Stirling's approximation gives that
$$ n!\sim \sqrt{2\pi n}(n/e)^n. $$ It has what I will call a quantitative version, namely $$ \sqrt{2\pi n}(n/e)^n\exp\left(\frac{1}{12n+1}\right) \leq n! \leq \sqrt{2\pi n}(n/e)^n\exp\left(\frac{1}{12n}\right), $$ i.e. rather than an asymptotic one gets upper and lower bounds.
The volume of the unit $\ell_p$ ball can be given by
$$\omega_n^{(p)} := 2^n \frac{\Gamma(1+p^{-1})^n}{\Gamma(1+np^{-1})},$$
where for $n\in\mathbb{N}$ $n! = \Gamma(n+1)$. It is straightforward to see that $\omega_n^{(1)}= \frac{2^n}{n!}$, and $\omega_n^{(\infty)} = 2^n$.
I am curious
- whether an asymptotic $\omega_n^{(p)}$ is known, and
- whether a quantitative form of this asymptotic is known.
There are some mild annoyances one runs into if they try to plug the quantitative form of Stirling's approximation into the above formula for $\omega_n^{(p)}$, namely that the terms $\exp(1/(12n))$ behave poorly. One can check that this term in the numerator is
$$\exp(1/(12p^{-1}))^n = \exp(np/12),$$
while in the denominator it is
$$\exp(1/(12np^{-1})) = \exp(p/(12n)),$$
i.e. they have significantly different behavior for large $n$. This leads to bounds that are somewhat annoying to try to interpret --- for example by comparison with the cases of $p = 1, \infty$ one should get an expression between $2^n/n!$ and $2^n$. The expression I get is difficult to interpret its precise order of growth. Explicitly plugging in the quantitative form of Stirling's approximation into the formula for $\omega_n^{(p)}$ (using the upper bound each time to slightly simplify notation), I find the volume is
$$\omega_n^{(p)} = 2^n \left(\frac{\sqrt{2\pi}\exp(p/12)}{pn^{1/p}}\right)^n \frac{1}{\sqrt{2\pi n/p}\exp(p/(12n))}.$$
I am wondering if there are "more enlightening" expressions than ones of this type.
Are there quantitative estimates for the volume of the unit $\ell_p$ ball, in a similar spirit to the above quantitative form of Stirling's approximation?
I prefer to use $$\log (\Gamma (x))=x (\log (x)-1)+\frac{1}{2} (\log (2 \pi )-\log (x))+\frac{1}{12 x}+O\left(\frac{1}{x^3}\right)$$ So, for $$\omega_n^{(p)} := 2^n \frac{\Gamma(1+p^{-1})^n}{\Gamma(1+np^{-1})}$$ we have $$\log\big[\omega_n^{(p)}\big]=n\,\log \left(\frac{2 e^{\frac{1}{p}} \left(\frac{p}{n}\right)^{\frac{1}{p}}}{\Gamma \left(1+\frac{1}{p}\right)}\right)+\frac{1}{2} \log \left(\frac p{2 \pi n}\right)-\frac{p}{12 n}+O\left(\frac{1}{n^2}\right)$$