Equivalence of Gamma Function Definitions

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

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

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle z \lim_{m \to \infty} \left({\prod_{n=1}^{m-1} \left({1 + \frac 1 n}\right)^{-z} \prod_{n=1}^m \left({1 + \frac z n}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle z \lim_{m \to \infty} \left({\left({1 + \frac 1 m}\right)^z \prod_{n=1}^m \left({1 + \frac 1 n}\right)^{-z} \left({1 + \frac z n}\right)}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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.