Kullback-Liebler divergence The Kullback-Liebler divergence between two distributions with pdfs $f(x)$ and $g(x)$ is defined
by
$$\mathrm{KL}(F;G) = \int_{-\infty}^{\infty} \ln \left(\frac{f(x)}{g(x)}\right)f(x)\,dx$$
Compute the Kullback-Lieber divergence when $F$ is the standard normal distribution and $G$
is the normal distribution with mean $\mu$ and variance $1$. For what value of $\mu$ is the divergence
minimized?
I was never instructed on this kind of divergence so I am a bit lost on how to solve this kind of integral.  I get that I can simplify my two normal equations in the natural log but my guess is that I should wait until after I take the integral. Any help is appreciated.

The pdf of the standard normal distribution is $
\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}$. Similarly, $g(x) = \frac{1}{\sqrt{2\pi}}e^{(x-\mu)^2/2}$.Therefore, $D_{KL}=\int_{-\infty}^{\infty} \frac{{2\mu-\mu}^2}{2\sqrt{2\pi}}e^{-x^{2}/2} dx$. This is equivalent to $\frac{{\mu}^2}{2}$.
This can be easily generalized to any two normal distributions with means $\mu_1, \mu_2$ and variances ${\sigma_1}^2, {\sigma_1}^2$.
The K-L divergence is obviously minimized at $\mu=0$ - where the distributions are the same!
EDIT: Sorry, I misread the mean as 1 originally - I have corrected it. 
