Simplifying the expression involving instances of Gamma function

Question

Is it possible to simplify the following expression involving instances of Gamma function:

$$E(p)=\frac{\frac{\Gamma(\frac{p+1}{2})}{\Gamma(\frac{p+2}{2})}} {\left(\frac{\Gamma(\frac{p+1}{p})^2}{\Gamma(\frac{p+2}{p})}\right)^{\frac{p+2}{2}}}$$

where $p$ is rational (or even real) and $p\geq2$. The bottom part of expression $E$ comes from the formula for the area of a superellipse, i.e., supercircle:

$$\mid x\mid ^p + \mid y \mid ^p =r^p,\ p\geq 2$$

and the rest is related to that also. Thanx in advance.

Answer 1

I guess it depends on what you mean by simplify. We could rewrite things in terms of (generalized) central binomial coefficients:

First the denominator: Notice that $$\frac{\Gamma\left(1+\frac{1}{p}\right)^{2}}{\Gamma\left(1+\frac{2}{p}\right)}=\binom{\frac{2}{p}}{\frac{1}{p}}^{-1}=\frac{1}{2p}\frac{\Gamma\left(\frac{1}{p}\right)^{2}}{\Gamma\left(\frac{2}{p}\right)}.$$ For the numerator $$\frac{\Gamma\left(\frac{p+1}{2}\right)}{\Gamma\left(\frac{p+2}{2}\right)}=\frac{\Gamma\left(\frac{p+1}{2}\right)^{2}}{\Gamma\left(\frac{p+1}{2}+\frac{1}{2}\right)\Gamma\left(\frac{p+1}{2}\right)}=\frac{\Gamma\left(\frac{p+1}{2}\right)^{2}}{\sqrt{\pi}2^{-p}\Gamma\left(p+1\right)}=\frac{2^{p}}{p\sqrt{\pi}}\binom{p-1}{\frac{p-1}{2}}^{-1}$$ so the fraction becomes $$\frac{2^{p}}{p\sqrt{\pi}}\binom{\frac{2}{p}}{\frac{1}{p}}^{\frac{p+2}{2}}\biggr/\binom{p-1}{\frac{p-1}{2}}.$$ You could also write it using the beta function, then it is $$\frac{2^{\frac{3p+2}{2}}p^{\frac{p+2}{2}}}{\sqrt{\pi}}\frac{\text{B}\left(\frac{1}{p},\frac{1}{p}\right)^{\frac{p+2}{2}}}{\text{B}\left(\frac{p+1}{2},\frac{p+1}{2}\right)}.$$ To clean it up, it feels like you need a nicer way to write $\Gamma\left(\frac{1}{p}\right)^{p}$. It seems to look like a multinomial coefficient.

Now, there is a way to rewrite everything as a multidimensional integral over a simplex, and I find this to be the cleanest way to rewrite it. This is related to a generalization of the Beta Function. Tell me if this interests you, and I can include it.

Answer 2

The only parts of this that can be simplified at all are $ \Gamma \left( {\frac {p+1}{p}} \right) ={\frac {\Gamma \left( \frac{1}{p} \right) }{p}}$, and similarly for $\Gamma\left(\frac{p+2}{p}\right)$ and $\Gamma\left(\frac{p+2}{2}\right)$