mr M.N july 1970 the question you posted earlier here goes the solution ∫


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Question Number 110910 by mathdave last updated on 31/Aug/20

$m r M . N j u l y 1970 t h e q u e s t i o n y o u p o s t e d e a r l i e r h e r e g o e s t h e s o l u t i o n$ $\int_{0}^{\frac{1}{2}} \frac{\ln^{2} (1 - x)}{x} d x$
	Answered by mathdave last updated on 31/Aug/20

	$s o l u t i o n$ $l e t I = \int_{0}^{\frac{1}{2}} \frac{\ln (1 - x) \ln (1 - x)}{x} d x$ $l e t u = \ln (1 - x)$ $, d u = - \frac{1}{1 - x} d x, \int d v = \int \frac{\ln (1 - x)}{x} d x, v = - {L i}_{2} (x)$ $u s i n g I B P$ $I = - \ln (1 - x) {L i}_{2} (x) ∣_{0}^{\frac{1}{2}} - \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{1 - x} d x$ $I = \ln (\frac{1}{2}) {L i}_{2} (\frac{1}{2}) - \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{1 - x} d x$ $n o w s o l v e \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{1 - x} d x (l e t$ $t = 1 - x, x = 1 - t, d x = - d t)$ $∵ \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (x)}{1 - x} d x = - \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (1 - t)}{t} d t = \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (1 - t)}{t} d t$ $r e c a l l t h a t$ ${L i}_{2} (z) + {L i}_{2} (1 - z) = \frac{π^{2}}{6} - \ln (z) \ln (1 - z)$ $∵ \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (1 - t)}{t} d t = \frac{π^{2}}{6} \int_{\frac{1}{2}}^{1} \frac{1}{t} d t - \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (t)}{t} d t - \int_{\frac{1}{2}}^{1} \frac{\ln (1 - t) \ln (t)}{t} d t$ $l e t$ $A = \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (t)}{t} d t = \int_{\frac{1}{2}}^{0} \frac{{L i}_{2} (t)}{t} d t + \int_{0}^{1} \frac{{L i}_{2} (t)}{t} d t$ $A = \int_{0}^{1} \frac{{L i}_{2} (t)}{t} d t - \int_{0}^{\frac{1}{2}} \frac{{L i}_{2} (t)}{t} d t$ $n o t e t h a t {L i}_{(n + 1)} (z) = \int_{b}^{a} \frac{{L i}_{n} (z)}{z} d z$ $A = {L i}_{3} (1) - {L i}_{3} (\frac{1}{2}) . . . . . . . . (x)$ $r e c a l l t h a t {L i}_{n} (z) = \underset{k = 1}{\sum^{\infty}} \frac{z^{k}}{k^{n}} w h e n n = 3, z = 1$ ${L i}_{3} (1) = \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{3}} = ζ (3)$ $a n d w h e n n = 3, z = \frac{1}{2} t h e n$ ${L i}_{3} (\frac{1}{2}) = \underset{k = 1}{\sum^{\infty}} \frac{1}{k^{3} 2^{k}} = \frac{1}{24} (21 ζ (3) + {4 \ln}^{3} (2) - 2 π^{2} \ln (2))$ $A = {L i}_{3} (1) - {L i}_{3} (\frac{1}{2}) = \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{6} + \frac{1}{12} π^{2} \ln 2 . . . . . . . . . (1)$ $a n d a g a i n l e t B = \int_{\frac{1}{2}}^{1} \frac{\ln (1 - t) \ln (t)}{t} d t$ $l e t$ $u = \ln t, d u = \frac{1}{t}, \int d v = \int \frac{\ln (1 - t)}{t} d t, v = - {L i}_{2} (t)$ $u s i n g I B P$ $B = - {L i}_{2} (t) \ln (t) ∣_{\frac{1}{2}}^{1} + \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (t)}{t} d t$ $B = {L i}_{2} (\frac{1}{2}) \ln (\frac{1}{2}) + \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (t)}{t} d t r e c a l l t h a t$ $\int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (t)}{t} d t = \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{6} + \frac{π^{2}}{12} \ln 2$ $∵ \int_{\frac{1}{2}}^{1} \frac{\ln (1 - t) \ln (t)}{t} d t = {L i}_{2} (\frac{1}{2}) \ln (\frac{1}{2}) + \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{6} + \frac{π^{2}}{12} \ln 2$ $B = - \frac{π^{2}}{12} \ln 2 + \frac{\ln^{3} (2)}{2} + \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{6} + \frac{π^{2}}{12} \ln 2$ $B = \int_{\frac{1}{2}}^{1} \frac{\ln (1 - t) \ln (t)}{t} d t = \frac{ζ (3)}{8} + \frac{\ln^{3} (2)}{3} . . . . . . (2)$ $b u t$ $I_{1} = \int_{\frac{1}{2}}^{1} \frac{{L i}_{2} (1 - t)}{t} d t = \frac{. π^{2}}{6} \int_{\frac{1}{2}}^{1} \frac{1}{t} d t - A - B$ $a n d I = - \ln (\frac{1}{2}) {L i}_{2} (\frac{1}{2}) - I_{1} s o$ $I_{1} = - \frac{π^{2}}{6} \ln (\frac{1}{2}) - \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{3} - \frac{π^{2}}{12} \ln 2 - \frac{ζ (3)}{8} - \frac{\ln^{3} (2)}{3}$ $I_{1} = \frac{π^{2}}{12} \ln 2 - \frac{ζ (3)}{4} - \frac{\ln^{3} (2)}{6}$ $b u t I = - \ln (\frac{1}{2}) {L i}_{2} (\frac{1}{2}) - I_{1} s o$ $I = \ln 2 (\frac{π^{2}}{12} - \frac{\ln^{2} (2)}{2}) - (\frac{π^{2}}{12} \ln 2 - \frac{ζ (3)}{4} - \frac{\ln^{3} (2)}{6})$ $I = \frac{π^{2}}{12} \ln 2 - \frac{\ln^{3} (2)}{2} - \frac{π^{2}}{12} \ln 2 + \frac{ζ (3)}{4} + \frac{\ln^{3} (2)}{6}$ $I = \frac{ζ (3)}{4} + \frac{\ln^{3} (2)}{6} - \frac{\ln^{3} (2)}{2} = \frac{ζ (3)}{4} + \frac{\ln^{3} (2) - {3 \ln}^{3} (2)}{6}$ $∵ I = \frac{ζ (3)}{4} - \frac{\ln^{3} (2)}{3}$ $∵ \int_{0}^{\frac{1}{2}} \frac{\ln^{2} (1 - x)}{x} d x = \frac{1}{4} ζ (3) - \frac{1}{3} \ln^{3} (2)$ $b y m a t h d a v e (31 / 08 / 2020)$
	Commented by mnjuly1970 last updated on 31/Aug/20

	$t a i e b a l l a h a n f a s a k o m . p e a c e$ $b e u p o n y o u . t h a n k y o u s i r . . .$
	Commented by I want to learn more last updated on 31/Aug/20

	$Weldone sir$
	Answered by mathmax by abdo last updated on 01/Sep/20
	$A =∫_0 ^(1/2) ((ln^2 (1−x))/x)dx ⇒A =_(1−x=u) ∫_1 ^(1/2) ((ln^2 u)/(1−u)) (−du) =∫_(1/2) ^1 ln^2 u(Σ_(n=0) ^∞ u^n )du =Σ_(n=0) ^∞ ∫_(1/2) ^1 u^n ln^2 u du U_n =∫_(1/2) ^1 u^n ln^2 (u)du =_(byparts) [(u^(n+1) /(n+1))ln^2 u]_(1/2) ^1 −∫_(1/2) ^1 (u^(n+1) /(n+1))×((2lnu)/u)du =−((ln^2 (2))/((n+1)2^(n+1) )) −(2/(n+1)) ∫_(1/2) ^1 u^n lnu du but ∫_(1/2) ^1 u^n lnu du =[(u^(n+1) /(n+1))lnu]_(1/2) ^1 −∫_(1/2) ^1 (u^(n+1) /(n+1))(du/u) =((ln2)/((n+1)2^(n+1) )) −(1/(n+1)) ∫_(1/2) ^(1 ) u^(n ) du =((ln(2))/((n+1)2^(n+1) ))−(1/((n+1)^2 )){1−(1/2^(n+1) )} ⇒ U_n =−((ln^2 2)/((n+1)2^(n+1) ))−((2ln2)/((n+1)^2 2^(n+1) )) +(2/((n+1)^3 ))(1−(1/2^(n+1) )) ⇒ A =−ln^2 2 Σ_(n=0) ^∞ (1/((n+1)2^(n+1) ))−2ln2 Σ_(n=0) ^∞ (1/((n+1)^2 2^(n+1) )) +2Σ_(n=0) ^∞ (1/((n+1)^3 )) −2 Σ_(n=0) ^∞ (1/((n+1)^3 2^(n+1) )) =−ln^2 2 Σ_(n=1) ^∞ (1/(n2^n ))−2ln2 Σ_(n=1) ^∞ (1/(n^2 2^n )) +2ξ(3)−2Σ_(n=1) ^∞ (1/(n^3 2^n )) =−ln^3 (2)+2ξ(3) −2ln(2) Σ_(n=1) ^∞ (1/(n^2 2^n )) −2 Σ_(n=1) ^∞ (1/(n^3 2^n )) we have Σ_(n=1) ^∞ (x^(n−1) /n) =−((ln(1−x))/x) ⇒ ∫Σ (x^(n−1) /n)dx =−∫((ln(1−x))/x)dx +c ⇒Σ_(n=1) ^∞ (x^n /(n^2 )) =−∫_0 ^x ((ln(1−t))/t) dt (c=0) ...be continued...$
	$A = \int_{0}^{\frac{1}{2}} \frac{\ln^{2} (1 - x)}{x} dx \Rightarrow A =_{1 - x = u} \int_{1}^{\frac{1}{2}} \frac{\ln^{2} u}{1 - u} (- du)$ $= \int_{\frac{1}{2}}^{1} \ln^{2} u (\sum_{n = 0}^{\infty} u^{n}) du = \sum_{n = 0}^{\infty} \int_{\frac{1}{2}}^{1} u^{n} \ln^{2} u du$ $U_{n} = \int_{\frac{1}{2}}^{1} u^{n} \ln^{2} (u) du =_{byparts} {[\frac{u^{n + 1}}{n + 1} \ln^{2} u]}_{\frac{1}{2}}^{1} - \int_{\frac{1}{2}}^{1} \frac{u^{n + 1}}{n + 1} \times \frac{2 lnu}{u} du$ $= - \frac{\ln^{2} (2)}{(n + 1) 2^{n + 1}} - \frac{2}{n + 1} \int_{\frac{1}{2}}^{1} u^{n} lnu du but$ $\int_{\frac{1}{2}}^{1} u^{n} lnu du = {[\frac{u^{n + 1}}{n + 1} lnu]}_{\frac{1}{2}}^{1} - \int_{\frac{1}{2}}^{1} \frac{u^{n + 1}}{n + 1} \frac{du}{u} = \frac{\ln 2}{(n + 1) 2^{n + 1}}$ $- \frac{1}{n + 1} \int_{\frac{1}{2}}^{1} u^{n} du = \frac{\ln (2)}{(n + 1) 2^{n + 1}} - \frac{1}{{(n + 1)}^{2}} {1 - \frac{1}{2^{n + 1}}} \Rightarrow$ $U_{n} = - \frac{\ln^{2} 2}{(n + 1) 2^{n + 1}} - \frac{2 \ln 2}{{(n + 1)}^{2} 2^{n + 1}} + \frac{2}{{(n + 1)}^{3}} (1 - \frac{1}{2^{n + 1}}) \Rightarrow$ $A = - \ln^{2} 2 \sum_{n = 0}^{\infty} \frac{1}{(n + 1) 2^{n + 1}} - 2 \ln 2 \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} 2^{n + 1}}$ $+ 2 \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{3}} - 2 \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{3} 2^{n + 1}}$ $= - \ln^{2} 2 \sum_{n = 1}^{\infty} \frac{1}{{n 2}^{n}} - 2 \ln 2 \sum_{n = 1}^{\infty} \frac{1}{n^{2} 2^{n}} + 2 ξ (3) - 2 \sum_{n = 1}^{\infty} \frac{1}{n^{3} 2^{n}}$ $= - \ln^{3} (2) + 2 ξ (3) - 2 \ln (2) \sum_{n = 1}^{\infty} \frac{1}{n^{2} 2^{n}} - 2 \sum_{n = 1}^{\infty} \frac{1}{n^{3} 2^{n}}$ $we have \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{n} = - \frac{\ln (1 - x)}{x} \Rightarrow$ $\int Σ \frac{x^{n - 1}}{n} dx = - \int \frac{\ln (1 - x)}{x} dx + c \Rightarrow \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}} = - \int_{0}^{x} \frac{\ln (1 - t)}{t} dt (c = 0)$ $. . . be continued . . .$
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