Differential entropy of a Gaussian law ($h_e(N(m,{\sigma }^2))=\frac{1}{2}\ln (2\pi e{\sigma }^2)$)

For a normal distribution with mean $m$ and positive variance ${\sigma }^2$, expand the logarithm of its probability density function:The second term has expected value $1/2$, hence $h_e(X)=\frac{1}{2}\ln (2\pi e{\sigma }^2)$. Translation of the mean has no effect, while shrinking the standard deviation lowers differential entropy. A standard normal distribution has approximately $1.41894$ nats. This connects the continuous spread calculation with entropy power, whose normalization matches a Gaussian distribution's variance.