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Use the product formula for $1/\Gamma(z)$ to prove that $$\Gamma'(1)=-\gamma$$


I know that for Euler constant $\gamma$,

$$\frac{1}{\Gamma(z)} =ze^{\gamma z}\prod _{k=1}^{\infty} (1+\frac{z}{k})e^{-z/k}$$

But I cannot prove it properly. Please show me explicitly. Thank you:)

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    $\begingroup$ related math.stackexchange.com/questions/461070 $\endgroup$ – mhd.math Dec 28 '13 at 10:44
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    $\begingroup$ Take the reciprocals of both sides of your identity, and then take the logarithmic derivative of both sides. Since $\Gamma(1)=1$, this gives you $\Gamma'(1)/\Gamma(1) = \Gamma'(1) = \dots$. $\endgroup$ – A l'Maeaux Dec 28 '13 at 10:45
  • $\begingroup$ I didnt see the question.thank you @hmedan.mnsh $\endgroup$ – user315 Dec 28 '13 at 10:49
  • $\begingroup$ Look up digamma function on wikipedia $\endgroup$ – Jonathan Mar 15 '17 at 0:16
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$$\log\Gamma(z)=-\log z-\gamma z-\sum_{k=1}^\infty\left[\log\left(1+\frac zk\right)-\frac zk\right]\implies$$

$$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{k=1}^\infty\left(\frac1k\frac k{k+z}-\frac1k\right)=-\frac1z-\gamma-\sum_{k=1}^\infty\left(\frac1{k+z}-\frac1k\right)\implies$$

$$\Gamma'(1)=\frac{\Gamma'(1)}{\Gamma(1)}=-1-\gamma-\left(\frac12-1+\frac13-\frac12+\ldots\right)=-1-\gamma-(-1)=-\gamma$$

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