Question Number 186735 by Rupesh123 last updated on 09/Feb/23
Answered by Ar Brandon last updated on 09/Feb/23
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2023}} {dx}\:,\:{t}=−\mathrm{ln}{x}\:\Rightarrow{x}={e}^{−{t}} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{2023}} {e}^{−{t}} {dt}=\Gamma\left(\mathrm{2024}\right)=\mathrm{2023}! \\ $$