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Question Number 137874 by mnjuly1970 last updated on 07/Apr/21

.......nice ... .... calculus.... evaluate :: Ω=∫_0 ^( 1) (((ln(1−x))/x))^3 =?....

.......nice.......calculus....evaluate::Ω=∫01(ln(1−x)x)3=?....

Answered by EnterUsername last updated on 07/Apr/21

Ω=∫_0 ^1 (((ln(1−x))/x))^3 dx =∫_0 ^1 ((ln^3 x)/((1−x)^3 ))dx=[((ln^3 x)/(2(1−x)^2 ))−(3/2)∫((ln^2 x)/(x(1−x)^2 ))dx]_0 ^1 =[((ln^3 x)/(2(1−x)^2 ))−(3/2)∫((1/x)+(1/(1−x))+(1/((1−x)^2 )))ln^2 xdx]_0 ^1 =[((ln^3 x)/(2(1−x)^2 ))−((ln^3 x)/2)−(3/2)∫((ln^2 x)/(1−x))dx−(3/2)∫((ln^2 x)/((1−x)^2 ))dx]_0 ^1 =−(3/2)(∫_0 ^1 ((ln^2 x)/(1−x))dx+∫_0 ^1 ((ln^2 x)/((1−x)^2 ))dx)=−(3/2)(−ψ′′(1)+2ζ(2)) =−(3/2)(2ζ(3)+2ζ(2))=−3(ζ(3)+ζ(2))

Ω=∫01(ln(1−x)x)3dx=∫01ln3x(1−x)3dx=[ln3x2(1−x)2−32∫ln2xx(1−x)2dx]01=[ln3x2(1−x)2−32∫(1x+11−x+1(1−x)2)ln2xdx]01=[ln3x2(1−x)2−ln3x2−32∫ln2x1−xdx−32∫ln2x(1−x)2dx]01=−32(∫01ln2x1−xdx+∫01ln2x(1−x)2dx)=−32(−ψ″(1)+2ζ(2))=−32(2ζ(3)+2ζ(2))=−3(ζ(3)+ζ(2))

Commented by EnterUsername last updated on 07/Apr/21

See Q137829 for ∫_0 ^1 ((ln^2 x)/((1−x)^2 ))dx=2ζ(2)

SeeQ137829for∫01ln2x(1−x)2dx=2ζ(2)

Commented by mnjuly1970 last updated on 07/Apr/21

very nice solution thanks alot master ....

verynicesolutionthanksalotmaster....

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