Question Number 131546 by mathmax by abdo last updated on 05/Feb/21
$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:=−\gamma \\ $$
Answered by Dwaipayan Shikari last updated on 06/Feb/21
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}} {dx}=\Gamma\left({a}+\mathrm{1}\right) \\ $$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}} {log}\left({x}\right){dx}=\Gamma'\left({a}+\mathrm{1}\right) \\ $$$${I}'\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {log}\left({x}\right){dx}=\Gamma'\left(\mathrm{1}\right)=\Gamma\left(\mathrm{1}\right)\psi\left(\mathrm{1}\right)=−\gamma \\ $$