方法1：因为积分值只与被积函数和积分域有关，与积分变量无关，所以
\[I^{2}=\left ( \int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x \right )^{2}=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x~~\cdot \int_{0}^{\infty }e^{-y^{2}}\mathrm{d}y=\int_{0}^{\infty }\int_{0}^{\infty }e^{-\left ( x^{2}+y^{2} \right )}\mathrm{d}x\mathrm{d}y
\]
用极坐标系下二重积分的计算法
\[I=\int_{0}^{\frac{\pi }{2}}\mathrm{d}\theta \int_{0}^{\infty }e^{-r^{2}}r\mathrm{d}r=\frac{\pi }{2}\left ( -\frac{1}{2}e^{-r^{2}} \right )\Bigg|_{0}^{\infty }=\frac{\pi }{4}
\]
而 \(e^{-r^{2}}\geq 0\), 则 \(I>0\). 即
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2}
\]
方法2：因为 \(\displaystyle \left ( 1+\frac{x^{2}}{n} \right )^{-n}\) 当 $n\rightarrow +\infty $ 时一致收敛于 \(e^{-x^{2}}\), 利用积分号下取极限，则有
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\int_{0}^{\infty }\left [ \lim_{n\rightarrow \infty }\left ( 1+\frac{x^{2}}{n} \right )^{-n} \right ]\mathrm{d}x=\lim_{n\rightarrow \infty }\int_{0}^{\infty }\left ( 1+\frac{x^{2}}{n} \right )^{-n}\mathrm{d}x
\]
令 \(x=\sqrt{nt}\), 则
\[I=\lim_{n\rightarrow \infty }\sqrt{n}\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )^{n}}\mathrm{d}t=\sqrt{n}I_n
\]
由于
\[\begin{align*}
I_{n-1}&=\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )^{n-1}}\mathrm{d}t=\frac{t}{\left ( 1+t^{2} \right )^{n-1}}\Bigg|_{0}^{\infty }+2\left ( n-1 \right )\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )}\mathrm{d}t\\
&=2\left ( n-1 \right )I_{n-1}-2\left ( n-1 \right )I_n
\end{align*}\]
所以 \(\displaystyle I_n=\frac{2n-3}{2n-2}I_{n-1}\), 而 \(\displaystyle I_1=\int_{0}^{\infty }\frac{1}{1+t^{2}}\mathrm{d}t=\frac{\pi }{2}\), 递推得
\[I_n=\frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2}
\]
因此 \(\displaystyle \int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\lim_{n\rightarrow \infty }\frac{\sqrt{n}\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2}\). 根据Wallis公式，有
\[\frac{\pi }{2}=\lim_{n\rightarrow \infty }\frac{\left [ \left ( 2n \right )!! \right ]^{2}}{\left ( 2n+1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}}=\lim_{n\rightarrow \infty }\frac{\left [ \left ( 2n-2 \right )!! \right ]^{2}}{\left ( 2n-1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}}
\]
所以
\[\begin{align*}
I&=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\frac{\pi }{2}\lim_{n\rightarrow \infty }\frac{\sqrt{n}\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}=\frac{\pi }{2}\lim_{n\rightarrow \infty }\frac{\left ( 2n-3 \right )!!\sqrt{2n-1}}{\left ( 2n-2 \right )!!}\cdot \sqrt{\frac{n}{2n-1}}\\
&=\frac{\pi }{2}\cdot \sqrt{\frac{2}{\pi }}\cdot \frac{1}{\sqrt{2}}=\frac{\sqrt{\pi }}{2}
\end{align*}\]
方法3：考虑两个含参变量积分
\[f\left ( x \right )=\left ( \int_{0}^{x}e^{-t^{2}}\mathrm{d}t \right )^{2}~~,~~g\left ( x \right )=\int_{0}^{1}\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u
\]
利用积分号下微分法，得
\[\begin{align*}
f'\left ( x \right )&= 2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}\mathrm{d}t\\
g'\left ( x \right )&=\int_{0}^{1}\frac{\partial }{\partial x}\left [ \frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}} \right ]\mathrm{d}u=-2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}u^{2}}\mathrm{d}u
\end{align*}\]
对后一积分，令\(xu=t\), 则
\[g'\left ( x \right )=-2xe^{-x^{2}}\int_{0}^{x}e^{-t^{2}}\mathrm{d}t=-f'\left ( x \right )~~~\left ( x\geq 0 \right )
\]
于是
\[\begin{align*}
f\left ( x \right )+g\left ( x \right )=c~~~\left ( x\geq 0 \right )\tag1
\end{align*}\]
由于 \(\displaystyle f(0)=0 , g\left ( 0 \right )=\frac{\pi }{4}\), 故 \(c=\dfrac{\pi }{4}\), 即
\[f(x)+g(x)=\dfrac{\pi }{4}~~~\left ( x\geq 0 \right )
\]
当 \(u\in \left [ 0,1 \right ]\), 有
\[0\leq \frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\leq e^{-x^{2}u^{2}}\leq e^{-x^{2}}~~~\left ( x\geq 0 \right )
\]
因此，当 $x\rightarrow \infty $ 时，函数 \(\displaystyle \frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\) 关于 \(u\in \left [ 0,1 \right ]\) 一致的趋于0.
\[\lim_{x\rightarrow \infty }g\left ( x \right )=\lim_{x\rightarrow \infty }\int_{0}^{1}\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u=\int_{0}^{1}\lim_{x\rightarrow \infty }\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u=0
\]
从而，由 \(f(x)\) 的定义及(1)，得
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\sqrt{\lim_{x\rightarrow \infty }f\left ( x \right )}=\sqrt{\lim_{x\rightarrow \infty }\frac{\pi }{4}-g\left ( x \right )}=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2}
\]
方法4：设 \(\displaystyle f\left ( t \right )=\int_{0}^{\infty }e^{-tx^{2}}\mathrm{d}x\), 对 \(f(t)\) 取拉普拉斯变换，得
\[\mathcal{L}\left ( \int_{0}^{\infty }e^{-tx^{2}}\mathrm{d}x \right )=\int_{0}^{\infty }\int_{0}^{\infty }e^{-tx^{2}}e^{-st}\mathrm{d}t\mathrm{d}x=\int_{0}^{\infty }\mathcal{L}\left ( e^{-tx^{2}} \right )\mathrm{d}x=\int_{0}^{\infty }\frac{\mathrm{d}x}{s+x^{2}}=\frac{\pi }{2\sqrt{s}}
\]
再取拉普拉斯逆变换，有 \(\displaystyle f\left ( t \right )=\int_{0}^{\infty }e^{-tx^{2}}\mathrm{d}x=\frac{\sqrt{\pi }}{2\sqrt{t}}\), 在上式中，令 \(t=1\), 则
\[I=f\left ( 1 \right )=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\frac{\sqrt{\pi }}{2}
\]
方法5：这种利用伽马函数的方法应该是高数中第一次接触的，出现在同济高数上册第五章最后，不过教材中打了星号，所以多数人都不了解，首先我们引入伽马函数的定义
\[\Gamma \left ( x \right )=\int_{0}^{\infty }t^{x-1}e^{-t}\mathrm{d}t
\]
所以，我们令 \(x^2=t\), 有
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\frac{1}{2}\int_{0}^{\infty }t^{-\frac{1}{2}}e^{-t}\mathrm{d}t=\frac{1}{2}\Gamma \left ( \frac{1}{2} \right )=\frac{\sqrt{\pi }}{2}
\]
其中 \(\displaystyle \Gamma \left ( \frac{1}{2} \right )=\sqrt{\pi }\) 可利用余元公式求得，这里不做证明.
方法6： 不难证明，函数 \((1+t)e^{-t}\) 在 \(t=0\) 时达到它的最大值1.因此当 \(t\neq 0\) 时，\((1+t)e^{-t}<1\), 令 \(t=\pm x^{2}\), 即得
\[\left ( 1-x^{2} \right )e^{x^{2}}<1~,~\left ( 1+x^{2} \right )e^{-x^{2}}<1
\]
或
\[\left ( 1-x^{2} \right )<e^{-x^{2}}<\frac{1}{1+x^{2}}~~~\left ( x>0 \right )
\]
假设限定第一个不等式中的 \(x\) 在(0,1)内变化，而第二个不等式中 \(x\) 则看作是任意的，把上式同 \(n\) 次方，有
\[\left ( 1-x^{2} \right )^{n}<e^{-nx^{2}}~~~\left ( 0<x<1 \right )
\]
\[e^{-nx^{2}}<\frac{1}{\left ( 1+x^{2} \right )^{n}}~~~\left ( x>0 \right )
\]
第一个不等式即从0到1积分，第二个不等式取从0到\(+\infty\)的积分，得
\[\int_{0}^{1}\left ( 1-x^{2} \right )^{n}\mathrm{d}x<\int_{0}^{1}e^{-nx^{2}}\mathrm{d}x<\int_{0}^{\infty }e^{-nx^{2}}\mathrm{d}x<\int_{0}^{\infty }\frac{1}{\left ( 1+x^{2} \right )^{n}}\mathrm{d}x
\]
在 \(\displaystyle \int_{0}^{1}\left ( 1-x^{2} \right )^{n}\mathrm{d}x\) 中，令 \(x=\cos t\), 则
\[\int_{0}^{1}\left ( 1-x^{2} \right )^{n}\mathrm{d}x=\int_{0}^{\frac{\pi }{2}}\sin^{2n+1}t\mathrm{d}t=\frac{\left ( 2n \right )!!}{\left ( 2n+1 \right )!!}
\]
在 \(\displaystyle \int_{0}^{\infty }\frac{1}{\left ( 1+x^{2} \right )^{n}}\mathrm{d}x\) 中，令 \(x=\cot t\), 则
\[\int_{0}^{\infty }\frac{1}{\left ( 1+x^{2} \right )^{n}}\mathrm{d}x=\int_{0}^{\frac{\pi }{2}}\sin^{2n-2}t\mathrm{d}t=\frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2}
\]
在 \(\displaystyle \int_{0}^{\infty }e^{-nx^{2}}\mathrm{d}x\) 中，令 \(\displaystyle x=\frac{t}{\sqrt{n}}\), 则
\[\int_{0}^{\infty }e^{-nx^{2}}\mathrm{d}x=\frac{1}{\sqrt{n}} \int_{0}^{\infty }e^{-t^{2}}\mathrm{d}t=\frac{1}{\sqrt{n}}I
\]
综上所述
\[\sqrt{n}\cdot \frac{\left ( 2n \right )!!}{\left ( 2n+1 \right )!!}<I<\sqrt{n}\cdot \frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2}
\]
取平方得
\[\frac{n}{2n+1}\cdot \frac{\left [ \left ( 2n \right )!! \right ]^{2}}{\left ( 2n+1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}}<I^2<\frac{n}{2n-1}\cdot \frac{\left [ \left ( 2n-3 \right )!! \right ]^{2}}{\left [ \left ( 2n-2 \right )!! \right ]^{2}}\cdot \left ( \frac{\pi }{2} \right )^{2}
\]
根据Wallis公式
\[\frac{\pi }{2}=\lim_{n\rightarrow \infty }\frac{\left [ \left ( 2n \right )!! \right ]^{2}}{\left ( 2n+1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}}
\]
不等式两边当 $n\rightarrow \infty $ 时的极限都是 \(\dfrac{\pi }{4}\), 所以
\[I^2=\frac{\pi }{4}\Rightarrow I=\frac{\sqrt{\pi }}{2}
\]
方法7：当然也可以利用三重积分
\[\begin{align*}
&8I^3 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\\
\Rightarrow &8I^3 = 4\pi\int_0^{\infty}\rho^2 e^{-\rho^2}\,\mathrm{d}\rho= 2\pi \int_0^{\infty} e^{-\rho^2}\,\mathrm{d}\rho=2\pi \cdot {I}
\Rightarrow 8I^3=2\pi I\Rightarrow I=\frac{\sqrt{\pi }}{2}
\end{align*}\]
方法8：注意到
\[n! =\int_0^\infty e^{-\sqrt[n]x} \mathrm{d}x\iff\frac1n! =\int_0^\infty e^{-x^n}\mathrm{d}x\rightarrow\frac12! = \int_0^\infty e^{-x^2}\mathrm{d}x
\]
\[\int_0^1\Big(1-\sqrt[n]x\Big)^m\,\mathrm{d}x = \int_0^1\Big(1-\sqrt[m]x\Big)^n\,\mathrm{d}x = \frac1{C_{m+n}^n} = \frac1{C_{m+n}^m} = \frac{m!\,n!}{(m+n)!}
\]
所以我们有
\[\frac\pi4 = \int_0^1\sqrt{1-x^2}\,\mathrm{d}x = \frac{\left(\dfrac12!\right)^2}{\left(\dfrac12 + \dfrac12\right)!} =\left(\frac12!\right)^2
\]
所以
\[I=\int_0^\infty e^{-x^2}\mathrm{d}x = \frac12! = \sqrt{\pi\over4} = \frac{\sqrt\pi}2
\]
方法9：利用
\[\int_0^\infty e^{-x^2}\mathrm{d}x=\sqrt \pi \int_0^\infty \frac{1}{\sqrt \pi}e^{-x^2}\mathrm{d}x=\sqrt \pi P(X\geq0)
\]
其中
\[X\sim N\left ( 0,\frac{1}{2} \right )~~,~~P(X>0)=P(X>E(X))=\frac{1}{2}
\]
所以
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\frac{\sqrt{\pi }}{2}
\]
方法10：利用
\[F(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} \exp\left(\frac{-t^2}{2}\right) \exp(- i \omega t) \mathrm{d}t
\]
所以
\[F(\omega) = \frac{1}{\pi} \int_{0}^{+\infty} \exp\left(\frac{-t^2}{2}\right) \cos( \omega t) \mathrm{d}t
\]
所以我们有
\[F'(\omega) = - \omega F(\omega)\Rightarrow F(\omega) = C \exp\left(\frac{-\omega^2}{2}\right)
\]
由
\[\exp\left(\frac{-x^2}{2}\right) = \int_{-\infty}^{+\infty} F(\omega) \exp( i \omega x) \mathrm{d}\omega
\]
可得 \(\displaystyle C = \frac{1}{\sqrt{2\pi}}.\) 令 \(\omega=0\), 有
\[F(0) = C = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} \exp\left(\frac{-t^2}{2}\right) \mathrm{d}t
\]
所以
\[\sqrt{2} \int_{-\infty}^{+\infty} \exp\left(-t^2\right) \mathrm{d}t = \sqrt{2\pi}\Rightarrow I=\frac{\sqrt{\pi }}{2}
\]
