TITLE: Prove $\int_{\mathbb{R}^n}e^{-\max\{|x_1|,\ldots,|x_n|\}}dx=2^nn!$
QUESTION [11 upvotes]: Prove $\int_{\mathbb{R}^n}e^{-\max\{|x_1|,\ldots,|x_n|\}}dx=2^nn!$
My attempt:
On the region where  $x_1$ has the largest absolute value
$\int_{\mathbb{R}^n}e^{-\max\{|x_1|,\ldots,|x_n|\}}=2^n\int_{\mathbb{R}^n}e^{-|x_1|}dx=2^n\int_0^{\infty}\int_{\mathbb{S}^{n-1}_r}e^{-|x_1|}dSdr$
But I got stuck here.

REPLY [5 votes]: I have another nice approach (hope you'll agree on that):
From symmetry on the sign:
$$\int_{\mathbb{R}^{n}}e^{-\max\{|x_1|,...,|x_n|\}}=
2^n\int_{\mathbb{R_{+}}^{n}}e^{-\max\{x_1,...,x_n\}}$$
By taking a permutation $0\leq x_1\leq x_2\leq...\leq x_n$, and taking the limit of the growing set $\lim_{k\to\infty}E_k = [0,k]^n$ we get
$$2^n\cdot n! \cdot \int_{\mathbb{R_{+}}^{n}\cap\{x\in\mathbb{R}^n|0\leq x_1\leq...\leq x_n\}}e^{-\max\{x_1,...,x_n\}} \\= 
2^nn!\int_{\{0\leq x_1\leq...\leq x_n\}}e^{-x_n} \\=
2^nn!\lim_{k\to\infty} \int_0^k \int_{x_1}^k...\int_{x_{n-1}}^k
e^{-x_n}$$
Im going to put $k=\infty$ just to make everything shorter. we get:
$$2^n\cdot n!\int_0^\infty \int_{x_1}^\infty...\int_{x_{n-1}}^\infty
e^{-x_n} = 2^n\cdot n! \int_0^\infty e^{-x_1} = 2^nn! $$