Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 166246 by mnjuly1970 last updated on 16/Feb/22

	Answered by Mathspace last updated on 18/Feb/22
	$I=_(by parts) [(1−(1/x))ln^2 (1−x^2 )]_0 ^1 −∫_0 ^1 (1−(1/x)).2ln(1−x^2 )×((−2x)/(1−x^2 ))dx =0−4∫_0 ^1 ((x−1)/x)×((xln(1−x^2 ))/((1−x)(1+x)))dx =4∫_0 ^1 ((ln(1−x^2 ))/(1+x))dx =4∫_0 ^1 ((ln(1−x)+ln(1+x))/(1+x))dx =4∫_0 ^1 ((ln(1−x))/(1+x)) +4∫_0 ^1 ((ln(1+x))/(1+x))dx we have ∫_0 ^1 ((ln(1+x))/(1+x))dx=_(1+x=t) ∫_1 ^2 ((ln(t))/t)dt =[(1/2)ln^2 (t)]_1 ^2 =(1/2)ln^2 (2) ∫_0 ^(1 ) ((ln(1−x))/(1+x))dx= ∫_0 ^1 ln(1−x)Σ_(n=0) ^∞ (−1)^n x^n dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n ln(1−x)dx =Σ_(n=0) ^∞ (−1)^n U_n U_n =[(x^(n+1) /(n+1))−(1/(n+1)))ln(1−x)]_0 ^1 −∫_0 ^1 (((x^(n+1) −1)/(n+1)))((−1)/(1−x))dx =0−(1/(n+1))∫_0 ^1 ((x^(n+1) −1)/(x−1))dx =−(1/(n+1))∫_0 ^1 (1+x+x^2 +...+x^n )dx =−(1/(n+1))[x+(x^2 /2)+....+(x^(n+1) /(n+1))]_0 ^1 =−(1/(n+1)){1+(1/2)+...+(1/(n+1))} =−(H_(n+1) /(n+1)) ⇒ ∫_0 ^1 ((ln(1−x))/(1+x))dx=−Σ_(n=0) ^∞ (−1)^n (H_(n+1) /(n+1)) =Σ_(n=1) ^∞ (−1)^n (H_n /n) ⇒ ∫_0 ^1 ((ln^2 (1−x^2 ))/x^2 )dx=2ln^2 (2)+4Σ_(n=1) ^∞ (−1)^n (H_n /n) rest to find the value of this serie...$
	$I =_{b y p a r t s} {[(1 - \frac{1}{x}) {l n}^{2} (1 - x^{2})]}_{0}^{1}$ $- \int_{0}^{1} (1 - \frac{1}{x}) . 2 l n (1 - x^{2}) \times \frac{- 2 x}{1 - x^{2}} d x$ $= 0 - 4 \int_{0}^{1} \frac{x - 1}{x} \times \frac{x l n (1 - x^{2})}{(1 - x) (1 + x)} d x$ $= 4 \int_{0}^{1} \frac{l n (1 - x^{2})}{1 + x} d x$ $= 4 \int_{0}^{1} \frac{l n (1 - x) + l n (1 + x)}{1 + x} d x$ $= 4 \int_{0}^{1} \frac{l n (1 - x)}{1 + x} + 4 \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x$ $w e h a v e \int_{0}^{1} \frac{l n (1 + x)}{1 + x} d x =_{1 + x = t} \int_{1}^{2} \frac{l n (t)}{t} d t$ $= {[\frac{1}{2} {l n}^{2} (t)]}_{1}^{2} = \frac{1}{2} {l n}^{2} (2)$ $\int_{0}^{1} \frac{l n (1 - x)}{1 + x} d x =$ $\int_{0}^{1} l n (1 - x) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n} d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{n} l n (1 - x) d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} U_{n}$ ${U_{n} = [\frac{x^{n + 1}}{n + 1} - \frac{1}{n + 1}) l n (1 - x)]}_{0}^{1} - \int_{0}^{1} (\frac{x^{n + 1} - 1}{n + 1}) \frac{- 1}{1 - x} d x$ $= 0 - \frac{1}{n + 1} \int_{0}^{1} \frac{x^{n + 1} - 1}{x - 1} d x$ $= - \frac{1}{n + 1} \int_{0}^{1} (1 + x + x^{2} + . . . + x^{n}) d x$ $= - \frac{1}{n + 1} {[x + \frac{x^{2}}{2} + . . . . + \frac{x^{n + 1}}{n + 1}]}_{0}^{1}$ $= - \frac{1}{n + 1} {1 + \frac{1}{2} + . . . + \frac{1}{n + 1}}$ $= - \frac{H_{n + 1}}{n + 1} \Rightarrow$ $\int_{0}^{1} \frac{l n (1 - x)}{1 + x} d x = - \sum_{n = 0}^{\infty} {(- 1)}^{n} \frac{H_{n + 1}}{n + 1}$ $= \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{H_{n}}{n} \Rightarrow$ $\int_{0}^{1} \frac{{l n}^{2} (1 - x^{2})}{x^{2}} d x = 2 {l n}^{2} (2) + 4 \sum_{n = 1}^{\infty} {(- 1)}^{n} \frac{H_{n}}{n}$ $r e s t t o f i n d t h e v a l u e o f t h i s s e r i e . . .$
	Commented by mnjuly1970 last updated on 19/Feb/22

	$t h a n k s a l o t$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum