I am trying to find $\mathbb{E}(e^{-X})$ where $X$ is a random variable with a general normal distribution. I end up with
$$(2\pi \sigma)^{-\frac{1}{2}} \int_{-\infty}^{\infty} e^{-x}e^{\left(\frac{-(\mu-x)^2}{2\sigma^2}\right)}dx $$
Tried some transformations and the polar coordinate trick used to derive values of $(2\pi \sigma)^{-\frac{1}{2}}\int e^{(\frac{-(\mu-x)^2}{2\sigma^2})}dx $ but haven't been able to come up with anything explicit. Can anyone point me in the right direction? I am in despair... do I need to use a numerical method?
If it helps, here is what the integral boils down to $$(2\pi \sigma)^{-\frac{1}{2}}\int \exp\left(-\frac{x(2\sigma - 2\mu +x)}{2\sigma^2}\right)dx $$
Question: Let $X \sim N(\mu, \sigma^2)$. Find $E[e^{-X}]$.
Answer: Let $Z = -X$, so $Z \sim N(m,\sigma^2)$, where $m = -\mu$. You seek $E[e^Z]$ which is just the mean of a Lognormal random variable, with Normal parent $N(m,\sigma^2)$. Thus:
$$E[e^{-X}] = E[e^Z] = e^{m +\frac{\sigma ^2}{2}} \quad \text{ where } m = -\mu$$
If you need help deriving the latter integral step-by-step, this is standard textbook material on the Lognormal.