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

1 Answer
Aug 8, 2018

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