find ∫_0 ^1 ((ln^2 (x))/((1−x^2 )^2 ))dx


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Question Number 60586 by Mr X pcx last updated on 22/May/19

$f i n d \int_{0}^{1} \frac{{l n}^{2} (x)}{{(1 - x^{2})}^{2}} d x$
Commented by maxmathsup by imad last updated on 23/May/19
$let A =∫_0 ^1 ((ln^2 x)/((1−x^2 )^2 ))dx we have for ∣x∣<1 Σ_(n=0) ^∞ x^n =(1/(1−x)) and Σ_(n=1) ^∞ nx^(n−1) = (1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ n x^(2n−2) =(1/((1−x^2 )^2 )) ⇒ A =∫_0 ^1 (Σ_(n=1) ^∞ nx^(n−2) )ln^2 x dx =Σ_(n=1) ^∞ n ∫_0 ^1 x^(n−2) ln^2 x dx =Σ_(n=1) ^∞ nw_n w_n =∫_0 ^1 x^(n−2) ln^2 x dx by parts u^′ =x^(n−2) and v =ln^2 x ⇒ w_n =[(1/(n−1))x^(n−1) ln^2 x]_0 ^1 −∫_0 ^1 (1/(n−1))x^(n−1) ((2lnx)/x)dx =−(2/(n−1)) ∫_0 ^1 x^(n−2) ln(x)dx by parts again u^′ =x^(n−2) and v =lnx ⇒ ∫_0 ^1 x^(n−2) ln(x)dx =[(1/(n−1))x^(n−1) lnx]_0 ^1 −∫_0 ^1 (1/(n−1)) x^(n−1) (dx/x) =−(1/(n−1)) ∫_0 ^1 x^(n−2) dx =−(1/(n−1))[(1/(n−1)) x^(n−1) ]_0 ^1 =−(1/((n−1)^2 )) ⇒w_n =(2/((n−1)^3 )) A =∫_0 ^1 (1+Σ_(n=2) ^∞ nx^(n−2) )ln^2 x dx =∫_0 ^1 ln^2 x dx +Σ_(n=2) ^∞ n ∫_0 ^1 x^(n−2) ln^2 xdx =∫_0 ^1 ln^2 x dx +Σ_(n=2) ^∞ n(2/((n−1)^3 )) =∫_0 ^1 ln^2 x dx + 2 Σ_(n=2) ^∞ (n/((n−1)^3 )) Σ_(n=2) ^∞ (n/((n−1)^3 )) =Σ_(n=1) ^∞ ((n+1)/n^3 ) =Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=1) ^∞ (1/n^3 ) =(π^2 /6) +ξ(3) ∫_0 ^1 ln^2 x dx =_(ln(x)=−t) ∫_(+∞) ^0 t^2 (−e^(−t) )dt =∫_0 ^∞ t^2 e^(−t) dt =[−t^2 e^(−t) ]_0 ^(+∞) −∫_0 ^(+∞) 2t (−e^(−t) )dt = 2 ∫_0 ^∞ t e^(−t) dt =2{ [−t e^(−t) ]_0 ^∞ −∫_0 ^∞ (−e^(−t) )dt} =2 ∫_0 ^∞ e^(−t) dt =2[−e^(−t) ]_0 ^(+∞) =2 ⇒ A = 2 +(π^2 /3) +2ξ(3)$
$l e t A = \int_{0}^{1} \frac{{l n}^{2} x}{{(1 - x^{2})}^{2}} d x w e h a v e f o r ∣ x ∣< 1 \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} a n d$ $\sum_{n = 1}^{\infty} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} n x^{2 n - 2} = \frac{1}{{(1 - x^{2})}^{2}} \Rightarrow$ $A = \int_{0}^{1} (\sum_{n = 1}^{\infty} {n x}^{n - 2}) {l n}^{2} x d x = \sum_{n = 1}^{\infty} n \int_{0}^{1} x^{n - 2} {l n}^{2} x d x = \sum_{n = 1}^{\infty} {n w}_{n}$ $w_{n} = \int_{0}^{1} x^{n - 2} {l n}^{2} x d x b y p a r t s u^{^{'}} = x^{n - 2} a n d v = {l n}^{2} x \Rightarrow$ $w_{n} = {[\frac{1}{n - 1} x^{n - 1} {l n}^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{1}{n - 1} x^{n - 1} \frac{2 l n x}{x} d x$ $= - \frac{2}{n - 1} \int_{0}^{1} x^{n - 2} l n (x) d x b y p a r t s a g a i n u^{^{'}} = x^{n - 2} a n d v = l n x \Rightarrow$ $\int_{0}^{1} x^{n - 2} l n (x) d x = {[\frac{1}{n - 1} x^{n - 1} l n x]}_{0}^{1} - \int_{0}^{1} \frac{1}{n - 1} x^{n - 1} \frac{d x}{x}$ $= - \frac{1}{n - 1} \int_{0}^{1} x^{n - 2} d x = - \frac{1}{n - 1} {[\frac{1}{n - 1} x^{n - 1}]}_{0}^{1} = - \frac{1}{{(n - 1)}^{2}} \Rightarrow w_{n} = \frac{2}{{(n - 1)}^{3}}$ $A = \int_{0}^{1} (1 + \sum_{n = 2}^{\infty} {n x}^{n - 2}) {l n}^{2} x d x = \int_{0}^{1} {l n}^{2} x d x + \sum_{n = 2}^{\infty} n \int_{0}^{1} x^{n - 2} {l n}^{2} x d x$ $= \int_{0}^{1} {l n}^{2} x d x + \sum_{n = 2}^{\infty} n \frac{2}{{(n - 1)}^{3}} = \int_{0}^{1} {l n}^{2} x d x + 2 \sum_{n = 2}^{\infty} \frac{n}{{(n - 1)}^{3}}$ $\sum_{n = 2}^{\infty} \frac{n}{{(n - 1)}^{3}} = \sum_{n = 1}^{\infty} \frac{n + 1}{n^{3}} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 1}^{\infty} \frac{1}{n^{3}} = \frac{π^{2}}{6} + ξ (3)$ $\int_{0}^{1} {l n}^{2} x d x =_{l n (x) = - t} \int_{+ \infty}^{0} t^{2} (- e^{- t}) d t = \int_{0}^{\infty} t^{2} e^{- t} d t$ $= {[- t^{2} e^{- t}]}_{0}^{+ \infty} - \int_{0}^{+ \infty} 2 t (- e^{- t}) d t = 2 \int_{0}^{\infty} t e^{- t} d t$ $= 2 {{[- t e^{- t}]}_{0}^{\infty} - \int_{0}^{\infty} (- e^{- t}) d t} = 2 \int_{0}^{\infty} e^{- t} d t = 2 {[- e^{- t}]}_{0}^{+ \infty} = 2 \Rightarrow$ $A = 2 + \frac{π^{2}}{3} + 2 ξ (3)$
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