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
$$
\frac{1}{\sqrt{2\pi\sigma^2}} \int_{-\infty}^{\infty} x^n e^{-\lambda x} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx.
$$

HINT
You have
$$
\begin{split}
\lambda x + \frac{(x - \mu)^2}{2\sigma^2}
&= \frac{x^2 + \mu^2 + 2x(\sigma^2\lambda-\mu)}{2\sigma^2} \\
&= \frac{x^2 + 2x(\sigma^2\lambda-\mu) +(\sigma^2\lambda-\mu)^2}
        {2\sigma^2}
  + \frac{\mu^2 - (\sigma^2\lambda-\mu)^2}{2\sigma^2}\\
&= \frac{(x + \sigma^2\lambda-\mu)^2}
        {2\sigma^2}
  + \frac{\mu^2 - (\sigma^2\lambda-\mu)^2}{2\sigma^2}\\
\end{split}
$$
and denote the last term on the RHS by $K$ and you have
$$
\int_\mathbb{R} x^n e^{\lambda x} e^{\frac{(x - \mu)^2}{2\sigma^2}} dx
= e^K\int_\mathbb{R} x^n \exp\left(\frac{(x + (\sigma^2\lambda-\mu))^2}{2\sigma^2}\right) dx
= e^K \mathbb{E}[Y]
$$
where $Y$ is also normal with same std deviation but shifted mean.
