# Why $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$

*This is based on a Mathematical Note written by George Osborn: link to DOME. ((I have made some modifications to the notation, treated limits with a little more care, and introduced a minor change in the line of reasoning, but otherwise this is entirely a gloss on Osborn’s work.))*

Start by defining a function, $G$, such that $G(n, 1+x) = \frac{n^x}{\Pi_{i=1}^{n}\left(1+\frac{x}{i}\right)}$.

For example, $G(3, 1 + 1) = \frac{3^1}{\left(1+ \frac{1}{1}\right)\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)}$, which is $\frac{3}{4)}$. In fact, there’s a lot of cancelling in the denominator, and in general $G(n, 1+1) = \frac{n}{n+1}$.

We’ll also let $\Gamma(1+x)$ be the limit of $G(n,1+x)$ as $n \to \infty$. This is the gamma function you’re possibly used to, but we’ve not proved that yet; you might have spotted that $\Gamma(1+1) = 1$.

### Is this $\Gamma$ our $\Gamma$?

We can look also at $G(n,x)$, which is $\frac{n^{x-1}}{\Pi_{i=1}^n 1 + \frac{x-1}{i}}$, and consider $\frac{G(n,1+x))}{xG(n,x)}$. That’s ((I’m taking the limits as read because I’m lazy)) $\frac{n^x \Pi 1 + \frac{x-1}{i}}{xn^{x-1} \Pi 1 + \frac{x}{i}}$; almost all of the $n$s cancel to give $\frac{n \Pi 1 + \frac{x-1}{i}}{x \Pi 1 + \frac{x}{i}}$.

Almost all of the terms in the products cancel, too! The product on top is $\frac{x}{1} \cdot \frac{1+x}{2} \dots \frac{n+x-1}{n}$; on the bottom, it’s $\frac{1+x}{1} \cdot \frac{2+x}{2} \dots \frac{n+x}{n}$. The denominators cancel, and all of the factors in the numerators vanish except for $x$ on top and $n+x$ on the bottom.

That leaves us with $\frac{G(n, 1+x)}{xG(n, x)} = \frac{nx}{x(n+x)} = \frac{n}{n+x}$.

As a result, when $n \to \infty$, $\Gamma(1+x) = x \Gamma(x)$. Also, $G(n, 1+0)$ is the ratio of two empty products, so $\Gamma(1) = 1$; by induction, you can show that $\Gamma(1+x) = x!$ ((Osborn’s note uses a sort of “L” shape around the $x$ to denote a factorial. Apparently that’s an old-fashioned thing.))

### $\Gamma\left(\frac{1}{2}\right)$

Now let’s consider $G(n, 1+x) \cdot G(n,1-x)$. Clearly the $n^x$s cancel, leaving just $\frac{1}{\left(\Pi 1 + \frac{x}{i}\right)\left(\Pi 1 - \frac{x}{i}\right)}$, so $\Gamma(1+x) \cdot \Gamma(1-x) = \Pi_{i=1}^{\infty} \left( 1 - \left(\frac{x}{i}\right)^2\right)^{-1}$.

What’s that product? It is undefined whenever $x$ is a positive integer, it’s 1 when $x=0$, and a little messing about will hopefully convince you that it’s equivalent to $\pi x \cosec(\pi x)$. ((Seriously, how neat is that?))

In particular, if $x= \frac{1}{2}$, then $\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right) = \piby 2$ (because $\cosec\left(\piby 2\right) = 1)$.

Also, $\Gamma\left(\frac{3}{2}\right) = \frac{1}{2}\Gamma\left(\frac{1}{2}\right)$, so $\left[\Gamma\left(\frac{1}{2}\right)\right]^2 = \pi$.

$\blacksquare$.