.....#advanced ............... calculus#..... prove that ::: 𝛗=∫_0 ^( 1) ((ln^2 (1−x))/x)dx=^? 2ζ(3) =^(1−x=t) ∫_0 ^( 1) ((ln^2 (t))/(1−t))dt=∫_0 ^( 1) Σ_(n=0) ^∞ ln^2 (t).t^n dt =Σ_(n=0) ^∞ {[(t^(n+1) /(n+1))ln^2 (t)]_0 ^1 −(2/(n+1))∫_0 ^( 1) t^n ln(t) =−2Σ_(n=0) ^∞ (1/(n+1)){[(t^(n+1) /(n+1))ln(t)]_0 ^1 −(1/(n+1))∫_0 ^( 1) t^n dt} =2Σ_(n=0) ^∞ (1/((n+1)^3 ))=2Σ_(n=1) ^∞ (1/n^3 )=2ζ(3) .................... 𝛗=2ζ(3) .................... ..........m.n.july.1970.........


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Question Number 133857 by mnjuly1970 last updated on 24/Feb/21
$.....#advanced ............... calculus#..... prove that ::: 𝛗=∫_0 ^( 1) ((ln^2 (1−x))/x)dx=^? 2ζ(3) =^(1−x=t) ∫_0 ^( 1) ((ln^2 (t))/(1−t))dt=∫_0 ^( 1) Σ_(n=0) ^∞ ln^2 (t).t^n dt =Σ_(n=0) ^∞ {[(t^(n+1) /(n+1))ln^2 (t)]_0 ^1 −(2/(n+1))∫_0 ^( 1) t^n ln(t) =−2Σ_(n=0) ^∞ (1/(n+1)){[(t^(n+1) /(n+1))ln(t)]_0 ^1 −(1/(n+1))∫_0 ^( 1) t^n dt} =2Σ_(n=0) ^∞ (1/((n+1)^3 ))=2Σ_(n=1) ^∞ (1/n^3 )=2ζ(3) .................... 𝛗=2ζ(3) .................... ..........m.n.july.1970.........$
$You can't use 'macro parameter character #' in math mode$ $p r o v e t h a t ::: ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x \overset{?}{=} 2 ζ (3)$ $\overset{1 - x = t}{=} \int_{0}^{1} \frac{{l n}^{2} (t)}{1 - t} d t = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {l n}^{2} (t) . t^{n} d t$ $= \underset{n = 0}{\sum^{\infty}} {{[\frac{t^{n + 1}}{n + 1} {l n}^{2} (t)]}_{0}^{1} - \frac{2}{n + 1} \int_{0}^{1} t^{n} l n (t)$ $= - 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{n + 1} {{[\frac{t^{n + 1}}{n + 1} l n (t)]}_{0}^{1} - \frac{1}{n + 1} \int_{0}^{1} t^{n} d t}$ $= 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = 2 ζ (3)$ $. . . . . . . . . . . . . . . . . . . . ϕ = 2 ζ (3) . . . . . . . . . . . . . . . . . . . .$ $. . . . . . . . . . m . n . j u l y . 1970 . . . . . . . . .$
	Answered by Dwaipayan Shikari last updated on 24/Feb/21

	$\underset{n = 1}{\sum^{\infty}} \int_{0}^{1} {l o g}^{2} (t) t^{n} d t$ $= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} \int_{0}^{\infty} u^{2} e^{- u} d t = \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{3}} = 2 ζ (3) u = - l o g t$
	Commented by mnjuly1970 last updated on 24/Feb/21

	$n i c e s o l u t i o n . .$
	Answered by Ñï= last updated on 04/Mar/21

	$ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x = {l n x l n}^{2} (1 - x) ∣_{0}^{1} + \int_{0}^{1} \frac{l n x}{1 - x} \cdot 2 l n (1 - x) d x$ $1 - x = t$ $ϕ = - \int_{1}^{0} \frac{l n (1 - t)}{t} \cdot 2 l n t d t = {L i}_{2} (t) 2 l n t ∣_{1}^{0} - 2 \int_{1}^{0} \frac{{L i}_{2} (t)}{t} d t$ $= - 2 {L i}_{3} (t) ∣_{1}^{0} = 2 {L i}_{3} (1) = 2 ζ (3)$
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