I want to derive some identities for $\Gamma(z)$ given \begin{equation*} \frac{1}{\Gamma(z)} = ze^{\gamma z} \prod_{k=1}^\infty \left(1 + \frac{z}{k} \right) e^{-z/k}. \end{equation*} I have already derived \begin{equation*} \frac{\Gamma'(z)}{\Gamma(z)} = -\frac{1}{z} - \gamma - \sum_{k=0}^\infty \frac{1}{z + k} - \frac{1}{k}. \end{equation*} From this, I would like to derive the relationship \begin{equation} \frac{\Gamma'(z+1)}{\Gamma(z+1)} - \frac{\Gamma'(z)}{\Gamma(z)} - \frac{1}{z} = 0. \end{equation} I would then like to use this to show that $\Gamma(z + 1) = Cz\Gamma(z)$ for some constant $C$.

From here it is easy (meaning I have done it) to show that $\lim_{z \to 0} z \Gamma(z) = 1$, which implies that $C = \Gamma(1)$. I would finally like to show that $\Gamma(1) = 1$.

I order to prove the first relationship, I have tried using the logarithmic derivative identity above , but to little avail. I have even tried just working from $\Gamma(1)z\Gamma(z)$ to arrive to $\Gamma(z+1)$ and I get close, but there is an extra term in my product that should not be there. Can anyone assist me?


1 Answer 1


Your first question is straightforward. You have $$\frac{\Gamma'(z)}{\Gamma(z)}=-\frac1z-\gamma-\sum_{k=1}^\infty \left(\frac1{z+k}-\frac1k\right).$$ Therefore $$\frac{\Gamma'(z+1)}{\Gamma(z+1)}=-\frac1{z+1}-\gamma-\sum_{k=1}^\infty \left(\frac1{z+1+k}-\frac1k\right).$$ Subtracting gives $$\frac{\Gamma'(z+1)}{\Gamma(z+1)}- \frac{\Gamma'(z)}{\Gamma(z)}=\frac1z-\frac1{z+1}+\sum_{k=1}^\infty \left(\frac{1}{z+k}-\frac1{z+k+1}\right).$$ The sum telescopes to $1/(z+1)$.

The logarithmic derivative of $z\Gamma(z)\Gamma(z+1)^{-1}$ vanishes, so that $z\Gamma(z)\Gamma(z+1)^{-1}$ is constant.

Finally $$\frac1{\Gamma(1)}=e^\gamma\prod_{k=1}^\infty\frac{k+1}{k}e^{-1/k} =\exp\left(\lim_{N\to\infty}\left(\gamma+\log(N+1)-\sum_{k=1}^N\frac1k\right)\right)$$ etc.

  • $\begingroup$ The problem with the last part is I am not sure how to show $\Gamma(1) = 1$ without using the fact that $\gamma$ is given by $\lim_{N \to \infty} \sum_{k=1}^N \frac{1}{k} - \log(N+1)$. Is there a way to do it without using that? $\endgroup$
    – swygerts
    Oct 4, 2017 at 11:37
  • $\begingroup$ @swygerts How do you propose to define $\gamma$ then? $\endgroup$ Oct 4, 2017 at 15:56
  • $\begingroup$ Not sure. Was thinking maybe there was another way to do it. Is there a way to prove the equivalence between this definition of the $\Gamma$ function and the integral definition without invoking $\gamma?$ $\endgroup$
    – swygerts
    Oct 5, 2017 at 12:35

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