方法1：因為積分值只與被積函數和積分域有關，與積分變量無關，所以
\[I^{2}=\left ( \int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x \right )^{2}=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x~~\cdot \int_{0}^{\infty }e^{-y^{2}}\mathrm{d}y=\int_{0}^{\infty }\int_{0}^{\infty }e^{-\left ( x^{2}+y^{2} \right )}\mathrm{d}x\mathrm{d}y \]
用極坐標系下二重積分的計算法
\[I=\int_{0}^{\frac{\pi }{2}}\mathrm{d}\theta \int_{0}^{\infty }e^{-r^{2}}r\mathrm{d}r=\frac{\pi }{2}\left ( -\frac{1}{2}e^{-r^{2}} \right )\Bigg|_{0}^{\infty }=\frac{\pi }{4} \]
而 \(e^{-r^{2}}\geq 0\), 則 \(I>0\). 即
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\sqrt{\frac{\pi }{4}}=\frac{\sqrt{\pi }}{2} \]
方法2：因為 \(\displaystyle \left ( 1+\frac{x^{2}}{n} \right )^{-n}\) 當 $n\rightarrow +\infty $ 時一致收斂於 \(e^{-x^{2}}\), 利用積分號下取極限，則有
\[I=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\int_{0}^{\infty }\left [ \lim_{n\rightarrow \infty }\left ( 1+\frac{x^{2}}{n} \right )^{-n} \right ]\mathrm{d}x=\lim_{n\rightarrow \infty }\int_{0}^{\infty }\left ( 1+\frac{x^{2}}{n} \right )^{-n}\mathrm{d}x \]
令 \(x=\sqrt{nt}\), 則
\[I=\lim_{n\rightarrow \infty }\sqrt{n}\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )^{n}}\mathrm{d}t=\sqrt{n}I_n \]
由於
\[\begin{align*} I_{n-1}&=\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )^{n-1}}\mathrm{d}t=\frac{t}{\left ( 1+t^{2} \right )^{n-1}}\Bigg|_{0}^{\infty }+2\left ( n-1 \right )\int_{0}^{\infty }\frac{1}{\left ( 1+t^{2} \right )}\mathrm{d}t\\ &=2\left ( n-1 \right )I_{n-1}-2\left ( n-1 \right )I_n \end{align*}\]
所以 \(\displaystyle I_n=\frac{2n-3}{2n-2}I_{n-1}\), 而 \(\displaystyle I_1=\int_{0}^{\infty }\frac{1}{1+t^{2}}\mathrm{d}t=\frac{\pi }{2}\), 遞推得
\[I_n=\frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2} \]
因此 \(\displaystyle \int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\lim_{n\rightarrow \infty }\frac{\sqrt{n}\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2}\). 根據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：考慮兩個含參變量積分
\[f\left ( x \right )=\left ( \int_{0}^{x}e^{-t^{2}}\mathrm{d}t \right )^{2}~~,~~g\left ( x \right )=\int_{0}^{1}\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u \]
利用積分號下微分法，得
\[\begin{align*} f'\left ( x \right )&= 2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}\mathrm{d}t\\ g'\left ( x \right )&=\int_{0}^{1}\frac{\partial }{\partial x}\left [ \frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}} \right ]\mathrm{d}u=-2xe^{-x^{2}}\int_{0}^{1}e^{-x^{2}u^{2}}\mathrm{d}u \end{align*}\]
對后一積分，令\(xu=t\), 則
\[g'\left ( x \right )=-2xe^{-x^{2}}\int_{0}^{x}e^{-t^{2}}\mathrm{d}t=-f'\left ( x \right )~~~\left ( x\geq 0 \right ) \]
於是
\[\begin{align*} f\left ( x \right )+g\left ( x \right )=c~~~\left ( x\geq 0 \right )\tag1 \end{align*}\]
由於 \(\displaystyle f(0)=0 , g\left ( 0 \right )=\frac{\pi }{4}\), 故 \(c=\dfrac{\pi }{4}\), 即
\[f(x)+g(x)=\dfrac{\pi }{4}~~~\left ( x\geq 0 \right ) \]
當 \(u\in \left [ 0,1 \right ]\), 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，當 $x\rightarrow \infty $ 時，函數 \(\displaystyle \frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\) 關於 \(u\in \left [ 0,1 \right ]\) 一致的趨於0.
\[\lim_{x\rightarrow \infty }g\left ( x \right )=\lim_{x\rightarrow \infty }\int_{0}^{1}\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u=\int_{0}^{1}\lim_{x\rightarrow \infty }\frac{e^{-x^{2}\left ( 1+u^{2} \right )}}{1+u^{2}}\mathrm{d}u=0 \]
從而，由 \(f(x)\) 的定義及(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：設 \(\displaystyle f\left ( t \right )=\int_{0}^{\infty }e^{-tx^{2}}\mathrm{d}x\), 對 \(f(t)\) 取拉普拉斯變換，得
\[\mathcal{L}\left ( \int_{0}^{\infty }e^{-tx^{2}}\mathrm{d}x \right )=\int_{0}^{\infty }\int_{0}^{\infty }e^{-tx^{2}}e^{-st}\mathrm{d}t\mathrm{d}x=\int_{0}^{\infty }\mathcal{L}\left ( e^{-tx^{2}} \right )\mathrm{d}x=\int_{0}^{\infty }\frac{\mathrm{d}x}{s+x^{2}}=\frac{\pi }{2\sqrt{s}} \]
再取拉普拉斯逆變換，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 \(t=1\), 則
\[I=f\left ( 1 \right )=\int_{0}^{\infty }e^{-x^{2}}\mathrm{d}x=\frac{\sqrt{\pi }}{2} \]
方法5：這種利用伽馬函數的方法應該是高數中第一次接觸的，出現在同濟高數上冊第五章最后，不過教材中打了星號，所以多數人都不了解，首先我們引入伽馬函數的定義
\[\Gamma \left ( x \right )=\int_{0}^{\infty }t^{x-1}e^{-t}\mathrm{d}t \]
所以，我們令 \(x^2=t\), 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，這里不做證明.
方法6： 不難證明，函數 \((1+t)e^{-t}\) 在 \(t=0\) 時達到它的最大值1.因此當 \(t\neq 0\) 時，\((1+t)e^{-t}<1\), 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
\[\left ( 1-x^{2} \right )<e^{-x^{2}}<\frac{1}{1+x^{2}}~~~\left ( x>0 \right ) \]
假設限定第一個不等式中的 \(x\) 在(0,1)內變化，而第二個不等式中 \(x\) 則看作是任意的，把上式同 \(n\) 次方，有
\[\left ( 1-x^{2} \right )^{n}<e^{-nx^{2}}~~~\left ( 0<x<1 \right ) \]
\[e^{-nx^{2}}<\frac{1}{\left ( 1+x^{2} \right )^{n}}~~~\left ( x>0 \right ) \]
第一個不等式即從0到1積分，第二個不等式取從0到\(+\infty\)的積分，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 \(x=\cos t\), 則
\[\int_{0}^{1}\left ( 1-x^{2} \right )^{n}\mathrm{d}x=\int_{0}^{\frac{\pi }{2}}\sin^{2n+1}t\mathrm{d}t=\frac{\left ( 2n \right )!!}{\left ( 2n+1 \right )!!} \]
在 \(\displaystyle \int_{0}^{\infty }\frac{1}{\left ( 1+x^{2} \right )^{n}}\mathrm{d}x\) 中，令 \(x=\cot t\), 則
\[\int_{0}^{\infty }\frac{1}{\left ( 1+x^{2} \right )^{n}}\mathrm{d}x=\int_{0}^{\frac{\pi }{2}}\sin^{2n-2}t\mathrm{d}t=\frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2} \]
在 \(\displaystyle \int_{0}^{\infty }e^{-nx^{2}}\mathrm{d}x\) 中，令 \(\displaystyle x=\frac{t}{\sqrt{n}}\), 則
\[\int_{0}^{\infty }e^{-nx^{2}}\mathrm{d}x=\frac{1}{\sqrt{n}} \int_{0}^{\infty }e^{-t^{2}}\mathrm{d}t=\frac{1}{\sqrt{n}}I \]
綜上所述
\[\sqrt{n}\cdot \frac{\left ( 2n \right )!!}{\left ( 2n+1 \right )!!}<I<\sqrt{n}\cdot \frac{\left ( 2n-3 \right )!!}{\left ( 2n-2 \right )!!}\cdot \frac{\pi }{2} \]
取平方得
\[\frac{n}{2n+1}\cdot \frac{\left [ \left ( 2n \right )!! \right ]^{2}}{\left ( 2n+1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}}<I^2<\frac{n}{2n-1}\cdot \frac{\left [ \left ( 2n-3 \right )!! \right ]^{2}}{\left [ \left ( 2n-2 \right )!! \right ]^{2}}\cdot \left ( \frac{\pi }{2} \right )^{2} \]
根據Wallis公式
\[\frac{\pi }{2}=\lim_{n\rightarrow \infty }\frac{\left [ \left ( 2n \right )!! \right ]^{2}}{\left ( 2n+1 \right )\left [ \left ( 2n-1 \right )!! \right ]^{2}} \]
不等式兩邊當 $n\rightarrow \infty $ 時的極限都是 \(\dfrac{\pi }{4}\), 所以
\[I^2=\frac{\pi }{4}\Rightarrow I=\frac{\sqrt{\pi }}{2} \]
方法7：當然也可以利用三重積分
\[\begin{align*} &8I^3 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}e^{-x^2 - y^2 - z^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\\ \Rightarrow &8I^3 = 4\pi\int_0^{\infty}\rho^2 e^{-\rho^2}\,\mathrm{d}\rho= 2\pi \int_0^{\infty} e^{-\rho^2}\,\mathrm{d}\rho=2\pi \cdot {I} \Rightarrow 8I^3=2\pi I\Rightarrow I=\frac{\sqrt{\pi }}{2} \end{align*}\]
方法8：注意到
\[n! =\int_0^\infty e^{-\sqrt[n]x} \mathrm{d}x\iff\frac1n! =\int_0^\infty e^{-x^n}\mathrm{d}x\rightarrow\frac12! = \int_0^\infty e^{-x^2}\mathrm{d}x \]
\[\int_0^1\Big(1-\sqrt[n]x\Big)^m\,\mathrm{d}x = \int_0^1\Big(1-\sqrt[m]x\Big)^n\,\mathrm{d}x = \frac1{C_{m+n}^n} = \frac1{C_{m+n}^m} = \frac{m!\,n!}{(m+n)!} \]
所以我們有
\[\frac\pi4 = \int_0^1\sqrt{1-x^2}\,\mathrm{d}x = \frac{\left(\dfrac12!\right)^2}{\left(\dfrac12 + \dfrac12\right)!} =\left(\frac12!\right)^2 \]
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
\[F(\omega) = \frac{1}{\pi} \int_{0}^{+\infty} \exp\left(\frac{-t^2}{2}\right) \cos( \omega t) \mathrm{d}t \]
所以我們有
\[F'(\omega) = - \omega F(\omega)\Rightarrow F(\omega) = C \exp\left(\frac{-\omega^2}{2}\right) \]
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
\[\sqrt{2} \int_{-\infty}^{+\infty} \exp\left(-t^2\right) \mathrm{d}t = \sqrt{2\pi}\Rightarrow I=\frac{\sqrt{\pi }}{2} \]
