Average norm of a N-dimensional vector given by a normal distribution

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 $$ \frac{N}{\sqrt{N+1}}\le \sigma^{-1}E(\|x\|)\le \sqrt{N} $$

The above answer contains mistakes, as has been noted in the comments. I needed recently to derive this so the general result is: $$\mathbb{E}\left[||x||_2^n\right] = 2^\frac{n-2}{2}\sigma^n N \frac{\Gamma\left(\frac{N+n}{2}\right)}{\Gamma\left(\frac{N+2}{2}\right)}$$

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.