Show that $\sum_{k=0}^n \binom{n}{k}\Gamma(k+\frac 1 2)\Gamma(n-k+\frac 12) = \pi n!$ 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
 A: 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,
$$\sum_{k=0}^n \binom{n}{k}\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(n-k+m+\frac{1}{2}\right)
=\frac{\pi(n+m)!}{4^m}\binom{2m}{m}.$$
A: 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
$$ n!\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}} = n!\cdot\Gamma\left(\tfrac{1}{2}\right)^2 = \color{red}{\pi\cdot n!}.$$
