I'm stuck with something that I think should be straightforward but can't see what to do.

I'm trying to find $\displaystyle E[e^X]$ where $\displaystyle X$~N($\displaystyle m$,$\displaystyle v^2$). I know that this is $\displaystyle \int^\infty_{-\infty}{e^{x}\cdot\frac{1}{\sqrt{2{\pi}v^2}}\cdot\ * e^{\frac{-(x-m)^2}{(2v^2)}}dx}$ i.e. the integral of the inner product of $\displaystyle e^{x}$ and the pdf of the Normal distribution. This is where it falls apart - how do I integrate this? I think it's done by substitution when dealing with $\displaystyle E(X)$ with the standard normal but this has the extra exponential and I'm not sure how to do it.

Help appreciated, thanks.