What is $\Gamma(1/2-n)\Gamma(1/2+n)$?

Question

I've been trying to work this out but am having little luck. $n$ here is an integer. Ive found that $$\Gamma(n+1/2) = \frac{1}{2}\Gamma(n-1/2)$$ But I am unsure how to handle $\Gamma$ when it holds a negative value.

Answer 1

Euler's reflection formula write $$\Gamma(z)\,\Gamma(1-z)=\frac \pi {\sin(\pi z)}$$ Making $z=\frac 12+n$ then gives $$\Gamma\left(\frac 12+n\right)\,\Gamma\left(\frac 12-n\right)=\frac \pi {\sin\left(\pi (\frac 12+n)\right)}=\pi \sec (\pi n)$$