expected value of squared infinity norm of vector of iid gaussians Given a random vector 
\begin{equation}
x=(x_1, \ldots, x_n)
\end{equation}
with independent and identically distributed entries $x_i \sim \mathcal{N}(0,\sigma^2)$, I would like to find a lower bound $f(n)$ 
\begin{equation}
\mathbb{E}[||x||^2_{\infty}] \geq f(n)
\end{equation}
which is reasonably tight. I know that the following equality for the non squared norm holds when $\sigma^2 =1$: 
\begin{equation}
E(\|x||_\infty)=\int_0^\infty(1-(2\Phi(x)-1)^n)dx, 
\end{equation}
where $\Phi$ is the CDF of $\mathcal{N}(0,1)$, see the comment to this question by @Did here. Unfortunately I am not even sure on how to (tightly) lower bound the right integral for this special case. 
Any help on solving the general case is much appreciated. 
 A: It is known that the max of i.i.d. subgaussian random variables with variance $\sigma^2$ is on the order of $\sigma \sqrt{\log n}$ so you can expect the squared max of the random variables to be roughly $\sigma^2 \log n$. A reference for this result is 'High Dimensional Probability' by Vershynin. In this case, Jensen's inequality immediately gives you that $$\mathbf{E}[||x||_{\infty}^2] \ge (\mathbf{E}[||x||_{\infty}])^2 = \Omega(\sigma^2 \log n).$$
We can also show that this is the right order of magnitude. Consider an arbitrary $\lambda > 0$ ($\lambda$ will have to satisfy a condition that we will address later). Then
\begin{align*}
\exp(\lambda \mathbf{E}(\max_i |x_i|)^2) &\le \mathbf{E}\exp(\lambda (\max_i |x_i|)^2) \\
&= \mathbf{E} \max_i \exp(\lambda |x_i|^2) \\
&\le \sum_{i=1}^n \mathbf{E}\exp(\lambda x_i^2).
\end{align*}
Now the last quantity is just the MGF of a chi-squared distribution which has an explicit form which is $1/(\sqrt{1-2\lambda \sigma^2})$. Then  by taking the logs we have
$$ \mathbf{E}(\max_i |x_i|)^2 \le \frac{\log n}{\lambda} + \frac{1}{\lambda \sqrt{1-2\lambda \sigma^2}}. $$
Optimizing this quantity in $\lambda$, we let $\lambda$ be such that $$\lambda \sigma^2 = \frac{1}2 - \frac{1}{2 \log(n)^2}.$$
(Note in the conference of the MGF above, we needed $\lambda \sigma^2 < 1/2$ which is satisfied here). Then plugging back in, we see that 
$$ \mathbf{E}(\max_i |x_i|)^2 \le \frac{4 \sigma^2 \log(n)}{1-1/\log(n)^2}$$
so $\sigma^2 \log n$ is the right order. 
A: Here's a general way how to obtain lower bounds in this case. The infinity norm $||\vec{X}||_\infty $ is defined on $\mathbb{R}^n$ as
$$ ||\vec{X}||_\infty =\max(|X_1|, ..., |X_n|)$$
Consequently, we also have
$$ ||\vec{X}||_\infty^2 = \max(X_1^2, ..., X_n^2)$$
as $X_i \leq X_j \implies X_i^2 \leq X_j^2$. Now, the maximum of several i.i.d variables, also known as the largest order statistic, has the distribution
$$ \max(X_1^2, ..., X_n^2) \sim \frac{d}{dx} F(x)^{n} = n \left(F(x)\right)^{n-1} f(x) $$
where $f(x)$ is the distribution of $X_i^2$ and $F(x)$ is its CDF. Finally, we therefore have
$$ \mathbb{E}[||\vec{X}||_\infty^2] = n \int_{-\infty}^{\infty} x F(x)^{n-1} f(x) \ dx = \int_{-\infty}^{\infty} x \frac{d}{dx} F(x)^{n} \ dx$$
Use the substitution $u = F(x)$, and we obtain
$$ \mathbb{E}[||\vec{X}||_\infty^2] = n \int_0^1 F^{-1}(u) u^{n-1} du $$
Now, for the random variables $X_i^2$ specifically, we have
$$ f(x) = \frac{1}{\sqrt{2 \pi x}\ \sigma} e^{- \frac{x}{2 \sigma^2}} \theta(x), \;\;\; F(x) = \text{erf}\left(\frac{\sqrt{x}}{\sqrt{2} \sigma}\right) $$
and therefore
$$F^{-1}(u) = 2 \sigma^2 \text{erf}^{-1}(u)^2 $$ .
Now, remarkably in this case, the function $\text{erf}^{-1}(u)$ has a Taylor series around $u = 0$ where all of the coefficients are positive. This means that any truncation of the series is strictly less than $\text{erf}^{-1}(u)$.
The first few terms are
$$ \text{erf}^{-1}(u) = \frac{\sqrt{\pi}}{2} u + \frac{\pi^{\frac{3}{2}}}{24} u^3 + O(u^5)$$
Therefore, if you replace $\text{erf}^{-1}(u)$ with the truncated series in the expression for $F^{-1}(u)$ and integrate the resulting polynomial expression, you obtain a lower bound.
As an example, take the lowest order term in the above expansion. We then must have
$$ \mathbb{E}[||\vec{X}||_\infty^2] \geq n \int_0^1 \sigma^2 \frac{\pi}{2} u^{n+1} du =  \sigma^2 \frac{\pi}{2} \frac{n}{n+2}$$
As you take more terms in the Taylor expansion and evaluate the integral, you will obtain increasingly tighter lower bounds.
