$E(x^k)$ under truncated $\mathcal{N}(\mu,1)$ There is a similar question in $E(x^k)$ under a Gaussian. However, it doesn't seem to be trivial when $\mu\ne0$. As mentioned in the previous question $k$ is not an integer.
The integral that I need to evaluate is as follows:
$$\int_0^\infty x^k\exp\left(-\frac{(x-\mu)^2}{2}\right)dx$$
If it helps for the case that $\mu=0$ the answer is $\frac{2^{(k-2)/2}}{\sqrt{\pi}}\Gamma(\frac{k+1}{2})$
 A: This is a Mellin transform. In general notation we have ($a>0, s>0$)
$$\int_{0}^{\infty}x^{s-1}\exp\left\{-ax^2-bx\right\}dx = (2a)^{-1/2}\Gamma(s)\exp\left\{\frac {b^2}{8a}\right\}D_{-s}\left(b(2a)^{-1/2}\right)$$
where $D_{-s}()$ is (Whittaker's) parabolic cylinder function.
For your integral we have
$$I=\int_0^\infty x^k\exp\left\{-\frac{(x-\mu)^2}{2}\right\}dx = \exp\left\{\frac {-\mu^2}{2}\right\}\int_0^\infty x^k\exp\left(-\frac{x^2}{2} +\mu x\right)dx  $$
Matching coefficients we get
$$s-1 = k \Rightarrow s=k+1,\;\; a=\frac 12,\;\; b=-\mu$$
Inserting into the general solution we have
$$I=\exp\left\{\frac {-\mu^2}{2}\right\}\left(2\frac 12\right)^{-1/2}\Gamma(k+1)\exp\left\{\frac {\mu^2}{8\frac 12}\right\}D_{-k-1}\left(-\mu\left(2\frac 12\right)^{-1/2}\right)$$
$$=\exp\left\{\frac {-\mu^2}{4}\right\}\Gamma(k+1)D_{-k-1}\left(-\mu\right)$$
Now set $k^* =k +\frac 12$. Then $D_{-k-1}\left(-\mu\right) = D_{-k^*-\frac 12}\left(-\mu\right)$ and for the second you can look up Abramowitz and Stegun p.687 and 686, starting with eq. $19.3.1$. You will indeed see that when $\mu\neq 0$ the situation is not trivial. The fact that $k$ is not an integer causes no special trouble.
