Integral Form of Gamma Function equivalent to Euler Form/Proof 1

Theorem

The following definitions of the concept of Gamma Function are equivalent:

Integral Form

The gamma function $\Gamma: \C \setminus \Z_{\le 0} \to \C$ is defined, for the open right half-plane, as:

$\ds \map \Gamma z = \map {\MM \set {e^{-t} } } z = \int_0^{\to \infty} t^{z - 1} e^{-t} \rd t$

where $\MM$ 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:

$\ds \map \Gamma z = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1} } = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \cdots \paren {z + m} }$

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

Proof

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

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

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

Let $n$ is a positive integer.

Then:

$\ds \frac {\log \map \Gamma {n - 1} - \log \map \Gamma n} {\paren {n - 1} - n} \le \frac {\log \map \Gamma {x + n} - \log \map \Gamma n} {\paren {x + n} - n} \le \frac {\log \map \Gamma {n + 1} - \log \map \Gamma n} {\paren {n + 1} - n}$

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

$\map \Gamma n = \paren {n - 1}!$

Simplifying, we get:

$\map \log {n - 1} \le \dfrac {\log \map \Gamma {x + n} - \map \log {\paren {n - 1}!} } x \le \map \log n$

We now make use of the identity:

$\ds \map \Gamma {x + n} = \prod_{k \mathop = 1}^n \paren {x + n - k} \map \Gamma x$

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

$\ds \paren {n - 1}^x \paren {n - 1}! \prod_{k \mathop = 1}^n \paren {x + n - k}^{-1} \le \map \Gamma x \le n^x \paren {n - 1}!\prod_{k \mathop = 1}^n \paren {x + n - k}^{-1}$

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

$\ds \map \Gamma x = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \paren {x + n - k}^{-1}$

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.

$\blacksquare$