无穷积分 ∫e^(-x^2)dx 的几种巧妙解法

  ${\int }_{-\infty }^{\infty }{e}^{-x^2}dx$ 是一个比较常见的无穷积分，在许多领域有着重要应用。

  在此介绍几种巧妙的解法，供读者欣赏。 为了您更好的阅读体验，请使用电脑浏览。

1. 极坐标变换

  $\begin{array}{rl}({\int }_{-\infty }^{\infty }{e}^{-x^2}dx{)}^2& ={\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\int }_{-\infty }^{\infty }{e}^{-y^2}dy\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-(x^2+y^2)}dxdy(\text{极坐标变换})\\ & ={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta \\ & =\left({\int }_0^{2\pi }d\theta \right)\left(\frac{1}{2}{\int }_0^{\infty }{e}^{-r^2}d{r}^2\right)\\ & =2\pi \frac{1}{2}=\pi \\ \therefore {\int }_{-\infty }^{\infty }{e}^{-x^2}dx& =\sqrt{\pi }.\end{array}$

2. 几何做法(一)

$$\begin{array}{rl}{S}_1& =\{(x,y)|x^2+y^2\le R^2\}\\ S_2& =\{(x,y)|0\le x\le R,0\le y\le R\}\\ S_3& =\{(x,y)|x^2+y^2\le 2R^2\}\\ & \end{array}$$    $\begin{array}{rl}({\int }_0^R{e}^{-x^2}dx{)}^2& ={\int }_0^R{e}^{-x^2}dx{\int }_0^R{e}^{-y^2}dy\\ & ={\int }_0^R{\int }_0^R{e}^{-(x^2+y^2)}dxdy\\ & ={\iint }_{S_2}{e}^{-(x^2+y^2)}dxdy\end{array}$

  $\begin{array}{rl}& \because e^{-(x^2+y^2)}\ge 0\\ & \therefore {\iint }_{S_1}{e}^{-(x^2+y^2)}dxdy\le {\iint }_{S_2}{e}^{-(x^2+y^2)}dxdy\le {\iint }_{S_3}{e}^{-(x^2+y^2)}dxdy\end{array}$

   $\begin{array}{rl}{\iint }_{S_1}{e}^{-(x^2+y^2)}dxdy& ={\int }_0^{\pi /2}{\int }_0^R{e}^{-r^2}rdrd\theta \\ & =\left({\int }_0^{\pi /2}d\theta \right)\left(\frac{1}{2}{\int }_0^R{e}^{-r^2}d{r}^2\right)\\ & =\frac{\pi }{4}(1-e^{-R^2})\end{array}$

$$\begin{gathered} \begin{array}{rl} \\ & \text{同理}{\iint }_{S_3}{e}^{-(x^2+y^2)}dxdy=\frac{\pi }{4}(1-e^{-2R^2}).\\ & \text{则}\frac{\pi }{4}(1-e^{-R^2})\le {\iint }_{S_2}{e}^{-(x^2+y^2)}dxdy\le \frac{\pi }{4}(1-e^{-2R^2}).\\ & \text{两边取极限}R\to \infty \text{，则}({\int }_0^{\infty }{e}^{-x^2}dx{)}^2={\iint }_{R^2}{e}^{-(x^2+y^2)}dxdy=\frac{\pi }{4}.\\ & \text{所以}{\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}.\end{array} \end{gathered}$$

3. 几何做法(二)

$\begin{array}{r}\text{考虑曲线}z=e^{-x^2}\text{绕}z\text{轴旋转一周生成的旋转体的体积，由高数知识}\end{array}$
$$\begin{array}{rl}V& ={\int }_0^1\pi x^{2}dz\\ & =\pi {\int }_0^1(-\ln z)dz\\ & =\pi (z-z\ln z){|}_0^1\\ & =\pi (1-\underset{z\to 0^{+}}{\lim }(z-z\ln z))\\ & =\pi \end{array}$$

$\begin{array}{rl}& \text{旋转体在几何上的表示如图所示，旋转生成的曲面方程为}z=e^{-(x^2+y^2)}.\\ & \text{旋转体的体积也可以理解为曲面}z=e^{-(x^2+y^2)}\text{与}xOy\text{之间部分的体积，即}\end{array}$
$$\pi =V={\iint }_{R^2}{e}^{-(x^2+y^2)}dxdy={\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\int }_{-\infty }^{\infty }{e}^{-y^2}dy=({\int }_{-\infty }^{\infty }{e}^{-x^2}dx{)}^2$$$\begin{array}{r}\text{则}{\int }_{-\infty }^{\infty }{e}^{-x^2}dx=\sqrt{\pi }.\end{array}$

4. 拉普拉斯变换

$\begin{array}{rl}& \text{令}f(t)={\int }_0^{\infty }{e}^{-x^{2}t}dx\text{，对}f(t)\text{作拉普拉斯变换，令}\end{array}$
  $\begin{array}{rl}F(s)=\mathcal{L}[f(t)]& ={\int }_0^{\infty }\mathcal{L}[e^{-x^{2}t}{]}_{t}dx\\ & ={\int }_0^{\infty }\frac{1}{s+x^2}dx\\ & =\frac{1}{\sqrt{s}}{\int }_0^{\infty }\frac{1}{1+(\frac{x}{\sqrt{s}}{)}^2}d(\frac{x}{\sqrt{s}})(\text{let}u=\frac{x}{\sqrt{s}})\\ & =\frac{1}{\sqrt{s}}\arctan u{|}_0^{\infty }\\ & =\frac{\pi }{2}{s}^{-\frac{1}{2}}\end{array}$

$\text{对}F(s)\text{作拉普拉斯反变换}$

  $\begin{array}{rl}f(t)={\mathcal{L}}^{-1}[F(s)]& =\frac{\pi }{2}{\mathcal{L}}^{-1}[\frac{1}{s^{-\frac{1}{2}+1}}]\\ & =\frac{\pi }{2}\frac{1}{\Gamma (-\frac{1}{2}+1)}{t}^{-\frac{1}{2}}\\ & =\frac{\pi }{2}\frac{1}{\Gamma (\frac{1}{2})}{t}^{-\frac{1}{2}}\end{array}$

$\begin{array}{rl}& \text{由余元公式：}\Gamma (s)\Gamma (1-s)=\frac{\pi }{\sin \pi s}.\\ & \text{令}s=\frac{1}{2}\text{，则}\Gamma (\frac{1}{2})=\sqrt{\pi }，\\ & \text{则}f(t)=\frac{\sqrt{\pi }}{2\sqrt{t}},{\int }_0^{\infty }{e}^{-x^2}dx=f(1)=\frac{\sqrt{\pi }}{2}.\end{array}$

5. 变量代换(一)

$\begin{array}{r}\text{令}t=x^2\text{，则}\frac{dt}{dx}=2x=2\sqrt{t},dx=\frac{dt}{2\sqrt{t}}.\end{array}$
$${\int }_0^{\infty }{e}^{-x^2}dx={\int }_0^{\infty }{e}^{-t}\frac{dt}{2\sqrt{t}}=\frac{1}{2}{\int }_0^{\infty }{e}^{-t}{t}^{\frac{1}{2}-1}dt=\frac{1}{2}\Gamma (\frac{1}{2})$$
$\begin{array}{rl}& \text{由上一种方法}\Gamma (\frac{1}{2})=\sqrt{\pi }\text{，则}{\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}.\end{array}$

6. 变量代换(二)

$\begin{array}{r}\text{记}I={\int }_0^{\infty }{e}^{-x^2}dx\text{，令}x=ut,u>0\text{为参数，则}\end{array}$ $$\begin{array}{rl}I& =u{\int }_0^{\infty }{e}^{-u^2{t}^2}dt\\ e^{-u^2}I& =e^{-u^2}u{\int }_0^{\infty }{e}^{-u^2{t}^2}dt\end{array}$$ $\begin{array}{rl}\text{两边对变量}u\text{积分，}I{\int }_0^{\infty }{e}^{-u^2}du& ={\int }_0^{\infty }\left(e^{-u^2}u{\int }_0^{\infty }{e}^{-u^2{t}^2}dt\right)du\end{array}$
$\begin{array}{r}\text{右边交换积分次序，}\end{array}$

$$\begin{array}{rl}{I}^2& =\frac{1}{2}{\int }_0^{\infty }\left({\int }_0^{\infty }{e}^{-(1+t^2)u^2}du\right)dt\\ & =\frac{1}{2}{\int }_0^{\infty }\frac{1}{1+t^2}dt\\ & =\frac{1}{2}\arctan t{|}_0^{\infty }\\ & =\frac{\pi }{4}\end{array}$$

$\begin{array}{r}\text{所以}I=\frac{\sqrt{\pi }}{2}.\end{array}$

7. 构造含参变量函数

$\begin{array}{r}\text{记}I={\int }_0^{\infty }{e}^{-x^2}dx.\text{令}f(x)={\int }_0^1\frac{e^{-x(1+t^2)}}{1+t^2}dt,x\ge 0\text{，则}\end{array}$
   $\begin{array}{rl}& f(0)={\int }_0^1\frac{1}{1+t^2}dt=\arctan t{|}_0^1=\frac{\pi }{4}\\ & f(\infty )=\underset{x\to \infty }{\lim }{\int }_0^1\frac{e^{-x(1+t^2)}}{1+t^2}dt=0\end{array}$

   $\begin{array}{rl}{f}^{\prime }(x)& =-{\int }_0^1\frac{\partial }{\partial x}\left(\frac{e^{-x(1+t^2)}}{1+t^2}\right)dt\\ & =-{\int }_0^1{e}^{-x(1+t^2)}dt\\ & =-e^{-x}{\int }_0^1{e}^{-x{t}^2}dt(letu=\sqrt{x}t,dt=du/\sqrt{x})\\ & =-\frac{e^{-x}}{\sqrt{x}}{\int }_0^{\sqrt{x}}{e}^{-u^2}du\end{array}$

$\begin{array}{rl}& \text{记}g(x)={\int }_0^x{e}^{-t^2}dt,\\ & \text{则}{f}^{\prime }(x)=-\frac{e^{-x}}{\sqrt{x}}g(\sqrt{x}),g^{\prime }(x)=e^{-x^2},g(0)=0,g(\infty )=I.\end{array}$

$\begin{array}{r}\text{由牛顿-莱布尼兹公式，}\end{array}$

   $\begin{array}{rl}f(\infty )-f(0)& ={\int }_0^{\infty }{f}^{\prime }(x)dx\\ & =-{\int }_0^{\infty }\frac{e^{-x}}{\sqrt{x}}g(\sqrt{x})dx(\text{let}u=\sqrt{x},du=\frac{dx}{2\sqrt{x}})\\ & =-2{\int }_0^{\infty }{e}^{-u^2}g(u)du\\ & =-2{\int }_0^{\infty }g(u)g^{\prime }(u)du\\ & =-2{\int }_0^{\infty }g(u)d(g(u))\\ & =-{\int }_0^{\infty }d(g(u{)}^2)\\ & =g(0{)}^2-g(\infty {)}^2\end{array}$

$\begin{array}{rl}\text{所以，}0-\frac{\pi }{4}& =0-I^2,I=\frac{\sqrt{\pi }}{2}.\end{array}$

8. Wallis 公式

$\begin{array}{r}\text{Wallis 公式内容：}\end{array}$
$$\underset{n\to \infty }{\lim }\frac{(2n-1)!!}{(2n)!!}\sqrt{2n}=\sqrt{\frac{2}{\pi }}$$ $\text{推导过程参见：}$Wallis公式_百度百科

  $\begin{array}{rl}\because \underset{n\to \infty }{\lim }(1+\frac{x^2}{n}{)}^{\frac{n}{x^2}}& =e\\ \therefore \underset{n\to \infty }{\lim }(1+\frac{x^2}{n}{)}^{-n}& =e^{-x^2}\end{array}$

 $\begin{array}{rl}{\int }_0^{\infty }{e}^{-x^2}dx=& {\int }_0^{\infty }\underset{n\to \infty }{\lim }(1+\frac{x^2}{n}{)}^{-n}dx=\underset{n\to \infty }{\lim }{\int }_0^{\infty }(1+\frac{x^2}{n}{)}^{-n}dx\end{array}$

 $$\begin{gathered} \begin{array}{rl} \\ & (\text{可以验证积分与极限次序是可交换的}).\end{array} \end{gathered}$$

$\begin{array}{r}\text{令}\tan t=\frac{x}{\sqrt{n}}\text{，则}\frac{dt}{dx}=\frac{1}{\sqrt{n}}(1+\frac{x^2}{n}{)}^{-1},dx=\sqrt{n}(1+\frac{x^2}{n})dt.\end{array}$
$\begin{array}{rl}\text{原式}& =\underset{n\to \infty }{\lim }{\int }_0^{\frac{\pi }{2}}\sqrt{n}(1+\frac{x^2}{n}{)}^{-n+1}dt\\ & =\underset{n\to \infty }{\lim }{\int }_0^{\frac{\pi }{2}}\sqrt{n}(1+(\tan t{)}^2{)}^{-(n-1)}dt\\ & =\underset{n\to \infty }{\lim }\sqrt{n}{\int }_0^{\frac{\pi }{2}}(\cos t{)}^{2n-2}dt({\int }_0^{\frac{\pi }{2}}(\cos t{)}^{2n}dt=\frac{(2n-1)!!}{(2n)!!}\frac{\pi }{2})\\ & =\underset{n\to \infty }{\lim }\sqrt{n}\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi }{2}\\ & =\frac{\pi }{2}\underset{n\to \infty }{\lim }\sqrt{(n-1)+1}\frac{(2(n-1)-1)!!}{(2(n-1))!!}(\text{let}{n}^{\prime }=n-1)\\ & =\frac{\pi }{2}\underset{n^{\prime }\to \infty }{\lim }\frac{\sqrt{n^{\prime }+1}}{\sqrt{2n^{\prime }}}\frac{(2n^{\prime }-1)!!}{(2n^{\prime })!!}\sqrt{2n^{\prime }}\\ & =\frac{\pi }{2\sqrt{2}}\underset{n^{\prime }\to \infty }{\lim }\frac{\sqrt{n^{\prime }+1}}{\sqrt{n^{\prime }}}\underset{n^{\prime }\to \infty }{\lim }\frac{(2n^{\prime }-1)!!}{(2n^{\prime })!!}\sqrt{2n^{\prime }}\\ & =\frac{\pi }{2\sqrt{2}}\cdot 1\cdot \sqrt{\frac{2}{\pi }}\\ & =\frac{\sqrt{\pi }}{2}\end{array}$

$\begin{array}{r}\text{所以}{\int }_0^{\infty }{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}\end{array}.$

$\text{若读者还有其他巧妙解法，请不吝赐教！}$

$\text{文末彩蛋：大家好，这是我的孪生兄弟：}$$\text{无穷积分}\int \frac{\sin x}{x}dx\text{的几种巧妙解法！}$

Plus: 如有错误、可以改进的地方、或任何想说的，请在评论区留言！