Gaussian integral

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The Gaussian integral, or probability integral, is an integral of the Gaussian function $\exp{(-x^2)}\,$ . It is named after the German mathematician and physicist Carl Friedrich Gauss.

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

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, due to the Risch algorithm, the Gaussian integral can be solved analytically through the tools of calculus.

1 The integral of a Gaussian function
2 Derivation by limits
3 Relation to the gamma function
4 n-dimensional and functional generalization
5 n-dimensional with linear term
6 References

[edit] The integral of a Gaussian function

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

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

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

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

Substituting y with cz gives

$ac\int_{-\infty}^\infty e^{-z^2}\,dz$

$=ac\sqrt{\pi}.$

[edit] Derivation by limits

To find a closed form of the Gaussian integral, start by defining an approximating function:

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

so that the integral may be found by

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

Taking the square of I yields

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

Using Fubini's theorem, the above double integral can be seen as an area integral $\int e^{-(x^2+y^2)}\,d(x,y)$ , 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 $I (a) 2$ , and similarly the integral taken over the square's circumcircle must be greater than $I (a) 2$ . The integrals over the two disks can easily be computed by switching from cartesian coordinates to polar coordinates: $x=r\,\cos \theta$ , $y= r\,\sin\theta$ , $d(x,y) = r\, d(r,\theta)$ :

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

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

Integrating,

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

By the squeeze theorem, this gives the Gaussian integral

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

[edit] Relation to the gamma function

Since the integrand is an even function,

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

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

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

where Γ is the gamma function. This shows why the factorial of a half-integer is a rational multiple of $\sqrt \pi$ . More generally,

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

[edit] n-dimensional and functional generalization

See main article multivariate normal distribution

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

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

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

Also,

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

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,

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

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 $(2\pi)^\infty$ is infinite and also, the functional determinant would also be infinite in general. This can be taken care of if we only consider ratios:

$\frac{\int f(x_1)\cdots f(x_{2N}) e^{-\iint \frac{A(x_{2N+1},x_{2N+2}) f(x_{2N+1}) f(x_{2N+2})}{2} d^dx_{2N+1} d^dx_{2N+2}} \mathcal{D}f}{\int e^{-\iint \frac{A(x_{2N+1},x_{2N+2}) f(x_{2N+1}) f(x_{2N+2})}{2} d^dx_{2N+1} d^dx_{2N+2}} \mathcal{D}f},$