Computing the expectation of $X^n e^{-\lambda X}$ Is there an elegant (probabilistic) way to compute $E(X^n e^{-\lambda X})$, where $n \in \mathbb{N}$, $\lambda > 0$ and $X$ is a random variable with normal distribution $N(\mu,\sigma^2)$?
Alternatively, my question is simply how to compute the integral
$$
\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty} x^n e^{-\lambda x} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx.
$$

$$\lambda x + \frac{(x-\mu)^2}{2\sigma^2} = \frac{x^2 - 2(\mu - \sigma^2 \lambda)x + \mu^2}{2\sigma^2} = \frac{(x-(\mu-\sigma^2 \lambda))^2}{2\sigma^2} + \frac{2 \lambda \mu -\sigma^2 \lambda^2}{2}$$
Ignoring constant factors, your integral becomes
$$\int x^n e^{-\frac{(x-(\mu-\sigma^2 \lambda))^2}{2\sigma^2}} \mathop{dx},$$
which can be computed from the [non-central] moments of a $N(\mu-\sigma^2 \lambda, \sigma^2)$ distribution.
