Let I_n = \mathbb E(\|G\|_\infty), i.e.

I_n = (2\pi)^{-\frac{n}{2}}\int_{x\in\mathbb R^n}\|x\|_\infty e^{-\frac{1}{2}\|x\|_2^2}\,dx.

I wonder if I can get its closed-form. By symmetry I got

I_n = 2n\sqrt{\frac{2}{\pi}}\int_0^\infty xe^{-x^2}\operatorname{erf}(x)^{n-1}\,dx,

and then by integration by parts, for n\ge2,

I_n = \frac{2\sqrt2}{\pi}n(n-1)\int_0^\infty e^{-2x^2}\operatorname{erf}(x)^{n-2}\,dx,

where \operatorname{erf} is the error function.These two formulas give me

I_1 = \sqrt{\frac{2}{\pi}},\quad I_2 = 2\sqrt{\frac{1}{\pi}},\quad I_3 = \frac{12}{\pi\sqrt\pi}\arctan\frac{\sqrt2}{2}.

In this step, I think a general closed-form is almost impossible, so I post here to see if anyone has a better approach (at least for I_4).

## Update

Series expansion of I_4:

I_4 = \frac{8\sqrt2}{\pi^2}\sum_{n=0}^{\infty}\left(\frac43\right)^n\frac{n!}{(2n+1)!}\,\Gamma(n+3/2)\,{}_2F_1(1/2,-n;3/2;1/4).By the way

I_n = \sqrt2n\int_0^1t^{n-1}\operatorname{erf}^{-1}(t)\,dt \,=\!\!\!?\; \sqrt2n\sum_{k=0}^\infty a_k \left(\frac{\sqrt\pi}{2}\right)^{2k+1}\frac1{2k+n+1},

where a_k is the k-th coefficient of the Maclaurin series of \operatorname{erf}^{-1}(2x/\sqrt\pi) (see InverseErf).Well, I don’t really know the behavior of (a_k), but numerically the series does converge. I don’t think this will lead to anything though.

Let me explain a little about this problem.

Imagine we have n points to throw at 0 at the real axis, and the resulted position of one point is determined by \mathcal N(0,1). We want to study the behavior of the farthest distance from 0.

This distance D = \|G\|_\infty is determined by the density function defined below

f:x \mapsto n\sqrt{\frac2\pi}\,\exp\left(-\frac{x^2}2\right) \operatorname{erf}^{n-1}\frac{x}{\sqrt2} \mathbb1_{x\ge0}.

(For fun one can check that \int_0^\infty f(x)\,dx=1.)

And now, what we want to know is, how to calculate \mathbb E(D) (at least when n=4)?

@YuriNegometyanov has given a formula for \mathbb E(\|G\|_2). Even though it’s not quite the topic, let’s write it down as well:

\mathbb E(\|G\|_2) =\sqrt2\,\frac{\Gamma\left(\dfrac{n+1}2\right)}{\Gamma\left(\dfrac n2\right)}.

A jupyter notebook to calculate numerical results.

So from the series expansion of I_4 mentioned above (and tons of calculation), I got:

I_4 = \frac{24}{\pi\sqrt\pi}\arctan\frac{1}{2\sqrt2}.

This is kind of interesting since the form is similar to I_3. Maybe a general closed-form is in fact possible?

**Answer**

Let J(a,n) = \int_0^\infty e^{-at}\operatorname{erf}^n\sqrt t\,dt for a>0.

Let J_n=J(1,n), we have then I_n = n\sqrt{\frac{2}{\pi}}J_{n-1}.

By some equalities, we have the recurrence relation below:

J(a,0)=\frac1a,\quad J(a,1)=\frac1{a\sqrt{a+1}},

J(a,n)=J(a,n-2)-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(1+s^2+a,n-2\right)\,ds.

Then by some calculation, we have

J(a,2)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac1{\sqrt{a+1}},

and

J(a,3)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},\text{ where }b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}.

(By the way, for fun one can prove that

2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2},

which shows up in I_4.)

We can also give an expression of J_4 (which gives I_5):

\begin{align}

J_4&=J_2-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(2+s^2,2\right)\,ds\\

&=J_2-\left(\frac{4}{\pi}\right)^2\int_0^1\frac{1}{1+s^2}\frac{1}{2+s^2}\frac{1}{\sqrt{3+s^2}}\arctan\frac{1}{\sqrt{3+s^2}}\,ds.

\end{align}

As you can see, these become more and more complicated. I really don’t think there’s a closed-form for I_n when n\ge5.

Alternatively, we also have

J_n=\sum_{k=0}^n\left(-1\right)^k\binom{n}{k}C_k,

where

C_k=\mathbb E\left[\phi(U)\right]=\pi^{-k}\int_{u\in\mathbb R^k}\phi(u)\prod_{i=1}^k\frac{1}{1+u_i^2}\,du,

\phi(u)=\frac{1}{1+\sum_{i=1}^k\left(1+u_i^2\right)},

and U=(U_i)_{1\le i\le k} is a random vector of independent \operatorname{Cauchy}(0,1) variables.

This might give us a global view of what happens in that recurrence relation (which I believe is unhelpful for a general closed-form).

(By the way, the formula of \mathbb E(\|G\|_2) given by @YuriNegometyanov can be easily found using \chi^2-distribution.)

A little simplification (see here).

J(a,3) = \frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).

**Attribution***Source : Link , Question Author : Aforest , Answer Author : Aforest*