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Suppose we defined the Gamma function $$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac zn\right)e^{-z/n}$$ where $\gamma$ is just a constant. I want to prove that $\Gamma(1)=1$, so I need to prove that $$e^{-\gamma}=\prod_{n=1}^\infty\left(1+\frac1n\right)e^{-1/n}\tag{*}$$

From definition, by taking logarithm and then differentiating, we arrive to $$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{n=1}^\infty\left(\frac1{z+n}-\frac1n\right)\tag{1}$$ which implies that $$\frac{\Gamma'(z+1)}{\Gamma(z+1)}-\frac{\Gamma'(z)}{\Gamma(z)}-\frac1z=0\tag{2}$$ taking integral we arrive to the following identity $$\Gamma(z+1)=Cz\Gamma(z)\tag{3}$$ where $C$ is constant and since $\lim_{z\to0}z\Gamma(z)=1$, so $$C=\Gamma(1)\tag{4}$$ I tried to from identities $(1-4)$ arrive to $(*)$, but failed, can anyone help me, please?

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    $\begingroup$ Shouldn't taking the logarithm and using $$\gamma = \sum_{k=1}^\infty (\frac{1}{k} - \log(\frac{k + 1}{k}))$$ suffice? $\endgroup$
    – user301452
    May 31, 2016 at 11:10
  • $\begingroup$ you get that $\Gamma(z+1) = z \Gamma(z)$ and $\Gamma(1) = 1$ but proving this is equal to $\int_0^\infty x^{z-1} e^{-x} dx$ requires more steps, I think you also need to show there is only one log-convex analytic function on $Re(z) > 1$ fulfilling $f(z+1) = zf(z)$, $f(1) = 1$ $\endgroup$
    – reuns
    May 31, 2016 at 11:18
  • $\begingroup$ @user1952009 Note that I want to prove that $\Gamma(1)=1$, we don't have it! $\endgroup$
    – user302007
    May 31, 2016 at 11:56
  • $\begingroup$ @user302007 : what is $\gamma$... as the other told you, the usual definition is the constant such that $f(z+1) = z f(z)$ with $f(z)$ your product $\endgroup$
    – reuns
    May 31, 2016 at 12:38
  • $\begingroup$ @user1952009 I don't know. This is a problem in Ablowitz-Fokas complex variables where the authors didn't give more information about $\gamma$! $\endgroup$
    – user302007
    May 31, 2016 at 13:28

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In order to prove that $C=1$, you need to prove that: $$ \gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) \tag{A}$$ but that is exactly the usual definition of the Euler-Mascheroni costant, since: $$ \sum_{n=1}^{N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) = H_N-\log(N+1).\tag{B}$$

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  • $\begingroup$ I don't know, but this is a problem in Ablowitz-Fokas complex variables where authors don't give more information about $\gamma$, page 171, problem 3.6.6.d? $\endgroup$
    – user302007
    May 31, 2016 at 11:51
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    $\begingroup$ @user302007: if you defined the $\Gamma$ function through its Weierstrass product you need some definition of $\gamma$, otherwise what is the meaning of $\gamma$ in you first line? You cannot ask to prove that $\Gamma(1)=1$ for any constant you put there, but the above lines show that $\Gamma(1)=1$ is equivalent to $$ \gamma = \lim_{N\to +\infty}\left(H_N-\log N\right).$$ $\endgroup$
    – Jack D'Aurizio
    May 31, 2016 at 11:52

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