Question Number 160895 by HongKing last updated on 08/Dec/21
![Prove that: ∫_( 0) ^( 1) ((ln(x))/(x^n + x^(n-1) + ... + 1)) dx = (1/n^2 ) [𝛙^((1)) ((2/n)) - 𝛙^((1)) ((1/n))]](Q160895.png)
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:+\:\mathrm{x}^{\boldsymbol{\mathrm{n}}-\mathrm{1}} \:+\:...\:+\:\mathrm{1}}\:\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\:\left[\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{\mathrm{n}}\right)\:-\:\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right] \\ $$
Answered by Kamel last updated on 08/Dec/21
![Ω_n =∫_0 ^1 ((Ln(x)(1−x))/(1−x^n ))dx=[(d/ds)∫_0 ^1 ((x^s −x^(s+1) )/(1−x^n ))dx]_(s=0) =^(t=x^n ) (1/n)[(d/ds_ )∫_0 ^1 ((t^(((s+1)/n)−1) −1+1−t^(((s+2)/n)−1) )/(1−t))dt]_(s=0) =(1/n^2 )(Ψ^((1)) ((2/n))−Ψ^((1)) ((1/n)))](Q160898.png)
$$\Omega_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{Ln}\left({x}\right)\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{{n}} }{dx}=\left[\frac{{d}}{{ds}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{s}} −{x}^{{s}+\mathrm{1}} }{\mathrm{1}−{x}^{{n}} }{dx}\right]_{{s}=\mathrm{0}} \\ $$$$\:\:\:\:\:\:\overset{{t}={x}^{{n}} } {=}\frac{\mathrm{1}}{{n}}\left[\frac{{d}}{{d}\underset{} {{s}}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{s}+\mathrm{1}}{{n}}−\mathrm{1}} −\mathrm{1}+\mathrm{1}−{t}^{\frac{{s}+\mathrm{2}}{{n}}−\mathrm{1}} }{\mathrm{1}−{t}}{dt}\right]_{{s}=\mathrm{0}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\Psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{2}}{{n}}\right)−\Psi^{\left(\mathrm{1}\right)} \left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$
Commented by HongKing last updated on 10/Dec/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Sir} \\ $$