Prove that $\Gamma(-m+\frac{1}{2})=\frac{(-1)^m 2^{2m}\sqrt{\pi}m!}{(2m)!}$

Question

I was able to get to this point \begin{eqnarray*} \Gamma(-m+\frac{1}{2})&=&\frac{\Gamma(\frac{1}{2})}{(-m+\frac{1}{2})_m}\\ &=&\frac{\sqrt{\pi}}{\frac{-2m+1}{2}\cdot\frac{-2m+3}{2}\cdots\frac{1}{2}}\\ &=&\frac{(-1)^m\cdot 2^m\cdot\sqrt{\pi}}{(2m-1)\cdot(2m-3)\cdots 2\cdot 1} \end{eqnarray*} How can I continue.

Answer 1

As you noticed, it is enough to apply the functional identity $\Gamma(z+1)=z\,\Gamma(z)$ multiple times.

$$\Gamma\left(-m+\frac{1}{2}\right)=\frac{\Gamma\left(-(m-1)+\frac{1}{2}\right)}{\left(-m+\frac{1}{2}\right)}=\frac{\Gamma\left(-(m-2)+\frac{1}{2}\right)}{\left(-m+\frac{1}{2}\right)\left(-(m-1)+\frac{1}{2}\right)}=\ldots $$

$$ \ldots = \frac{\Gamma\left(\frac{1}{2}\right)}{\left(-m+\frac{1}{2}\right)\cdots\left(-1+\frac{1}{2}\right)}=\frac{(-1)^m \sqrt{\pi}}{\left(m-\frac{1}{2}\right)\cdots\left(1-\frac{1}{2}\right)}=\frac{(-1)^m 2^m\sqrt{\pi}}{(2m-1)!!}$$ By multiplying the last ratio by $1=\frac{(2m)!!}{(2m)!!}=\frac{2^m m!}{(2m)!!}$ we get $$ \Gamma\left(-m+\frac{1}{2}\right) = \frac{(-1)^m 2^{2m}m!\sqrt{\pi}}{(2m)!} $$ as wanted.

Answer 2

Thank you I have seen what to do. When you multiply the expression by $\frac{2^m\cdot m!}{2^m\cdot m!}$ then I derived the equation.