Equivalence of Definitions of Gamma Function

Theorem
The various versions of the Gamma function given on the definition page are equivalent.

Proof
We first show that the Weierstrass form is equivalent to the Euler form. By the Weierstrass form, we have


 * $\displaystyle \frac 1 {\Gamma(z)} = z e^{\gamma z} \prod_{n=1}^\infty \left({\left({1 + \frac z n}\right) e^{\frac{-z} n} }\right) = z \left({\lim_{m \to \infty} \exp \left({\left({1 + \frac 1 2 + \dots + \frac 1 m - \log(m)}\right) z}\right)}\right) \left({\lim_{m \to \infty} \prod_{n=1}^\infty \left({\left({1 + \frac z n}\right) e^{-z/n} }\right)}\right)$

by the definition of Euler's constant.

Combining the limits, we get

But:
 * $\displaystyle m = \frac{m!}{(m-1)!} = \frac 2 1 \cdot \frac 3 2 \dots \frac{x+1} x \dots \frac m {m-1}$

and each term here is just $\dfrac{x+1} x = 1 + \dfrac 1 x$, so:
 * $\displaystyle m = \prod_{n=1}^{m-1} \left({1 + \frac 1 n}\right)$

and so our expression for $\displaystyle \frac 1 {\Gamma(z)}$ becomes:

a limit which is easily evaluated to be:
 * $\displaystyle z \prod_{n=1}^\infty \left({1 + \frac 1 n}\right)^{-z} \left({1 + \frac z n}\right)$

and hence:
 * $\displaystyle \Gamma(z) = \frac 1 z \prod_{n=1}^{\infty} \left({1 + \frac 1 n}\right)^z \left({1 + \frac z n}\right)^{-1}$

which is precisely the Euler form of the Gamma function.

It remains to be shown that the integral form is equivalent to these definitions.

The Integral Form
Here, we take it for granted that the Gamma function increases monotonically for real numbers greater than or equal to 1. We begin with an inequality that can easily be verified using cross multiplication. x is a real number between 0 and 1 whereas n is a positive integer:


 * $\displaystyle\frac{\log\Gamma(n-1)-\log\Gamma(n)}{(n-1)-n}\leq\frac{\log\Gamma(x+n)-\log\Gamma(n)}{(x+n)-n}\leq\frac{\log\Gamma(n+1)-\log\Gamma(n)}{(n+1)-n}$

Since n is a positive integer, we can make use of the identity $\displaystyle \Gamma(n)=(n-1)!$. Simplifying, we get


 * $\displaystyle\log(n-1)\leq\frac{\log\Gamma(x+n)-\log((n-1)!)}{x}\leq\log(n)$

We now make use of the identity $\displaystyle \Gamma(x+n)=\prod_{k=1}^{n}(x+n-k)\Gamma(x)$, along with the fact that the Gamma function is log-convex, to simplify the inequality:


 * $\displaystyle(n-1)^x(n-1)!\prod_{k=1}^{n}(x+n-k)^{-1}\leq\Gamma(x)\leq n^x(n-1)!\prod_{k=1}^{n}(x+n-k)^{-1}$

Taking the limit as n goes to infinity and using the Squeeze Theorem, we get


 * $\displaystyle \Gamma(x)=\lim_{n\to\infty}n^{x}n!\prod_{k=0}^{n}(x+n-k)^{-1}$

which is another representation of Euler's form. This product can also be shown to be equivalent to the Weierstrass form.

We proved the theorem for x between 0 and 1. If this satisfies the Gamma functional equation, then our proof is valid for all reals but the simple poles of the Gamma function. It turns out this indeed satisfies the functional equation. The proof of this is trivial and hence not presented.