I was tasked to evaluate the Beta function $$B\left(\frac{2021}{2},\frac{1}{2}\right) \tag{1}$$
I've expanded this to $$\frac{\Gamma\left(\dfrac{2021}{2}\right)\Gamma\left(\dfrac{1}{2}\right)}{\Gamma(1011)} \tag{2}$$
How do I simplify $\Gamma\left(\dfrac{2021}{2}\right)$ and then $(2)$, knowing that $\Gamma(1011) = 1010!$ and $\Gamma\left(\frac{1}{2}\right) = \sqrt \pi$ ?
Check the duplication formula here https://brilliant.org/wiki/gamma-function/#eulers-reflection-formula , with $z=2020/2$, so that:
$\Gamma(2020/2+1/2)\Gamma(1010)=\frac{\sqrt{\pi}}{2^{2019}}\Gamma(2020)$ [1]
Note that from here one can get: $\Gamma(2020/2+1/2)=\frac{\sqrt{\pi}}{2^{2019}}\frac{2019!}{1009!}$.
Therefore, using your expansion and the above:
$B(2021/2,1/2)=\frac{\pi}{2^{2019}}\frac{2019!}{1010!1009!}=\frac{\pi}{2^{2019}}\binom{2019}{1010}$
Is the last form acceptable ? Using the same methodology one could check the general identity:
$$B\left(k+\frac{1}{2},\frac{1}{2}\right)=\frac{\pi}{2^{2k-1}}\binom{2k-1}{k}$$
for every $k$ natural. For $k=1010$ we have the OP expression.
Legendre's duplication formula: $$\Gamma(2z)=\frac{2^{2z-1}\Gamma(z)\Gamma(z+1/2)}{\sqrt{\pi}}$$ Using $z=2021/2$, $$\Gamma(2021)=\frac{2^{2\frac{2021}{2}-1}\Gamma(2021/2)\Gamma(2021/2+1/2)}{\sqrt{\pi}}$$ $$2020!=\frac{2^{2020}\Gamma(2021/2)\cdot1010!}{\sqrt{\pi}}$$ $$\Gamma(2021/2)=\sqrt{\pi}\frac{2020!}{2^{2020}1010!}$$
