Average norm of a N-dimensional vector given by a normal distribution I'm interested in knowing what is the expected value of the norm of a vector obtained from a gaussian distribution in function of the number of dimensions $N$ and $\sigma$, i.e:
$$E[\|x\|_2],\quad x\sim\mathcal{N}(0,\sigma I_N)$$
I tried to search for this but didn't find anything. Can I get some help from you? 
 A: The above answer contains mistakes, as has been noted in the comments. I needed recently to derive this so the general result is:
$$\mathbb{E}\left[||x||_2^n\right] = 2^\frac{n-2}{2}\sigma^n N \frac{\Gamma\left(\frac{N+n}{2}\right)}{\Gamma\left(\frac{N+2}{2}\right)}$$
A: When $\sigma=1$, this is the first moment of Chi-distribution. Furthermore, \begin{equation}
\mathbb{E}_{x\sim\mathcal{N}(0,\sigma^2I)}\{ \|x\|_2^k \} = \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|\sigma x\|_2^k \}= \sigma^k \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}, 
\end{equation}
where $\mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \| x\|_2^k \}$ is the $k$th moment of Chi-distribution, which has value 
\begin{equation}
    \mathbb{E}_{x\sim\mathcal{N}(0,I)}\{ \|x\|_2^k \} = 2^{k/2} \frac{\Gamma((N+k)/2)}{\Gamma(N/2)}. 
\end{equation}
Two comments:
1. I think $x\sim\mathcal{N}(0,\sigma I_N)$ should be replaced by $x\sim\mathcal{N}(0,\sigma^2I_N)$ in the original post. 
2. This result is different from that of Botev's. 
Let me know if I made any mistakes. 
A: This amounts to integration in spherical coordinates $(r=\|x\|)$: 
$$
E(\|x\|) = \frac{1}{(\sqrt{2\pi} \sigma)^N } \frac{N\pi^{N/2}}{\Gamma\big(\frac{N}{2}+1\big)}\int_0^\infty e^{-r^2/(2\sigma^2)} r^{N-1} \,dr 
\tag1$$
This is not so bad: substitute $t=r^2/(2\sigma^2)$, so that $dt = r/\sigma^2$. The resulting integral gives Euler's gamma function $\Gamma$. I'll skip the boring cancellations and get to the result: 
$$
E(\|x\|) = \frac{\sqrt{2}\, \Gamma\big(\frac{N+1}{2}\big)}{\Gamma\big(\frac{N }{2}\big)}\,\sigma
$$
As stated in this paper, where you can also find the inequalities
$$
\frac{N}{\sqrt{N+1}}\le \sigma^{-1}E(\|x\|)\le \sqrt{N}
$$
