Show that $\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+\frac 12) = \pi n!$

Question

Could someone please provide a detailed derivation of the following result? $$\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+\frac 12) = \pi n!$$ Also, is it possible to generalize the result for the following sum, given a generic integer $m$? $$\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+m+\frac 12).$$

Thanks in advance for your help.

Graziano

Answer 1

Since $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(1/2)=\sqrt{\pi}$, we have that for any non-negative integer $N$, $$\Gamma\left(N+\frac{1}{2}\right)=\Gamma\left(\frac{1}{2}\right)\prod_{k=0}^{N-1}\left(k+\frac{1}{2}\right)=\sqrt{\pi}\,\frac{(2N)!}{4^N N!}.$$ Hence $$\begin{align} \sum_{k=0}^n \binom{n}{k}\left(k+\frac{1}{2}\right)\Gamma\left(n-k+\frac{1}{2}\right)&=\sum_{k=0}^n \frac{n!}{k!(n-k)!}\cdot\sqrt{\pi}\,\frac{(2k)!}{4^k k!}\cdot\sqrt{\pi}\, \frac{(2(n-k))!}{4^{n-k} (n-k)!}\\ &=\frac{\pi n!}{4^n}\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=\pi n! \end{align}$$ where at the last step we used the Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ .

In a similar way you should be able to show the more general identity, $$\sum_{k=0}^n \binom{n}{k}\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(n-k+m+\frac{1}{2}\right) =\frac{\pi(n+m)!}{4^m}\binom{2m}{m}.$$

Answer 2

In an alternative way,

$$ \Gamma(k+1/2)\Gamma(n-k+1/2) = n! B(k+1/2,n-k+1/2)=n!\int_{0}^{1}(1-x)^{k-1/2}x^{n-k-1/2}\,dx $$ hence by multiplying both sides by $\binom{n}{k}$ and summing over $k=0,1,\ldots,n$ we get $$ \sum_{k=0}^{n}\binom{n}{k}\Gamma(k+1/2)\Gamma(n-k+1/2)=n!\int_{0}^{1}\frac{x^n}{\sqrt{x(1-x)}}\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1-x}{x}\right)^k\,dx $$ and the RHS equals $$ n!\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}} = n!\cdot\Gamma\left(\tfrac{1}{2}\right)^2 = \color{red}{\pi\cdot n!}.$$