# How to prove that $e^{-\gamma}=\prod_{n=1}^\infty\left(1+\frac1n\right)e^{-1/n}$

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?

• Shouldn't taking the logarithm and using $$\gamma = \sum_{k=1}^\infty (\frac{1}{k} - \log(\frac{k + 1}{k}))$$ suffice?
– user301452
May 31, 2016 at 11:10
• 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 $f(z+1) = zf(z)$, $f(1) = 1$ May 31, 2016 at 11:18
• @user1952009 Note that I want to prove that $\Gamma(1)=1$, we don't have it!
– user302007
May 31, 2016 at 11:56
• @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 May 31, 2016 at 12:38
• @user1952009 I don't know. This is a problem in Ablowitz-Fokas complex variables where the authors didn't give more information about $\gamma$!
– user302007
May 31, 2016 at 13:28

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: $$\sum_{n=1}^{N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) = H_N-\log(N+1).\tag{B}$$
• 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?
• @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 $$\gamma = \lim_{N\to +\infty}\left(H_N-\log N\right).$$ May 31, 2016 at 11:52