Question Number 212470 by Spillover last updated on 14/Oct/24
Answered by Ar Brandon last updated on 14/Oct/24
![Ω=∫_0 ^∞ (((1−x)lnx)/(1−x^6 ))dx =∫_0 ^1 ((1−x)/(1−x^6 ))lnxdx+∫_1 ^∞ ((1−x)/(1−x^6 ))lnxdx =∫_0 ^1 ((1−x)/(1−x^6 ))lnxdx−∫_0 ^1 ((1−(1/x))/(x^2 −(1/x^4 )))lnxdx =∫_0 ^1 ((1−x−x^3 +x^4 )/(1−x^6 ))lnxdx , ψ′(t)=−∫_0 ^1 ((x^(t−1) lnx)/(1−x))dx =(1/(36))∫_0 ^1 ((x^(−(5/6)) −x^(−(2/3)) −x^(−(1/3)) +x^(−(1/6)) )/(1−x))ln(x)dx =−(1/(36))(ψ′((1/6))−ψ′((1/3))−ψ′((2/3))+ψ′((5/6))) =−(1/(36))((ψ′((1/6))+ψ′((5/6)))−(ψ′((1/3))+ψ′((2/3))) [ ψ′(x)+ψ′(1−x)=π^2 cosec^2 (πx) ] =−(1/(36))(π^2 cosec^2 ((π/6))−π^2 cosec^2 ((π/3))) =−(π^2 /(36))(2^2 −((2/( (√3))))^2 )=−(π^2 /(36))(((12−4)/3))=−((2π^2 )/(27))](https://www.tinkutara.com/question/Q212476.png)
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\left(\mathrm{1}−{x}\right)\mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{\mathrm{6}} }\mathrm{ln}{xdx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{\mathrm{6}} }\mathrm{ln}{xdx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}−{x}^{\mathrm{6}} }\mathrm{ln}{xdx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}\mathrm{ln}{xdx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}−{x}^{\mathrm{3}} +{x}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{6}} }\mathrm{ln}{xdx}\:,\:\psi'\left({t}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{t}−\mathrm{1}} \mathrm{ln}{x}}{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{−\frac{\mathrm{5}}{\mathrm{6}}} −{x}^{−\frac{\mathrm{2}}{\mathrm{3}}} −{x}^{−\frac{\mathrm{1}}{\mathrm{3}}} +{x}^{−\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{1}−{x}}\mathrm{ln}\left({x}\right){dx} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{36}}\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{6}}\right)−\psi'\left(\frac{\mathrm{1}}{\mathrm{3}}\right)−\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\psi'\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right) \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{36}}\left(\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\psi'\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)−\left(\psi'\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\psi'\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right)\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:\psi'\left({x}\right)+\psi'\left(\mathrm{1}−{x}\right)=\pi^{\mathrm{2}} \mathrm{cosec}^{\mathrm{2}} \left(\pi{x}\right)\:\right] \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{36}}\left(\pi^{\mathrm{2}} \mathrm{cosec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)−\pi^{\mathrm{2}} \mathrm{cosec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$\:\:\:\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\left(\mathrm{2}^{\mathrm{2}} −\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\left(\frac{\mathrm{12}−\mathrm{4}}{\mathrm{3}}\right)=−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{27}} \\ $$