....advanced .... calculus.... prove that:: 𝛗=∫_0 ^( 1) ((ln(x).ln(1−x))/(1+x))dx=((13)/8)ζ(3)−(π^2 /4)ln(2)....


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Question Number 137501 by mnjuly1970 last updated on 03/Apr/21

$. . . . a d v a n c e d . . . . c a l c u l u s . . . .$ $p r o v e t h a t ::$ $ϕ = \int_{0}^{1} \frac{l n (x) . l n (1 - x)}{1 + x} d x = \frac{13}{8} ζ (3) - \frac{π^{2}}{4} l n (2) . . . .$
	Answered by Ñï= last updated on 04/Apr/21
	$φ=∫_0 ^1 ((lnx ln(1−x))/(1+x))dx =∫_0 ^1 ((lnxln(1−x^2 ))/(1+x))−∫_0 ^1 ((lnxln(1+x))/(1+x))dx =∫_0 ^1 ((lnxln(1−x^2 ))/(1−x^2 ))dx−∫_0 ^1 ((xln xln (1−x^2 ))/(1−x^2 ))dx−∫_0 ^1 ((ln xln (1+x))/(1+x))dx =(∂/(∂a∂b))∣_(a=0,b=−1) ∫_0 ^1 x^a (1−x^2 )^b dx−(∂/(∂a∂b))∣_(a=1,b=−1) ∫_0 ^1 x^a (1−x^2 )^b dx−{∫((ln(1+x){ln(1+x)+ln(x/(1+x))})/(1+x))}_0 ^1 =(∂/(∂a∂b))∣_(a=0,b=−1) (1/2)∫_0 ^1 u^((a−1)/2) (1−u)^b du−(∂/(∂a∂b))∣_(a=1,b=−1) (1/2)∫_0 ^1 u^((a−1)/2) (1−u)^b du−{∫((ln^2 (1+x))/(1+x))dx+∫((ln(1−(1/(1+x))))/(1+x))ln (1+x)dx}_0 ^1 =(∂/(∂a∂b))∣_(a=0,b=−1) (1/2)B(((a+1)/2),b+1)−(∂/(∂a∂b))∣_(a=1,b=−1) (1/2)B(((a+1)/2),b+1)−(1/3)ln^3 2+Σ_(n=1) ^∞ ∫_0 ^1 (((1+x)^(−n) )/n)∙((ln (1+x))/(1+x))dx =(3/2)ζ(3)−(π^2 /4)ln2−(1/3)ln^3 2+Σ_(n=1) ^∞ (1/n){∫_0 ^1 (1+x)^(−n−1) ln(1+x)dx} =(3/2)ζ(3)−(π^2 /4)ln2−(1/3)ln^3 2+Σ_(n=1) ^∞ (1/n){(((1+x)^(−n) )/(−n))ln(1+x)−∫(((1+x)^(−n−1) )/(−n))dx}_0 ^1 =(3/2)ζ(3)−(π^2 /4)ln2−(1/3)ln^3 2−Σ_(n=1) ^∞ (((1+x)^(−n) ln(1+x))/n^2 )∣_0 ^1 +Σ_(n=1) ^∞ (((1+x)^(−n) )/n^3 )∣_0 ^1 =(3/2)ζ(3)−(π^2 /4)ln2−(1/3)ln^3 2−Li_2 ((1/(1+x)))ln(1+x)∣_0 ^1 −Li_3 ((1/(1+x)))∣_0 ^1 =(3/2)ζ(3)−(π^2 /4)ln2−(1/3)ln^3 2−ln2Li_2 ((1/2))−Li_3 ((1/2))+Li_3 (1) =ans...$
	$ϕ = \int_{0}^{1} \frac{l n x l n (1 - x)}{1 + x} d x$ $= \int_{0}^{1} \frac{l n x l n (1 - x^{2})}{1 + x} - \int_{0}^{1} \frac{l n x l n (1 + x)}{1 + x} d x$ $= \int_{0}^{1} \frac{l n x l n (1 - x^{2})}{1 - x^{2}} d x - \int_{0}^{1} \frac{x l n x l n (1 - x^{2})}{1 - x^{2}} d x - \int_{0}^{1} \frac{l n x l n (1 + x)}{1 + x} d x$ $= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \int_{0}^{1} x^{a} {(1 - x^{2})}^{b} d x - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \int_{0}^{1} x^{a} {(1 - x^{2})}^{b} d x - {\int \frac{l n (1 + x) {l n (1 + x) + l n \frac{x}{1 + x}}}{1 + x}}_{0}^{1}$ $= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \frac{1}{2} \int_{0}^{1} u^{(a - 1) / 2} {(1 - u)}^{b} d u - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \frac{1}{2} \int_{0}^{1} u^{(a - 1) / 2} {(1 - u)}^{b} d u - {\int \frac{{l n}^{2} (1 + x)}{1 + x} d x + \int \frac{l n (1 - \frac{1}{1 + x})}{1 + x} l n (1 + x) d x}_{0}^{1}$ $= \frac{\partial}{\partial a \partial b} ∣_{a = 0, b = - 1} \frac{1}{2} B (\frac{a + 1}{2}, b + 1) - \frac{\partial}{\partial a \partial b} ∣_{a = 1, b = - 1} \frac{1}{2} B (\frac{a + 1}{2}, b + 1) - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \int_{0}^{1} \frac{{(1 + x)}^{- n}}{n} \cdot \frac{l n (1 + x)}{1 + x} d x$ $= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} {\int_{0}^{1} {(1 + x)}^{- n - 1} l n (1 + x) d x}$ $= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} {\frac{{(1 + x)}^{- n}}{- n} l n (1 + x) - \int \frac{{(1 + x)}^{- n - 1}}{- n} d x}_{0}^{1}$ $= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - \underset{n = 1}{\sum^{\infty}} \frac{{(1 + x)}^{- n} l n (1 + x)}{n^{2}} ∣_{0}^{1} + \underset{n = 1}{\sum^{\infty}} \frac{{(1 + x)}^{- n}}{n^{3}} ∣_{0}^{1}$ $= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - {L i}_{2} (\frac{1}{1 + x}) l n (1 + x) ∣_{0}^{1} - {L i}_{3} (\frac{1}{1 + x}) ∣_{0}^{1}$ $= \frac{3}{2} ζ (3) - \frac{π^{2}}{4} l n 2 - \frac{1}{3} {l n}^{3} 2 - l n 2 {L i}_{2} (\frac{1}{2}) - {L i}_{3} (\frac{1}{2}) + {L i}_{3} (1)$ $= a n s . . .$
	Commented by mnjuly1970 last updated on 04/Apr/21

	$g r e a t . . . t h a n k s a l o t . . .$
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