Gamma function and product of $\int_0^\pi \sin^k x\,dx$

Question

I want to prove that: \begin{equation} \pi \prod_{k=1}^{n-2}\int_0^\pi\sin^kx\,dx=\frac{\pi^{n/2}}{\Gamma(n/2)},\quad\forall n\in\mathbb{N}. \end{equation} I have set: $$ I_k:=\int_0^\pi\sin^kx\,dx,\quad\forall k\in\mathbb{N},$$ then, by integration by part: $$ I_k:=\frac{k-1}{k}I_{k-2},\quad\forall k\in\mathbb{N}. $$ At this points i haven't any idea to go on. Any help would be appreciated.

Answer 1

We use the trigonometric form of the beta function, which reads $$2\int_0^{\pi/2}\sin^{2m-1}x\cos^{2n-1}x\:dx=\frac{\Gamma(m)\:\Gamma(n)}{\Gamma(m+n)}$$ Setting $m=\frac{k+1}{2}$ and $n=\frac{1}{2}$, we obtain $$2\int_0^{\pi/2} \sin^k x\:dx=\frac{\Gamma\left(\frac{k+1}{2}\right)\:\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)}$$ Since $\sin(\pi-x) = \sin x$, we have $$I_k\stackrel{\text{def}}{=}\int_0^{\pi}\sin^kx\:dx=2\int_0^{\pi/2}\sin^k x\:dx=\frac{\Gamma\left(\frac{k+1}{2}\right)\:\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)}$$ Now, we would like to determine the product $$\begin{align} \pi\:\prod_{k=1}^{n-2}I_k &= \pi\:\prod_{k=1}^{n-2}\frac{\Gamma\left(\frac{k+1}{2}\right)\:\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{k+2}{2}\right)} \\ &= \pi\:\frac{\Gamma(1)\:\Gamma\left(\frac{1}{2}\right)^{n-2}}{\Gamma\left(\frac{n}{2}\right)} \\ &=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)} \end{align} $$ The problem is solved.