Gaussian integral

From Free net encyclopedia

The integral of any Gaussian function (named after Carl Friedrich Gauss) is quickly reducible to the Gaussian integral

<math>\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}.</math>

This integral cannot be computed by elementary means since the function has no simple antiderivative. The solution is, however, readily apparent when one examines the square of the integral, and converts to polar coordinates. Let the value of the integral be <math>s</math>. Then, by applying Fubini's theorem,

<math>s^2 = \left(\int_{-\infty}^\infty e^{-x^2}\,dx\right) \left(\int_{-\infty}^\infty e^{-y^2}\,dy\right) = \int_{\mathbb{R}^2}\!\!e^{-(x^2 + y^2)}\,dA. </math>

In polar coordinates we have:

<math>s^2 = \int_0^{2\pi}\!\!\!\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi\int_0^\infty e^{-r^2}r\,dr=2\pi\left[\frac{-e^{-r^{2}}}{2}\right]_0^\infty= \pi.</math>

(The factor of <math>r</math> that makes the integral straightforward comes from the change of variables.)

It then follows that

<math>s = \int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}.</math>

n-dimensional and functional generalization

Suppose A is a symmetric invertible covariant tensor of rank two. Then,

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

where the integral is understood to be over Rⁿ. This fact is applied in the study of the multivariate normal distribution.

Also,

<math>\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)}}</math>

where σ is a permutation of {1, ..., 2N} and the extra factor on the right-hand side is the sum over all combinatorial pairings of {1, ..., 2N} of N copies of A⁻¹.

Alternatively,

<math>\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}</math>

for some analytic function f, provided it satisfies some appropriate bounds on its growth and some other technical criteria. (It works for some functions and fails for others. Polynomials are fine.) The exponential over a differential operator is understood as a power series.

While functional integrals have no rigorous definition (or even a nonrigorous computational one in most cases), we can define a Gaussian functional integral in analogy to the finite-dimensional case. There is still the problem, though, that <math>(2\pi)^\infty</math> is infinite and also, the functional determinant would also be infinite in general. This can be taken care of if we only consider ratios:

<math>\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},</math>

<math>=\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)}).</math>

In the deWitt notation, the equation looks identical to the finite-dimensional case.

n-dimensional with linear term

Once again A is a symetric matrix, then

<math>\int e^{-A_{ij} x^i x^j+B_i x_i} d^nx=\sqrt{ \frac{\pi^n}{\det{A}} }e^{\frac{1}{n}B^TA^{-1}B}</math>it:Integrale di Gauss