高斯积分与正态密度归一化

(1) 令 $I=\int_{-\infty}^{\infty}e^{-x^2}dx$，按极坐标计算可得 $I^2=\pi$，因此 $\boxed{I=\sqrt\pi}$。

(2) 标准正态密度为

$$\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$$

令 $u=x/\sqrt2$，则

$$\int_{-\infty}^{\infty}\varphi(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2/2}dx =\frac{1}{\sqrt{2\pi}}\cdot\sqrt2\int_{-\infty}^{\infty}e^{-u^2}du =\frac{1}{\sqrt{2\pi}}\cdot\sqrt2\cdot\sqrt\pi=1.$$

一般 $N(\mu,\sigma^2)$ 密度用代换 $z=(x-\mu)/\sigma$ 同理可证积分为 1。

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英文解析

(1) Let $I=\int_{-\infty}^{\infty}e^{-x^2}dx$. Computing via polar coordinates yields $I^2=\pi$, therefore $\boxed{I=\sqrt\pi}$.

(2) The standard normal density is

$$\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}.$$

Let $u=x/\sqrt2$, then

$$\int_{-\infty}^{\infty}\varphi(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^2/2}dx =\frac{1}{\sqrt{2\pi}}\cdot\sqrt2\int_{-\infty}^{\infty}e^{-u^2}du =\frac{1}{\sqrt{2\pi}}\cdot\sqrt2\cdot\sqrt\pi=1.$$

Generally, for $N(\mu,\sigma^2)$, the density integrates to 1 by the same substitution $z=(x-\mu)/\sigma$.