calculate-0-1-1-x-ln-1-x-1-x-dx-

Question Number 33984 by abdo imad last updated on 28/Apr/18

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

Commented by prof Abdo imad last updated on 29/Apr/18

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

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

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

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

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

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

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

= - (\frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}})

= \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} - \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

= \frac{π^{2}}{8} - \frac{1}{4} \frac{π^{2}}{6} = \frac{3 π^{2}}{24} - \frac{π^{2}}{24} = \frac{π^{2}}{12} . l e t c a l c u l a t e

\int_{0}^{1} \frac{1}{x} l n (1 - x) d x w e h a v e

{l n}^{^{'}} (1 - x) = \frac{- 1}{1 - x} = - \sum_{n = 0}^{\infty} x^{n} \Rightarrow

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

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

= - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n - 1} d x = - \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = - \frac{π^{2}}{6} \Rightarrow

I = \frac{π^{2}}{12} + \frac{π^{2}}{6} = \frac{3 π^{2}}{12} = \frac{π^{2}}{4}

★ I = \frac{π^{2}}{4} ★

Answered by hknkrc46 last updated on 29/Apr/18

\ln (\frac{1 + x}{1 - x}) = \ln (\frac{(1 + x) (1 + x)}{(1 - x) (1 + x)}) = \ln (\frac{{(1 + x)}^{2}}{1 - x^{2}})

= \ln {(\frac{1 + x}{\sqrt{1 - x^{2}}})}^{2} = 2 \ln \frac{1 + x}{\sqrt{1 - x^{2}}}

\int \frac{1}{x} \ln (\frac{1 + x}{1 - x}) d x = 2 \int \frac{1 + x}{x \sqrt{1 - x^{2}}} d x \to {\begin{cases} x = \cos 2 ψ \Rightarrow d x = - 2 \sin 2 ψ d ψ \\ x \to 0, ψ \to \frac{π}{4} \land x \to 1, ψ \to 0 \end{cases}

- 2 \int \frac{(1 + \cos 2 ψ) \sin 2 ψ}{\cos 2 ψ \sqrt{1 - \cos^{2} 2 ψ}} d ψ = - 2 \int \frac{2 \cos^{2} ψ \sin 2 ψ}{\cos 2 ψ \sin 2 ψ} d ψ

= - 4 \int \frac{\cos^{2} ψ}{\cos 2 ψ} d ψ = - 4 \int \frac{1 - \sin^{2} ψ}{\cos 2 ψ} d ψ

= - 4 \int \sec 2 ψ d ψ + 4 \int \frac{\sin^{2} ψ}{\cos 2 ψ} d ψ

= - 4 \int \sec 2 ψ d ψ + 4 \int \frac{\frac{1 - \cos 2 ψ}{2}}{\cos 2 ψ} d ψ

= - 4 \int \sec 2 ψ d ψ + 2 \int \sec 2 ψ d ψ - 2 \int d ψ

= - 2 \int \sec 2 ψ d ψ - 2 \int d ψ = - 2 \int \frac{d ψ}{\cos 2 ψ} - 2 \int d ψ

= - 2 \int \frac{d ψ}{\cos 2 ψ} - 2 \int d ψ = - \ln ∣ \sec 2 ψ + \tan 2 ψ ∣ - 2 ψ + C

\int_{0}^{1} \frac{1}{x} l n (\frac{1 + x}{1 - x}) d x = {[- \ln ∣ \sec 2 ψ + \tan 2 ψ ∣ - 2 ψ]}_{\frac{π}{4}}^{0} = - \infty

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