For a given probability distribution $\mu$, the associated rate function is $I_\mu(x) = sup \left \lbrace x \lambda - ln \left( \int e^{\lambda t} \mu (dx) \right) \right \rbrace$, and if there happens to be a probability distribution function $f(x)$ for $\mu$ this becomes $I_\mu(x) = sup \left \lbrace x \lambda - ln \left( \int f(x) e^{\lambda t} dx \right) \right \rbrace$. That's all well and good, but how does one actually calculate what this function is for a particular distribution?
For example, if $\mu$ has the normal distribution $N(m, \sigma^2)$, then its distribution function is $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-(x-m)^2/(2\sigma^2)}$ (at least according to wikipedia it is, anyhow). Is there some known way to then explicitly calculate what $I_\mu(x)$ is as a basic function of x, that is, to render it in terms of something that doesn't involve a supremum over all real numbers? When I tried just taking the definition, putting the distribution function in, and chugging along, I got something that requires a calculation of $\int e^{((2\sigma^2 \lambda + 2m)t - t^2)/(2\sigma^2)} dt$, and even fudging this integral's value using wolfram alpha (which is obviously not a satisfactory way to solve the problem) I wound up with something that I don't really know how to take a general supremum of.