Question Number 160895 by HongKing last updated on 08/Dec/21
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}-\mathrm{1}} \:+\:…\:+\:\mathrm{1}}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:\left[\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{n}}\right)\:-\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right] \\ $$
Answered by Kamel last updated on 08/Dec/21
$$\Omega_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left({x}\right)\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{{n}} }{dx}=\left[\frac{{d}}{{ds}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} −{x}^{{s}+\mathrm{1}} }{\mathrm{1}−{x}^{{n}} }{dx}\right]_{{s}=\mathrm{0}} \\ $$$$\:\:\:\:\:\:\overset{{t}={x}^{{n}} } {=}\frac{\mathrm{1}}{{n}}\left[\frac{{d}}{{d}\underset{} {{s}}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{s}+\mathrm{1}}{{n}}−\mathrm{1}} −\mathrm{1}+\mathrm{1}−{t}^{\frac{{s}+\mathrm{2}}{{n}}−\mathrm{1}} }{\mathrm{1}−{t}}{dt}\right]_{{s}=\mathrm{0}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\Psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{{n}}\right)−\Psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$
Commented by HongKing last updated on 10/Dec/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Sir} \\ $$