无穷积分 ∫e^(-x^2)dx 的几种巧妙解法

  $\displaystyle\int_{-\infin}^{\infin}e^{-x^2}dx$∫−∞∞​e−x2dx 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。

1. 极坐标变换

  $\begin{aligned}\big(\int_{-\infin}^{\infin}e^{-x^2}dx\big)^2&=\int_{-\infin}^{\infin}e^{-x^2}dx\int_{-\infin}^{\infin}e^{-y^2}dy\\&=\int_{-\infin}^{\infin}\int_{-\infin}^{\infin}e^{-(x^2+y^2)}dxdy \quad(\textbf{极坐标变换})\\&=\int_{0}^{2\pi}\int_{0}^{\infin}e^{-r^2}rdrd\theta\\&=\big(\int_{0}^{2\pi}d\theta\big)\big(\frac{1}{2}\int_{0}^{\infin}e^{-r^2}dr^2\big)\\&=2\pi\frac{1}{2}=\pi\\ \therefore \,\,\int_{-\infin}^{\infin}e^{-x^2}dx&=\sqrt{\pi}\,.\end{aligned}$(∫−∞∞​e−x2dx)2∴∫−∞∞​e−x2dx​=∫−∞∞​e−x2dx∫−∞∞​e−y2dy=∫−∞∞​∫−∞∞​e−(x2+y2)dxdy(极坐标变换)=∫02π​∫0∞​e−r2rdrdθ=(∫02π​dθ)(21​∫0∞​e−r2dr2)=2π21​=π=π​.​

2. 几何做法(一)

$\begin{aligned}S_1&=\{(x,y)\,|\,x^2+y^2\leq R^2\}\\S_2&=\{(x,y)\,|\,0\leq x \leq R,\,0\leq y \leq R\}\\S_3&=\{(x,y)\,|\,x^2+y^2\leq 2R^2\}\\\,\end{aligned}$S1​S2​S3​​={(x,y)∣x2+y2≤R2}={(x,y)∣0≤x≤R,0≤y≤R}={(x,y)∣x2+y2≤2R2}​ $\begin{aligned}\big(\int_{0}^{R}e^{-x^2}dx\big)^2&=\int_{0}^{R}e^{-x^2}dx\int_{0}^{R}e^{-y^2}dy\\&=\int_{0}^{R}\int_{0}^{R}e^{-(x^2+y^2)}dxdy\\&=\iint_{S_2}e^{-(x^2+y^2)}dxdy\end{aligned}$(∫0R​e−x2dx)2​=∫0R​e−x2dx∫0R​e−y2dy=∫0R​∫0R​e−(x2+y2)dxdy=∬S2​​e−(x2+y2)dxdy​

$\begin{aligned}&\because e^{-(x^2+y^2)}\geq 0 \\ &\therefore \iint_{S_1}e^{-(x^2+y^2)}dxdy\leq\iint_{S_2}e^{-(x^2+y^2)}dxdy\leq\iint_{S_3}e^{-(x^2+y^2)}dxdy\end{aligned}$​∵e−(x2+y2)≥0∴∬S1​​e−(x2+y2)dxdy≤∬S2​​e−(x2+y2)dxdy≤∬S3​​e−(x2+y2)dxdy​

  $\begin{aligned}\iint_{S_1}e^{-(x^2+y^2)}dxdy&=\int_{0}^{\pi/2}\int_{0}^{R}e^{-r^2}rdrd\theta\\&=\big(\int_{0}^{\pi/2}d\theta\big)\big(\frac{1}{2}\int_{0}^{R}e^{-r^2}dr^2\big)\\&=\frac{\pi}{4}(1-e^{-R^2})\end{aligned}$∬S1​​e−(x2+y2)dxdy​=∫0π/2​∫0R​e−r2rdrdθ=(∫0π/2​dθ)(21​∫0R​e−r2dr2)=4π​(1−e−R2)​

$\begin{aligned}\\&\textbf{同理}\,\iint_{S_3}e^{-(x^2+y^2)}dxdy=\frac{\pi}{4}(1-e^{-2R^2})\,.\\&\textbf{则}\,\, \frac{\pi}{4}(1-e^{-R^2})\leq\iint_{S_2}e^{-(x^2+y^2)}dxdy\leq\frac{\pi}{4}(1-e^{-2R^2})\,.\\&\textbf{两边取极限}\,R\to\infin,\,\textbf{则}\,\Big(\int_{0}^{R}e^{-x^2}dx\Big)^2=\iint_{S_2}e^{-(x^2+y^2)}dxdy=\frac{\pi}{4}.\\&\textbf{所以，}\int_{0}^{\infin}e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \,.\end{aligned}$​同理∬S3​​e−(x2+y2)dxdy=4π​(1−e−2R2).则4π​(1−e−R2)≤∬S2​​e−(x2+y2)dxdy≤4π​(1−e−2R2).两边取极限R→∞,则(∫0R​e−x2dx)2=∬S2​​e−(x2+y2)dxdy=4π​.所以，∫0∞​e−x2dx=2π​​.​

3. 几何做法(二)

$\begin{aligned}\textbf{考虑曲线}\, z=e^{-x^2}\, \textbf{绕}\, z\, \textbf{轴旋转一周生成的旋转体的体积，由高数知识}\end{aligned}$考虑曲线z=e−x2绕z轴旋转一周生成的旋转体的体积，由高数知识​
$\begin{aligned}V&= \int_0^{1}\pi x^2dz\\&=\pi \int_0^{1}(-\ln z)dz\\&=\pi (z-z\ln z)\,\Big|_0^1\\&=\pi (1-\lim_{z\to 0^{+}}(z-z\ln z))\\&=\pi\end{aligned}$V​=∫01​πx2dz=π∫01​(−lnz)dz=π(z−zlnz)∣∣∣​01​=π(1−z→0+lim​(z−zlnz))=π​

$\begin{aligned}&\textbf{旋转体在几何上的表示如图所示，旋转生成的曲面方程为} \,z=e^{-(x^2+y^2)}. \\ &\textbf{旋转体的体积也可以理解为曲面} \,z=e^{-(x^2+y^2)} \textbf{与}\,xOy\,\textbf{之间部分的体积，即}\end{aligned}$​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−(x2+y2).旋转体的体积也可以理解为曲面z=e−(x2+y2)与xOy之间部分的体积，即​
$\pi=V=\iint_{R^2}e^{-(x^2+y^2)}dxdy=\int_{-\infin}^{\infin}e^{-x^2}dx\int_{-\infin}^{\infin}e^{-y^2}dy=\big(\int_{-\infin}^{\infin}e^{-x^2}dx\big)^2$π=V=∬R2​e−(x2+y2)dxdy=∫−∞∞​e−x2dx∫−∞∞​e−y2dy=(∫−∞∞​e−x2dx)2$\begin{aligned} \textbf{则} \int_{-\infin}^{\infin}e^{-x^2}dx=\sqrt{\pi}.\end{aligned}$则∫−∞∞​e−x2dx=π​.​

4. 拉普拉斯变换

$\begin{aligned}&\textbf{令}\,f(t)=\int_0^{\infin}e^{-x^2t}dx\textbf{，对}\,f(t)\,\textbf{作拉普拉斯变换，令}\end{aligned}$​令f(t)=∫0∞​e−x2tdx，对f(t)作拉普拉斯变换，令​
  $\begin{aligned}F(s)=\mathscr{L}[f(t)]&=\int_0^{\infin}\mathscr{L}[e^{-x^2t}]_t\,dx\\&=\int_0^{\infin}\frac{1}{s+x^2}dx\\&=\frac{1}{\sqrt{s}}\int_{0}^{\infin}\frac{1}{1+(\displaystyle \frac{x}{\sqrt{s}})^2}d(\frac{x}{\sqrt{s}}) \quad (\textrm{let}\,\, u=\frac{x}{\sqrt{s}})\\&=\frac{1}{\sqrt{s}}\arctan u\,\Big|_0^{\infin}\\&=\frac{\pi}{2}s^{-\frac{1}{2}}\end{aligned}$F(s)=L[f(t)]​=∫0∞​L[e−x2t]t​dx=∫0∞​s+x21​dx=s​1​∫0∞​1+(s​x​)21​d(s​x​)(letu=s​x​)=s​1​arctanu∣∣∣​0∞​=2π​s−21​​

$\textbf{对}\,F(s)\,\textbf{作拉普拉斯反变换}$对F(s)作拉普拉斯反变换

  $\begin{aligned}f(t)=\mathscr{L}^{-1}[F(s)]&=\frac{\pi}{2}\mathscr{L}^{-1}[\frac{1}{s^{-\frac{1}{2}+1}}]\\&=\frac{\pi}{2}\frac{1}{\Gamma(-\frac{1}{2}+1)}t^{-\frac{1}{2}}\\&=\frac{\pi}{2}\frac{1}{\Gamma(\frac{1}{2})}t^{-\frac{1}{2}}\end{aligned}$f(t)=L−1[F(s)]​=2π​L−1[s−21​+11​]=2π​Γ(−21​+1)1​t−21​=2π​Γ(21​)1​t−21​​

$\begin{aligned}&\textbf{由余元公式：}\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s} .\\ &\textbf{令}\,s=\frac{1}{2}\textbf{，则}\,\Gamma(\frac{1}{2})=\sqrt{\pi}，\\&\textbf{则}\, f(t)=\frac{\sqrt{\pi}}{2\sqrt{t}}\,,\,\int_0^{\infin}e^{-x^2}dx=f(1)=\frac{\sqrt{\pi}}{2}\,.\end{aligned}$​由余元公式：Γ(s)Γ(1−s)=sinπsπ​.令s=21​，则Γ(21​)=π​，则f(t)=2t​π​​,∫0∞​e−x2dx=f(1)=2π​​.​

5. 变量代换(一)

$\begin{aligned}\textbf{令}\,t=x^2,\textbf{则}\,\frac{dt}{dx}=2x=2\sqrt{t} ,dx=\frac{dt}{2\sqrt{t}}\,.\end{aligned}$令t=x2,则dxdt​=2x=2t​,dx=2t​dt​.​
$\int_{0}^{\infin}e^{-x^2}dx=\int_{0}^{\infin}e^{-t}\frac{dt}{2\sqrt{t}}=\frac{1}{2}\int_{0}^{\infin}e^{-t}t^{\frac{1}{2}-1}dt=\frac{1}{2}\,\Gamma(\frac{1}{2})$∫0∞​e−x2dx=∫0∞​e−t2t​dt​=21​∫0∞​e−tt21​−1dt=21​Γ(21​)
$\begin{aligned}&\textbf{由上一种方法}\,\Gamma(\frac{1}{2})=\sqrt{\pi}\textbf{，则}\int_{0}^{\infin}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}.\end{aligned}$​由上一种方法Γ(21​)=π​，则∫0∞​e−x2dx=2π​​.​

6. 变量代换(二)

$\begin{aligned}\textbf{记}\,I=\int_0^{\infin}e^{-x^2}dx,\,\textbf{令}\,x=ut,\,u>0\,\textbf{为参数，则}\end{aligned}$记I=∫0∞​e−x2dx,令x=ut,u>0为参数，则​ $\begin{aligned}I&=u\int_0^{\infin}e^{-u^2t^2}dt\\e^{-u^2}I&=e^{-u^2}u\int_0^{\infin}e^{-u^2t^2}dt\end{aligned}$Ie−u2I​=u∫0∞​e−u2t2dt=e−u2u∫0∞​e−u2t2dt​ $\begin{aligned}\textbf{两边对变量}\,u\,\textbf{积分，} I\int_0^{\infin}e^{-u^2}du&=\int_0^{\infin}\big(e^{-u^2}u\int_0^{\infin}e^{-u^2t^2}dt\big)du\\ \end{aligned}$两边对变量u积分，I∫0∞​e−u2du​=∫0∞​(e−u2u∫0∞​e−u2t2dt)du​
$\begin{aligned}\textbf{右边交换积分次序，} \end{aligned}$右边交换积分次序，​

$\begin{aligned}I^2&=\frac{1}{2}\int_0^{\infin}\big(\int_0^{\infin}e^{-(1+t^2)u^2}du\big)dt\\&=\frac{1}{2}\int_0^{\infin}\frac{1}{1+t^2}dt \\&=\frac{1}{2}\arctan t\,\Big|_0^{\infin}\\&=\frac{\pi}{4}\end{aligned}$I2​=21​∫0∞​(∫0∞​e−(1+t2)u2du)dt=21​∫0∞​1+t21​dt=21​arctant∣∣∣​0∞​=4π​​

$\begin{aligned}\textbf{所以}\,I=\frac{\sqrt{\pi}}{2}\,.\end{aligned}$所以I=2π​​.​

7. 构造含参变量函数

$\begin{aligned}\textbf{记}\,I=\int_0^{\infin}e^{-x^2}dx\,. \,\textbf{令}\,f(x)=\int_0^{1}\frac{e^{-x(1+t^2)}}{1+t^2}dt,x\geq 0\textbf{，则}\end{aligned}$记I=∫0∞​e−x2dx.令f(x)=∫01​1+t2e−x(1+t2)​dt,x≥0，则​
   $\begin{aligned} &f(0)=\int_0^{1}\frac{1}{1+t^2}dt=\arctan t\,\Big|_0^1=\frac{\pi}{4} \\ &f(\infin)=\lim_{x\to\infin}\int_0^{1}\frac{e^{-x(1+t^2)}}{1+t^2}dt=0\end{aligned}$​f(0)=∫01​1+t21​dt=arctant∣∣∣​01​=4π​f(∞)=x→∞lim​∫01​1+t2e−x(1+t2)​dt=0​

   $\begin{aligned}f'(x)&=-\int_0^{1}e^{-x(1+t^2)}dt \\ &=-e^{-x}\int_0^{1}e^{-xt^2}dt \quad (\textrm{let }u=\sqrt{x}t,\,dt=du/\sqrt{x}\,) \\&=-\frac{e^{-x}}{\sqrt{x}}\int_0^{\sqrt{x}}e^{-u^2}du\end{aligned}$f′(x)​=−∫01​e−x(1+t2)dt=−e−x∫01​e−xt2dt(let u=x​t,dt=du/x​)=−x​e−x​∫0x​​e−u2du​

$\begin{aligned}&\textbf{记}\,\,g(x)=\int_0^{x}e^{-t^2}dt\,,\\&\textbf{则} \, f'(x)=-\frac{e^{-x}}{\sqrt{x}}g(\sqrt{x}),\,g'(x)=e^{-x^2},\,g(0)=0,\,g(\infin)=I.\end{aligned}$​记g(x)=∫0x​e−t2dt,则f′(x)=−x​e−x​g(x​),g′(x)=e−x2,g(0)=0,g(∞)=I.​

$\begin{aligned}\textbf{由牛顿-莱布尼兹公式，}\end{aligned}$由牛顿-莱布尼兹公式，​

   $\begin{aligned}f(\infin)-f(0)&=\int_0^{\infin}f'(x)dx\\&=-\int_0^{\infin}\frac{e^{-x}}{\sqrt{x}}g(\sqrt{x})dx\quad (\textrm{let} \,u=\sqrt{x},\,du=\frac{dx}{2\sqrt{x}} )\\&=-2\int_0^{\infin}e^{-u^2}g(u)du\\&=-2\int_0^{\infin}g(u)g'(u)du\\&=-2\int_0^{\infin}g(u)d(g(u))\\&=-\int_0^{\infin}d\big(g(u)^2\big)\\&=g(0)^2-g(\infin)^2 \end{aligned}$f(∞)−f(0)​=∫0∞​f′(x)dx=−∫0∞​x​e−x​g(x​)dx(letu=x​,du=2x​dx​)=−2∫0∞​e−u2g(u)du=−2∫0∞​g(u)g′(u)du=−2∫0∞​g(u)d(g(u))=−∫0∞​d(g(u)2)=g(0)2−g(∞)2​

$\begin{aligned}\textbf{所以，} 0-\frac{\pi}{4}&=0-I^2,\,I=\frac{\sqrt{\pi}}{2}\,.\end{aligned}$所以，0−4π​​=0−I2,I=2π​​.​

8. Wallis 公式

$\begin{aligned}\textbf{Wallis 公式内容：}\end{aligned}$Wallis 公式内容：​
$\lim_{n\to\infin}\frac{(2n-1)!!}{(2n)!!}\sqrt{2n}=\sqrt{\frac{2}{\pi}}$n→∞lim​(2n)!!(2n−1)!!​2n​=π2​​ $\textbf{推导过程参见：}$推导过程参见：Wallis公式_百度百科

  $\begin{aligned}\because \,\lim_{n\to\infin} \Big(1+\frac{x^2}{n}\Big)^{\frac{n}{x^2}}&=e\\ \therefore \,\lim_{n\to\infin} \Big(1+\frac{x^2}{n}\Big)^{-n}&=e^{-x^2}\end{aligned}$∵n→∞lim​(1+nx2​)x2n​∴n→∞lim​(1+nx2​)−n​=e=e−x2​

 $\begin{aligned}\int_0^{\infin}e^{-x^2}dx=&\int_0^{\infin}\lim_{n\to\infin} \Big(1+\frac{x^2}{n}\Big)^{-n}dx =\lim_{n\to\infin} \int_0^{\infin}\Big(1+\frac{x^2}{n}\Big)^{-n}dx \end{aligned}$∫0∞​e−x2dx=​∫0∞​n→∞lim​(1+nx2​)−ndx=n→∞lim​∫0∞​(1+nx2​)−ndx​

 $\begin{aligned}\\&\textbf{可以验证积分与极限次序是可交换的}\end{aligned}$​可以验证积分与极限次序是可交换的​

$\begin{aligned}\textbf{令}\,\tan t=\frac{x}{\sqrt{n}} \textbf{，则}\,\frac{dt}{dx}=\frac{1}{\sqrt{n}}(1+\frac{x^2}{n})^{-1},\, dx=\sqrt{n}(1+\frac{x^2}{n})dt.\end{aligned}$令tant=n​x​，则dxdt​=n​1​(1+nx2​)−1,dx=n​(1+nx2​)dt.​
$\begin{aligned} \textbf{原式}&=\lim_{n\to\infin} \int_0^{\frac{\pi}{2}}\sqrt{n}(1+\frac{x^2}{n})^{-n+1}dt \\&=\lim_{n\to\infin} \int_0^{\frac{\pi}{2}}\sqrt{n}(1+(\tan t)^2)^{-(n-1)}dt\\&=\lim_{n\to\infin} \sqrt{n}\int_0^{\frac{\pi}{2}}(\cos t)^{2n-2}dt \quad \Big( \int_0^{\frac{\pi}{2}}(\cos t)^{2n}dt=\frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}\,\Big)\\&=\lim_{n\to\infin} \sqrt{n}\,\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2} \\&=\frac{\pi}{2}\lim_{n\to\infin} \sqrt{(n-1)+1}\,\frac{(2(n-1)-1)!!}{(2(n-1))!!} \quad (\textrm{let}\,\,n'=n-1)\\&=\frac{\pi}{2}\lim_{n'\to\infin} \frac{\sqrt{n'+1}}{\sqrt{2n'}}\,\frac{(2n'-1)!!}{(2n')!!}\sqrt{2n'}\\&=\frac{\pi}{2\sqrt{2}}\lim_{n'\to\infin} \frac{\sqrt{n'+1}}{\sqrt{n'}}\lim_{n'\to\infin} \frac{(2n'-1)!!}{(2n')!!}\sqrt{2n'}\\&=\frac{\pi}{2\sqrt{2}}\cdot1\cdot\sqrt{\frac{2}{\pi}}\\&=\frac{\sqrt{\pi}}{2}\end{aligned}$原式​=n→∞lim​∫02π​​n​(1+nx2​)−n+1dt=n→∞lim​∫02π​​n​(1+(tant)2)−(n−1)dt=n→∞lim​n​∫02π​​(cost)2n−2dt(∫02π​​(cost)2ndt=(2n)!!(2n−1)!!​2π​)=n→∞lim​n​(2n−2)!!(2n−3)!!​2π​=2π​n→∞lim​(n−1)+1​(2(n−1))!!(2(n−1)−1)!!​(letn′=n−1)=2π​n′→∞lim​2n′​n′+1​​(2n′)!!(2n′−1)!!​2n′​=22​π​n′→∞lim​n′​n′+1​​n′→∞lim​(2n′)!!(2n′−1)!!​2n′​=22​π​⋅1⋅π2​​=2π​​​

$\begin{aligned}\textbf{所以}\,\int_0^{\infin}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\end{aligned}.$所以∫0∞​e−x2dx=2π​​​.

$\begin{aligned}\small \textbf{如读者还有其他巧妙解法，请不吝赐教！}\end{aligned}$如读者还有其他巧妙解法，请不吝赐教！​