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How do you prove that $$ \Gamma'(1)=-\gamma, $$ where $\gamma$ is the Euler-Mascheroni constant?

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    $\begingroup$ What definition the the gamma function are you using? $\endgroup$ Apr 22, 2015 at 15:22
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    $\begingroup$ And what definition of $\gamma$? $\endgroup$
    – Chappers
    Apr 22, 2015 at 15:32
  • $\begingroup$ Would this answer your question ? $\endgroup$
    – Lucian
    Apr 22, 2015 at 16:01

3 Answers 3

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Consider the integral form of the Gamma function, \begin{align} \Gamma(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt \end{align} taking the derivative with respect to $x$ yields \begin{align} \Gamma'(x) = \int_{0}^{\infty} e^{-t} \, t^{x-1} \, \ln(t) \, dt. \end{align} Setting $x=1$ leads to \begin{align} \Gamma'(1) = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align} This is one of the many definitions of the Euler-Mascheroni constant. Hence, \begin{align} \Gamma'(1) = - \gamma = \int_{0}^{\infty} e^{-t} \, \ln(t) \, dt. \end{align}

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  • $\begingroup$ How to take derivative with respect to x of$ \int_{0}^{\infty} e^{-t} \, t^{x-1} \, dt$? $\endgroup$
    – Jonathen
    Oct 7, 2022 at 2:36
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    $\begingroup$ @Jonathen Look up "differentiation under the integral sign". $\endgroup$
    – asrxiiviii
    Oct 18, 2022 at 6:36
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The Weierstrass product for the $\Gamma$ function gives: $$\Gamma(z+1)=e^{-\gamma z}\cdot\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}\tag{1}$$ hence by considering $\frac{d}{dz}\log(\cdot)$ of both terms we get: $$ \psi(z+1)=\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{n+z}\right) \tag{2}$$ and by evaluating the previous identity in $z=0$ it follows that: $$ \psi(1) = \Gamma'(1) = -\gamma.\tag{3}$$

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I was wrong I cannot delete my post because I having trouble singing in sorry for my lapse in judgement and failed math skills I will try to be better the solutions above work just fine. $$\Gamma^{\prime}(z) = \frac{d}{dz} \int^{\infty}_{0} e^{-t}t^{z-1}dt = \int^{\infty}_{0} \frac{d}{dz} e^{-t}t^{z-1}dt = \int^{\infty}_{0} e^{-t} \frac{d}{dz} t^{z-1} dt = \int^{\infty}_{0} e^{-t} ln(t) t^{z-1} dt$$ $$\Gamma^{\prime}(1) = \int^{\infty}_{0} e^{-t} ln(t) t^{1-1} dt = \int^{\infty}_{0} e^{-t} ln(t) dt$$ this integral can be solved numerically to show that it comes out to $$-\gamma_{\,_\mathrm{EM}}$$

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