KL divergence between two univariate Gaussians I need to determine the KL-divergence between two Gaussians. I am comparing my results to these, but I can't reproduce their result. My result is obviously wrong, because the KL is not 0 for KL(p, p).
I wonder where I am doing a mistake and ask if anyone can spot it.
Let $p(x) = N(\mu_1, \sigma_1)$ and $q(x) = N(\mu_2, \sigma_2)$. From Bishop's
PRML I know that
$$KL(p, q) = - \int p(x) \log q(x) dx + \int p(x) \log p(x) dx$$
where integration is done over all real line, and that
$$\int p(x) \log p(x) dx = -\frac{1}{2} (1 + \log 2 \pi \sigma_1^2),$$
so I restrict myself to $\int p(x) \log q(x) dx$, which I can write out as
$$-\int p(x) \log \frac{1}{(2 \pi \sigma_2^2)^{(1/2)}} e^{-\frac{(x-\mu_2)^2}{2 \sigma_2^2}} dx,$$
which can be separated into
$$\frac{1}{2} \log (2 \pi \sigma_2^2) - \int p(x) \log e^{-\frac{(x-\mu_2)^2}{2 \sigma_2^2}} dx.$$
Taking the log I get
$$\frac{1}{2} \log (2 \pi \sigma_2^2) - \int p(x) \bigg(-\frac{(x-\mu_2)^2}{2 \sigma_2^2} \bigg) dx,$$
where I separate the sums and get $\sigma_2^2$ out of the integral.
$$\frac{1}{2} \log (2 \pi \sigma^2_2) + \frac{\int p(x) x^2 dx - \int p(x) 2x\mu_2 dx + \int p(x) \mu_2^2 dx}{2 \sigma_2^2}$$
Letting $\langle \rangle$ denote the expectation operator under $p$, I can rewrite this as
$$\frac{1}{2} \log (2 \pi \sigma_2^2) + \frac{\langle x^2 \rangle - 2 \langle x \rangle \mu_2 + \mu_2^2}{2 \sigma_2^2}.$$
We know that $var(x) = \langle x^2 \rangle - \langle x \rangle ^2$. Thus
$$\langle x^2 \rangle = \sigma_1^2 + \mu_1^2$$
and therefore 
$$\frac{1}{2} \log (2 \pi \sigma_2^2) + \frac{\sigma_1^2 + \mu_1^2 - 2 \mu_1 \mu_2 + \mu_2^2}{2 \sigma_2^2},$$
which I can put as
$$\frac{1}{2} \log (2 \pi \sigma_2^2) + \frac{\sigma_1^2 + (\mu_1 - \mu_2)^2}{2 \sigma_2^2}.$$
Putting everything together, I get to
\begin{align*}
KL(p, q) &= - \int p(x) \log q(x) dx + \int p(x) \log p(x) dx\\\\
&= \frac{1}{2} \log (2 \pi \sigma_2^2) + \frac{\sigma_1^2 + (\mu_1 - \mu_2)^2}{2 \sigma_2^2} - \frac{1}{2} (1 + \log 2 \pi \sigma_1^2)\\\\
&= \log \frac{\sigma_2}{\sigma_1} + \frac{\sigma_1^2 + (\mu_1 - \mu_2)^2}{2 \sigma_2^2}.
\end{align*}
Which is wrong since it equals $1$ for two identical Gaussians.
Can anyone spot my error?
Update
Thanks to mpiktas for clearing things up. The correct answer is:
$KL(p, q) = \log \frac{\sigma_2}{\sigma_1} + \frac{\sigma_1^2 + (\mu_1 - \mu_2)^2}{2 \sigma_2^2} - \frac{1}{2}$

I did not have a look at your calculation but here is mine with a lot of details.
Suppose $p$ is the density of a normal random variable with mean $\mu_1$ and variance $\sigma^2_1$, and that $q$ is the density of a normal random variable with mean $\mu_2$ and variance $\sigma^2_2$.
The Kullback-Leibler distance from $q$ to $p$ is:
$$\int \left[\log( p(x)) - \log( q(x)) \right] p(x) dx$$
\begin{align}&=\int \left[ -\frac{1}{2} \log(2\pi) - \log(\sigma_1) - \frac{1}{2} \left(\frac{x-\mu_1}{\sigma_1}\right)^2 + \frac{1}{2}\log(2\pi) + \log(\sigma_2) + \frac{1}{2} \left(\frac{x-\mu_2}{\sigma_2}\right)^2  \right]\times \frac{1}{\sqrt{2\pi}\sigma_1} \exp\left[-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2\right] dx\\&=\int \left\{\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2} \left[ \left(\frac{x-\mu_2}{\sigma_2}\right)^2 - \left(\frac{x-\mu_1}{\sigma_1}\right)^2 \right] \right\}\times \frac{1}{\sqrt{2\pi}\sigma_1} \exp\left[-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2\right] dx\\&
=E_{1} \left\{\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2} \left[ \left(\frac{x-\mu_2}{\sigma_2}\right)^2 - \left(\frac{x-\mu_1}{\sigma_1}\right)^2 \right]\right\}\\&=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2} E_1 \left\{(X-\mu_2)^2\right\} - \frac{1}{2\sigma_1^2} E_1 \left\{(X-\mu_1)^2\right\}\\ &=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2} E_1 \left\{(X-\mu_2)^2\right\} - \frac{1}{2};\end{align}
(Now note that $(X - \mu_2)^2 = (X-\mu_1+\mu_1-\mu_2)^2 = (X-\mu_1)^2 + 2(X-\mu_1)(\mu_1-\mu_2) + (\mu_1-\mu_2)^2$)
\begin{align}&=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2}
\left[E_1\left\{(X-\mu_1)^2\right\} + 2(\mu_1-\mu_2)E_1\left\{X-\mu_1\right\} + (\mu_1-\mu_2)^2\right] - \frac{1}{2}\\&=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{\sigma_1^2 + (\mu_1-\mu_2)^2}{2\sigma_2^2} - \frac{1}{2}.\end{align}
