Editing Gaussian integral

{{Use American English|date = January 2019}}
{{Short description|Integral of the Gaussian function, equal to sqrt(π)}}
{{hatnote|This integral from statistics and physics is not to be confused with [[Gaussian quadrature]], a method of numerical integration.}}
[[Image:Gaussian Integral.svg|thumb|right|A graph of the function <math>f(x) = e^{-x^2}</math> and the area between it and the <math>x</math>-axis, (i.e. the entire real line) which is equal to <math>\sqrt{\pi}</math>.]]

The '''Gaussian integral''', also known as the '''Euler–Poisson integral''', is the [[integral]] of the [[Gaussian function]] <math>f(x) = e^{-x^2}</math> over the entire real line. Named after the German mathematician [[Carl Friedrich Gauss]], the integral is
<math display="block">\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.</math>

[[Abraham de Moivre]] originally discovered this type of integral in 1733, while Gauss published the precise integral in 1809,<ref name="The Evolution of the Normal Distribution">{{cite web |url=https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/stahl96.pdf |title=The Evolution of the Normal Distribution |work=MAA.org |first=Saul|last=Stahl|date=April 2006|access-date=May 25, 2018}}</ref> attributing its discovery to [[Laplace]]. The integral has a wide range of applications.  For example, with a slight change of variables it is used to compute the [[normalizing constant]] of the [[normal distribution]].  The same integral with finite limits is closely related to both the [[error function]] and the [[cumulative distribution function]] of the [[normal distribution]]. In physics this type of integral appears frequently, for example, in [[quantum mechanics]], to find the probability density of the ground state of the harmonic oscillator. This integral is also used in the path integral formulation, to find the propagator of the harmonic oscillator, and in [[statistical mechanics]], to find its [[partition function (statistical mechanics)|partition function]].

Although no [[elementary function]] exists for the error function, as can be proven by the [[Risch algorithm]],<ref>{{cite journal |first=G. W. |last=Cherry |title=Integration in Finite Terms with Special Functions: the Error Function |journal=Journal of Symbolic Computation |volume=1 |issue=3 |year=1985 |pages=283–302 |doi=10.1016/S0747-7171(85)80037-7 |doi-access=free }}</ref> the Gaussian integral can be solved analytically through the methods of [[multivariable calculus]]. That is, there is no elementary ''[[indefinite integral]]'' for
<math display="block">\int e^{-x^2}\,dx,</math>
but the [[definite integral]]
<math display="block">\int_{-\infty}^\infty e^{-x^2}\,dx</math>
can be evaluated. The definite integral of an arbitrary [[Gaussian function]] is
<math display="block">\int_{-\infty}^{\infty}  e^{-a(x+b)^2}\,dx= \sqrt{\frac{\pi}{a}}.</math>

==Computation==

===By polar coordinates===
A standard way to compute the Gaussian integral, the idea of which goes back to Poisson,<ref name="york.ac.uk">{{cite web |title=The Probability Integral | last=Lee | first=Peter M. |url=https://www.york.ac.uk/depts/maths/histstat/normal_history.pdf }}</ref> is to make use of the property that:

<math display="block">\left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \int_{-\infty}^{\infty} e^{-y^2}\,dy = \int_{-\infty}^{\infty}  \int_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\, dx\,dy. </math>

Consider the function <math>e^{-\left(x^2 + y^2\right)} = e^{-r^{2}}</math>on the plane <math>\mathbb{R}^2</math>, and compute its integral two ways:
# on the one hand, by [[double integration]] in the [[Cartesian coordinate system]], its integral is a square: <math display="block">\left(\int e^{-x^2}\,dx\right)^2;</math>
# on the other hand, by [[shell integration]] (a case of double integration in [[polar coordinates]]), its integral is computed to be <math>\pi</math>

Comparing these two computations yields the integral, though one should take care about the [[improper integral]]s involved.

<math display="block">\begin{align}
   \iint_{\R^2} e^{-\left(x^2 + y^2\right)}dx\,dy
   &= \int_0^{2\pi} \int_0^{\infty} e^{-r^2}r\,dr\,d\theta\\[6pt]
   &= 2\pi \int_0^\infty re^{-r^2}\,dr\\[6pt]
   &= 2\pi \int_{-\infty}^0 \tfrac{1}{2} e^s\,ds && s = -r^2\\[6pt]
   &= \pi \int_{-\infty}^0 e^s\,ds \\[6pt]
   &= \pi \, \left[ e^s\right]_{-\infty}^{0} \\[6pt]
   &= \pi \,\left(e^0 - e^{-\infty}\right) \\[6pt]
   &= \pi \,\left(1 - 0\right) \\[6pt]
   &=\pi,
 \end{align}</math>
where the factor of {{mvar|r}} is the [[Jacobian determinant]] which appears because of the [[list of canonical coordinate transformations|transform to polar coordinates]] ({{math|''r'' ''dr'' ''dθ''}} is the standard measure on the plane, expressed in polar coordinates [[Wikibooks:Calculus/Polar Integration#Generalization]]), and the substitution involves taking {{math|1=''s'' = −''r''<sup>2</sup>}}, so {{math|1=''ds'' = −2''r'' ''dr''}}.

Combining these yields
<math display="block">\left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right )^2=\pi,</math>
so
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}.</math>

====Complete proof====
To justify the improper double integrals and equating the two expressions, we begin with an approximating function:
<math display="block">I(a) = \int_{-a}^a e^{-x^2}dx.</math>

If the integral
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx</math>
were [[absolutely convergent]] we would have that its [[Cauchy principal value]], that is, the limit
<math display="block">\lim_{a\to\infty} I(a) </math>
would coincide with
<math display="block">\int_{-\infty}^\infty e^{-x^2}\,dx.</math>
To see that this is the case, consider that

<math display="block">\int_{-\infty}^\infty \left|e^{-x^2}\right| dx < \int_{-\infty}^{-1} -x e^{-x^2}\, dx + \int_{-1}^1 e^{-x^2}\, dx+ \int_{1}^{\infty} x e^{-x^2}\, dx < \infty .</math>

So we can compute
<math display="block">\int_{-\infty}^\infty e^{-x^2} \, dx</math>
by just taking the limit
<math display="block">\lim_{a\to\infty} I(a).</math>

Taking the square of <math>I(a)</math> yields

<math display="block">\begin{align}
I(a)^2 & = \left ( \int_{-a}^a e^{-x^2}\, dx \right ) \left ( \int_{-a}^a e^{-y^2}\, dy \right ) \\[6pt]
& = \int_{-a}^a \left ( \int_{-a}^a e^{-y^2}\, dy \right )\,e^{-x^2}\, dx \\[6pt]
& = \int_{-a}^a \int_{-a}^a e^{-\left(x^2+y^2\right)}\,dy\,dx.
\end{align}</math>

Using [[Fubini's theorem]], the above double integral can be seen as an area integral
<math display="block">\iint_{[-a, a] \times [-a, a]} e^{-\left(x^2+y^2\right)}\,d(x,y),</math>
taken over a square with vertices {{math|{(−''a'', ''a''), (''a'', ''a''), (''a'', −''a''), (−''a'', −''a'')}<nowiki/>}} on the ''xy''-[[Cartesian plane|plane]].

Since the exponential function is greater than 0 for all real numbers, it then follows that the integral taken over the square's [[incircle]] must be less than <math>I(a)^2</math>, and similarly the integral taken over the square's [[circumcircle]] must be greater than <math>I(a)^2</math>. The integrals over the two disks can easily be computed by switching from Cartesian coordinates to [[list of canonical coordinate transformations|polar coordinates]]:

<math display="block">\begin{align}
x &= r \cos \theta, &
y &= r \sin\theta
\end{align}</math>
<math display="block">
\mathbf J(r, \theta) = 
\begin{bmatrix}
  \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial\theta}\\[1em]
  \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial\theta} \end{bmatrix}
= \begin{bmatrix}
  \cos\theta &           -  r\sin \theta \\
  \sin\theta & \hphantom{-} r\cos \theta
\end{bmatrix}
</math>
<math display="block">d(x,y) = \left|J(r, \theta)\right| d(r,\theta) = r\, d(r,\theta).</math>
<math display="block">\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.</math>

(See [[list of canonical coordinate transformations|to polar coordinates from Cartesian coordinates]] for help with polar transformation.)

Integrating,
<math display="block">\pi \left(1-e^{-a^2}\right) <  I^2(a) < \pi \left(1 - e^{-2a^2}\right). </math>

By the [[squeeze theorem]], this gives the Gaussian integral
<math display="block">\int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}.</math>

===By Cartesian coordinates===
A different technique, which goes back to Laplace (1812),<ref name="york.ac.uk" /> is the following. Let
<math display="block">\begin{align}
y & = xs \\
dy & = x\,ds.
\end{align}</math>

Since the limits on {{mvar|s}} as {{math|''y'' → ±∞}} depend on the sign of {{mvar|x}}, it simplifies the calculation to use the fact that {{math|''e''<sup>−''x''<sup>2</sup></sup>}} is an [[even function]], and, therefore, the integral over all real numbers is just twice the integral from zero to infinity. That is,

<math display="block">\int_{-\infty}^{\infty} e^{-x^2} \, dx = 2\int_{0}^{\infty} e^{-x^2}\,dx.</math>

Thus, over the range of integration, {{math|''x'' ≥ 0}}, and the variables {{mvar|y}} and {{mvar|s}} have the same limits. This yields:
<math display="block">\begin{align}
I^2 &= 4 \int_0^\infty \int_0^\infty e^{-\left(x^2 + y^2\right)} dy\,dx \\[6pt]
&= 4 \int_0^\infty \left( \int_0^\infty e^{-\left(x^2 + y^2\right)} \, dy \right) \, dx \\[6pt]
&= 4 \int_0^\infty \left( \int_0^\infty e^{-x^2\left(1+s^2\right)} x\,ds \right) \, dx \\[6pt]
\end{align}</math>
Then, using [[Fubini's theorem]] to switch the [[order of integration (calculus)|order of integration]]:
<math display="block">\begin{align}
I^2 &= 4 \int_0^\infty \left( \int_0^\infty e^{-x^2\left(1 + s^2\right)} x \, dx \right) \, ds  \\[6pt]
&= 4 \int_0^\infty \left[ \frac{e^{-x^2\left(1+s^2\right)} }{-2 \left(1+s^2\right)} \right]_{x=0}^{x=\infty} \, ds \\[6pt]
&= 4 \left (\frac{1}{2} \int_0^\infty \frac{ds}{1+s^2} \right) \\[6pt]
&= 2 \arctan(s)\Big |_0^\infty \\[6pt]
&= \pi.
\end{align}</math>

Therefore, <math>I = \sqrt{\pi}</math>, as expected.

=== By [[Laplace's method]] ===

In Laplace approximation, we deal only with up to second-order terms in Taylor expansion, so we consider <math>e^{-x^2}\approx 1-x^2 \approx (1+x^2)^{-1}</math>.  

In fact, since <math>(1+t)e^{-t} \leq 1</math> for all <math>t</math>, we have the exact bounds:<math display="block">1-x^2 \leq e^{-x^2} \leq (1+x^2)^{-1}</math>Then we can do the bound at Laplace approximation limit:<math display="block">\int_{[-1, 1]}(1-x^2)^n dx \leq \int_{[-1, 1]}e^{-nx^2} dx \leq \int_{[-1, 1]}(1+x^2)^{-n} dx</math>  

That is,
<math display="block">2\sqrt n\int_{[0, 1]}(1-x^2)^n dx \leq \int_{[-\sqrt n, \sqrt n]}e^{-x^2} dx \leq 2\sqrt n\int_{[0, 1]}(1+x^2)^{-n} dx</math>  

By trigonometric substitution, we exactly compute those two bounds: <math>2\sqrt n(2n)!!/(2n+1)!!</math> and <math>2\sqrt n (\pi/2)(2n-3)!!/(2n-2)!!</math>  

By taking the square root of the [[Wallis formula]], <math display="block">\frac \pi 2 = \prod_{n=1} \frac{(2n)^2}{(2n-1)(2n+1)}</math>we have <math>\sqrt \pi = 2 \lim_{n\to \infty} \sqrt{n} \frac{(2n)!!}{(2n+1)!!}</math>, the desired lower bound limit. Similarly we can get the desired upper bound limit.
Conversely, if we first compute the integral with one of the other methods above, we would obtain a proof of the Wallis formula.

==Relation to the gamma function==

The integrand is an [[even function]],

<math display="block">\int_{-\infty}^{\infty} e^{-x^2} dx = 2 \int_0^\infty e^{-x^2} dx</math>

Thus, after the change of variable <math display="inline">x = \sqrt{t}</math>, this turns into the Euler integral

<math display="block">2 \int_0^\infty e^{-x^2} dx = 2\int_0^\infty \frac{1}{2}\ e^{-t} \ t^{-\frac{1}{2}} dt = \Gamma{\left(\frac{1}{2}\right)} = \sqrt{\pi}</math>

where <math display="inline"> \Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt </math> is the [[gamma function]]. This shows why the [[factorial]] of a half-integer is a rational multiple of <math display="inline">\sqrt \pi</math>. More generally,
<math display="block">\int_0^\infty x^n e^{-ax^b} dx = \frac{\Gamma{\left((n+1)/b\right)}}{b a^{(n+1)/b}}, </math>
which can be obtained by substituting <math>t=a x^b</math> in the integrand of the gamma function  to get <math display="inline"> \Gamma(z) = a^z b \int_0^{\infty} x^{bz-1} e^{-a x^b} dx </math>.

==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]].

==See also==
{{Portal|Mathematics|Physics}}
* [[List of integrals of Gaussian functions]]
* [[Common integrals in quantum field theory]]
* [[Normal distribution]]
* [[List of integrals of exponential functions]]
* [[Error function]]
* [[Berezin integral]]

== References ==
=== Citations ===
{{Reflist}}

=== Sources ===
{{refbegin}}
* {{MathWorld |title = Gaussian Integral |urlname = GaussianIntegral }}
* {{cite book |first=David |last=Griffiths |title=Introduction to Quantum Mechanics |edition=2nd }}
* {{cite book |last1=Abramowitz |first1=M. |last2=Stegun |first2=I. A. |title = Handbook of Mathematical Functions |publisher=Dover Publications |location=New York }}
{{refend}}

{{-}}
{{integral}}

[[Category:Integrals]]
[[Category:Articles containing proofs]]
[[Category:Gaussian function]]
[[Category:Theorems in mathematical analysis]]