= Differential entropy of a Gaussian law
{title2=$h_e(N(m,\sigma^2))=\frac12\ln(2\pi e\sigma^2)$}
For a <normal distribution> with <mean> $m$ and positive <variance> $\sigma^2$, expand the <logarithm> of its <probability density function>:
$$
-\ln f(X)=\frac12\ln(2\pi\sigma^2)+\frac{(X-m)^2}{2\sigma^2}.
$$
The second term has <expected value> $1/2$, hence $h_e(X)=\tfrac12\ln(2\pi e\sigma^2)$. 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>.
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