Since there is no definite answer to this question here but many good tips, I tried to compute $E(e^x)$ myselfe and this is what I got: $$ E(e^X)=\displaystyle{\int_{-\infty}^\infty e^x \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}}dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^x e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{x-\frac{(x-\mu)^2}{2\sigma^2}}dx = $$
$$ = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{\frac{2x\sigma^2-(x-\mu)^2}{2\sigma^2}}dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{\frac{2x\sigma^2-(x^2-2x\mu+\mu^2)}{2\sigma^2}}dx = \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{\frac{-(x^2-2x(\sigma^2+\mu)+\mu^2)}{2\sigma^2}}dx $$
$$ \frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^\infty e^{\frac{-(x-(\sigma^2+\mu))^2+2\sigma^2\mu+\sigma^4}{2\sigma^2}}dx =e^{{\frac{2\sigma^2\mu+\sigma^4}{2\sigma^2}}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma} e^{\frac{-(x-(\sigma^2+\mu))^2}{2\sigma^2}}dx = e^{\mu}e^{\frac{\sigma^2}{2}} $$
Is this ok? But does this also prove that $E(|e^X|)$ is finite? If not, how can I show that?
