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

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

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.