Calculating $\Gamma (n+5/2)$

Question

Hey I'm pretty new to the gamma function and was trying to calculate $\Gamma (n+\frac52)$. I got to the integer $\int_0^\infty t^{n+\frac32}e^{-t}\mathrm dt\\$, and I really don't know how to go on from here. Any help is greatly appreciated.

Answer 1

There is the easy way of using the identity $z\Gamma(z)=\Gamma(z+1)$.

However, we will use the duplication formula \begin{align} \Gamma\left(z+\frac{1}{2}\right) = 2^{1-2z}\sqrt{\pi}\frac{\Gamma(2z)}{\Gamma(z)}. \end{align}

In particular, when $z = n+\frac{5}{2}$ we see that \begin{align} \Gamma\left(n+2+\frac{1}{2}\right) = 2^{-2n-4}\sqrt{\pi} \frac{(2n+3)!}{(n+1)!}. \end{align}

Answer 2

$$\Gamma(\frac{1}{2})=\sqrt\pi$$ $$\Gamma(\frac{3}{2})=\Gamma(\frac{1}{2})\frac{1}{2}$$ $$\Gamma(\frac{5}{2})=\Gamma(\frac{3}{2})\frac{3}{2}=\Gamma(\frac{1}{2})\ \frac{1}{2}\ \frac{3}{2}$$ Following this pattern: $$\Gamma(\frac{1}{2}+n)=\Gamma(\frac{1}{2})\ \frac{1}{2}\ \frac{3}{2}...\frac{2n-1}{2}=\Gamma(\frac{1}{2})\frac{1\cdot3\cdot5...(2n-1)}{2^n}$$ $$1\cdot3\cdot5...(2n-1)=\frac{2n!}{2\cdot4\cdot...2n}=\frac{(2n)!}{2^nn!}$$ Therefore $$\Gamma(\frac{1}{2}+n)=\Gamma(\frac{1}{2})\frac{(2n)!}{4^nn!}=\sqrt\pi\frac{(2n)!}{4^nn!}$$ I know it is not rigorous, but the result is proper ;).