I know the expression for $\Gamma(\frac{1}{2}+n)$ as found in here Proving that $\Gamma \left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$. . To me, even an expression for $\Gamma(\frac{n}{2}+1)$, $n\in \mathbb{N}$ it gets complicated. Any idea how to understand the simpler one in order to extend it to the case where both $m$ and $n$ are involved?
If I started this $\Gamma(\frac{n}{2}+1)=(\frac{n}{2})!=(\frac{n}{2}-1) (\frac{n}{2} -2)\cdots \sqrt{\pi}$ this gives me no simple expression or something .
$$\Gamma\left(\frac{m+n}{2}+1\right)\\ =\Gamma\left(\frac{m+n+2}{2}\right)$$ Wikipedia gives the formula: $$\Gamma\left(\frac{x}{2}\right)=\frac{\sqrt{\pi}\cdot\left(x-2\right)!!}{2^{\frac{x+1}{2}}},\ for\ x\in\mathbb{N}$$ For some positive integer $x$, where $!!$ is the double factorial. We can now substitute $x=m+n+2$ to expand our original equation into the form: $$\frac{\sqrt{\pi}\cdot\left(m+n+2-2\right)!!}{2^{\frac{m+n+2+1}{2}}},\ for\ m+n+2\in \mathbb{N}\\ =\frac{\left(m+n\right)!!}{2}\cdot\sqrt{\frac{\pi}{2^{m+n+1}}},\ for\ m+n+2\in \mathbb{N}$$ However, simplifying the expression further requires splitting the equation into two. WolframAlpha gives the formula: $$\frac{x!}{x!!}=(x-1)!!$$ As well as: $$(2x)!!=2^x\cdot x!,\ for\ x=0,1,\cdots$$ Together, we can say that for odd $x$: $$x!!=\frac{(x+1)!}{(x+1)!!}=\frac{(x+1)!}{2^\frac{x+1}{2}\cdot \left(\frac{x+1}{2}\right)!}=\frac{(x+1)!}{2^\frac{x+1}{2}\cdot \left(\frac{\sqrt{\pi}\cdot\left(x+1-2\right)!!}{2^{\frac{x+1+1}{2}}}\right)}=\frac{(x+1)!}{\left(x-1\right)!!}\cdot\sqrt{\frac{2}{\pi}}$$ Reiterating this process $x-1$ times gives the resulting formula: $$x!!=(x+1)!\cdot\left(\frac{2}{\pi}\right)^\frac{x}{2},\ for\ \ \frac{x-1}{2}\in\mathbb{N}$$ Returning once again to our original equation with this newfound knowledge, we can further expand our equation for neither odd $m+n$ nor even $m+n$. This is because both expressions can be rewritten in terms of $\left(\frac{m+n}{2}\right)!$ which is our original equation. So, with this in mind we have obtained the following result: $$\Gamma\left(\frac{m+n}{2}+1\right)=\frac{\left(m+n\right)!!}{2}\cdot\sqrt{\frac{\pi}{2^{m+n+1}}},\ for\ m+n+2\in \mathbb{N}$$ However, you might be able to come up with a similar result by using another formula from WolframAlpha : $$\Gamma\left(n+\frac{1}{2}\right)=\frac{\left(2n-1\right)!!}{2^n}\cdot\sqrt{\pi}$$