The good old factorial is defined line $n!:=\prod_{k=1}^{n}k$.

Given this definition it is fairly easy to arrive to some recursive formulas such as \begin{equation} n!=(n-m)!\prod_{k=1}^{m}(n-m+k)\text{, } \end{equation} or, equivalently, $(n+m)!=n!\prod_{k=1}^{m}(n+k)$.

If $n$ is large, it is intuitively true that $\prod_{k=1}^{m}(n+k)\sim n^m$, written in precise terms \begin{equation} \lim_{n\to\infty}\frac{n^m}{\prod_{k=1}^{m}(n+k)}=1\text{. } \end{equation} Then, multiplying both sides by $m!$ we have that \begin{equation} m!=\lim_{n\to\infty}\frac{n^mm!}{\prod_{k=1}^{m}(n+k)}=\lim_{n\to\infty}\frac{n^mm!}{\frac{(m+n)!}{n!}}=\lim_{n\to\infty}\frac{n^mm!n!}{m!\prod_{k=1}^{n}(m+k)}=\lim_{n\to\infty}\frac{n^mn!}{\prod_{k=1}^{n}(m+k)}\text{, } \end{equation} so, to sum up, we have the following expression, which allows us to potentially compute the "factorial" of any real number $x$ (except for the negative integers) \begin{equation} x!=\lim_{n\to\infty}\frac{n^xn!}{\prod_{k=1}^{n}(x+k)}\text{. } \end{equation}

It is known that this expression is a shifted version of the Gamma function, but why? And I mean, really why? I have read a proof here, but it is pretty magical and non informative, the appropiate function just appears out of thin air.

My main goal is trying to derive the Gamma function integral formula from scratch.

  • $\begingroup$ Which definition of the Gamma function would you like to show this is equivalent to? $\endgroup$
    – Jack M
    Commented Oct 19, 2021 at 11:42
  • $\begingroup$ @JackM The classical $\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}dx$ $\endgroup$ Commented Oct 19, 2021 at 11:59

1 Answer 1


Here is a hint:

Theorem. (Bohr-Mollerup)

If a function $f(x)$ satisfies the following three conditions, then it is identical in its domain of definition with the Gamma function:

  1. $f(x+1) = xf(x)$
  2. The domain of definition of $f(x)$ contains all $x>0$ and is log-convex for these $x$.
  3. $f(1) = 1$.

It can be shown that

\begin{equation} f(x)=\lim_{n\to\infty}\frac{n^xn!}{\prod_{k=1}^{n}(x+k)}\text{. } \end{equation}

Satisfy these conditions and hence is identical to the Gamma function.

It is important to note that not all the functions that satisfy the functional equation

$$ f(x+1) = xf(x), \;\; f(1)=1$$ can be called "Gamma". The property that determine a Gamma function is the log-convexity.

Please see the proof of the Bohr-Mollerup theorem here (is a little bit extensive) and through it you will discover that $\Gamma(x) = f(x)$

  • $\begingroup$ Thank you, I knew about that theorem, but I was looking for a more elementary proof, which could be understood by early undergraduate students without prior knowledge of complex analysis or the classical integral form of the Gamma function. $\endgroup$ Commented Oct 19, 2021 at 13:05
  • $\begingroup$ Take a look at the history of how Euler developed the formula here math.stackexchange.com/questions/840227/…. Also you can take a look on page 1 and 2 of this document citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – Bertrand87
    Commented Oct 19, 2021 at 13:32

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