Let $m,n\in\mathbb N$ with $m<n$ and $0<s<1$, $1\leq p<\infty$. Is the following holds $$\frac{\Gamma(\frac{sp+p+n-2}{2})\Gamma(\frac{n-m+sp}{2})}{\Gamma(\frac{n+sp}{2})\Gamma(\frac{sp+p+n-m-2}{2})}=1.$$ Where $\Gamma$ is the usual gamma function.
Clearly for $p=2$, its true.
I'm trying to show this for any $p$ with the above range but couldn't do.
For example with $m=1$ , $n=2$ , $s=\frac12$ , $p=4$
$\Gamma(\frac{sp+p+n-2}{2})= 2$
$\Gamma(\frac{n-m+sp}{2}) = \frac{\sqrt{\pi}}{2}$
$\Gamma(\frac{n+sp}{2}) = 1$
$\Gamma(\frac{sp+p+n-m-2}{2}) = \frac{3\sqrt{\pi}}{4}$
$$\frac{\Gamma(\frac{sp+p+n-2}{2}) \Gamma(\frac{n-m+sp}{2}) }{ \Gamma(\frac{n+sp}{2}) \Gamma(\frac{sp+p+n-m-2}{2})} = \frac{2\frac{\sqrt{\pi}}{2}}{1\frac{3\sqrt{\pi}}{4}} = \frac43$$
The result is different from $1$.
Thus, $\frac{\Gamma(\frac{sp+p+n-2}{2}) \Gamma(\frac{n-m+sp}{2}) }{ \Gamma(\frac{n+sp}{2}) \Gamma(\frac{sp+p+n-m-2}{2})} $ is not always equal to $1$.