# 高斯积分

$\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}$

## 高斯函数的积分

$\int_{-\infty}^{\infty} ae^{\frac{-(x-b)^2}{c^2}}\,dx.$

$a\int_{-\infty}^\infty e^{-\frac{y^2}{c^2}}\,dy.$

$ac\int_{-\infty}^\infty e^{-z^2}\,dz=ac\sqrt{\pi}.$

## 通过极限求解

$I(a)=\int_{-a}^a e^{-x^2}dx.$

$\lim_{a\to\infty} I(a) = \int_{-a}^a e^{-x^2}\, dx.$

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

$\int_0^{2\pi}\int_0^a re^{-r^2}\,dr\,d\theta < I^2(a) < \int_0^{2\pi}\int_0^{a\sqrt{2}} re^{-r^2}\,dr\,d\theta.$

$\pi (1-e^{-a^2}) < I^2(a) < \pi (1 - e^{-2a^2}).$

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

## 与Γ函数的关系

$\int_{-\infty}^{\infty} e^{-x^2} dx = 2 \int_0^\infty e^{-x^2} dx$

$\int_0^\infty e^{-t} \ t^{-\frac{1}{2}} dt \, = \, \Gamma\left(\frac{1}{2}\right)$

$b\int_0^\infty e^{-ax^b} dx = a^{-\frac{1}{b}} \, \Gamma\left(1-\frac{1}{b}\right).$

## n维和一般化

$A$为一个对称的、正的、可逆的、二维协变的张量，则

$\int e^{-\frac{A_{ij} x^i x^j}{2}} d^nx=\frac{(2\pi)^{\frac{n}{2}}}{\sqrt{\det{A}}}$

$\int x^{k_1}\cdots x^{k_{2N}} e^{-\frac{A_{ij} x^i x^j}{2}} d^nx=\frac{(2\pi)^{n/2}}{\sqrt{\det{A}}}\frac{1}{2^N N!}\sum_{\sigma \in S_{2N}}(A^{-1})^{k_{\sigma(1)}k_{\sigma(2)}}\cdots (A^{-1})^{k_{\sigma(2N-1)}k_{\sigma(2N)}}$

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

$\frac{\int f(x_1)\cdots f(x_{2N}) e^{-\iint \frac{A(x_{2N+1},x_{2N+2}) f(x_{2N+1}) f(x_{2N+2})}{2} d^dx_{2N+1} d^dx_{2N+2}} \mathcal{D}f}{\int e^{-\iint \frac{A(x_{2N+1},x_{2N+2}) f(x_{2N+1}) f(x_{2N+2})}{2} d^dx_{2N+1} d^dx_{2N+2}} \mathcal{D}f},$
$=\frac{1}{2^N N!}\sum_{\sigma \in S_{2N}}A^{-1}(x_{\sigma(1)},x_{\sigma(2)})\cdots A^{-1}(x_{\sigma(2N-1)},x_{\sigma(2N)}).$

## 带线性项的n维

A依然是一个对称矩阵，则

$\int e^{-A_{ij} x^i x^j+B_i x_i} d^nx=\sqrt{ \frac{\pi^n}{\det{A}} }e^{\frac{1}{4}B^TA^{-1}B}.$