Closed-form of E(‖\mathbb E(\|G\|_\infty) where G\sim\mathcal N(0,\mathbf{Id}_n)G\sim\mathcal N(0,\mathbf{Id}_n).

Let $$I_n = \mathbb E(\|G\|_\infty)I_n = \mathbb E(\|G\|_\infty)$$, i.e.
$$I_n = (2\pi)^{-\frac{n}{2}}\int_{x\in\mathbb R^n}\|x\|_\infty e^{-\frac{1}{2}\|x\|_2^2}\,dx.I_n = (2\pi)^{-\frac{n}{2}}\int_{x\in\mathbb R^n}\|x\|_\infty e^{-\frac{1}{2}\|x\|_2^2}\,dx.$$
I wonder if I can get its closed-form. By symmetry I got
$$I_n = 2n\sqrt{\frac{2}{\pi}}\int_0^\infty xe^{-x^2}\operatorname{erf}(x)^{n-1}\,dx,I_n = 2n\sqrt{\frac{2}{\pi}}\int_0^\infty xe^{-x^2}\operatorname{erf}(x)^{n-1}\,dx,$$
and then by integration by parts, for $$n\ge2n\ge2$$,
$$I_n = \frac{2\sqrt2}{\pi}n(n-1)\int_0^\infty e^{-2x^2}\operatorname{erf}(x)^{n-2}\,dx,I_n = \frac{2\sqrt2}{\pi}n(n-1)\int_0^\infty e^{-2x^2}\operatorname{erf}(x)^{n-2}\,dx,$$
where $$\operatorname{erf}\operatorname{erf}$$ is the error function.

These two formulas give me

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

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 $$I_4I_4$$).

Update

Series expansion of $$I_4I_4$$:
$$I_4 = \frac{8\sqrt2}{\pi^2}\sum_{n=0}^{\infty}\left(\frac43\right)^n\frac{n!}{(2n+1)!}\,\Gamma(n+3/2)\,{}_2F_1(1/2,-n;3/2;1/4).I_4 = \frac{8\sqrt2}{\pi^2}\sum_{n=0}^{\infty}\left(\frac43\right)^n\frac{n!}{(2n+1)!}\,\Gamma(n+3/2)\,{}_2F_1(1/2,-n;3/2;1/4).$$

By the way
$$I_n = \sqrt2n\int_0^1t^{n-1}\operatorname{erf}^{-1}(t)\,dt \,=\!\!\!?\; \sqrt2n\sum_{k=0}^\infty a_k \left(\frac{\sqrt\pi}{2}\right)^{2k+1}\frac1{2k+n+1},I_n = \sqrt2n\int_0^1t^{n-1}\operatorname{erf}^{-1}(t)\,dt \,=\!\!\!?\; \sqrt2n\sum_{k=0}^\infty a_k \left(\frac{\sqrt\pi}{2}\right)^{2k+1}\frac1{2k+n+1},$$
where $$a_ka_k$$ is the $$kk$$-th coefficient of the Maclaurin series of $$\operatorname{erf}^{-1}(2x/\sqrt\pi)\operatorname{erf}^{-1}(2x/\sqrt\pi)$$ (see InverseErf).

Well, I don’t really know the behavior of $$(a_k)(a_k)$$, but numerically the series does converge. I don’t think this will lead to anything though.

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

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

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

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

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

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

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

A jupyter notebook to calculate numerical results.

So from the series expansion of $$I_4I_4$$ mentioned above (and tons of calculation), I got:
$$I_4 = \frac{24}{\pi\sqrt\pi}\arctan\frac{1}{2\sqrt2}.I_4 = \frac{24}{\pi\sqrt\pi}\arctan\frac{1}{2\sqrt2}.$$
This is kind of interesting since the form is similar to $$I_3I_3$$. Maybe a general closed-form is in fact possible?

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

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

By some equalities, we have the recurrence relation below:

$$J(a,0)=\frac1a,\quad J(a,1)=\frac1{a\sqrt{a+1}},J(a,0)=\frac1a,\quad J(a,1)=\frac1{a\sqrt{a+1}},$$
$$J(a,n)=J(a,n-2)-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(1+s^2+a,n-2\right)\,ds.J(a,n)=J(a,n-2)-\frac{4}{\pi}\int_0^1\frac{1}{1+s^2}J\left(1+s^2+a,n-2\right)\,ds.$$

Then by some calculation, we have

$$J(a,2)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac1{\sqrt{a+1}},J(a,2)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac1{\sqrt{a+1}},$$
and
$$J(a,3)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},\text{ where }b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}.J(a,3)=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},\text{ where }b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}.$$

(By the way, for fun one can prove that
$$2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2},2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2},$$
which shows up in $$I_4I_4$$.)

We can also give an expression of $$J_4J_4$$ (which gives $$I_5I_5$$):
\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}\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 $$I_nI_n$$ when $$n\ge5n\ge5$$.

Alternatively, we also have
$$J_n=\sum_{k=0}^n\left(-1\right)^k\binom{n}{k}C_k,J_n=\sum_{k=0}^n\left(-1\right)^k\binom{n}{k}C_k,$$
where
$$C_k=\mathbb E\left[\phi(U)\right]=\pi^{-k}\int_{u\in\mathbb R^k}\phi(u)\prod_{i=1}^k\frac{1}{1+u_i^2}\,du,C_k=\mathbb E\left[\phi(U)\right]=\pi^{-k}\int_{u\in\mathbb R^k}\phi(u)\prod_{i=1}^k\frac{1}{1+u_i^2}\,du,$$
$$\phi(u)=\frac{1}{1+\sum_{i=1}^k\left(1+u_i^2\right)},\phi(u)=\frac{1}{1+\sum_{i=1}^k\left(1+u_i^2\right)},$$
and $$U=(U_i)_{1\le i\le k}U=(U_i)_{1\le i\le k}$$ is a random vector of independent $$\operatorname{Cauchy}(0,1)\operatorname{Cauchy}(0,1)$$ 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 $$\mathbb E(\|G\|_2)\mathbb E(\|G\|_2)$$ given by @YuriNegometyanov can be easily found using $$\chi^2\chi^2$$-distribution.)

A little simplification (see here).
$$J(a,3) = \frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).J(a,3) = \frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).$$