高斯积分

（重定向自高斯積分

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

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

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}})

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

利用沃利斯积分计算

${\displaystyle I_{n}=\int _{0}^{\frac {\pi }{2}}\sin ^{n}x\mathrm {d} x={\begin{cases}{\dfrac {n-1}{n}}\cdot {\dfrac {n-3}{n-2}}\cdots {\dfrac {3}{4}}\cdot {\dfrac {1}{2}}\cdot {\dfrac {\pi }{2}}&2|n\\{\dfrac {n-1}{n}}\cdot {\dfrac {n-3}{n-2}}\cdots {\dfrac {4}{5}}\cdot {\dfrac {2}{3}}\cdot 1&2\nmid n\end{cases}}}$

{\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {(n+1)I_{n+1}}{nI_{n}}}&=\lim _{n\to \infty }{\frac {(n+1)I_{n}I_{n+1}}{nI_{n}^{2}}}=\lim _{n\to \infty }{\frac {\pi }{2nI_{n}^{2}}}=1\\\lim _{n\to \infty }{\sqrt {n}}I_{n}&={\sqrt {\frac {\pi }{2}}}\end{aligned}}}

${\displaystyle {\frac {1}{1-t}}\geqslant \mathrm {e} ^{t}\geqslant 1+t}$

${\displaystyle t=x^{2}}$ 并且不等式各边取倒数之后，变成：

${\displaystyle 1-x^{2}\leqslant \mathrm {e} ^{-x^{2}}\leqslant {\frac {1}{1+x^{2}}}}$

${\displaystyle \int _{0}^{1}(1-x^{2})^{n}\mathrm {d} x\leqslant \int _{0}^{1}\mathrm {e} ^{-nx^{2}}\mathrm {d} x\leqslant \int _{0}^{\infty }{\frac {\mathrm {d} x}{(1+x^{2})^{n}}}}$

${\displaystyle \int _{0}^{\frac {\pi }{2}}\cos ^{2n+1}\theta \mathrm {d} \theta \leqslant {\frac {1}{\sqrt {n}}}\int _{0}^{\sqrt {n}}\mathrm {e} ^{-y^{2}}\mathrm {d} y\leqslant \int _{0}^{\frac {\pi }{2}}\cos ^{2n-2}\theta \mathrm {d} \theta }$

${\displaystyle {\sqrt {n}}\int _{0}^{\frac {\pi }{2}}\sin ^{2n+1}\theta \mathrm {d} \theta \leqslant \int _{0}^{\sqrt {n}}\mathrm {e} ^{-y^{2}}\mathrm {d} y\leqslant {\sqrt {n}}\int _{0}^{\frac {\pi }{2}}\sin ^{2n-2}\theta \mathrm {d} \theta }$

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.