....nice calculus.... prove that:::: I:=∫_0 ^(1/2) {((log^2 (1−x))/x)}dx = ((ζ(3))/4) − (1/3)log^3 (2)✓


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Question Number 123547 by mnjuly1970 last updated on 26/Nov/20
$....nice calculus.... prove that:::: I:=∫_0 ^(1/2) {((log^2 (1−x))/x)}dx = ((ζ(3))/4) − (1/3)log^3 (2)✓$
$. . . . n i c e c a l c u l u s . . . .$ $p r o v e t h a t ::::$ $I := \int_{0}^{\frac{1}{2}} {\frac{{l o g}^{2} (1 - x)}{x}} d x$ $= \frac{ζ (3)}{4} - \frac{1}{3} {l o g}^{3} (2) ✓$
	Answered by mnjuly1970 last updated on 26/Nov/20
	$solution: special values:: i:: li_2 ((1/2))[ =_(try it.) ^(why??) ](π^2 /(12)) −(1/2)log^2 (2) ✓ ii:: li_3 ((1/2))[ =_(prove it.) ^(why??) ](7/8)ζ(3)+(1/6)log^3 (2)−(π^2 /(12))log(2) ✓ iii:: li_3 (1) =ζ(3)✓ I=^(1−x=t) ∫_(1/2) ^( 1) ((log^2 (t))/(1−t))dt=^(i.b.p) [−log(1−t)log^2 (t)]_(1/2) ^1 +2∫_(1/2) ^( 1) ((log(t)log(1−t))/t)dt =log((1/2))log^2 ((1/2))+2∫_(1/2) ^( 1) log(t)d(−li_2 (t)) =−log^3 (2)+2{[log(t)li_2 (t)]_((1/2) ) ^1 +∫_(1/2) ^( 1) ((li_2 (t))/t)dt} =−log^3 (2)+2[log((1/2))li_2 ((1/2))]+2li_3 (1)−2li_3 ((1/2)) =−log^3 (2)−2log(2)((π^2 /(12))−(1/2)log^2 (2))+_↫ −^↬ (7/4)ζ(3)−(1/3)log^3 (2)+(π^2 /6)log(2)+2ζ(3) =((ζ (3 ))/4) −(1/3)log^3 (2) ✓ ✓ nice.calculus. m.n.july.1970$
	$s o l u t i o n :$ $s p e c i a l v a l u e s ::$ $i :: {l i}_{2} (\frac{1}{2}) [\underset{t r y i t .}{\overset{w h y ? ?}{=}}] \frac{π^{2}}{12} - \frac{1}{2} {l o g}^{2} (2) ✓$ $i i :: {l i}_{3} (\frac{1}{2}) [\underset{p r o v e i t .}{\overset{w h y ? ?}{=}}] \frac{7}{8} ζ (3) + \frac{1}{6} {l o g}^{3} (2) - \frac{π^{2}}{12} l o g (2) ✓$ $i i i :: {l i}_{3} (1) = ζ (3) ✓$ $I \overset{1 - x = t}{=} \int_{\frac{1}{2}}^{1} \frac{{l o g}^{2} (t)}{1 - t} d t \overset{i . b . p}{=} {[- l o g (1 - t) {l o g}^{2} (t)]}_{\frac{1}{2}}^{1} + 2 \int_{\frac{1}{2}}^{1} \frac{l o g (t) l o g (1 - t)}{t} d t$ $= l o g (\frac{1}{2}) {l o g}^{2} (\frac{1}{2}) + 2 \int_{\frac{1}{2}}^{1} l o g (t) d (- {l i}_{2} (t))$ $= - {l o g}^{3} (2) + 2 {{[l o g (t) {l i}_{2} (t)]}_{\frac{1}{2}}^{1} + \int_{\frac{1}{2}}^{1} \frac{{l i}_{2} (t)}{t} d t}$ $= - {l o g}^{3} (2) + 2 [l o g (\frac{1}{2}) {l i}_{2} (\frac{1}{2})] + 2 {l i}_{3} (1) - 2 {l i}_{3} (\frac{1}{2})$ $= - {l o g}^{3} (2) - 2 l o g (2) (\frac{π^{2}}{12} - \frac{1}{2} {l o g}^{2} (2)) \underset{↫}{+}$ $\overset{↬}{-} \frac{7}{4} ζ (3) - \frac{1}{3} {l o g}^{3} (2) + \frac{π^{2}}{6} l o g (2) + 2 ζ (3)$ $= \frac{ζ (3)}{4} - \frac{1}{3} {l o g}^{3} (2) ✓ ✓$ $n i c e . c a l c u l u s . m . n . j u l y . 1970$
	Answered by mathmax by abdo last updated on 26/Nov/20
	$I =∫_0 ^(1/2) ((log^2 (1−x))/x)dx ⇒I =_(x=(t/2)) 2∫_0 ^1 ((log^2 (1−(t/2)))/t)(dt/2) =∫_0 ^1 ((log^2 (1−(t/2)))/t) dt let f(a) =∫_0 ^1 ((log^2 (1−at))/t)dt[with 0<a<1 f^′ (a) =2∫_0 ^1 ((−t)/(1−at))((log(1−at))/t)dt =−2 ∫_0 ^1 ((log(1−at))/(1−at))dt =−2 ∫_0 ^1 log(1−at)Σ_(n=0) ^∞ a^n t^n dt =−2 Σ_(n=0) ^∞ a^n ∫_0 ^1 t^n log(1−at)dt =−2Σ_(n=0) ^∞ a^n U_n U_n =∫_0 ^1 t^n log(1−at)dt =[(t^(n+1) /(n+1))log(1−at)]_0 ^1 =((log(1−a))/(n+1)) −∫_0 ^1 (t^(n+1) /(n+1))×((−a)/(1−at))dt =((log(1−a))/(n+1)) +(a/(n+1))∫_0 ^1 (t^(n+1) /(1−at))dt and ∫_0 ^1 (t^(n+1) /(1−at))dt =_(1−at=z) ∫_1 ^(1−a) (((((1−z)/a))^(n+1) )/z)(−(1/a))dz =(1/a^(n+2) )∫_(1−a) ^a (((1−z)^(n+1) )/z)dz =(1/a^(n+2) )∫_(1−a) ^a (1/z)Σ_(k=0) ^(n+1) C_(n+1) ^k (−z)^k dz =(1/a^(n+2) )∫_(1−a) ^a Σ_(k=0) ^(n+1) C_(n+1) ^k (−1)^k z^(k−1) dz =(1/a^(n+2) )Σ_(k=0) ^(n+1) C_(n+1) ^k (−1)^k [(z^k /k)]_(1−a) ^a =(1/a^(n+2) )Σ_(k=0) ^(n+1) C_(n+1) ^k (((−1)^k )/k){a^k −(1−a)^k }....be continued...$
	$I = \int_{0}^{\frac{1}{2}} \frac{\log^{2} (1 - x)}{x} dx \Rightarrow I =_{x = \frac{t}{2}} 2 \int_{0}^{1} \frac{\log^{2} (1 - \frac{t}{2})}{t} \frac{dt}{2}$ $= \int_{0}^{1} \frac{\log^{2} (1 - \frac{t}{2})}{t} dt let f (a) = \int_{0}^{1} \frac{\log^{2} (1 - at)}{t} dt [with 0 < a < 1$ $f^{^{'}} (a) = 2 \int_{0}^{1} \frac{- t}{1 - at} \frac{\log (1 - at)}{t} dt = - 2 \int_{0}^{1} \frac{\log (1 - at)}{1 - at} dt$ $= - 2 \int_{0}^{1} \log (1 - at) \sum_{n = 0}^{\infty} a^{n} t^{n} dt$ $= - 2 \sum_{n = 0}^{\infty} a^{n} \int_{0}^{1} t^{n} \log (1 - at) dt = - 2 \sum_{n = 0}^{\infty} a^{n} U_{n}$ $U_{n} = \int_{0}^{1} t^{n} \log (1 - at) dt = {[\frac{t^{n + 1}}{n + 1} \log (1 - at)]}_{0}^{1} = \frac{\log (1 - a)}{n + 1}$ $- \int_{0}^{1} \frac{t^{n + 1}}{n + 1} \times \frac{- a}{1 - at} dt = \frac{\log (1 - a)}{n + 1} + \frac{a}{n + 1} \int_{0}^{1} \frac{t^{n + 1}}{1 - at} dt and$ $\int_{0}^{1} \frac{t^{n + 1}}{1 - at} dt =_{1 - at = z} \int_{1}^{1 - a} \frac{{(\frac{1 - z}{a})}^{n + 1}}{z} (- \frac{1}{a}) dz$ $= \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \frac{{(1 - z)}^{n + 1}}{z} dz = \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \frac{1}{z} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- z)}^{k} dz$ $= \frac{1}{a^{n + 2}} \int_{1 - a}^{a} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- 1)}^{k} z^{k - 1} dz$ $= \frac{1}{a^{n + 2}} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} {(- 1)}^{k} {[\frac{z^{k}}{k}]}_{1 - a}^{a}$ $= \frac{1}{a^{n + 2}} \sum_{k = 0}^{n + 1} C_{n + 1}^{k} \frac{{(- 1)}^{k}}{k} {a^{k} - {(1 - a)}^{k}} . . . . be continued . . .$
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