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Question Number 186735 by Rupesh123 last updated on 09/Feb/23

Answered by Ar Brandon last updated on 09/Feb/23

$$I={\int }_0^1{\left(\ln \left(\frac{1}{x}\right)\right)}^{2023}dx,t=-\ln x\Rightarrow x=e^{-t}$$$$={\int }_0^{\mathrm{\infty }}{t}^{2023}{e}^{-t}dt=\mathrm{\Gamma }\left(2024\right)=2023!$$

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