# Integral Form of Gamma Function equivalent to Euler Form

## Theorem

The following definitions of the concept of 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:

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

### 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 1

It is taken for granted that the Gamma function increases monotonically on $\R_{\ge 1}$.

We begin with an inequality that can easily be verified using cross multiplication.

Let $x$ be a real number between $0$ and $1$.

Let $n$ is a positive integer.

Then:

$\displaystyle \frac {\log \Gamma \left({n - 1}\right) - \log \Gamma \left({n}\right)} {\left({n - 1}\right) - n} \le \frac {\log \Gamma \left({x + n}\right) - \log \Gamma \left({n}\right)} {\left({x + n}\right) - n} \le \frac {\log \Gamma \left({n + 1}\right) - \log \Gamma \left({n}\right)}{\left({n + 1}\right) - n}$

Since n is a positive integer, we can make use of the identity:

$\Gamma \left({n}\right) = \left({n - 1}\right)!$

Simplifying, we get:

$\log \left({n - 1}\right) \le \dfrac {\log \Gamma \left({x + n}\right) - \log \left({\left({n - 1}\right)!}\right)} x \le \log \left({n}\right)$

We now make use of the identity:

$\displaystyle \Gamma \left({x + n}\right) = \prod_{k \mathop = 1}^n \left({x + n - k}\right) \Gamma \left({x}\right)$

along with the fact that the Gamma Function is Log-Convex, to simplify the inequality:

$\displaystyle \left({n - 1}\right)^x \left({n - 1}\right)! \prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1} \le \Gamma \left({x}\right) \le n^x \left({n - 1}\right)!\prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1}$

Taking the limit as $n$ goes to infinity and using the Squeeze Theorem:

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

which is another representation of Euler's form.

This proves equivalence for $x$ between $0$ and $1$.

The result follows from the Gamma Difference Equation.

$\blacksquare$

## Proof 2

First we present a lemma:

Let $0 \le t \le m$.

Then:

$0 \le e^{-t} - \left({1 - \dfrac t m}\right)^m \le t^2 \dfrac {e^{-t} } m$

Recall the definition of the partial Gamma function:

$\displaystyle \Gamma_m \left({x}\right) := \frac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$

We have that:

 $\displaystyle$  $\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)$ $\displaystyle$ $=$ $\displaystyle \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)$ $\displaystyle$ $=$ $\displaystyle \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \int_0^m \left({1 - \frac t m}\right)^m t^{x - 1} \rd t$ Partial Gamma Function expressed as Integral $\displaystyle$ $=$ $\displaystyle \int_m^\infty e^{-t} t^{x - 1} \rd t + \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t$

We have that for large $t$:

$t^{x - 1} < t^{t / 2}$

and so as $m \to \infty$:

$\displaystyle \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$

Then:

 $\displaystyle$  $\displaystyle \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t$ $\displaystyle$ $\le$ $\displaystyle \int_0^m \frac {t^2 e^{-t} } m e^{-t} t^{x - 1} \rd t$ Lemma $\displaystyle$ $=$ $\displaystyle \frac 1 m \int_0^m t^{x + 1} e^{-t} \rd t$ $\displaystyle$ $<$ $\displaystyle \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t$

Now we have that as $m \to \infty$:

$\displaystyle \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$

so:

$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right) = 0$

$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$
$\blacksquare$