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

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.