Define $f:\mathbb{R}^n \to \mathbb{R}$ as $f(x) = \|x\|_\infty^{-n-1}$, where for $x = (x_1, x_2, \cdots, x_n)$, $\|x\|_\infty = \max\{|x_i|, i = 1,\cdots,n\}$ (the usual coordinate-wise $\infty$-norm). Let $E = \{ x \in \mathbb{R}^n: \|x\|_\infty \geq 1 \}$; show that $\int_E f = n2^n$.
Using Fubini, I have the tried to compute iterated integral \begin{align*}\int_E f &= \int_{E^*} \int_{E_\mathbf{x}}\|(\mathbf{x},x_n)\|_\infty^{-n-1}dx_n d\mathbf{x} \\ &= 2\int_1^\infty \int_{E_\mathbf{x},\|\mathbf{x}\|_\infty < 1}\|(\mathbf{x},x_n)\|_\infty^{-n-1}dx_n d\mathbf{x} + 2\int_0^\infty \int_{E_\mathbf{x},\|\mathbf{x}\|_\infty \geq 1}\|(\mathbf{x},x_n)\|_\infty^{-n-1}dx_n d\mathbf{x} \end{align*} where the latter expression is my attempt to exclude the region $(-1,1)^n$ and take advantage of symmetry.
Am I on the right track? Is there a simpler way?
Using the symmetry, I get
$$\begin{align} \int_E f(x)\, dx &= 2n \int_1^\infty \left(\int_{\lVert \mathbf{x}\rVert_\infty \leqslant x_n} \,d\mathbf{x}\right) \frac{1}{x_n^{n+1}}\,dx_n\\ &= 2n \int_1^\infty (2x_n)^{n-1}\frac{1}{x_n^{n+1}}\,dx_n\\ &= n2^n \int_1^\infty \frac{dx_n}{x_n^2}\\ &= n2^n. \end{align}$$
The space $\mathbb{R}^n$ can be partitioned in the $n$ regions where $\lVert x\rVert_\infty = \lvert x_k\rvert$, which are disjoint modulo null sets (lower dimensional intersections), and congruent; that gives the factor $n$. Each of these region consists of two disjoint congruent parts, the part where $x_k < 0$ and where $x_k > 0$; that gives the factor $2$. The remaining integral is simple, the integrand depends only on one component.