高斯分布熵的推导

本文主要进行了一元高斯分布熵的推导：

$$\begin{array}{}\text{(1)}& p(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2\sigma }}\end{array}$$

$$\begin{array}{rl}H(x)& =-\int p(x)\ln p(x)\mathrm{d}x\\ & =-\int \frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\cdot \ln \frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\mathrm{d}x\\ & =-\int \frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\cdot (-\ln \sqrt{2\pi {\sigma }^2}-\frac{(x-\mu {)}^2}{2{\sigma }^2})\mathrm{d}x\\ & =\int \frac{\ln \sqrt{2\pi {\sigma }^2}}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\cdot \mathrm{d}x+\int \frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}\cdot \frac{(x-\mu {)}^2}{2{\sigma }^2}\mathrm{d}x\\ & =\ln \sqrt{2\pi {\sigma }^2}+\frac{1}{\sqrt{\pi }}\int e^{-t^2}\cdot t^2\mathrm{d}t\\ & =\ln \sqrt{2\pi {\sigma }^2}+\frac{1}{2}\\ \text{(2)}& & =\frac{1}{2}(\ln (2\pi {\sigma }^2)+1)\end{array}$$ 其中： $$\begin{array}{}\text{(3)}& \int e^{-x^2}\mathrm{d}x=\sqrt{\pi }\end{array}$$

$$\begin{array}{rl}\int x^2\cdot e^{-x^2}\mathrm{d}x& =-\frac{1}{2}\int x(-2x)e^{-x^2}\mathrm{d}x\\ & =-\frac{1}{2}\int x\mathrm{d}{e}^{-x^2}\\ & =-\frac{1}{2}(x\cdot e^{-x^2}{|}_{-\mathrm{\infty }}^{+\mathrm{\infty }}-\int e^{-x^2}\mathrm{d}x)\\ & =\frac{1}{2}\int e^{-x^2}\mathrm{d}x)\\ \text{(4)}& & =\frac{\sqrt{\pi }}{2}\end{array}$$