Beta Function and the Legendre Duplication Formula

Consider the fact that,

$$m!n!=\lim_{a^2 \rightarrow \infty} \int^{a^2}_0 e^{-u}u^m du \int^{a^2}_0e^{-v}v^n dv$$ (3.1)

When we transform to polar coordinates, we have,

$$m!n! = (m+n+1)!2\int^{\pi/2}_0 \cos^{2m+1}\theta \sin^{2n+1}\theta d\theta$$ (3.2)

The beta function is so defined as,

$$B(m+1,n+1)\equiv 2 \int^{\pi/2}_0 \cos^{2m+1}\theta \sin^{2n+1}\theta d\theta = \frac{m!n!}{(m+n+1)!}$$ (3.3)

With $t=\cos^2\theta$, we have

$$\frac{m!n!}{(m+n+1)!} = \int^1_0 t^m(1-t)^n dt \underbrace{=}_{t = x^2} 2 \int^1_0 x^{2m+1}(1-x^2)^n dx$$ (3.4)

From these we can obtain the Legendre duplication formula, where we start with,

$$\frac{z!z!}{(2z+1)!} = \int^1_0 t^z(1-t)^z dt \underbrace{=}_{t = (1+s)/2}2^{-2z}\int^1_0 (1-s^2)^z ds$$ (3.5)

From the definition of the Beta function, we have,

$$\frac{z!z!}{(2z+1)!}=2^{-2z-1}\frac{z!(-\frac{1}{2})!}{(z+\frac{1}{2})!}$$ (3.6)

It is easy to show that $(-\frac{1}{2})!=\sqrt{\pi}$. Multiplying through equation 3.6 by $z+\frac{1}{2}$, we have,

$$z!(z-\frac{1}{2})! = 2^{-2z}\sqrt{\pi}(2z)!$$ (3.7)