d t

3) d e v e l o p p f (x) a t i n t e g r s e r i e i f ∣ x ∣< 1

Commented by maxmathsup by imad last updated on 17/May/19

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

f^{^{'}} (x) =_{e^{t} = u} \int_{1}^{e} \frac{1}{u + x} \frac{d u}{u} = \frac{1}{x} \int_{1}^{e} (\frac{1}{u} - \frac{1}{u + x}) d u (f o r x \neq 0)

= \frac{1}{x} {[l n ∣ \frac{u}{u + x} ∣]}_{1}^{e} = \frac{1}{x} {l n (\frac{e}{e + x}) - l n (\frac{1}{1 + x})}

= \frac{1}{x} {1 - l n (e + x) + l n (1 + x)} = \frac{1}{x} - \frac{l n (x + e)}{x} + \frac{l n (1 + x)}{x} \Rightarrow

f (x) = l n ∣ x ∣ - \int \frac{l n (x + e)}{x} d x + \int \frac{l n (1 + x)}{x} d x + C \dots . b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 17/May/19

l e t t a k e a n o t h e r e s s a y b y p a r t s u^{^{'}} = 1 a n d v = l n (1 + {x e}^{- t})

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

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

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

= \frac{1}{x} \int_{1}^{e} l n u {\frac{1}{u} - \frac{1}{u + x}} d x = \frac{1}{x} \int_{1}^{e} \frac{l n (u)}{u} d u - \frac{1}{x} \int_{1}^{e} \frac{l n (u)}{u + x} d x

\int_{1}^{e} \frac{l n (u)}{u} d u =_{b y p a r t s} {[{l n}^{2} (u)]}_{1}^{e} - \int_{1}^{e} \frac{l n u}{u} d u \Rightarrow 2 \int_{1}^{e} \frac{l n (u)}{u} d u = 1 \Rightarrow

\int_{1}^{e} \frac{l n (u)}{u} d u = \frac{1}{2} a l s o b y p a r t s

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

i f w e s u p p o s e x > 0 \dots . b e c o n t i n u e d \dots

Commented by maxmathsup by imad last updated on 17/May/19

2) l e t I = \int_{0}^{1} l n (1 + 2 e^{- t}) d t c h a n g . 2 e^{- t} = x g i v e e^{- t} = \frac{x}{2} \Rightarrow t = - l n (\frac{x}{2})

3) w e h a v e {l n}^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow l n (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} u^{n + 1}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} u^{n} \Rightarrow l n (1 + {x e}^{- t}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {({x e}^{- t})}^{n}

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

= \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{x^{n}}{n^{2}} {e^{- n} - 1} \Rightarrow f (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} (e^{- n} - 1)}{n^{2}} x^{n} .