# Gaussian integral

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The **Gaussian integral**, or **probability integral**, is the improper integral of the Gaussian function **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): e^{{-x}^2}**
over the entire real line. It is named after the German mathematician and physicist Carl Friedrich Gauss, and the equation is:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}**

This integral has wide applications including normalization in probability theory and continuous Fourier transform. It also appears in the definition of the error function.

Although no elementary function exists for the error function, as can be proven by the Risch algorithm, the Gaussian integral can be solved analytically through the tools of calculus. That is, there is no elementary *indefinite integral* for **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int e^{-x^2}\,dx**
, but the *definite integral* **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^\infty e^{-x^2}\,dx**
can be evaluated.

## Computation

A standard way to compute the Gaussian integral is

- consider the function
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): e^{-(x^2+y^2)}=e^{-r^2}**on the plane**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \mathbf{R}^2**, and compute its integral two ways: - on the one hand, by double integration in the Cartesian coordinate system, its integral is a square:
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \left(\int e^{-x^2}\,dx\right)^2**; - on the other hand, by shell integration (a case of double integration in polar coordinates), its integral is computed to be
**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \pi**.

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

### Brief proof

Briefly, one computes that on the one hand,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \begin{align} \int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dx\,dy &= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy\\ &= \left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right ) \cdot \left ( \int_{-\infty}^\infty e^{-y^2}\,dy \right )\\ &= \left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right )^2. \end{align}**

On the other hand,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \begin{align} \int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dx\,dy &= \int_0^{2\pi} \int_0^{\infin} re^{-r^2}\,dr\,d\theta\\ &= 2\pi \int_0^\infty re^{-r^2}\,dr\\ &= 2\pi \int_{-\infty}^0 \frac{1}{2} e^s\,ds = \pi \int_{-\infty}^0 e^s\,ds = \pi (e^0 - e^{-\infty}) = \pi (1 - 0) = \pi, \end{align}**

where the factor of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): r**
comes from the transform to polar coordinates (**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): r\,dr\,d\theta**
is the standard measure on the plane, expressed in polar coordinates), and the substitution involves taking **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): s=-r^2**
, so **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): ds=-2r\,dr**
.

Combining these yields **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right )^2=\pi**
, so **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}**
.

### Careful proof

To justify the improper double integrals and equating the two expressions, we begin with an approximating function:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): I(a)=\int_{-a}^a e^{-x^2}dx.**

so that the integral may be found by

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \lim_{a\to\infty} I(a) = \int_{-\infty}^{\infty} e^{-x^2}\, dx.**

Taking the square of *I* yields

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): 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.**

Using Fubini's theorem, the above double integral can be seen as an area integral **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int e^{-(x^2+y^2)}\,d(x,y)**
, taken over a square with vertices {(−*a*, *a*), (*a*, *a*), (*a*, −*a*), (−*a*, −*a*)} on the *xy*-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 **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): I(a)^2**
, and similarly the integral taken over the square's circumcircle must be greater than **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): I(a)^2**
. The integrals over the two disks can easily be computed by switching from cartesian coordinates to
polar coordinates: **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): x=r\,\cos \theta**
, **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): y= r\,\sin\theta**
, **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): d(x,y) = r\, d(r,\theta)**
:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \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.**

(See to polar coordinates from cartesian coordinates for help with polar transformation.)

Integrating,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \pi (1-e^{-a^2}) < I^2(a) < \pi (1 - e^{-2a^2}).**

By the squeeze theorem, this gives the Gaussian integral

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}.**

## Relation to the gamma function

Since the integrand is an even function,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^{\infty} e^{-x^2} dx = 2 \int_0^\infty e^{-x^2} dx**

which, after a change of variable, turns into the Euler integral

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_0^\infty e^{-t} \ t^{-1/2} dt \, = \, \Gamma\left(\frac{1}{2}\right)**

where Γ is the gamma function. This shows why the factorial of a half-integer is a rational multiple of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \sqrt \pi**
. More generally,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_0^\infty e^{-ax^b} dx = a^{-1/b} \, \Gamma\left(1+\frac{1}{b}\right).**

## Generalizations

### The integral of a Gaussian function

The integral of any Gaussian function is reducible in terms of the Gaussian integral

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^{\infty} ae^{-(x+b)^2/c^2}\,dx.**

The constant *a* can be factored out of the integral. Replacing *x* with *y* - *b* yields

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): a\int_{-\infty}^\infty e^{-y^2/c^2}\,dy.**

Substituting *y* with *cz* gives

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): a |c| \int_{-\infty}^\infty e^{-z^2}\,dz****Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): =a |c| \sqrt{\pi}.**

*n*-dimensional and functional generalization

Suppose *A* is a symmetric positive-definite invertible covariant tensor of rank two. Then,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int \exp\left( - \frac 1 2 A_{ij} x^i x^j \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}}**

where the integral is understood to be over **R**^{n}. This fact is applied in the study of the multivariate normal distribution.

Also,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int x^{k_1}\cdots x^{k_{2N}} \, \exp\left( - \frac 1 2 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)}}**

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

Alternatively,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int f(\vec x) \, \exp\left( - \frac 1 2 A_{ij} x^i x^j \right) 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}**

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 **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): (2\pi)^\infty**
is infinite and also, the functional determinant would also be infinite in general. This can be taken care of if we only consider ratios:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \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},**

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): =\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)}).**

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

*n*-dimensional with linear term

Once again A is a symmetric matrix, then

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \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}.**

### Integrals of similar form

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \begin{align} &\int_0^\infty x^{2n} e^{-x^2/a^2}\,dx &&= \sqrt{\pi} \frac{(2n)!}{n!} \left(\frac{a}{2}\right)^{2n+1} \\ &\int_0^\infty x^{2n+1}e^{-x^2/a^2}\,dx &&= \frac{n!}{2} a^{2n+2} \end{align}**

## Beyond Gaussian Integrals

Exponentials of other even polynomials can easily be solved using series. For example the solution to the integral of the exponential of a quartic polynomial is:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://api.formulasearchengine.com/v1/":): \int_{-\infty}^{\infty}{\exp({a x^4+b x^3+c x^2+d x+f})dx} = e^f\sum_{n,m,p=0}^{\infty}{ \frac{ b^{4n}}{(4n)!}\frac{c^{2m}}{(2m)!}\frac{d^{4p}}{(4p)!} \frac{ \Gamma(3n+m+p+\frac14) }{a^{3n+m+p+\frac14} } } **

These integrals turn up in subjects such as quantum field theory.

## References

- Eric W. Weisstein,
*Gaussian Integral*at MathWorld. - David Griffiths. Introduction to Quantum Mechanics. 2nd Edition back cover.ca:Integral de Gauß