Question Number 206674 by 073 last updated on 22/Apr/24
Answered by Frix last updated on 22/Apr/24
$$\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}} \mathrm{e}^{−{t}} {dt} \\ $$$$\frac{{d}\Gamma\left({x}\right)}{{dx}}=\Gamma'\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}} \mathrm{e}^{−{t}} \mathrm{ln}\:{t}\:{dt} \\ $$$$\Rightarrow \\ $$$$\Gamma'\left(\mathrm{1}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}} \mathrm{ln}\:{t}\:{dt} \\ $$$$\mathrm{We}\:\mathrm{know}\:\Gamma'\left(\mathrm{1}\right)=\psi\left(\mathrm{1}\right)=−\gamma \\ $$