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Question Number 122572 by rs4089 last updated on 18/Nov/20

Answered by Olaf last updated on 18/Nov/20

$$\mathrm{With}{\mathrm{Euler}}^{\prime }\mathrm{s}\mathrm{Gamma}\mathrm{function}\mathrm{\Gamma }:$$$${\int }_0^{\mathrm{\infty }}{e}^{-x}\ln xdx={\mathrm{\Gamma }}^{\prime }\left(1\right)=-\gamma$$$$\mathrm{where}\gamma \mathrm{is}\mathrm{the}\mathrm{Euler}-\mathrm{Mascheroni}$$$$\mathrm{constant}:\gamma \approx 0.5772156649$$

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