By Long Luo

$f(x) = \mathrm{e}^{-cx^2}, \, c > 0 \tag{1} \label{1}$

$\int_{- \infty}^{+ \infty} f(x) \mathrm{d}x = 1 \tag{2} \label{2}$

$f(x) = {\frac {1}{\sigma {\sqrt {2 \pi }}}}\;e^{-{\frac {\left(x - \mu \right)^{2}}{2 \sigma ^{2}}}} \tag{3} \label{3}$

# 什么是高斯积分？

$F(x) = \int_{0}^{x} e^{-t^2} \mathrm{d}t \tag{4} \label{4}$

$\int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} = \sqrt{\pi} \tag{5} \label{5}$

# 极坐标系法

$I = \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} \tag{6} \label{6}$

$I = \int_{-\infty}^{+\infty} \mathrm{e} ^{-y^2} \,\mathrm{d}{y} \tag{7} \label{7}$

\begin{aligned} I^2 & = \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} \int_{-\infty}^{+\infty} \mathrm{e} ^{-y^2} \,\mathrm{d}{y} \\ & = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \mathrm{e} ^{-(x^2 + y^2)} \,\mathrm{d}{x} \,\mathrm{d}{y} \end{aligned}

## 转换坐标系

$\begin{cases} x = r \cos \theta \\ y = r \sin \theta \end{cases}$

$\begin{cases} -\infty < x < \infty \\ -\infty < y < \infty \end{cases}$

$\begin{cases} 0< r < \infty \\ 0< \theta < 2 \pi \end{cases}$

## 为什么 $$\mathrm{d}x \mathrm{d}y = r \mathrm{d}r\mathrm{d}\theta$$ ？

\begin{aligned} \mathrm{d} \sigma & = \frac{1}{2}\left(r + \mathrm{d}r \right)^2\cdot \mathrm{d}\theta - \frac{1}{2}r^2\cdot \mathrm{d}\theta \\ & = \frac{1}{2}\left[r^2 + 2r \mathrm{d}r + \left(\mathrm{d} r\right)^2-r^2\right] \mathrm{d}\theta \\ & = \frac{1}{2}\left[2r \mathrm{d} r + \left(\mathrm{d} r \right)^2 \right] \cdot \mathrm{d}\theta \end{aligned}

$\mathrm{d} \sigma = \frac{1}{2}(2r \mathrm{d}r) \mathrm{d}\theta = r \mathrm{d}r \mathrm{d}\theta$

$\mathrm{d}x \mathrm{d}y = r \mathrm{d}r\mathrm{d}\theta$

## 求解积分

\begin{aligned} I^2 & = \int_0^{2\pi} \int_0^{+\infty} \mathrm{e} ^{-r^2} r \,\mathrm{d}{r} \,\mathrm{d}{\theta} \\ & = \int_{\theta = 0}^{2\pi} \mathrm{d}{\theta} \int_{r=0}^{\infty} \mathrm{e} ^{-r^2} r \,\mathrm{d}{r} \\ & = 2 \pi \int_0^{+\infty} r \mathrm{e} ^{-r^2} \,\mathrm{d}{r} \end{aligned}

$$u = r^2$$ ，则有：

$\mathrm{d}u = 2r \mathrm{d}{r}$

$r \mathrm{d}r = \frac{1}{2}\mathrm{d}u$

\begin{aligned} I^2 & = \pi \int_0^{+\infty} \mathrm{e} ^{-u} \,\mathrm{d}{u} \\ & = \pi \left (-\mathrm{e} ^{-u}) \right\rvert _0^{+\infty} \\ & = \pi \end{aligned}

$I = \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} = \sqrt{\pi}$

# 3D 钟形曲面下的体积

$$1$$ 维的高斯函数为：

$f_1(x) = \mathrm{e}^{-x^2}$

$f_2(x, y) = \mathrm{e}^{-x^2} \mathrm{e}^{-y^2} = \mathrm{e}^{- (x^2 + y^2)}$

$$1$$ 维的高斯函数表现为钟形曲线， $$2$$ 维的高斯函数则是 $$1$$ 维的高斯函数沿着 $$z$$ 轴旋转，如下图 8 所示：

$$2$$ 维的高斯函数表达式也可以写成：

$f_2(x, y) = \mathrm{e}^{-r^2}$

$\mathrm{d}V = 2 \pi r \mathrm{e}^{-r^2}\mathrm{d}r$

\begin{aligned} \mathbb{V} & = \int_0^{+\infty} \mathrm{d}V \\ & = \int_0^{+\infty} 2 \pi r \mathrm{e}^{-r^2}\mathrm{d}r \\ & = \pi \int_0^{+\infty} 2r \mathrm{e}^{-r^2}\mathrm{d}r \\ & = \pi \left (-e^{-\infty^2} -(-e^0) \right ) \\ & = \pi \end{aligned}

# 沿着 y 轴方向进行切片

$f_2(x, y) = \mathrm{e}^{-x^2} \mathrm{e}^{-y^2} = \mathrm{e}^{-x^2}$

$S = \mathrm{e}^{-x^2} \int_{-\infty}^{+\infty} \mathrm{e} ^{-y^2} \,\mathrm{d}{y}$

$\mathrm{d}V = \mathrm{e}^{-y^2} \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x}$

$\mathbb{V} = \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} \int_{-\infty}^{+\infty} \mathrm{e} ^{-y^2} \,\mathrm{d}{y}$

$\mathbb{V} = \left ( \int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} \right )^2$

$\int_{-\infty}^{+\infty} \mathrm{e} ^{-x^2} \,\mathrm{d}{x} = \sqrt {\pi}$