How to calculate the moments of a normal random variable with mean $\mu$ and variance $\sigma^2$?
Using integration by parts we get the recurrence relation (calling $a_n = E(X^n)$)
$$\begin{cases} a_{n+2} = \mu a_{n+1} + \sigma^2(n+1)a_n \\ a_1 = \mu\\ a_2 = \mu + \sigma^2 \end{cases}$$
If $\mu = 0$, $\sigma^2 = 1$ then we find $a_{2n+1} = 0$, $\displaystyle a_{2n} = \frac{(2n)!}{2^nn!}= (2n-1)!! $ (double factorial)
But in the general case? How do I solve the recurrence relation written above?
One approach is to do a transformation to standardize the distribution, giving you the recurrence you already have, then undo the transformation: e.g., $$X = \mu + \sigma Z$$ implies that $$\operatorname{E}[X^n] = \operatorname{E}[(\mu + \sigma Z)^n] = \sum_{k=0}^n \binom{n}{k} \sigma^k \mu^{n-k} \operatorname{E}[Z^k] .$$