The proof I am dealing with is worded exactly as follows:
Prove $\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$.
The proof itself can be done easily with induction, I assume. However, my issue is with the domain of the given $n$; granted, the factorial operator is only defined for positive integer values. However, the gamma function, as far as I know, is defined for all complex numbers bar $\mathbb{Z}^-$. I don't think induction will suffice for generalizing the proof for said domain. Would that be necessary, or should I be looking for other methods to tackle the proof?
Best Answer
We can also exploit the fact that $\Gamma(x+1)=x\Gamma(x)$ to obtain \begin{align*} \Gamma\left(n + \frac{1}{2}\right) & =\left(n-1+\frac{1}{2}\right)\Gamma\left(n-1+\frac{1}{2}\right) \\ & =\left(n-1+\frac{1}{2}\right)\left(n-2+\frac{1}{2}\right)\Gamma\left(n-2+\frac{1}{2}\right) \\ & = \ldots = \left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\dots \ \frac{1}{2}\ \Gamma\left(\frac{1}{2}\right) \\ & = \frac{(2n-1)(2n-3)\cdots1}{2^n} \cdot \Gamma\left( \frac{1}{2}\right) \\ & =\frac{(2n-1)(2n-2)(2n-3)(2n-4)\cdots1}{2^n(2n-2)(2n-4)\dots2} \cdot\Gamma\left( \frac{1}{2}\right) \\ & =\frac{(2n-1)!}{2^{2n-1}(n-1)!}\Gamma\left( \frac{1}{2}\right)\\ & =\frac{2n!}{2^{2n}n!}\sqrt{\pi}. \end{align*}