Editing Gaussian integral (section)

==Generalizations==

===The integral of a Gaussian function===
{{Main|Integral of a Gaussian function}}
The integral of an arbitrary [[Gaussian function]] is
<math display="block">\int_{-\infty}^{\infty}  e^{-a(x+b)^2}\,dx= \sqrt{\frac{\pi}{a}}.</math>

An alternative form is
<math display="block">\int_{-\infty}^{\infty}e^{- (a x^2 + b x + c)}\,dx=\sqrt{\frac{\pi}{a}}\,e^{\frac{b^2}{4a}-c}.</math>

This form is useful for calculating expectations of some continuous probability distributions related to the normal distribution, such as the [[log-normal distribution]], for example.

=== Complex form ===
{{main|Fresnel integral}}
<math display="block">\int_{-\infty}^{\infty} e^{\frac 12 it^2} dt = e^{i\pi/4} \sqrt{2\pi}</math>and more generally,<math display="block">\int_{\mathbb{R}^N} e^{\frac{1}{2} i \mathbf{x}^T A \mathbf{x}}dx = \det(A)^{-\frac{1}{2}} {\left(e^{i\pi/4} \sqrt{2\pi}\right)}^N</math>for any positive-definite symmetric matrix <math>A</math>.

===''n''-dimensional and functional generalization===
{{main|multivariate normal distribution}}
Suppose ''A'' is a symmetric positive-definite (hence invertible) {{math|''n'' × ''n''}} [[precision matrix]], which is the matrix inverse of the [[covariance matrix]]. Then,

<math display="block">\begin{align}
\int_{\mathbb{R}^n} \exp{\left(-\frac 1 2 \mathbf{x}^\mathsf{T} A \mathbf{x} \right)} \, d^n \mathbf{x}
&= \int_{\mathbb{R}^n} \exp{\left(-\frac 1 2 \sum\limits_{i,j=1}^{n} A_{ij} x_i x_j \right)} \, d^n \mathbf{x} \\[1ex]
&= \sqrt{\frac{{\left(2\pi\right)}^n}{\det A}}
= \sqrt{\frac{1}{\det \left(A / 2\pi\right)}} \\[1ex]
&= \sqrt{\det \left(2 \pi A^{-1}\right)}
\end{align}</math>By completing the square, this generalizes to<math display="block">\int_{\mathbb{R}^n} \exp{\left(-\tfrac 1 2 \mathbf{x}^\mathsf{T} A \mathbf{x} + \mathbf{b}^\mathsf{T} \mathbf{x} + c\right)} \, d^n \mathbf{x} = \sqrt{\det \left(2 \pi A^{-1}\right)} \exp\left(\tfrac{1}{2} \mathbf{b}^\mathsf{T} A^{-1} \mathbf{b} + c\right)</math>

This fact is applied in the study of the [[multivariate normal distribution]].

Also,
<math display="block">\int x_{k_1}\cdots x_{k_{2N}} \, \exp{\left( -\frac{1}{2} \sum\limits_{i,j=1}^{n}A_{ij} x_i x_j \right)} \, d^nx =\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)}}</math>
where ''σ'' is a [[permutation]] of {{math|{1, …, 2''N''}<nowiki/>}} and the extra factor on the right-hand side is the sum over all combinatorial pairings of {{math|{1, …, 2''N''}<nowiki/>}} of ''N'' copies of ''A''<sup>−1</sup>.

Alternatively,<ref name="Central identity explanation">{{cite web |title=Reference for Multidimensional Gaussian Integral |date=March 30, 2012 |work=[[Stack Exchange]] |url=https://math.stackexchange.com/q/126227 }}</ref>

<math display="block">\int f(\mathbf x) \exp{\left( - \frac 1 2 \sum_{i,j=1}^n A_{ij} x_i x_j \right)} d^n\mathbf{x}
= \sqrt{\frac{{\left(2\pi\right)}^n}{\det A}} \, \left. \exp\left(\frac{1}{2} \sum_{i,j=1}^{n}\left(A^{-1}\right)_{ij}{\partial \over \partial x_i}{\partial \over \partial x_j}\right) f(\mathbf{x})\right|_{\mathbf{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 integral]]s 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. {{Citation needed|date=June 2011}} 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 display="block">\begin{align}
& \frac{\displaystyle\int f(x_1)\cdots f(x_{2N}) \exp\left[{-\iint \frac{1}{2}A(x_{2N+1},x_{2N+2}) f(x_{2N+1}) f(x_{2N+2}) \, d^dx_{2N+1} \, d^dx_{2N+2}}\right] \mathcal{D}f}{\displaystyle\int \exp\left[{-\iint \frac{1}{2} A(x_{2N+1}, x_{2N+2}) f(x_{2N+1}) f(x_{2N+2}) \, d^dx_{2N+1} \, d^dx_{2N+2}}\right] \mathcal{D}f} \\[6pt]
= {} & \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)}).
\end{align}</math>

In the [[DeWitt notation]], the equation looks identical to the finite-dimensional case.

===''n''-dimensional with linear term===
If ''A'' is again a symmetric positive-definite matrix, then (assuming all are column vectors)
<math display="block">\begin{align}
\int \exp\left(-\frac{1}{2}\sum_{i,j=1}^n A_{ij} x_i x_j+\sum_{i=1}^n b_i x_i\right) d^n \mathbf{x}
&= \int \exp\left(-\tfrac{1}{2} \mathbf{x}^\mathsf{T} A \mathbf{x} + \mathbf{b}^\mathsf{T} \mathbf{x}\right) d^n \mathbf{x} \\
&= \sqrt{ \frac{(2\pi)^n}{\det A} } \exp\left(\tfrac{1}{2} \mathbf{b}^\mathsf{T} A^{-1} \mathbf{b}\right).
\end{align}</math>

===Integrals of similar form===
<math display="block">\int_0^\infty x^{2n}  e^{-{x^2}/{a^2}}\,dx = \sqrt{\pi}\frac{a^{2n+1} (2n-1)!!}{2^{n+1}}</math>
<math display="block">\int_0^\infty x^{2n+1} e^{-{x^2}/{a^2}} \, dx = \frac{n!}{2} a^{2n+2}</math>
<math display="block">\int_0^\infty x^{2n}e^{-bx^2}\,dx = \frac{(2n-1)!!}{b^n 2^{n+1}} \sqrt{\frac{\pi}{b}}</math>
<math display="block">\int_0^\infty x^{2n+1}e^{-bx^2}\,dx = \frac{n!}{2b^{n+1}}</math>
<math display="block">\int_0^\infty x^{n}e^{-bx^2}\,dx = \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}}</math>
where <math>n</math> is a positive integer

An easy way to derive these is by [[Leibniz integral rule#Evaluating definite integrals|differentiating under the integral sign]].

<math display="block">\begin{align}
\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 \\[1ex]
&= \left(-1\right)^n\frac{\partial^n}{\partial \alpha^n} \int_{-\infty}^\infty e^{-\alpha x^2}\,dx\\[1ex]
&= \sqrt{\pi} \left(-1\right)^n\frac{\partial^n}{\partial \alpha^n}\alpha^{-\frac{1}{2}} \\[1ex]
&= \sqrt{\frac{\pi}{\alpha}}\frac{(2n-1)!!}{\left(2\alpha\right)^n}
\end{align}</math>

One could also integrate by parts and find a [[recurrence relation]] to solve this.

===Higher-order polynomials===

Applying a linear change of basis shows that the integral of the exponential of a homogeneous polynomial in ''n'' variables may depend only on [[SL(n)|SL(''n'')]]-invariants of the polynomial. One such invariant is the [[discriminant]],
zeros of which mark the singularities of the integral. However, the integral may also depend on other invariants.<ref name="morozov2009">{{cite journal | last1 = Morozov | first1 = A. | last2 = Shakirove | first2= Sh. | journal = Journal of High Energy Physics | pages = 002 | title = Introduction to integral discriminants | doi = 10.1088/1126-6708/2009/12/002 | volume = 2009 | year = 2009 | issue = 12 | arxiv = 0903.2595 | bibcode = 2009JHEP...12..002M }}</ref>

Exponentials of other even polynomials can numerically be solved using series. These may be interpreted as [[formal calculation]]s when there is no convergence. For example, the solution to the integral of the exponential of a quartic polynomial is{{citation needed|date=August 2015}}

<math display="block">\int_{-\infty}^{\infty} e^{a x^4+b x^3+c x^2+d x+f}\,dx
= \frac{1}{2} e^f \sum_{\begin{smallmatrix}n,m,p=0 \\ n+p=0 \bmod 2\end{smallmatrix}}^{\infty} \frac{b^n}{n!} \frac{c^m}{m!} \frac{d^p}{p!} \frac{\Gamma{\left (\frac{3n+2m+p+1}{4} \right)}}{{\left(-a\right)}^{\frac{3n+2m+p+1}4}}.</math>

The {{math|1=''n'' + ''p'' = 0}} mod 2 requirement is because the integral from −∞ to 0 contributes a factor of {{math|(−1)<sup>''n''+''p''</sup>/2}} to each term, while the integral from 0 to +∞ contributes a factor of 1/2 to each term. These integrals turn up in subjects such as [[quantum field theory]].