Prove $\int_{\mathbb{R}^n}e^{-\max\{|x_1|,\ldots,|x_n|\}}dx=2^nn!$ 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.
 A: 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:
$$2^n\cdot n!\int_0^\infty \int_{x_1}^\infty...\int_{x_{n-1}}^\infty
e^{-x_n} = 2^n\cdot n! \int_0^\infty e^{-x_1} = 2^nn! $$
A: There are $n$ cases, all of them mutually exclusive (except on a set of measure $0$):
$$E_k=\{|x_k|=\max\{|x_j|:1\leq j\leq n\}, \quad k=1,\ldots,n$$
One each $E_k$, by Fubini's theorem yields
$$\int_{E_k}e^{-\max_{1\leq j\leq n}|x_j|}=\int_{\mathbb{R}}e^{-|x|}\Big(\int^{|x|}_{-|x|}\,dx_2\ldots\int^{|x|}_{-|x|}\,dx_n\Big)\,dx$$
Thus, thee integral of interest becomes
$$n\int^\infty_{-\infty}e^{-|x|}2^{n-1}|x|^{n-1}\,dx=n2^n\int
^\infty_0x^{n-1}e^{-x}\,dx=n2^n\Gamma(n)=2^n n!$$

For another more geometric approach, similar to polar coordinates, notice that $\max_{1\leq j\leq n}|x_j|=\|x\|_\infty$ is a norm in $\mathbb{R}^n$.  It is not difficult to check (via monotone class arguments) that for any norm $\rho$ on $\mathbb{R}^n$ and any nonnegative measurable function $f$ on $\mathbb{R}$
$$ \int_{\mathbb{R}^n}f(\rho(x))\,dx=n \lambda_n(\{x:\rho(x)\leq1\})\int^\infty_0 f(r) r^{n-1}\,dr$$
where $\lambda_n$ is the Lebesgue measure on $\mathbb{R}^n$.
For the particular norm $\|\,\|_\infty$, the volume of the $\|\;\|_\infty$-unit ball is $2^n$
