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Question-122572




Question Number 122572 by rs4089 last updated on 18/Nov/20
Answered by Olaf last updated on 18/Nov/20
With Euler′s Gamma function Γ :  ∫_0 ^∞ e^(−x) lnxdx = Γ′(1) = −γ  where γ is the Euler−Mascheroni  constant : γ ≈ 0.5772156649
$$\mathrm{With}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{Gamma}\:\mathrm{function}\:\Gamma\:: \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} \mathrm{ln}{xdx}\:=\:\Gamma'\left(\mathrm{1}\right)\:=\:−\gamma \\ $$$$\mathrm{where}\:\gamma\:\mathrm{is}\:\mathrm{the}\:\mathrm{Euler}−\mathrm{Mascheroni} \\ $$$$\mathrm{constant}\::\:\gamma\:\approx\:\mathrm{0}.\mathrm{5772156649} \\ $$

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