Question Number 127775 by Bird last updated on 02/Jan/21
$${prove}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {lnxdx}=−\gamma \\ $$
Answered by Dwaipayan Shikari last updated on 02/Jan/21
$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {logx}\:{dx}\:=\Gamma'\left(\mathrm{1}\right)=\psi\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}\right)=−\gamma \\ $$