Working out the expectation of a function of iid random variables I have found the maximum likelihood estimator $\hat{\sigma}$ of a iid r.vs $X_1, ..., X_n$ which all have normal distribution with known mean $\mu$ and unknown variance $\sigma^2$. 
So $\hat{\sigma}$ turns out to be $\sqrt{\frac{1}{n} \displaystyle\sum_{i=1}^n (X_i - \mu)^2} $. Now if it wasn't for the square root sign I'd have no problem working out $\mathbb{E}[\hat{\sigma}]$.
Could someone please help?
 A: The random variable $Q:=\sum_{i=1}^n {\left(\frac{X_i - \mu}{\sigma}\right)}^2$ has a Chi-squared distribution with $n$ degrees of freedom. Denote by $f$ its pdf, which you can find on Wikipedia or many other places. It is given by
$$
f(x) = C x^{\frac{n}{2}-1}\exp\bigl(-\frac{x}{2}\bigr)
$$
where $C=\frac{1}{2^{\frac{n}{2}}\Gamma\bigl(\frac{n}{2}\bigr)}$ is a constant (not depending on $x$).
Your are looking for the expectation of the random variable 
$$
R= \sqrt{\frac{1}{n} \displaystyle\sum_{i=1}^n (X_i - \mu)^2}.
$$
Once you get the pdf of $R$, say $g$, you can get the expectation of $R$ by calculating $E(R) = \int yg(y)\mathrm{d}y$. 
The random variable $R$ is a function of $Q$, namely $R=h(Q)$ where $h(x)=\sigma \sqrt{\frac{1}{n}x}$. This function is one-to-one map from $[0, \infty[$ to $[0, \infty[$, therefore you can use the change of variables formula to get the pdf of $R$. Denoting by $g$ this pdf, it is given by
$$
g(y) = {(h^{-1})}'(y)\times f\bigl(h^{-1}(y)\bigr).
$$
The inverse of $h$ is $h^{-1}(y) = \frac{n}{\sigma^2}y^2$. 
Setting $\lambda=\frac{n}{\sigma^2}$ for notational simplicity,  one gets (for $y > 0$)
$$
\begin{align*}
g(y) & = 2 \lambda y \times f\bigl(\lambda y^2\bigr) \\
& = 2 C\lambda y^{n-1} \exp \bigl(-\frac{\lambda y^2}{2}\bigr)
\end{align*},
$$
and one finally has to calculate 
$$
E(R) = 2 C\lambda^{\frac{n}{2}} \int_0^\infty y^{n} \exp \bigl(-\frac{\lambda y^2}{2}\bigr)\mathrm{d}y.
$$
By the change of variables $x=\alpha y^2$, 
$$
\begin{align*}
\int_0^\infty y^{n} \exp \bigl(-\alpha y^2\bigr)\mathrm{d}y & = \frac{1}{2} \alpha^{-\frac{n+1}{2}} \int_0^\infty x^{\frac{n}{2}-1} \exp \bigl(-x\bigr)\mathrm{d}x \\ & = \frac{1}{2} \alpha^{-\frac{n+1}{2}} \Gamma\Bigl(\frac{n+1}{2} \Bigr). 
\end{align*}
$$
Thus one finally gets
$$
\begin{align*}
E(R) & = \frac{1}{2^{\frac{n}{2}}\Gamma\bigl(\frac{n}{2}\bigr)} {\left(\frac{n}{\sigma^2}\right)}^{\frac{n}{2}} 
{\left(\frac{1}{2}\frac{n}{\sigma^2}\right)}^{-\frac{n+1}{2}}\Gamma\Bigl(\frac{n+1}{2} \Bigr) \\
& = \frac{\sqrt{2}\sigma}{\sqrt{n}}\frac{\Gamma\Bigl(\frac{n+1}{2} \Bigr)}{\Gamma\bigl(\frac{n}{2}\bigr)}.
\end{align*}
$$
Checking:
> n <- 5
> sigma <- 2
> Q <- rchisq(10000, n)
> h <- function(x) sigma*sqrt(x/n)
> R <- h(Q)
> mean(R)
[1] 1.903887
> sqrt(2)*sigma/sqrt(n)*gamma((n+1)/2)/gamma(n/2)
[1] 1.903066

