# 高斯积分

f(x) = ex2 的图像，这个函数与 x 轴之间的面积等于 ${\displaystyle \scriptstyle {\sqrt {\pi }}}$
${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx={\sqrt {\pi }}}$

${\displaystyle \int _{-\infty }^{\infty }e^{-a(x+b)^{2}}\,dx={\sqrt {\frac {\pi }{a}}}.}$

## 计算方式

### 通过极限计算

${\displaystyle I(a)=\int _{-a}^{a}e^{-x^{2}}dx}$

${\displaystyle \lim _{a\to \infty }I(a)=\int _{-\infty }^{+\infty }e^{-x^{2}}\,dx.}$

${\displaystyle I}$ 取平方获得

${\displaystyle I^{2}(a)=\left(\int _{-a}^{a}e^{-x^{2}}\,dx\right)\cdot \left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)=\int _{-a}^{a}\left(\int _{-a}^{a}e^{-y^{2}}\,dy\right)\,e^{-x^{2}}\,dx=\int _{-a}^{a}\int _{-a}^{a}e^{-(x^{2}+y^{2})}\,dx\,dy.}$

${\displaystyle \int _{0}^{2\pi }\int _{0}^{a}re^{-r^{2}}\,dr\,d\theta

${\displaystyle \pi (1-e^{-a^{2}})

${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}.}$

## 与Γ函数的关系

${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}dx=2\int _{0}^{\infty }e^{-x^{2}}dx}$

${\displaystyle \int _{0}^{\infty }e^{-t}\ t^{-{\frac {1}{2}}}dt\,=\,\Gamma \left({\frac {1}{2}}\right)}$

${\displaystyle b\int _{0}^{\infty }e^{-ax^{b}}dx=a^{-{\frac {1}{b}}}\,\Gamma \left({\frac {1}{b}}\right).}$

## 推广

### 高斯函数的积分

${\displaystyle \int _{-\infty }^{\infty }e^{-a(x+b)^{2}}\,dx={\sqrt {\frac {\pi }{a}}}}$

 ${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}+bx+c}\,dx={\sqrt {\frac {\pi }{a}}}\,e^{{\frac {b^{2}}{4a}}+c}}$

### n维和泛函推广

${\displaystyle A}$ 为一个对称的、正定的（因而可逆${\displaystyle n\times n}$  精密矩阵英语precision matrix（即协方差矩阵的逆矩阵），则

${\displaystyle \int _{-\infty }^{\infty }e^{\left(-{\frac {1}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}\,d^{n}x=\int _{-\infty }^{\infty }e^{\left(-{\frac {1}{2}}x^{T}Ax\right)}\,d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}={\sqrt {\frac {1}{\det(A/2\pi )}}}={\sqrt {\det(2\pi A^{-1})}}}$

${\displaystyle \int x^{k_{1}}\cdots x^{k_{2N}}\,e^{\left(-{\frac {1}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}\,d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\,{\frac {1}{2^{N}N!}}\,\sum _{\sigma \in S_{2N}}(A^{-1})_{k_{\sigma (1)}k_{\sigma (2)}}\cdots (A^{-1})_{k_{\sigma (2N-1)}k_{\sigma (2N)}}}$

${\displaystyle \int f({\vec {x}})e^{\left(-{\frac {1}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}\right)}d^{n}x={\sqrt {(2\pi )^{n} \over \det A}}\,\left.e^{\left({1 \over 2}\sum \limits _{i,j=1}^{n}(A^{-1})_{ij}{\partial \over \partial x_{i}}{\partial \over \partial x_{j}}\right)}f({\vec {x}})\right|_{{\vec {x}}=0}}$

${\displaystyle {\frac {\int f(x_{1})\cdots f(x_{2N})e^{-\iint {\frac {1}{2}}A(x_{2N+1},x_{2N+2})f(x_{2N+1})f(x_{2N+2})d^{d}x_{2N+1}d^{d}x_{2N+2}}{\mathcal {D}}f}{\int e^{-\iint {\frac {1}{2}}A(x_{2N+1},x_{2N+2})f(x_{2N+1})f(x_{2N+2})d^{d}x_{2N+1}d^{d}x_{2N+2}}{\mathcal {D}}f}}={\frac {1}{2^{N}N!}}\sum _{\sigma \in S_{2N}}A^{-1}(x_{\sigma (1)},x_{\sigma (2)})\cdots A^{-1}(x_{\sigma (2N-1)},x_{\sigma (2N)}).}$

### 带线性项的n维

${\displaystyle \int e^{-{\frac {1}{2}}\sum \limits _{i,j=1}^{n}A_{ij}x_{i}x_{j}+\sum \limits _{i=1}^{n}B_{i}x_{i}}d^{n}x=\int e^{-{\frac {1}{2}}{\vec {x}}^{T}\mathbf {A} {\vec {x}}+{\vec {B}}^{T}{\vec {x}}}d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det {A}}}}e^{{\frac {1}{2}}{\vec {B}}^{T}\mathbf {A} ^{-1}{\vec {B}}}.}$

### 形式相似的积分

${\displaystyle \int _{0}^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}{\frac {a^{2n+1}(2n-1)!!}{2^{n+1}}}}$
${\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\frac {n!}{2}}a^{2n+2}}$
${\displaystyle \int _{0}^{\infty }x^{2n}e^{-ax^{2}}\,dx={\frac {(2n-1)!!}{a^{n}2^{n+1}}}{\sqrt {\frac {\pi }{a}}}}$
${\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}}$
${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,dx={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{\frac {n+1}{2}}}}}$

{\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }x^{2n}e^{-\alpha x^{2}}\,dx&=\left(-1\right)^{n}\int _{-\infty }^{\infty }{\frac {\partial ^{n}}{\partial \alpha ^{n}}}e^{-\alpha x^{2}}\,dx~=\left(-1\right)^{n}{\frac {\partial ^{n}}{\partial \alpha ^{n}}}\int _{-\infty }^{\infty }e^{-\alpha x^{2}}\,dx\\&={\sqrt {\pi }}\left(-1\right)^{n}{\frac {\partial ^{n}}{\partial \alpha ^{n}}}\alpha ^{-{\frac {1}{2}}}~={\sqrt {\frac {\pi }{\alpha }}}{\frac {(2n-1)!!}{\left(2\alpha \right)^{n}}}\end{aligned}}}

## 参考资料

• Griffiths, David. Introduction to Quantum Mechanics 2nd.
• Abramowitz, M.; Stegun, I. A. Handbook of Mathematical Functions. New York: Dover Publications.