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- The Gaussian integral, or probability integral, is the integral of the Gaussian function e over the entire real line. It is named after the German mathematician and physicist Carl Friedrich Gauss. The integral is: \int_{-\infty}^\infty e^{-x^2}\,dx \sqrt{\pi}. This integral has wide applications. When normalized so that its value is 1, it is the density function of the normal distribution. It is an eigenfunction of the continuous Fourier transform. 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 <math>\int e^{-x^2}\,dx, but the definite integral <math>\int_{-\infty}^\infty e^{-x^2}\,dx can be evaluated. Computation By polar coordinates A standard way to compute the Gaussian integral is consider the function e + y e on the plane R, and compute its integral two ways: on the one hand, by double integration in the Cartesian coordinate system, its integral is a square: \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 <math>\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, \begin{align} \int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dA & \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, \begin{align} \int_{\mathbf{R}^2} e^{-(x^2+y^2)}\,dA & \int_0^{2\pi} \int_0^{\infin} e^{-r^2}r\,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 r comes from the transform to polar coordinates (r dr dθ is the standard measure on the plane, expressed in polar coordinates), and the substitution involves taking s −r, so ds −2r dr. Combining these yields \left (\int_{-\infty}^\infty e^{-x^2}\,dx \right)^2\pi, so \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: 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, since \int_{-\infty}^\infty e^{-x^2}\, 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. Taking the square of I yields \begin{align} I(a)^2 & \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. \end{align} 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 <math>I(a)^2, and similarly the integral taken over the square's circumcircle must be greater than <math>I(a)^2. The integrals over the two disks can easily be computed by switching from cartesian coordinates to polar coordinates: \begin{align} x & r \cos \theta \\ y & r \sin\theta \\ d(x,y) & r\, d(r,\theta). \end{align} \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}. By Cartesian coordinates Georgakis wrote that the following is "a better alternative to the usual method of reduction to polar coordinates". Let \begin{align} y & xs \\ dy & x\,ds. \end{align} Since the limits on <math>s as <math>y goes to <math>\pm\infty depend on the sign of <math>x, it simplifies the calculation to use the fact that <math>e^{-x^2} is an even function, and, therefore, the integral over all real numbers is just twice the integral from zero to infinity. That is, <math>\int_{-\infty}^{\infty} e^{-x^2}\,dx 2\int_{0}^{\infty} e^{-x^2}\,dx. Thus, over the range of integration, <math>x\ge 0, and the variables <math>y and <math>s have the same limits. This yields: I^2 4 \int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} dy\,dx . Then \begin{align} \frac{I^2}{4} & \int_0^\infty \left(\int_0^\infty e^{-} \, dy \right) \, dx \int_0^\infty \left(\int_0^\infty e^{-x^2} x\,ds \right) dx \\[5pt] & \int_0^\infty \left(\int_0^\infty e^{-x^2} x \, dx \right) \, ds \\[5pt] & \int_0^\infty \left[ \frac{1}{-2(1+s^2)} e^{-x^2(1+s^2)} \right]_0^\infty \, ds \frac{1}{2} \int_0^\infty \frac{ds}{1+s^2} \\[5pt] & \frac{1}{2} \left. \arctan s \frac{}{} \right|_0^\infty \frac{\pi}{4}. \end{align} Finally, <math> I \sqrt\pi, as expected. Relation to the gamma function The integrand is an even function, \int_{-\infty}^{\infty} e^{-x^2} dx 2 \int_0^\infty e^{-x^2} dx Thus, after the change of variable <math>x\sqrt{t}, this 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 <math>\sqrt \pi. More generally, \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 an arbitrary Gaussian function is \int_{-\infty}^{\infty} e^{-(x+b)^2/c^2}\,dxc \sqrt{\pi}. An alternative form is \int_{-\infty}^{\infty}d\,e^{-f x^2 + g x + h}\,dxd\,\sqrt{\frac{\pi}{f}}\,\exp\left(g^2/4f + h\right), where f must be strictly positive for the integral to converge. n-dimensional and functional generalization Suppose A is a symmetric positive-definite invertible covariant tensor of rank two. Then, \quad \int_{-\infty}^\infty \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. This fact is applied in the study of the multivariate normal distribution. Also, \begin{align} & {} \quad \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)}} \end{align} 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. Alternatively, \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}^{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 <math>(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}, \frac{1}{2^N N!}\sum_{\sigma \in S_{2N}}A^{-1}(x_{\sigma},x_{\sigma})\cdots A^{-1}(x_{\sigma},x_{\sigma}). In the deWitt notation, the equation looks identical to the finite-dimensional case. n-dimensional with linear term If A is again a symmetric matrix, then \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 \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} Higher-order polynomials 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 \begin{align} & \int_{-\infty}^{\infty} e^{a x^4+b x^3+c x^2+d x+f}\,dx \\ & {} \quad \frac12 e^f \!\!\!\!\!\!\!\! \sum_{\begin{smallmatrix}n,m,p0 \\ n+p0 \mod 2\end{smallmatrix}}^{\infty} \!\!\!\! \frac{b^n}{n!} \frac{c^m}{m!} \frac{d^p}{p!} \frac{\Gamma(\frac{3n+2m+p+1}4)}{(-a)^{\frac{3n+2m+p+1}4}}. \end{align} The n + p 0 mod 2 requirement is because the integral from −∞ to 0 contributes a factor of (−1)/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. References Weisstein, Eric W. , "Integral. html urlnameGaussianIntegral" from MathWorld. David Griffiths. Introduction to Quantum Mechanics. 2nd Edition back cover. See also Common integrals in quantum field theory
- Das gaußsche Fehlerintegral wird auch Wahrscheinlichkeitsverteilungsfunktion <math>\Phi</math> genannt. Es ist das Integral von <math>-\infty</math> bis <math>z</math> über die Normalverteilung (hier mit <math>\mu = 0</math> und <math>\sigma = 1</math>). Da die gesamte Fläche unterhalb der Normalverteilung (auch Gauß-Glocke genannt) 1 ist, ist der Wert des Fehlerintegrals für <math>z\rightarrow\infty</math> ebenfalls 1. Das Fehlerintegral ist durch <math>\Phi(z)=\frac 1{\sqrt{2\pi}} \int_{-\infty}^z e^{-\frac 12 t^2} \mathrm dt</math> definiert. Lässt man das Integral erst bei <math>0</math> statt bei <math>-\infty</math> beginnen, so spricht man von <math>\Phi_0</math>: <math>\Phi_0(z)=\frac 1{\sqrt{2\pi}} \int_0^z e^{-\frac 12 t^2} \mathrm dt = \Phi(z)-\frac 12. </math>
- La integral de Gauß és una integral definida, que fou calculada per primera vegada per Gauß. És la base de la distribució normal (o distribució gaussiana). És un element fonamental de la teoria de la probabilitat. La integral s'expressa habitualment com <math>\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}~,</math> o, de forma equivalent, com <math>\int_{0}^\infty e^{-x^2}\,dx= \frac {\sqrt{\pi}} 2. </math> La demostració d'aquesta integral està basada en el Teorema de Fubini.
- Gaussův integrál je integrál typu <math>I_p(\lambda) = \int_0^\infty x^p \mathrm{e}^{-\lambda x^2} \mathrm{d}x</math>, kde <math>\Re (p)>-1, \Re(\lambda) \geq 0</math>.
- La integral de Gauss, integral gaussiana o integral de probabilidad, es la integral impropia de la función gaussiana <math>e^{{-x}^2} sobre toda la recta de los reales. Debe su nombre al matemático y físico alemán Carl Friedrich Gauss, y su ecuación es: \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi} Esta integral tiene amplias aplicaciones, incluyendo normalización, en teoría de la probabilidad y transformada continua de Fourier. También aparece en la definición de la función error. A pesar de que no existe ninguna función elemental para la función error, como se puede demostrar mediante el algoritmo de Risch, la integral Gaussiana puede ser resuelta analíticamente con las herramientas del cálculo. O sea, no existe una integral indefinida elemental para <math>\int e^{-x^2}\,dx, pero si es posible evaluar la integral definida<math>\int_{-\infty}^\infty e^{-x^2}\,dx.
- Cette intégrale est appelée intégrale de Gauss. Elle intervient dans la définition de la loi de probabilité appelée loi gaussienne, ou loi normale. La valeur de cette intégrale fut donnée pour la première fois par Pierre-Simon Laplace.
- L'integrale di Gauss è un integrale definito, calcolato per la prima volta da Gauss. È alla base della distribuzione normale (detta pure gaussiana), mattone fondamentale della teoria della probabilità. La funzione integranda, normalizzata affinché l'area dell'integrale da -∞ a ∞ sia 1, è detta anche funzione gaussiana. La forma solitamente usata per l'integrale di Gauss è: <math>\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}~. </math> o l'equivalente <math>\int_{0}^\infty e^{-x^2}\,dx= \frac {\sqrt{\pi}} 2. </math>
- ガウス積分（がうす-せきぶん、Gaussian integral）はガウス関数 exp(−x) の（両側無限）積分。名称は、数学・物理学者のカール・フリードリヒ・ガウスに由来する。 {{Indent|<math>\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi} この積分は、確率論における正規化や連続フーリエ変換など広い応用を持ち、また、誤差関数 erf の定義などにも用いられる。ガウス関数の原始関数である誤差関数を初等関数で表すことはできないが、リッシュのアルゴリズムにより、定積分としてのガウス積分の値は微分積分学の道具を用いて解析的に求められる。
- 高斯积分（Gauss integration），有时也被称为概率积分，是高斯函数（<math>f(x) = e^{-x^2}</math>）的积分。它是以德国数学家和物理学家卡爾·弗里德里希·高斯命名的。 <math>\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}</math> 高斯积分在概率论和连续傅里叶变换等的统一化等计算中有广泛的应用。在误差函数的定义中它也出现。虽然误差函数没有初等函数，但是高斯积分可以通过微积分学的手段解析求解。
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