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1-1-ln-x-dx-




Question Number 153164 by alisiao last updated on 05/Sep/21
∫_(−1) ^( 1) ln(𝚪(x)) dx
$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\right)\:\boldsymbol{{dx}} \\ $$
Answered by Ar Brandon last updated on 05/Sep/21
Ω=∫_(−1) ^1 ln(Γ(x))dx=∫_(−1) ^0 ln(Γ(x))dx_(Ω_1 ) +∫_0 ^1 ln(Γ(x))dx_(Ω_2 )   Ω_2 =∫_0 ^1 ln(Γ(x))dx=∫_0 ^1 ln(Γ(1−x))dx        =(1/2)∫_0 ^1 {ln(Γ(x))+ln(Γ(1−x))dx}dx        =(1/2)∫_0 ^1 ln(Γ(x)Γ(1−x))dx=(1/2)∫_0 ^1 ln((π/(sin(πx))))dx        =((lnπ)/2)−(1/2)∫_0 ^1 ln(sin(πx))dx=((lnπ)/2)−(1/(2π))∫_0 ^π ln(sin(u))du_(=0)
$$\Omega=\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}=\underset{\Omega_{\mathrm{1}} } {\underbrace{\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}}}+\underset{\Omega_{\mathrm{2}} } {\underbrace{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}}} \\ $$$$\Omega_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{ln}\left(\Gamma\left({x}\right)\right)+\mathrm{ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\frac{\pi}{\mathrm{sin}\left(\pi{x}\right)}\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{ln}\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{sin}\left(\pi{x}\right)\right){dx}=\frac{\mathrm{ln}\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{=\mathrm{0}} {\underbrace{\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}\left({u}\right)\right){du}}} \\ $$

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