Expected value for $\min\{|x_j|\}$ of normal distribution sample Let $\{x_1, x_2, \ldots, x_n\} $ is sample of the size $n$ of  normally distributed random value:
$$X \sim N(0,1).\tag{1}$$
How to find expected value for $$m_n = \min_{1\le j \le n}\{|x_j|\}.\tag{2}$$
I have just experimental data in the table:
\begin{array}{|l|l|}
\hline
n & E(m_n) \\
\hline
1 & \approx 0.795 \\
2 & \approx 0.467 \\
5 & \approx 0.215 \\
10 & \approx 0.115 \\
100 & \approx 0.0124 \\
1000 & \approx 0.00126 \\
\hline
\end{array}
from which we can derive that
$$E(m_n) = \frac{c(n)}{n},\tag{3}$$
where roughly $0.5<c(n)<1.5$; or 
$$E(m_n) = \frac{c(n)}{n+1},\tag{3'}$$
where roughly $1.2<c(n)<1.6$.  
So is there any formula (or more accurate estimation) for numerator $c(n)$ in $(3)$ or $(3')$?
 A: We can find the expected value of a nonnegative random variable $X$ with $\mathbb{E}(X)=\int_{0}^{\infty}\mathbb{P}(X>x)\mathrm{d}x.$ The event $\{m_{n}>x\}$ is just the intersection of the events $\{|x_{i}|>x\},$ so assuming this is an iid sample, we have $\mathbb{P}(m_{n}>x)=\prod_{i=1}^{n}\mathbb{P}(|x_{i}|>x)=\mathbb{P}(|x_{1}|>x)^{n}=(1-\mathrm{erf}(x/\sqrt{2}))^{n}.$
Then one way to write this expected value is $$E(m_{n})=\int_{0}^{\infty}(1-\mathrm{erf}(x/\sqrt{2}))^{n}\,\mathrm{d}x=\int_{0}^{\infty}\mathrm{erfc}(x/\sqrt{2})^{n}\,\mathrm{d}x.$$
Using the bound $\mathrm{erfc}(x)\leq e^{-x^{2}}$ for all $x>0,$ we see that $E(m_{n})\leq\int_{0}^{\infty}e^{-nx^{2}/2}\mathrm{d}x$, and by symmetry of this function, this equals $\frac{1}{2}\int_{-\infty}^{\infty}e^{-nx^{2}/2}\mathrm{d}x$. Recognizing the integrand as the density of a $\mathcal{N}(0,1/n)$ random variable, we obtain $$E(m_{n})\leq\frac{1}{2}\sqrt{\frac{2\pi}{n}}\int_{-\infty}^{\infty}\sqrt{\frac{n}{2\pi}}e^{-x^{2}/(2/n)}\,\mathrm{d}x=\sqrt{\frac{\pi}{2n}}.$$ This is weaker than the estimate you made from your calculations, but has been rigorously proved. There are many bounds on the complementary error function that could be brought to bear to obtain tighter estimates and lower bounds, but this is a simple one that gives some evidence in favor of what you've already stated. 
