The integral of any Gaussian function (named after Carl Friedrich Gauss) is quickly reducible to the Gaussian integral

$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-x^2\}dx=\backslash sqrt\{\backslash pi\}.$

This integral cannot be computed by elementary means since the function has no simple antiderivative.

Evaluation using double integrals

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The value of the Gaussian integral can be determined by considering its square as a double integral. Let the value of the integral be $s$. Then, by applying Fubini's theorem,

$s^2\; =\; \backslash left(\backslash int\_\{-\backslash infty\}^\backslash infty\; e^\{-x^2\}\backslash ,dx\backslash right)\; \backslash left(\backslash int\_\{-\backslash infty\}^\backslash infty\; e^\{-y^2\}\backslash ,dy\backslash right)\; =\; \backslash int\_\{\backslash mathbb\{R\}^2\}\backslash !\backslash !e^\{-(x^2\; +\; y^2)\}\backslash ,dA.$

In polar coordinates we have:

$s^2\; =\; \backslash int\_0^\{2\backslash pi\}\backslash !\backslash !\backslash !\backslash int\_0^\backslash infty\; e^\{-r^2\}r\backslash ,dr\backslash ,d\backslash theta\; =\; 2\backslash pi\backslash int\_0^\backslash infty\; e^\{-r^2\}r\backslash ,dr=2\backslash pi\backslash left*\_0^\backslash infty=\; \backslash pi.$

(The factor of $r$ that makes the integral straightforward comes from the change of variables.)

It then follows that

$s\; =\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-x^2\}dx\; =\; \backslash sqrt\{\backslash pi\}.$

If there is a multiplicative constant, the result changes slightly.

$s\; =\; \backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-a\; x^2\}dx\; =\; \backslash sqrt\{\backslash pi/a\}.$

Relation to the gamma function

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Since the integrand is an even function,

$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-x^2\}\; dx\; =\; 2\; \backslash int\_0^\backslash infty\; e^\{-x^2\}\; dx$

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

$\backslash int\_0^\backslash infty\; e^\{-t\}\; \backslash \; t^\{-1/2\}\; dt\; \backslash ,\; =\; \backslash ,\; \backslash Gamma\backslash left(\backslash frac\{1\}\{2\}\backslash right)$

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

$\backslash int\_0^\backslash infty\; e^\{-ax^b\}\; dx\; =\; a^\{-1/b\}\; \backslash ,\; \backslash Gamma\backslash left(1+\backslash frac\{1\}\{b\}\backslash right).$

n-dimensional and functional generalization

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

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

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

Also,

$\backslash int\; x^\{k\_1\}\backslash cdots\; x^\{k\_\{2N\}\}\; e^\{-\backslash frac\{A\_\{ij\}\; x^i\; x^j\}\{2\}\}\; d^nx=\backslash frac\{(2\backslash pi)^\{n/2\}\}\{\backslash sqrt\{\backslash det\{A\}\}\}\backslash frac\{1\}\{2^N\; N!\}\backslash sum\_\{\backslash sigma\; \backslash in\; S\_\{2N\}\}(A^\{-1\})^\{k\_\{\backslash sigma(1)\}k\_\{\backslash sigma(2)\}\}\backslash cdots\; (A^\{-1\})^\{k\_\{\backslash sigma(2N-1)\}k\_\{\backslash sigma(2N)\}\}$

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,

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

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

$\backslash frac\{\backslash int\; f(x\_1)\backslash cdots\; f(x\_\{2N\})\; e^\{-\backslash iint\; \backslash frac\{A(x\_\{2N+1\},x\_\{2N+2\})\; f(x\_\{2N+1\})\; f(x\_\{2N+2\})\}\{2\}\; d^dx\_\{2N+1\}\; d^dx\_\{2N+2\}\}\; \backslash mathcal\{D\}f\}\{\backslash int\; e^\{-\backslash iint\; \backslash frac\{A(x\_\{2N+1\},x\_\{2N+2\})\; f(x\_\{2N+1\})\; f(x\_\{2N+2\})\}\{2\}\; d^dx\_\{2N+1\}\; d^dx\_\{2N+2\}\}\; \backslash mathcal\{D\}f\},$

$=\backslash frac\{1\}\{2^N\; N!\}\backslash sum\_\{\backslash sigma\; \backslash in\; S\_\{2N\}\}A^\{-1\}(x\_\{\backslash sigma(1)\},x\_\{\backslash sigma(2)\})\backslash cdots\; A^\{-1\}(x\_\{\backslash sigma(2N-1)\},x\_\{\backslash sigma(2N)\}).$

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

n-dimensional with linear term

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Once again A is a symmetric matrix, then

$\backslash int\; e^\{-A\_\{ij\}\; x^i\; x^j+B\_i\; x\_i\}\; d^nx=\backslash sqrt\{\; \backslash frac\{\backslash pi^n\}\{\backslash det\{A\}\}\; \}e^\{\backslash frac\{1\}\{4\}B^TA^\{-1\}B\}$

Exponentials | Integrals

Integrale di Gauss