prove-that-0-1-dx-x-e-x-n-0-1-n-n-1-n-1-A-n-with-A-n-0-n-1-t-n-e-t-dt-

Question Number 26756 by abdo imad last updated on 28/Dec/17

p r o v e t h a t \int_{0}^{1} \frac{d x}{x + e^{x}} = \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n}}{{(n + 1)}^{n + 1}} A_{n}

w i t h A_{n} = \int_{0}^{n + 1} t^{n} e^{- t} d t .

Commented by abdo imad last updated on 01/Jan/18

l e t p u t I = \int_{0}^{1} \frac{d x}{x + e^{x}}

I = \int_{0}^{1} \frac{e^{- x}}{1 + x e^{- x}} d x b u t / {x e}^{- x} / ⩽ 1

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

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

a n d b y t h e c h s n g e m e n t (n + 1) x = t

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

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

a n d f i n a l l y ? I = \sum_{n = 0}^{n =\propto} \frac{{(- 1)}^{n}}{{(n + 1)}^{n + 1}} A_{n}

w i t h A_{n} = \int_{0}^{1} t^{n} e^{- t} d t .

Commented by abdo imad last updated on 01/Jan/18

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

= \frac{A_{n}}{{(n + 1)}^{n + 1}} w i t h A_{n} = \int_{0}^{n + 1} t^{n} e^{- t} d t .