TITLE: Hypervolume of a $N$-dimensional ball in $p$-norm
QUESTION [7 upvotes]: Suppose I have a N-dimensional ball with radius R in p-norm:
$$ \sum_{n=1}^N |x_n|^p = R^p $$
Is there a closed formula for its (hyper)volume? I can't find anything.
If there isn't, can we at least know how it varies with $R$?

REPLY [9 votes]: Wiki do has an entry for Volume of n-ball. 
The hypervolume of $n$-dim ball in p-norm with radius R is given by:
$$V_n^p(R) = \frac{(2\Gamma(\frac{1}{p}+1)R)^n}{\Gamma(\frac{n}{p}+1)}$$
It is actually pretty easy to prove this ourselves.
Observe
$$\begin{align}V_n^p(1)
&= \int_{-1}^{1}dx_1\int_{|x_2|^p+\ldots+|x_n|^p\le 1-|x_1|^p}\prod_{i=2}^n dx_i\\
&=\int_{-1}^{1} V_{n-1}^p((1 - |x|^p)^\frac{1}{p}) dx\\
&= V_{n-1}^p(1) \left[2\int_0^1 (1-x^p)^\frac{n-1}{p} dx\right]\\
&= V_{n-1}^p(1) \left[\frac{2}{p} \int_{0}^1 y^{\frac{1}{p}-1}(1-y)^{\frac{n-1}{p}} dy\right]\\
&= V_{n-1}^p(1) \left[ \frac{2}{p}\frac{\Gamma(\frac{1}{p})\Gamma(\frac{n-1}{p}+1)}{\Gamma(\frac{n}{p}+1)}\right]\\
&= V_{n-1}^p(1) \left[2\Gamma(\frac{1}{p}+1) \frac{\Gamma(\frac{n-1}{p}+1)}{\Gamma(\frac{n}{p}+1)}\right]
\end{align}$$
Since $V_1^p(1) = 2$, we conclude:
$$ V_n^p(1) = V_1^p(1) (2\Gamma(\frac{1}{p}+1))^{n-1}\frac{\Gamma(\frac{1}{p}+1)}{\Gamma(\frac{n}{p}+1)} = \frac{(2\Gamma(\frac{1}{p}+1))^n}{\Gamma(\frac{n}{p}+1)}$$