find ∫_0 ^1 ((ln(1−x^2 )ln(x))/x^2 )dx prove first the convergence.


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Question Number 78270 by msup trace by abdo last updated on 15/Jan/20

$f i n d \int_{0}^{1} \frac{l n (1 - x^{2}) l n (x)}{x^{2}} d x$ $p r o v e f i r s t t h e c o n v e r g e n c e .$
Commented by mathmax by abdo last updated on 15/Jan/20
$let I =∫_0 ^1 ((ln(1−x^2 )ln(x))/x^2 )dx the convergence is assured by sir <mind is power> we have ln^′ (1−u)=−(1/(1−u))=−Σ_(n=0) ^∞ u^n for ∣u∣<1 ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) +c (c=0)=−Σ_(n=1) ^∞ (u^n /n) ⇒ ln(1−x^2 ) =−Σ_(n=1) ^∞ (x^(2n) /n) ⇒ ((ln(1−x^2 )ln(x))/x^2 )=−Σ_(n=1) ^∞ (x^(2n−2) /n)ln(x) ⇒I =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 x^(2n−2) ln(x)dx by parts ∫_0 ^1 x^(2n−2) ln(x)dx =[(1/(2n−1))x^(2n−1) ln(x)]_0 ^1 −∫_0 ^1 (1/(2n−1))x^(2n−1) (dx/x) =−(1/(2n−1))∫_0 ^1 x^(2n−2) dx =−(1/((2n−1)^2 ))[x^(2n−1) ]_0 ^1 =((−1)/((2n−1)^2 )) ⇒ I =Σ_(n=1) ^∞ (1/(n(2n−1)^2 )) let decompose F(x)=(1/(x(2x−1)^2 )) F(x)=(a/x) +(b/(2x−1)) +(c/((2x−1)^2 )) a=xF(x)∣_(x=0) =1 c =(2x−1)^2 F(x)∣_(x=(1/2)) =2 ⇒F(x)=(1/x) +(b/(2x−1)) +(2/((2x−1)^2 )) F(1)=1 =1 +b +2 ⇒b=−2 ⇒F(x)=(1/x)−(2/(2x−1)) +(2/((2x−1)^2 )) ⇒ I =Σ_(n=1) ^∞ F(n) =Σ_(n=1) ^∞ ((1/n)−(2/(2n−1))) +2Σ_(n=1) ^∞ (1/((2n−1)^2 )) Σ_(n=1) ^∞ ((1/n)−(2/(2n−1))) =lim s_n with s_n =Σ_(k=1) ^n ((1/k)−(2/(2k−1))) =H_n −2Σ_(k=1) ^n (1/(2k−1)) but Σ_(k=1) ^n (1/(2k−1)) =1 +(1/3) +(1/5) +...+(1/(2n−1)) =1+(1/2) +(1/3) +(1/4) +... +(1/(2n−1)) +(1/(2n)) −(1/2)−(1/4)−...−(1/(2n)) =H_(2n) −(1/2) H_n ⇒ s_n =H_n −2{H_(2n) −(1/2) H_n } =2H_n −2H_(2n) =2( ln(n)+γ +o((1/n))−ln(2n)−γ+o((1/n))) =2( ln((n/(2n))) +o((1/n))) ⇒lim s_n =−2ln(2) let findA= Σ_(n=1) ^∞ (1/((2n−1)^2 )) A =Σ_(n=0) ^∞ (1/((2n+1)^2 )) we have (π^2 /6) =Σ_(n=1) ^∞ (1/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1−(1/4))(π^2 /6) =(3/4)×(π^2 /6) =(π^2 /8) ⇒ I =−2ln(2)+2×(π^2 /8) =(π^2 /4)−2ln(2)$
$l e t I = \int_{0}^{1} \frac{l n (1 - x^{2}) l n (x)}{x^{2}} d x t h e c o n v e r g e n c e i s a s s u r e d b y s i r$ $< m i n d i s p o w e r > w e h a v e {l n}^{^{'}} (1 - u) = - \frac{1}{1 - u} = - \sum_{n = 0}^{\infty} u^{n} f o r$ $∣ u ∣< 1 \Rightarrow l n (1 - u) = - \sum_{n = 0}^{\infty} \frac{u^{n + 1}}{n + 1} + c (c = 0) = - \sum_{n = 1}^{\infty} \frac{u^{n}}{n} \Rightarrow$ $l n (1 - x^{2}) = - \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n} \Rightarrow \frac{l n (1 - x^{2}) l n (x)}{x^{2}} = - \sum_{n = 1}^{\infty} \frac{x^{2 n - 2}}{n} l n (x)$ $\Rightarrow I = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{2 n - 2} l n (x) d x b y p a r t s$ $\int_{0}^{1} x^{2 n - 2} l n (x) d x = {[\frac{1}{2 n - 1} x^{2 n - 1} l n (x)]}_{0}^{1} - \int_{0}^{1} \frac{1}{2 n - 1} x^{2 n - 1} \frac{d x}{x}$ $= - \frac{1}{2 n - 1} \int_{0}^{1} x^{2 n - 2} d x = - \frac{1}{{(2 n - 1)}^{2}} {[x^{2 n - 1}]}_{0}^{1} = \frac{- 1}{{(2 n - 1)}^{2}} \Rightarrow$ $I = \sum_{n = 1}^{\infty} \frac{1}{n {(2 n - 1)}^{2}} l e t d e c o m p o s e F (x) = \frac{1}{x {(2 x - 1)}^{2}}$ $F (x) = \frac{a}{x} + \frac{b}{2 x - 1} + \frac{c}{{(2 x - 1)}^{2}}$ $a = x F (x) ∣_{x = 0} = 1$ $c = {(2 x - 1)}^{2} F (x) ∣_{x = \frac{1}{2}} = 2 \Rightarrow F (x) = \frac{1}{x} + \frac{b}{2 x - 1} + \frac{2}{{(2 x - 1)}^{2}}$ $F (1) = 1 = 1 + b + 2 \Rightarrow b = - 2 \Rightarrow F (x) = \frac{1}{x} - \frac{2}{2 x - 1} + \frac{2}{{(2 x - 1)}^{2}} \Rightarrow$ $I = \sum_{n = 1}^{\infty} F (n) = \sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{2}{2 n - 1}) + 2 \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}}$ $\sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{2}{2 n - 1}) = l i m s_{n} w i t h s_{n} = \sum_{k = 1}^{n} (\frac{1}{k} - \frac{2}{2 k - 1})$ $= H_{n} - 2 \sum_{k = 1}^{n} \frac{1}{2 k - 1} b u t$ $\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \frac{1}{5} + . . . + \frac{1}{2 n - 1}$ $= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + . . . + \frac{1}{2 n - 1} + \frac{1}{2 n} - \frac{1}{2} - \frac{1}{4} - . . . - \frac{1}{2 n}$ $= H_{2 n} - \frac{1}{2} H_{n} \Rightarrow s_{n} = H_{n} - 2 {H_{2 n} - \frac{1}{2} H_{n}}$ $= 2 H_{n} - 2 H_{2 n} = 2 (l n (n) + γ + o (\frac{1}{n}) - l n (2 n) - γ + o (\frac{1}{n}))$ $= 2 (l n (\frac{n}{2 n}) + o (\frac{1}{n})) \Rightarrow l i m s_{n} = - 2 l n (2) l e t f i n d A = \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{2}}$ $A = \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} w e h a v e \frac{π^{2}}{6} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow$ $\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = (1 - \frac{1}{4}) \frac{π^{2}}{6} = \frac{3}{4} \times \frac{π^{2}}{6} = \frac{π^{2}}{8} \Rightarrow$ $I = - 2 l n (2) + 2 \times \frac{π^{2}}{8} = \frac{π^{2}}{4} - 2 l n (2)$
	Answered by mind is power last updated on 15/Jan/20
	$cv in zero ((ln(1−x^2 ))/x^2 )∼1 ((ln(1−x^2 ))/x^2 ).ln(x)∼ln(x) log is integrabl in zero in 1 ((ln(1−x^2 ).ln(x))/x^2 )=g(x) lim_(x→1) (√(1−x )) g(x)=0⇒g(x)=o((1/(√(1−x)))) since (1/(√(1−x))) is integral in 1^− ⇒g also since g is continus ]0,1[ ⇒is integrable ∫_0 ^1 ((ln(1−x^2 )ln(x))/x^2 )dx ln(1−x^2 )=−Σ_(k≥1) (x^(2k) /k) =−∫_0 ^1 Σ_(k≥1) ((x^(2k) ln(x))/k)dx =−Σ_(k≥1) ∫_0 ^1 ((ln(x)x^(2k) )/k)dx u=−ln(x)⇒dx=−e^(−u) du =Σ_(k≥1) ∫_0 ^(+∞) (((−u).(−e^(−u) )e^(−2ku) )/k)du =Σ_(k≥1) ∫_0 ^(+∞) ((ue^(−(1+2k)u) )/k)du,w=(1+2k)u⇒du=(dw/(1+2k)) =Σ_(k≥1) ∫_0 ^(+∞) ((we^(−w) )/(k(1+2k)^2 ))dw =Σ_(k≥1) (1/(k(1+2k)^2 )).∫_0 ^(+∞) we^(−w) dw=Σ_(k≥1) (1/(k(1+2k)^2 )) =Σ((1/k)−(2/((1+2k)^2 ))−(2/((1+2k)))) =Σ_(k≥1) ((2/(2k))−(2/(1+2k)))−2Σ_(k≥1) (1/((1+2k)^2 )) =2{Σ_(k≥1) (((−1)^k )/k)+1}−2((3ζ(2))/4) =2{1−ln(2))−(3/2).(π^2 /6) =2−2ln(2)−(π^2 /4)$
	$cv$ $in zero$ $\frac{\ln (1 - x^{2})}{x^{2}} \sim 1$ $\frac{\ln (1 - x^{2})}{x^{2}} . \ln (x) \sim \ln (x) \log is integrabl in zero$ $in 1$ $\frac{\ln (1 - x^{2}) . \ln (x)}{x^{2}} = g (x)$ $\lim_{x \to 1} \sqrt{1 - x} g (x) = 0 \Rightarrow g (x) = o (\frac{1}{\sqrt{1 - x}})$ $since \frac{1}{\sqrt{1 - x}} is integral in 1^{-} \Rightarrow g also$ $since g is continus] 0, 1 [\Rightarrow is integrable$ $\int_{0}^{1} \frac{\ln (1 - x^{2}) \ln (x)}{x^{2}} dx$ $\ln (1 - x^{2}) = - \sum_{k ⩾ 1} \frac{x^{2 k}}{k}$ $= - \int_{0}^{1} \sum_{k ⩾ 1} \frac{x^{2 k} \ln (x)}{k} dx$ $= - \sum_{k ⩾ 1} \int_{0}^{1} \frac{\ln (x) x^{2 k}}{k} dx$ $u = - \ln (x) \Rightarrow dx = - e^{- u} du$ $= \sum_{k ⩾ 1} \int_{0}^{+ \infty} \frac{(- u) . (- e^{- u}) e^{- 2 ku}}{k} du$ $= \sum_{k ⩾ 1} \int_{0}^{+ \infty} \frac{{ue}^{- (1 + 2 k) u}}{k} du, w = (1 + 2 k) u \Rightarrow du = \frac{dw}{1 + 2 k}$ $= \sum_{k ⩾ 1} \int_{0}^{+ \infty} \frac{{we}^{- w}}{k {(1 + 2 k)}^{2}} dw$ $= \sum_{k ⩾ 1} \frac{1}{k {(1 + 2 k)}^{2}} . \int_{0}^{+ \infty} {we}^{- w} dw = \sum_{k ⩾ 1} \frac{1}{k {(1 + 2 k)}^{2}}$ $= Σ (\frac{1}{k} - \frac{2}{{(1 + 2 k)}^{2}} - \frac{2}{(1 + 2 k)})$ $= \sum_{k ⩾ 1} (\frac{2}{2 k} - \frac{2}{1 + 2 k}) - 2 \sum_{k ⩾ 1} \frac{1}{{(1 + 2 k)}^{2}}$ $= 2 {\sum_{k ⩾ 1} \frac{{(- 1)}^{k}}{k} + 1} - 2 \frac{3 ζ (2)}{4}$ $= 2 {1 - \ln (2)) - \frac{3}{2} . \frac{π^{2}}{6}$ $= 2 - 2 \ln (2) - \frac{π^{2}}{4}$
	Commented by msup trace by abdo last updated on 15/Jan/20

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 15/Jan/20

	$y^{'} re Welcom$
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