Gaussian integral

By polar coordinates A standard way to compute the Gaussian integral, the idea of which goes back to Poisson, is to make use of the property that: \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. Consider the function e^{-\left(x^2 + y^2\right)} = e^{-r^{2}}on the plane \mathbb{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: \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 \pi Comparing these two computations yields the integral, though one should take care about the improper integrals involved. \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} where the factor of is the Jacobian determinant which appears because of the transform to polar coordinates ( is the standard measure on the plane, expressed in polar coordinates Wikibooks:Calculus/Polar Integration#Generalization), and the substitution involves taking , so . Combining these yields \left ( \int_{-\infty}^\infty e^{-x^2}\,dx \right )^2=\pi, so \int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}. Complete 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. If the integral \int_{-\infty}^\infty e^{-x^2} \, dx were absolutely convergent we would have that its Cauchy principal value, that is, the limit \lim_{a\to\infty} I(a) would coincide with \int_{-\infty}^\infty e^{-x^2}\,dx. To see that this is the case, consider that \int_{-\infty}^\infty \left|e^{-x^2}\right| dx So we can compute \int_{-\infty}^\infty e^{-x^2} \, dx by just taking the limit \lim_{a\to\infty} I(a). Taking the square of I(a) yields \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} Using Fubini's theorem, the above double integral can be seen as an area integral \iint_{[-a, a] \times [-a, a]} e^{-\left(x^2+y^2\right)}\,d(x,y), taken over a square with vertices {{math|{(−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: \begin{align} x &= r \cos \theta, & y &= r \sin\theta \end{align} \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} d(x,y) = \left|J(r, \theta)\right| d(r,\theta) = r\, d(r,\theta). \int_0^{2\pi} \int_0^a re^{-r^2} \, dr \, d\theta (See to polar coordinates from Cartesian coordinates for help with polar transformation.) Integrating, \pi \left(1-e^{-a^2}\right) By the squeeze theorem, this gives the Gaussian integral \int_{-\infty}^\infty e^{-x^2}\, dx = \sqrt{\pi}. By Cartesian coordinates A different technique, which goes back to Laplace (1812), ==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 x = \sqrt{t}, this turns into the Euler integral 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} where \Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt is the gamma function. More generally, \int_0^\infty x^n e^{-ax^b} dx = \frac{\Gamma{\left((n+1)/b\right)}}{b a^{(n+1)/b}}, which can be obtained by substituting t=a x^b in the integrand of the gamma function to get \Gamma(z) = a^z b \int_0^{\infty} x^{bz-1} e^{-a x^b} dx . ==Generalizations==

The integral of a Gaussian function The integral of an arbitrary Gaussian function is \int_{-\infty}^{\infty} e^{-a(x+b)^2}\,dx= \sqrt{\frac{\pi}{a}}. An alternative form is \int_{-\infty}^{\infty}e^{- (a x^2 - b x + c)}\,dx=\sqrt{\frac{\pi}{a}}\,e^{\frac{b^2}{4a}-c}. 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 \int_{-\infty}^{\infty} e^{\frac 12 it^2} dt = e^{i\pi/4} \sqrt{2\pi}and more generally,\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)}^Nfor any positive-definite symmetric matrix A. n-dimensional and functional generalization Suppose A is a symmetric positive-definite (hence invertible) precision matrix, which is the matrix inverse of the covariance matrix. Then, \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} By completing the square, this generalizes to\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) This fact is applied in the study of the multivariate normal distribution. Also, \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)}} where σ is a permutation of {{math|{1, …, 2N}}} and the extra factor on the right-hand side is the sum over all combinatorial pairings of {{math|{1, …, 2N}}} of N copies of A−1. Alternatively, \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} 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: \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} 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) \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} Integrals of similar form \int_0^\infty x^{2n} e^{-{x^2}/{a^2}}\,dx = \sqrt{\pi}\frac{a^{2n+1} (2n-1)!!}{2^{n+1}} \int_0^\infty x^{2n+1} e^{-{x^2}/{a^2}} \, dx = \frac{n!}{2} a^{2n+2} \int_0^\infty x^{2n}e^{-bx^2}\,dx = \frac{(2n-1)!!}{b^n 2^{n+1}} \sqrt{\frac{\pi}{b}} \int_0^\infty x^{2n+1}e^{-bx^2}\,dx = \frac{n!}{2b^{n+1}} \int_0^\infty x^{n}e^{-bx^2}\,dx = \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}} where n is a positive integer An easy way to derive these is by differentiating under the integral sign. \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} 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)-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. Exponentials of other even polynomials can numerically be solved using series. These may be interpreted as formal calculations when there is no convergence. For example, the solution to the integral of the exponential of a quartic polynomial is \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}}. The mod 2 requirement is because the integral from −∞ to 0 contributes a factor of 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==