I have a Bank Account with Indian Bank.
Is it possible for me to transfer money from my Indian bank account via atm to Citibank Account? 

If Yes what the steps to do this.


Question: KL divergence between two univariate Gaussians
Answer: 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:

$\int \left[\log( p(x)) - log( q(x)) \right] p(x) dx$

$=\int \left[ -\frac{1}{2} \log(2\pi) - \log(\sigma_1) - \frac{1}{2} \left(\frac{x-\mu_1}{\sigma_1}\right)^2 + \frac{1}{2}\log(2\pi) + \log(\sigma_2) + \frac{1}{2} \left(\frac{x-\mu_2}{\sigma_2}\right)^2  \right]$
$\times \frac{1}{\sqrt{2\pi}\sigma_1} \exp\left[-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2\right] dx$

$=\int \left\{\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2} \left[ \left(\frac{x-\mu_2}{\sigma_2}\right)^2 - \left(\frac{x-\mu_1}{\sigma_1}\right)^2 \right] \right\}$
$\times \frac{1}{\sqrt{2\pi}\sigma_1} \exp\left[-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2\right] dx$

$=E_{1} \left\{\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2} \left[ \left(\frac{x-\mu_2}{\sigma_2}\right)^2 - \left(\frac{x-\mu_1}{\sigma_1}\right)^2 \right]\right\}$

$=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2} E_1 \left\{(X-\mu_2)^2\right\} - \frac{1}{2\sigma_1^2} E_1 \left\{(X-\mu_1)^2\right\}$ 

$=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2} E_1 \left\{(X-\mu_2)^2\right\} - \frac{1}{2}$

(Now note that $(X - \mu_2)^2 = (X-\mu_1+\mu_1-\mu_2)^2 = (X-\mu_1)^2 + 2(X-\mu_1)(\mu_1-\mu_2) + (\mu_1-\mu_2)^2$)

$=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{1}{2\sigma_2^2}
\left[E_1\left\{(X-\mu_1)^2\right\} + 2(\mu_1-\mu_2)E_1\left\{X-\mu_1\right\} + (\mu_1-\mu_2)^2\right] - \frac{1}{2}$

$=\log\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{\sigma_1^2 + (\mu_1-\mu_2)^2}{2\sigma_2^2} - \frac{1}{2}$