Question Number 137874 by mnjuly1970 last updated on 07/Apr/21
$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:…….{nice}\:…\:….\:{calculus}…. \\ $$$$\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\right)^{\mathrm{3}} =?…. \\ $$
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))](https://www.tinkutara.com/question/Q137883.png)
$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\right)^{\mathrm{3}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{3}} {x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }{dx}=\left[\frac{{ln}^{\mathrm{3}} {x}}{\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} {x}}{{x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\frac{{ln}^{\mathrm{3}} {x}}{\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\right){ln}^{\mathrm{2}} {xdx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\frac{{ln}^{\mathrm{3}} {x}}{\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{{ln}^{\mathrm{3}} {x}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx}−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{ln}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {x}}{\mathrm{1}−{x}}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}\right)=−\frac{\mathrm{3}}{\mathrm{2}}\left(−\psi''\left(\mathrm{1}\right)+\mathrm{2}\zeta\left(\mathrm{2}\right)\right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{2}\zeta\left(\mathrm{3}\right)+\mathrm{2}\zeta\left(\mathrm{2}\right)\right)=−\mathrm{3}\left(\zeta\left(\mathrm{3}\right)+\zeta\left(\mathrm{2}\right)\right) \\ $$
Commented by EnterUsername last updated on 07/Apr/21
$${See}\:{Q}\mathrm{137829}\:{for}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }{dx}=\mathrm{2}\zeta\left(\mathrm{2}\right) \\ $$
Commented by mnjuly1970 last updated on 07/Apr/21
$${very}\:{nice}\:{solution}\:{thanks}\:{alot} \\ $$$${master}\:…. \\ $$