How to derive $\mathbb E(e^X)$ if $X$ is normally distributed?

Question

Assume that $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$. How do I derive $\mathbb E(e^X)$? Note I'm not asking for a mere answer, but an approach how to solve it. I end up messing with integrals, and change of variables, and it's not working.

Answer 1

As always, $$\mathbb E[f(X)]=\int_{\mathbb R} f(x)\mu_X(\,\mathrm d x),$$ where $\mu_X$ is the measure on $\mathbb R$ induced by $X$. In the case where $X\sim \mathcal N(\mu,\sigma ^2)$, $$\mathbb E[f(X)]=\frac{1}{\sigma \sqrt{2\pi}}\int_{\mathbb R}f(x)e^{-\frac{(x-\mu)^2}{2\sigma ^2}}\,\mathrm d x.$$