Question Number 185517 by Noorzai last updated on 23/Jan/23
Answered by Ar Brandon last updated on 23/Jan/23
$$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{4}} \mathrm{ln}{x}}{\mathrm{1}−{x}^{\mathrm{6}} }{dx}=−\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{−\frac{\mathrm{5}}{\mathrm{6}}} +{x}^{−\frac{\mathrm{1}}{\mathrm{6}}} }{\mathrm{1}−{x}}\mathrm{ln}{xdx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{36}}\left(\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{36}}\mathrm{cosec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{6}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{9}} \\ $$