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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.

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    $\begingroup$ Where did $2^n$ come from? I ask this for learning, not for challenging your working. How does one get that from $\exp(-\max\{\cdots\})$? $\endgroup$
    – FShrike
    Aug 8, 2021 at 11:24
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    $\begingroup$ For actual question helping, I believe that assuming $x_1$ has the largest absolute value is false, since saying that restricts you to a subset of $\Bbb{R}^n$ instead of the whole set $\endgroup$
    – FShrike
    Aug 8, 2021 at 11:25
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    $\begingroup$ @FShrike $2^n$ probably comes from the symmetry of the integrand and the fact that there are $2^n$ "quadrants". For example if $n=2$, there are the four quadrants, for $n=3$ there are the eight octants and so on. The integral over each piece is equal due to the symmetry of the integrand under change of $x_i\mapsto -x_i$. So, if we let $A_{+}=\{x\in\Bbb{R}^n\,:\, x_1\geq0,\dots x_n\geq 0\}$, then we have to show the integral over $A_+$ is $n!$. $\endgroup$
    – peek-a-boo
    Aug 8, 2021 at 11:30
  • $\begingroup$ Ah yes, thank you @peek-a-boo. This now reminds me of a Gaussian integral $\endgroup$
    – FShrike
    Aug 8, 2021 at 11:32
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    $\begingroup$ @ReneMorningstar $dx$ here means integration with respect to the $n$ variables $dx_1\dots dx_n$. This is an integral in $\Bbb{R}^n$, and $dx$ is just a short way of writing it. $\endgroup$
    – peek-a-boo
    Aug 8, 2021 at 12:43

2 Answers 2

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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 $2^n$

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  • $\begingroup$ Well done @Oliver Diaz :) $\endgroup$
    – convxy
    Aug 8, 2021 at 11:43
  • $\begingroup$ I've figured where I was wrong thank you :) $\endgroup$
    – convxy
    Aug 8, 2021 at 12:00
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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:

$$2^n\cdot n!\int_0^\infty \int_{x_1}^\infty...\int_{x_{n-1}}^\infty e^{-x_n} = 2^n\cdot n! \int_0^\infty e^{-x_1} = 2^nn! $$

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