# 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.

• Where did $2^n$ come from? I ask this for learning, not for challenging your working. How does one get that from $\exp(-\max\{\cdots\})$? Aug 8, 2021 at 11:24
• 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 Aug 8, 2021 at 11:25
• @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 $n!$. Aug 8, 2021 at 11:30
• Ah yes, thank you @peek-a-boo. This now reminds me of a Gaussian integral Aug 8, 2021 at 11:32
• @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. Aug 8, 2021 at 12:43

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 $$2^n$$

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

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:

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