Ok, I've been searching for ages for the Kullback-Liebler divergence between two Normal distributions and didn't find it, but RS's answer enabled me to calculate it quite simply. Here's my derivation. I've also derived it for Beta distributions.
Kullback-Liebler divergence of Normal Distributions
Suppose we have two Normal distributions $F\sim N(\mu_{f},\sigma_{f})$; $G\sim N(\mu_{g},\sigma_{g})$. The Kullback-Liebler divergence is defined as:
$ \mathrm{KL}(F||G)=\int f(x)\ln\left(\frac{f(x)}{g(x)}\right)dx\quad\mathrm{nats} $
Divide by $\ln(2)$ to get the answer in bits. The Gaussian PDF is:
$ f(x)=\frac{1}{\sigma_{f}\sqrt{2\pi}}\, e^{\dfrac{-(x-\mu_{f})^{2}}{2\sigma_{f}^{2}}} $
Substituting we get
\begin{eqnarray*} \mathrm{KL}(F||G) & = & \int f(x)\ln\left(\frac{e^{\frac{-(x-\mu_{f})^{2}}{2\sigma_{f}^{2}}}}{\sigma_{f}\sqrt{2\pi}}\frac{\sigma_{g}\sqrt{2\pi}}{e^{\frac{-(x-\mu_{g})^{2}}{2\sigma_{g}^{2}}}}\right)dx\\ & = & \int f(x)\ln\left(\frac{e^{\frac{-(x-\mu_{f})^{2}}{2\sigma_{f}^{2}}}}{\sigma_{f}}\frac{\sigma_{g}}{e^{\frac{-(x-\mu_{g})^{2}}{2\sigma_{g}^{2}}}}\right)dx\\ & = & \int f(x)\left[\ln\left(\frac{e^{\frac{-(x-\mu_{f})^{2}}{2\sigma_{f}^{2}}}}{e^{\frac{-(x-\mu_{g})^{2}}{2\sigma_{g}^{2}}}}\right)+\ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)\right]dx\\ & = & \int f(x)\left[\frac{-(x-\mu_{f})^{2}}{2\sigma_{f}^{2}}-\frac{-(x-\mu_{g})^{2}}{2\sigma_{g}^{2}}+\ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)\right]dx \end{eqnarray*}
Then via a tedious and error-prone but straightforward expansion we get
$ =\left[\ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)+\frac{-\mu_{f}^{2}}{2\sigma_{f}^{2}}-\frac{-\mu_{g}^{2}}{2\sigma_{g}^{2}}\right]\int f(x)dx+\left[\frac{2\mu_{f}}{2\sigma_{f}^{2}}-\frac{2\mu_{g}}{2\sigma_{g}^{2}}\right]\int x\, f(x)dx+\left[\frac{-1}{2\sigma_{f}^{2}}-\frac{-1}{2\sigma_{g}^{2}}\right]\int x^{2}f(x)dx $
Then we have the following properties:
\begin{eqnarray*} \int f(x)dx & = & 1\\ \int x\, f(x)dx & = & \mu_{f}\\ \int x^{2}f(x)dx & = & \mu_{f}^{2}+\sigma_{f}^{2} \end{eqnarray*}
Which gives:
\begin{eqnarray*} \mathrm{KL}(F||G) & = & \left[\ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)+\frac{-\mu_{f}^{2}}{2\sigma_{f}^{2}}-\frac{-\mu_{g}^{2}}{2\sigma_{g}^{2}}\right]+\left[\frac{2\mu_{f}}{2\sigma_{f}^{2}}-\frac{2\mu_{g}}{2\sigma_{g}^{2}}\right]\mu_{f}+\left[\frac{-1}{2\sigma_{f}^{2}}-\frac{-1}{2\sigma_{g}^{2}}\right]\left(\mu_{f}^{2}+\sigma_{f}^{2}\right)\\ & = & \ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)+\frac{-\mu_{f}^{2}}{2\sigma_{f}^{2}}+\frac{\mu_{g}^{2}}{2\sigma_{g}^{2}}+\frac{2\mu_{f}^{2}}{2\sigma_{f}^{2}}+\frac{-2\mu_{g}\mu_{f}}{2\sigma_{g}^{2}}+\frac{-\mu_{f}^{2}-\sigma_{f}^{2}}{2\sigma_{f}^{2}}+\frac{\mu_{f}^{2}+\sigma_{f}^{2}}{2\sigma_{g}^{2}}\\ & = & \ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)+\frac{\mu_{g}^{2}-2\mu_{g}\mu_{f}+\mu_{f}^{2}+\sigma_{f}^{2}}{2\sigma_{g}^{2}}+\frac{-\mu_{f}^{2}+2\mu_{f}^{2}-\mu_{f}^{2}-\sigma_{f}^{2}}{2\sigma_{f}^{2}}\\ & = & \ln\left(\frac{\sigma_{g}}{\sigma_{f}}\right)+\frac{(\mu_{f}-\mu_{g})^{2}+\sigma_{f}^{2}-\sigma_{g}^{2}}{2\sigma_{g}^{2}} \end{eqnarray*}
I verified this numerically in Matlab, after fixing many sign errors! If anyone wants to do this for the Beta distribution I'd be greatful!
Kullback-Liebler divergence of Beta Distributions
Suppose we have two Beta distributions $F\sim\mathrm{Beta}(\alpha_{f},\beta_{f})$; $G\sim\mathrm{Beta}(\alpha_{g},\beta_{g})$. The Kullback-Liebler divergence is defined as:
$\mathrm{KL}(F||G)=\int f(x)\ln\left(\frac{f(x)}{g(x)}\right)dx\quad\mathrm{nats}$
Divide by $\ln(2)$ to get the answer in bits. The Beta PDF is:
$f(x)=\frac{\Gamma(\alpha_{f}+\beta_{f})}{\Gamma(\alpha_{f})\Gamma(\beta_{f})}x^{\alpha_{f}-1}(1-x)^{\beta_{f}-1}$
Where $\Gamma()$ is the Gamma function. Substituting gives:
\begin{eqnarray*} \mathrm{KL}(F||G) & = & \int_{0}^{1}f(x)\ln\left(\frac{\frac{\Gamma(\alpha_{f}+\beta_{f})}{\Gamma(\alpha_{f})\Gamma(\beta_{f})}x^{\alpha_{f}-1}(1-x)^{\beta_{f}-1}}{\frac{\Gamma(\alpha_{g}+\beta_{g})}{\Gamma(\alpha_{g})\Gamma(\beta_{g})}x^{\alpha_{g}-1}(1-x)^{\beta_{g}-1}}\right)dx\\ & = & \int_{0}^{1}f(x)\ln\left(\frac{\frac{\Gamma(\alpha_{f}+\beta_{f})}{\Gamma(\alpha_{f})\Gamma(\beta_{f})}x^{\alpha_{f}-\alpha_{g}}(1-x)^{\beta_{f}-\beta_{g}}}{\frac{\Gamma(\alpha_{g}+\beta_{g})}{\Gamma(\alpha_{g})\Gamma(\beta_{g})}}\right)dx\\ & = & \int_{0}^{1}f(x)\ln\left(\frac{\Gamma(\alpha_{f}+\beta_{f})\Gamma(\alpha_{g})\Gamma(\beta_{g})}{\Gamma(\alpha_{g}+\beta_{g})\Gamma(\alpha_{f})\Gamma(\beta_{f})}x^{\alpha_{f}-\alpha_{g}}(1-x)^{\beta_{f}-\beta_{g}}\right)dx\\ & = & \int_{0}^{1}f(x)\left[\ln\frac{\Gamma(\alpha_{f}+\beta_{f})\Gamma(\alpha_{g})\Gamma(\beta_{g})}{\Gamma(\alpha_{g}+\beta_{g})\Gamma(\alpha_{f})\Gamma(\beta_{f})}+\ln\left(x^{\alpha_{f}-\alpha_{g}}\right)+\ln\left((1-x)^{\beta_{f}-\beta_{g}}\right)\right]dx\\ & = & \ln\frac{\Gamma(\alpha_{f}+\beta_{f})\Gamma(\alpha_{g})\Gamma(\beta_{g})}{\Gamma(\alpha_{g}+\beta_{g})\Gamma(\alpha_{f})\Gamma(\beta_{f})}+(\alpha_{f}-\alpha_{g})\int_{0}^{1}f(x)\ln x\, dx+(\beta_{f}-\beta_{g})\int_{0}^{1}f(x)\ln(1-x)dx \end{eqnarray*}
In terms of expectations this is:
$\ln\frac{\Gamma(\alpha_{f}+\beta_{f})\Gamma(\alpha_{g})\Gamma(\beta_{g})}{\Gamma(\alpha_{g}+\beta_{g})\Gamma(\alpha_{f})\Gamma(\beta_{f})}+(\alpha_{f}-\alpha_{g})\mathrm{E}\left(\ln F\right)+(\beta_{f}-\beta_{g})\mathrm{E}\left(\ln(1-F)\right)$
From Wikipedia we have:
$\mathrm{E}(\ln F)=\psi(\alpha_{f})-\psi(\alpha_{f}+\beta_{f})$
Where $\psi(x)=\frac{d}{dx}\ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ is the digamma function (also known as the polygamma function; it is \texttt{psi} in Matlab). By swapping variables it is easy to show that
$\mathrm{E}(\ln(1-F))=\psi(\beta_{f})-\psi(\alpha_{f}+\beta_{f})$
Therefore the final solution is
$\mathrm{KL}(F||G)=\ln\frac{\Gamma(\alpha_{f}+\beta_{f})\Gamma(\alpha_{g})\Gamma(\beta_{g})}{\Gamma(\alpha_{g}+\beta_{g})\Gamma(\alpha_{f})\Gamma(\beta_{f})}+(\alpha_{f}-\alpha_{g})\left(\psi(\alpha_{f})-\psi(\alpha_{f}+\beta_{f})\right)+(\beta_{f}-\beta_{g})\left(\psi(\beta_{f})-\psi(\alpha_{f}+\beta_{f})\right)$