# Equivalence of Definitions of Gamma Function

## Theorem

The following definitions of the concept of the gamma function are equivalent:

### Integral Form

The Gamma function $\Gamma: \C \to \C$ is defined, for the open right half-plane, as:

$\displaystyle \map \Gamma z = \map {\mathcal M \set {e^{-t} } } z = \int_0^{\to \infty} t^{z - 1} e^{-t} \rd t$

where $\mathcal M$ is the Mellin transform.

For all other values of $z$ except the non-positive integers, $\map \Gamma z$ is defined as:

$\map \Gamma {z + 1} = z \, \map \Gamma z$

### Weierstrass Form

The Weierstrass form of the Gamma function is:

$\displaystyle \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$

where $\gamma$ is the Euler-Mascheroni constant.

The Weierstrass form is valid for all $\C$.

### Hankel Form

The Hankel form of the Gamma function is:

$\displaystyle \frac 1 {\Gamma \left({z}\right)} = \dfrac 1 {2 \pi i} \oint_{\mathcal H} \frac {e^t \, \mathrm d t} {t^z}$

where $\mathcal H$ is the contour starting at $-\infty$, circling the origin in an anticlockwise direction, and returning to $-\infty$.

The Hankel form is valid for all $\C$.

### Euler Form

The Euler form of the Gamma function is:

$\displaystyle \Gamma \left({z}\right) = \frac 1 z \prod_{n \mathop = 1}^\infty \left({\left({1 + \frac 1 n}\right)^z \left({1 + \frac z n}\right)^{-1}}\right) = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \left({z + 1}\right) \left({z + 2}\right) \cdots \left({z + m}\right)}$

which is valid except for $z \in \left\{{0, -1, -2, \ldots}\right\}$.

## Proof

### Weierstrass Form equivalent to Euler Form

First it is shown that the Weierstrass form is equivalent to the Euler form.

 $\displaystyle \frac 1 {\map \Gamma z}$ $=$ $\displaystyle z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} }$ Weierstrass Form of $\Gamma$ Function $\displaystyle$ $=$ $\displaystyle z \paren {\lim_{m \mathop \to \infty} \exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} } \paren {\lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} } }$ Definition of Euler-Mascheroni Constant

Combining the limits:

 $\displaystyle \frac 1 {\map \Gamma z}$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {\paren {1 + \frac z n} e^{-z/n} } }$ $\displaystyle$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \exp \paren {\frac {-z} 1 + \frac {-z} 2 + \cdots + \frac {-z} m} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }$ $\displaystyle$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 - 1 + \frac 1 2 - \frac 1 2 + \cdots + \frac 1 m - \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }$ Exponential of Sum $\displaystyle$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {m^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }$

But:

$(1): \quad m = \dfrac {m!} {\paren {m - 1}!} = \dfrac 2 1 \cdot \dfrac 3 2 \cdots \dfrac {x + 1} x \cdots \dfrac m {m - 1}$

Each term in $(1)$ is just $\dfrac {x + 1} x = 1 + \dfrac 1 x$, so:

$\displaystyle m = \prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}$

Thus the expression for $\dfrac 1 {\map \Gamma z}$ becomes:

 $\displaystyle$  $\displaystyle z \lim_{m \mathop \to \infty} \paren {\prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }$ $\displaystyle$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {\paren {1 + \frac 1 m}^z \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n} }$ $\displaystyle$ $=$ $\displaystyle z \lim_{m \mathop \to \infty} \paren {1 + \frac 1 m}^z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n}$ Product Rule for Complex Sequences $\displaystyle$ $=$ $\displaystyle z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n}$

Hence:

$\displaystyle \map \Gamma z = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1}$

which is the Euler form of the Gamma function.

$\Box$

### Integral Form equivalent to Euler Form

This is proved in the page:

Integral Form of Gamma Function equivalent to Euler Form

$\blacksquare$