Factorial of Half/Proof 2

Theorem

$\left({\dfrac 1 2}\right)! = \dfrac {\sqrt \pi} 2$

Proof

$\displaystyle \prod_{n \mathop \ge 1} \dfrac {\left({n + \alpha_1}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \cdots \left({n + \beta_k}\right)} = \dfrac {\Gamma \left({1 + \beta_1}\right) \cdots\Gamma \left({1 + \beta_1}\right)} {\Gamma \left({1 + \alpha_1}\right) \cdots\Gamma \left({1 + \alpha_k}\right)}$

where:

$\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
none of the $\beta$s is a negative integer.

Setting:

$k = 2$
$\alpha_1 = \alpha_2 = 0$
$\beta_1 = -\dfrac 1 2, \beta_2 = \dfrac 1 2$

we see that:

$\alpha_1 + \alpha_2 = \beta_1 + \beta_2$

So this reduces to:

 $\ds \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2}$ $=$ $\ds \dfrac {\Gamma \left({1 - \frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)} {\Gamma \left({1}\right) \Gamma \left({1}\right)}$ $\ds$ $=$ $\ds \Gamma \left({\frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)$ as $\Gamma \left({1}\right) = 0! = 1$ $\ds$ $=$ $\ds 2 \Gamma \left({1 + \frac 1 2}\right)^2$ Gamma Difference Equation

We then note that by Wallis's Product:

$\displaystyle \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2} = \frac \pi 2$

Thus:

 $\ds \frac \pi 2$ $=$ $\ds 2 \Gamma \left({1 + \frac 1 2}\right)^2$ $\ds \leadsto \ \$ $\ds \frac \pi 4$ $=$ $\ds \left({\left({\frac 1 2}\right)!}\right)^2$ Gamma Function Extends Factorial $\ds \leadsto \ \$ $\ds \left({\frac 1 2}\right)!$ $=$ $\ds \frac {\sqrt \pi} 2$

$\blacksquare$