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Question Number 65290 by mathmax by abdo last updated on 27/Jul/19

f (x) = \int_{0}^{1} \frac{d t}{x + e^{t}} w i t h 0 ⩽ x ⩽ 1

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} \frac{d t}{2 + e^{t}}

3) f i n d g (x) = \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{2}}

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

5) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s

6) d e v e l o p p f a t i n t e g r s e r i e .

Commented by mathmax by abdo last updated on 28/Jul/19

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

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

f (x) = \frac{l n (1 + x)}{x} - \frac{l n (1 + {x e}^{- 1})}{x} i f x \neq 0

2) \int_{0}^{1} \frac{d t}{2 + e^{t}} = f (2) = \frac{l n (3)}{2} - \frac{l n (1 + 2 e^{- 1})}{2}

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

f^{^{'}} (x) = - \frac{1}{x^{2}} l n (1 + x) + \frac{1}{x (1 + x)} + \frac{1}{x^{2}} l n (1 + {x e}^{- 1}) - \frac{1}{x} \frac{e^{- 1}}{1 + {x e}^{- 1}} \Rightarrow

g (x) = \frac{1}{x^{2}} l n (1 + x) - \frac{1}{x (1 + x)} - \frac{1}{x^{2}} l n (1 + {x e}^{- 1}) + \frac{e^{- 1}}{x (1 + {x e}^{- 1})}

4) \int_{0}^{1} \frac{d t}{{(1 + e^{t})}^{2}} = g (1) = l n (2) - \frac{1}{2} - l n (1 + e^{- 1}) + \frac{e^{- 1}}{1 + e^{- 1}}

Commented by mathmax by abdo last updated on 28/Jul/19

5) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{x + e^{t}} \Rightarrow f^{(n)} (x) = \int_{0}^{1} \frac{{(- 1)}^{n} n!}{{(x + e^{t})}^{n + 1}} d t

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

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

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

= - \frac{{(- 1)}^{n} n!}{n + 1} {e^{- (n + 1)} - 1} \Rightarrow f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{n + 1} {e^{- (n + 1)} - 1}

\Rightarrow f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} {1 - e^{- (n + 1)}} .

Commented by mathmax by abdo last updated on 28/Jul/19

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

Answered by Eminem last updated on 28/Jul/19

f (x) = \int_{0}^{1} \frac{e^{- t}}{{x e}^{- t} + 1} d t = - \frac{1}{x} \int_{0}^{1} \frac{- {x e}^{- t}}{1 + {x e}^{- t}} d t \forall x \in {I R}^{*}

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

2 ⟩ f (2) = \frac{1}{2} l n (\frac{3}{1 + 2 e^{- 1}})

3) f i n d g (x} = \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{2}}

\forall x \in {I R}^{*} f (x) = \frac{1}{x} l n (\frac{1 + x}{1 + {x e}^{- 1}})

f^{^{'}} (x) = \int_{0}^{1} \frac{d}{d x} (\frac{1}{x + e^{t}}) d t = \int_{0}^{1} \frac{- 1}{{(x + e^{t})}^{2}} d t = - g (x) ==>

g (x) = - f^{^{'}} (x) = - \frac{d}{d x} {\frac{1}{x} l n (\frac{1 + x}{1 + {x e}^{- 1}})}

= - {\frac{- 1}{x^{2}} l n (\frac{1 + x}{1 + {x e}^{- 1}}) + \frac{1}{x} \times (\frac{1 + {x e}^{- 1} - e^{- 1} (1 + x)}{\frac{{(1 + {x e}^{- 1})}^{2}}{\frac{(1 + x)}{(1 + {x e}^{- 1})}}})}

= \frac{1}{x^{2}} l n (\frac{1 + x}{1 + {x e}^{- 1}}) - \frac{1}{x} \times \frac{(1 - e^{- 1})}{(1 + x) (1 + {x e}^{- 1})} = g (x)

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

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

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

= \frac{1}{x} {\sum_{k = 0} ({(- 1)}^{k} \frac{x^{k + 1}}{k + 1} - {(- 1)}^{k} \frac{{(e^{- 1} x)}^{k + 1}}{k + 1})

= \sum_{k = 0} {(- 1)}^{k} (1 - e^{- (k + 1)}) \frac{x^{k}}{k + 1}

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