Question

Prove that `((1/2)!)=sqrt(pi)/2`?

Answer 1

We are going to begin by switching to a much more useful form: the Gamma function, which is defined thus:

`Gamma(z) = int_0^infty x^-z e^-x dx = (z+1)!`
The last property technically means that this is an analytical continuation/extension of the factorial function, which arguably only makes sense on integers.

Anyway, by the above,
`(1/2)! = Gamma(-1/2) = int_0^infty x^(1/2)e^-x dx`
This is a bit difficult (it involves three u-substitutions versus our 2), so let's apply another property of factorial:
`(1/2)! = 1/2 * (-1/2)! implies (-1/2)! = 2 * (1/2)!`
hence
`(1/2)! = 1/2 * int_0^infty x^(-1/2) e^-x dx`
While this seems more complicated, consider the `u`-sub, `u = sqrt(x)`. Thus,
`(1/2)! = int_0^inftye^(-u^2) du`

We can do the Gaussian integral by hand (or you can just know its value):
`I = int_0^infty e^(-u^2)du implies I^2 =int_0^infty int_0^infty e^(-(u^2+v^2))dudv`
`I^2 = int_0^(pi/2) int_0^infty e^(-r^2) cdot r dr d theta = pi/2 * int_0^infty e^-q 1/2 dq = pi/4`
`I = sqrt(pi)/2`

Therefore,
`(1/2)! = sqrt(pi)/2`