calculate ∫_0 ^1 (1/x)ln(((1+x)/(1−x)))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	$ln (((1+x)/(1−x)))=ln ((((1+x)(1+x))/((1−x)(1+x))))=ln ((((1+x)^2 )/(1−x^2 ))) =ln (((1+x)/(√(1−x^2 ))))^2 =2ln ((1+x)/(√(1−x^2 ))) ∫(1/x)ln (((1+x)/(1−x)))dx=2∫((1+x)/(x(√(1−x^2 ))))dx→ { ((x=cos 2ψ⇒dx=−2sin2 ψdψ)),((x→0,ψ→(π/4)∧x→1,ψ→0)) :} −2∫(((1+cos 2ψ)sin 2ψ)/(cos 2ψ(√(1−cos^2 2ψ))))dψ=−2∫((2cos^2 ψsin 2ψ)/(cos 2ψsin 2ψ))dψ =−4∫((cos^2 ψ)/(cos 2ψ))dψ=−4∫((1−sin^2 ψ)/(cos 2ψ))dψ =−4∫sec 2ψdψ+4∫((sin^2 ψ)/(cos 2ψ))dψ =−4∫sec 2ψdψ+4∫(((1−cos 2ψ)/2)/(cos 2ψ))dψ =−4∫sec 2ψdψ+2∫sec 2ψdψ−2∫dψ =−2∫sec 2ψdψ−2∫dψ=−2∫(dψ/(cos 2ψ))−2∫dψ =−2∫(dψ/(cos 2ψ))−2∫dψ=−ln ∣sec 2ψ+tan 2ψ∣−2ψ+C ∫_0 ^1 (1/x)ln(((1+x)/(1−x)))dx=[−ln ∣sec 2ψ+tan 2ψ∣−2ψ]_(π/4) ^0 =−∞$
	$\ln (\frac{1 + x}{1 - x}) = \ln (\frac{(1 + x) (1 + x)}{(1 - x) (1 + x)}) = \ln (\frac{{(1 + x)}^{2}}{1 - x^{2}})$ $= \ln {(\frac{1 + x}{\sqrt{1 - x^{2}}})}^{2} = 2 \ln \frac{1 + x}{\sqrt{1 - x^{2}}}$ $\int \frac{1}{x} \ln (\frac{1 + x}{1 - x}) d x = 2 \int \frac{1 + x}{x \sqrt{1 - x^{2}}} d x \to {\begin{cases} x = \cos 2 ψ \Rightarrow d x = - 2 \sin 2 ψ d ψ \\ x \to 0, ψ \to \frac{π}{4} \land x \to 1, ψ \to 0 \end{cases}$ $- 2 \int \frac{(1 + \cos 2 ψ) \sin 2 ψ}{\cos 2 ψ \sqrt{1 - \cos^{2} 2 ψ}} d ψ = - 2 \int \frac{2 \cos^{2} ψ \sin 2 ψ}{\cos 2 ψ \sin 2 ψ} d ψ$ $= - 4 \int \frac{\cos^{2} ψ}{\cos 2 ψ} d ψ = - 4 \int \frac{1 - \sin^{2} ψ}{\cos 2 ψ} d ψ$ $= - 4 \int \sec 2 ψ d ψ + 4 \int \frac{\sin^{2} ψ}{\cos 2 ψ} d ψ$ $= - 4 \int \sec 2 ψ d ψ + 4 \int \frac{\frac{1 - \cos 2 ψ}{2}}{\cos 2 ψ} d ψ$ $= - 4 \int \sec 2 ψ d ψ + 2 \int \sec 2 ψ d ψ - 2 \int d ψ$ $= - 2 \int \sec 2 ψ d ψ - 2 \int d ψ = - 2 \int \frac{d ψ}{\cos 2 ψ} - 2 \int d ψ$ $= - 2 \int \frac{d ψ}{\cos 2 ψ} - 2 \int d ψ = - \ln ∣ \sec 2 ψ + \tan 2 ψ ∣ - 2 ψ + C$ $\int_{0}^{1} \frac{1}{x} l n (\frac{1 + x}{1 - x}) d x = {[- \ln ∣ \sec 2 ψ + \tan 2 ψ ∣ - 2 ψ]}_{\frac{π}{4}}^{0} = - \infty$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum