# Find a formula for $\Gamma(\frac{n}{2})$ for positive integer n.

Find a formula for $\Gamma(\frac{n}{2})$ for positive integer n.

I know the following relations; $\Gamma (z+1)=z\Gamma (z)$ and $\Gamma(n+1)=n!$

Please give me a way how to show this. Thank you.

• If you know $\Gamma(\frac12)$, this is trivial. If you don't, you should. – Did Dec 29 '13 at 0:06
• Yes $\Gamma(1/2)=\sqrt{\pi}$ @Did – user315 Dec 29 '13 at 0:07
• – lhf Dec 29 '13 at 0:07

$$\Gamma(n)=(n-1)!$$ and \begin{align} \Gamma\left(n+\frac12\right) &=\Gamma\left(\frac12\right)\frac12\frac32\frac52\cdots\frac{2n-1}{2}\\ &=\sqrt\pi\frac{1\cdot\color{#A0A0A0}{2}\cdot3\cdot\color{#A0A0A0}{4}\cdot5\cdot\color{#A0A0A0}{6}\cdots(2n-1)\cdot\color{#A0A0A0}{2n}}{2^n(\color{#A0A0A0}{2}\cdot\color{#A0A0A0}{4}\cdot\color{#A0A0A0}{6}\cdots\color{#A0A0A0}{2n})}\\ &=\sqrt\pi\frac{(2n)!}{4^nn!} \end{align}
For even $n$, $$\Gamma\left(\frac n2\right)=\left(\frac n2-1\right)!$$ For odd $n$, $$\Gamma\left(\frac n2\right)=\sqrt\pi\frac{(n-1)!}{2^{n-1}\frac{n-1}{2}!}$$
If $n$ is even, use that second identity. Otherwise, use that first identity to relate $\Gamma\left(\frac{n}{2}\right)$ to $\Gamma\left(\frac{1}{2}\right)=\sqrt\pi$.