关于Euler-Poisson积分的几种解法

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} ]