How to prove this gamma identity?

Question

How to prove this? $$2^n \ \Gamma(n+\frac{1}{2})\ =\ 1.3.5...(2n-1)\ \sqrt{\pi}$$ I tried rewriting the right-hand side as $$\frac{(2n-1)!}{2(n-1/2)}\ \sqrt{\pi}=\frac{\Gamma(2n)}{2\Gamma(n+1/2)}\sqrt{\pi}$$ But I have no idea what to do after that.

Answer 1

Use the functional equation of the Gamma function $$\Gamma\left(n+\frac{1}{2}\right)= (n-\frac{1}{2})\Gamma\left(n-\frac{1}{2}\right) =\Gamma\left(n-\frac{1}{2}\right)\frac{2n-1}{2} $$ Repeat this to get $$\Gamma\left(n+\frac{1}{2}\right)= \Gamma\left(\frac{1}{2}\right)\prod_{k=1}^n\frac{2k-1}{2} = \Gamma\left(\frac{1}{2}\right)\frac{\prod_{k=1}^n(2k-1)}{2^n} $$

Now multiply with ${2^n}$ and use $\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}.$

Answer 2

First, from the integral expression of $\Gamma(x) = \int\limits_0^\infty t^{x-1} e^{-t} dt$ you can prove by integration by parts that $\Gamma(x+1)=x\Gamma(x)$. Applying this recursivelly to $2^n \Gamma(n+1)$ yields $$2^n \Gamma(n+\frac{1}{2}) = (2n-1)(2n-2) \ldots 3 \cdot 1 \cdot \Gamma(\frac{1}{2}).$$ Finally, $\Gamma(\frac{1}{2})=\sqrt{\pi}$, which can be seen substituting $t \rightarrow t^2$ in the integral expression for $\Gamma(\frac{1}{2})$ and using the well known $\int\limits_0^\infty e^{-t^2} dt = \frac{\sqrt{\pi}}{2}$.