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Question Number 206674 by 073 last updated on 22/Apr/24

Answered by Frix last updated on 22/Apr/24

$$\mathrm{\Gamma }\left(x\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{t}^{x-1}{\mathrm{e}}^{-t}dt$$$$\frac{d\mathrm{\Gamma }\left(x\right)}{dx}={\mathrm{\Gamma }}^{\prime }\left(x\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{t}^{x-1}{\mathrm{e}}^{-t}\ln tdt$$$$\Rightarrow$$$${\mathrm{\Gamma }}^{\prime }\left(1\right)=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-t}\ln tdt$$$$\mathrm{We}\mathrm{know}{\mathrm{\Gamma }}^{\prime }\left(1\right)=\psi \left(1\right)=-\gamma$$

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