find-J-x-0-dt-x-e-t-

Question Number 29971 by abdo imad last updated on 14/Feb/18

f i n d J (x) = \int_{0}^{\infty} \frac{d t}{x + e^{t}} ? .

Commented by abdo imad last updated on 16/Feb/18

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

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

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

J (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 1} \Rightarrow x J (x) = \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} l e t d e r i v a t e

J (x) + {x J}^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} s o J i s s o l u t i o n o f t h e

d . e y + {x y}^{^{'}} = \frac{1}{1 - x} e h \Rightarrow {x y}^{^{'}} = - y \Rightarrow \frac{y^{^{'}}}{y} = \frac{- 1}{x} \Rightarrow

l n ∣ y ∣= - l n ∣ x ∣ + c \Rightarrow y = \frac{λ}{x} m v c m e t h o d g i v e

y^{^{'}} = \frac{λ^{^{'}} x - λ}{x^{2}} \Rightarrow \frac{λ}{x} + λ^{^{'}} - \frac{λ}{x} = \frac{1}{1 - x} \Rightarrow λ (x) = \int \frac{d x}{1 - x} + κ

= - l n ∣ 1 - x ∣ + k b u t k = λ (0) = 0

y (x) = - \frac{1}{x} l n ∣ 1 - x ∣ \Rightarrow J (x) = - \frac{1}{x} l n ∣ 1 - x ∣ .

Commented by abdo imad last updated on 16/Feb/18

t h e Q . i s f i n d J (x) = \int_{0}^{\infty} \frac{d t}{x + e^{t}} w i t h ∣ x ∣< 1 .