Suppose $X_1, \ldots, X_n \sim N(0,\sigma^2)$ are iid. Find the expected value of $M$, the median of $\vert X_1 \vert, \ldots \vert X_n \vert$
What I have so far:
The density of $\vert X_i \vert$ is given by: $$f_{\vert X \vert}(x) = \frac{2}{\sqrt{2\pi \sigma^2}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$$
If $n = 2m - 1$, the median $M$ is the $m^{th}$ order statistic of $\vert X_1 \vert, \ldots, \vert X_n \vert$. This has density $$f_{\vert X \vert_{(m)}} (x) = \frac{n!}{(m-1)!(n-m)!} f_{\vert X \vert}(x) (F_{\vert X \vert})^{m-1}(1 - F_{\vert X \vert})^{n-m}$$ where $F_{\vert X \vert}(x) = F_X(x) - F_X(-x)$ is the cdf of $\vert X \vert$. So the expected value of the median is $$\mathbb E M = \int_0^\infty x f_{\vert X \vert_{(m)}} (x) dx$$
Is there a better way to approach this, rather than having to compute this integral?