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
ight )^{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}
ight )}mathrm{d}xmathrm{d}y
]
用极坐标系下二重积分的计算法
[I=int_{0}^{frac{pi }{2}}mathrm{d} heta int_{0}^{infty }e^{-r^{2}}rmathrm{d}r=frac{pi }{2}left ( -frac{1}{2}e^{-r^{2}}
ight )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}
ight )^{-n}) 当 $n
ightarrow +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
ightarrow infty }left ( 1+frac{x^{2}}{n}
ight )^{-n}
ight ]mathrm{d}x=lim_{n
ightarrow infty }int_{0}^{infty }left ( 1+frac{x^{2}}{n}
ight )^{-n}mathrm{d}x
]
令 (x=sqrt{nt}), 则
[I=lim_{n
ightarrow infty }sqrt{n}int_{0}^{infty }frac{1}{left ( 1+t^{2}
ight )^{n}}mathrm{d}t=sqrt{n}I_n
]
由于
[egin{align*}
I_{n-1}&=int_{0}^{infty }frac{1}{left ( 1+t^{2}
ight )^{n-1}}mathrm{d}t=frac{t}{left ( 1+t^{2}
ight )^{n-1}}Bigg|_{0}^{infty }+2left ( n-1
ight )int_{0}^{infty }frac{1}{left ( 1+t^{2}
ight )}mathrm{d}t\
&=2left ( n-1
ight )I_{n-1}-2left ( n-1
ight )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
ight )!!}{left ( 2n-2
ight )!!}cdot frac{pi }{2}
]
因此 (displaystyle int_{0}^{infty }e^{-x^{2}}mathrm{d}x=lim_{n
ightarrow infty }frac{sqrt{n}left ( 2n-3
ight )!!}{left ( 2n-2
ight )!!}cdot frac{pi }{2}). 根据Wallis公式，有
[frac{pi }{2}=lim_{n
ightarrow infty }frac{left [ left ( 2n
ight )!!
ight ]^{2}}{left ( 2n+1
ight )left [ left ( 2n-1
ight )!!
ight ]^{2}}=lim_{n
ightarrow infty }frac{left [ left ( 2n-2
ight )!!
ight ]^{2}}{left ( 2n-1
ight )left [ left ( 2n-1
ight )!!
ight ]^{2}}
]
所以
[egin{align*}
I&=int_{0}^{infty }e^{-x^{2}}mathrm{d}x=frac{pi }{2}lim_{n
ightarrow infty }frac{sqrt{n}left ( 2n-3
ight )!!}{left ( 2n-2
ight )!!}=frac{pi }{2}lim_{n
ightarrow infty }frac{left ( 2n-3
ight )!!sqrt{2n-1}}{left ( 2n-2
ight )!!}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
[fleft ( x
ight )=left ( int_{0}^{x}e^{-t^{2}}mathrm{d}t
ight )^{2}~~,~~gleft ( x
ight )=int_{0}^{1}frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{1+u^{2}}mathrm{d}u
]
利用积分号下微分法，得
[egin{align*}
f'left ( x
ight )&= 2e^{-x^{2}}int_{0}^{x}e^{-t^{2}}mathrm{d}t\
g'left ( x
ight )&=int_{0}^{1}frac{partial }{partial x}left [ frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{1+u^{2}}
ight ]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
ight )=-2xe^{-x^{2}}int_{0}^{x}e^{-t^{2}}mathrm{d}t=-f'left ( x
ight )~~~left ( xgeq 0
ight )
]
于是
[egin{align*}
fleft ( x
ight )+gleft ( x
ight )=c~~~left ( xgeq 0
ight ) ag1
end{align*}]
由于 (displaystyle f(0)=0 , gleft ( 0
ight )=frac{pi }{4}), 故 (c=dfrac{pi }{4}), 即
[f(x)+g(x)=dfrac{pi }{4}~~~left ( xgeq 0
ight )
]
当 (uin left [ 0,1
ight ]), 有
[0leq frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{1+u^{2}}leq e^{-x^{2}u^{2}}leq e^{-x^{2}}~~~left ( xgeq 0
ight )
]
因此，当 $x
ightarrow infty $ 时，函数 (displaystyle frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{1+u^{2}}) 关于 (uin left [ 0,1
ight ]) 一致的趋于0.
[lim_{x
ightarrow infty }gleft ( x
ight )=lim_{x
ightarrow infty }int_{0}^{1}frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{1+u^{2}}mathrm{d}u=int_{0}^{1}lim_{x
ightarrow infty }frac{e^{-x^{2}left ( 1+u^{2}
ight )}}{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
ightarrow infty }fleft ( x
ight )}=sqrt{lim_{x
ightarrow infty }frac{pi }{4}-gleft ( x
ight )}=sqrt{frac{pi }{4}}=frac{sqrt{pi }}{2}
]
方法4：设 (displaystyle fleft ( t
ight )=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
ight )=int_{0}^{infty }int_{0}^{infty }e^{-tx^{2}}e^{-st}mathrm{d}tmathrm{d}x=int_{0}^{infty }mathcal{L}left ( e^{-tx^{2}}
ight )mathrm{d}x=int_{0}^{infty }frac{mathrm{d}x}{s+x^{2}}=frac{pi }{2sqrt{s}}
]
再取拉普拉斯逆变换，有 (displaystyle fleft ( t
ight )=int_{0}^{infty }e^{-tx^{2}}mathrm{d}x=frac{sqrt{pi }}{2sqrt{t}}), 在上式中，令 (t=1), 则
[I=fleft ( 1
ight )=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
ight )=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}
ight )=frac{sqrt{pi }}{2}
]
其中 (displaystyle Gamma left ( frac{1}{2}
ight )=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
eq 0) 时，((1+t)e^{-t}<1), 令 (t=pm x^{2}), 即得
[left ( 1-x^{2}
ight )e^{x^{2}}<1~,~left ( 1+x^{2}
ight )e^{-x^{2}}<1
]
或
[left ( 1-x^{2}
ight )<e^{-x^{2}}<frac{1}{1+x^{2}}~~~left ( x>0
ight )
]
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}
ight )^{n}<e^{-nx^{2}}~~~left ( 0<x<1
ight )
]
[e^{-nx^{2}}<frac{1}{left ( 1+x^{2}
ight )^{n}}~~~left ( x>0
ight )
]
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}
ight )^{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}
ight )^{n}}mathrm{d}x
]
在 (displaystyle int_{0}^{1}left ( 1-x^{2}
ight )^{n}mathrm{d}x) 中，令 (x=cos t), 则
[int_{0}^{1}left ( 1-x^{2}
ight )^{n}mathrm{d}x=int_{0}^{frac{pi }{2}}sin^{2n+1}tmathrm{d}t=frac{left ( 2n
ight )!!}{left ( 2n+1
ight )!!}
]
在 (displaystyle int_{0}^{infty }frac{1}{left ( 1+x^{2}
ight )^{n}}mathrm{d}x) 中，令 (x=cot t), 则
[int_{0}^{infty }frac{1}{left ( 1+x^{2}
ight )^{n}}mathrm{d}x=int_{0}^{frac{pi }{2}}sin^{2n-2}tmathrm{d}t=frac{left ( 2n-3
ight )!!}{left ( 2n-2
ight )!!}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
ight )!!}{left ( 2n+1
ight )!!}<I<sqrt{n}cdot frac{left ( 2n-3
ight )!!}{left ( 2n-2
ight )!!}cdot frac{pi }{2}
]
取平方得
[frac{n}{2n+1}cdot frac{left [ left ( 2n
ight )!!
ight ]^{2}}{left ( 2n+1
ight )left [ left ( 2n-1
ight )!!
ight ]^{2}}<I^2<frac{n}{2n-1}cdot frac{left [ left ( 2n-3
ight )!!
ight ]^{2}}{left [ left ( 2n-2
ight )!!
ight ]^{2}}cdot left ( frac{pi }{2}
ight )^{2}
]
根据Wallis公式
[frac{pi }{2}=lim_{n
ightarrow infty }frac{left [ left ( 2n
ight )!!
ight ]^{2}}{left ( 2n+1
ight )left [ left ( 2n-1
ight )!!
ight ]^{2}}
]
不等式两边当 $n
ightarrow 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
[egin{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 = 4piint_0^{infty}
ho^2 e^{-
ho^2}\,mathrm{d}
ho= 2pi int_0^{infty} e^{-
ho^2}\,mathrm{d}
ho=2pi cdot {I}
Rightarrow 8I^3=2pi IRightarrow I=frac{sqrt{pi }}{2}
end{align*}]
方法8：注意到
[n! =int_0^infty e^{-sqrt[n]x} mathrm{d}xifffrac1n! =int_0^infty e^{-x^n}mathrm{d}x
ightarrowfrac12! = int_0^infty e^{-x^2}mathrm{d}x
]
[int_0^1Big(1-sqrt[n]xBig)^m\,mathrm{d}x = int_0^1Big(1-sqrt[m]xBig)^n\,mathrm{d}x = frac1{C_{m+n}^n} = frac1{C_{m+n}^m} = frac{m!\,n!}{(m+n)!}
]
所以我们有
[fracpi4 = int_0^1sqrt{1-x^2}\,mathrm{d}x = frac{left(dfrac12!
ight)^2}{left(dfrac12 + dfrac12
ight)!} =left(frac12!
ight)^2
]
所以
[I=int_0^infty e^{-x^2}mathrm{d}x = frac12! = sqrt{piover4} = frac{sqrtpi}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(Xgeq0)
]
其中
[Xsim Nleft ( 0,frac{1}{2}
ight )~~,~~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} expleft(frac{-t^2}{2}
ight) exp(- i omega t) mathrm{d}t
]
所以
[F(omega) = frac{1}{pi} int_{0}^{+infty} expleft(frac{-t^2}{2}
ight) cos( omega t) mathrm{d}t
]
所以我们有
[F'(omega) = - omega F(omega)Rightarrow F(omega) = C expleft(frac{-omega^2}{2}
ight)
]
由
[expleft(frac{-x^2}{2}
ight) = int_{-infty}^{+infty} F(omega) exp( i omega x) mathrm{d}omega
]
可得 (displaystyle C = frac{1}{sqrt{2pi}}.) 令 (omega=0), 有
[F(0) = C = frac{1}{2 pi} int_{-infty}^{+infty} expleft(frac{-t^2}{2}
ight) mathrm{d}t
]
所以
[sqrt{2} int_{-infty}^{+infty} expleft(-t^2
ight) mathrm{d}t = sqrt{2pi}Rightarrow I=frac{sqrt{pi }}{2}
]