Common integrals in quantum field theory: Difference between revisions

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There are '''common integrals in quantum field theory''' that appear repeatedly.<ref>{{cite book| author=A. Zee| title=Quantum Field Theory in a Nutshell| publisher= Princeton University| year=2003 | isbn=0-691-01019-6}} pp. 13-15</ref> These integrals are all variations and generalizations of [[gaussian integral]]s to the complex plane and to multiple dimensions. Other integrals can be approximated by versions of the gaussian integral. Fourier integrals are also considered.

'''Common integrals in quantum field theory''' are all variations and generalizations of [[Gaussian integral|Gaussian integrals]] to the [[complex plane]] and to multiple dimensions.<ref name="Zee">{{cite book|author=A. Zee|title=Quantum Field Theory in a Nutshell|publisher=Princeton University|year=2003|isbn=0-691-01019-6}}</ref>{{rp|pp=13–15}} Other integrals can be approximated by versions of the Gaussian integral. Fourier integrals are also considered.

== Variations on a simple gaussianGaussian integral ==

=== Gaussian integral ===

The first integral, with broad application outside of quantum field theory, is the [[gaussianGaussian integral]].

<math display="block"> G \equiv \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx</math>

:<math> G \equiv \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx</math>

In physics the factor of 1/2 in the argument of the exponential is common.

Note that, if we let <math>r=\sqrt{x^2+y^2}</math> be the radius, then we can use the usual polar coordinate change of variables (which in particular renders <math>dx\,dy=r\,dr\,d\theta</math>) to get

Note:

<math display="block"> G^2 = \left ( \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx \right ) \cdot \left ( \int_{-\infty}^{\infty} e^{-{1 \over 2} y^2}\,dy \right ) = 2\pi \int_{0}^{\infty} r e^{-{1 \over 2} r^2}\,dr = 2\pi \int_{0}^{\infty} e^{- w}\,dw = 2 \pi.</math>

:<math> G^2 = \left ( \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx \right ) \cdot \left ( \int_{-\infty}^{\infty} e^{-{1 \over 2} y^2}\,dy \right ) = 2\pi \int_{0}^{\infty} r e^{-{1 \over 2} r^2}\,dr = 2\pi \int_{0}^{\infty} e^{- w}\,dw = 2 \pi.</math>

Thus we obtain

<math display="block"> \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx = \sqrt{2\pi}. </math>

=== Slight generalization of the Gaussian integral ===

:<math> \int_{-\infty}^{\infty} e^{-{1 \over 2} x^2}\,dx = \sqrt{2\pi}. </math>

<math display="block"> \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = \sqrt{2\pi \over a} </math>

===Slight generalization of the gaussian integral===

:<math> \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = \sqrt{2\pi \over a} </math>

where we have scaled

<math display="block"> x \to {x \over \sqrt{a}}. </math>

=== Integrals of exponents and even powers of ''x'' ===

:<math> x \to {x \over \sqrt{a}} </math>.

<math display="block"> \int_{-\infty}^{\infty} x^2 e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \left ( {2\pi \over a } \right ) ^{1\over 2} = \left ( {2\pi \over a } \right ) ^{1\over 2} {1\over a}</math>

===Integrals of exponents and even powers of ''x''===

:<math> \int_{-\infty}^{\infty} x^2 e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = -2{d\over da} \left ( {2\pi \over a } \right ) ^{1\over 2} = \left ( {2\pi \over a } \right ) ^{1\over 2} {1\over a}</math>

and

<math display="block"> \int_{-\infty}^{\infty} x^4 e^{-{1 \over 2} a x^2}\,dx = \left ( -2{d\over da} \right) \left ( -2{d\over da} \right) \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = \left ( -2{d\over da} \right) \left ( -2{d\over da} \right) \left ( {2\pi \over a } \right ) ^{1\over 2} = \left ( {2\pi \over a } \right ) ^{1\over 2} {3\over a^2}</math>

:<math> \int_{-\infty}^{\infty} x^4 e^{-{1 \over 2} a x^2}\,dx = \left ( -2{d\over da} \right) \left ( -2{d\over da} \right) \int_{-\infty}^{\infty} e^{-{1 \over 2} a x^2}\,dx = \left ( -2{d\over da} \right) \left ( -2{d\over da} \right) \left ( {2\pi \over a } \right ) ^{1\over 2} = \left ( {2\pi \over a } \right ) ^{1\over 2} {3\over a^2}</math>

In general

<math display="block"> \int_{-\infty}^{\infty} x^{2n} e^{-{1 \over 2} a x^2}\,dx = \left ( {2\pi \over a } \right ) ^{1\over {2}} {1\over a^{n}} \left ( 2n -1 \right ) \left ( 2n -3 \right ) \cdots 5 \cdot 3 \cdot 1 = \left ( {2\pi \over a } \right ) ^{1\over {2}} {1\over a^{n}} \left ( 2n -1 \right )!! </math>

:<math> \int_{-\infty}^{\infty} x^{2n} e^{-{1 \over 2} a x^2}\,dx = \left ( {2\pi \over a } \right ) ^{1\over {2}} {1\over a^{n}} \left ( 2n -1 \right ) \left ( 2n -3 \right ) \cdots 5 \cdot 3 \cdot 1 = \left ( {2\pi \over a } \right ) ^{1\over {2}} {1\over a^{n}} \left ( 2n -1 \right )!! </math>

Note that the integrals of exponents and odd powers of x are 0, due to [[odd function|odd]] symmetry.

=== Integrals with a linear term in the argument of the exponent ===

:<math display="block"> \int_{-\infty}^{\infty} \exp\left( -{1 \overfrac 1 2} a x^2 + Jx\right ) dx </math>

This integral can be performed by [[completing the square]]:

<math display="block"> \left( -{1 \over 2} a x^2 + Jx\right ) = -{1 \over 2} a \left ( x^2 - { 2 Jx \over a } + { J^2 \over a^2 } - { J^2 \over a^2 } \right ) = -{1 \over 2} a \left ( x - { J \over a } \right )^2 + { J^2 \over 2a } </math>

:<math> \left( -{1 \over 2} a x^2 + Jx\right ) = -{1 \over 2} a \left ( x^2 - { 2 Jx \over a } + { J^2 \over a^2 } - { J^2 \over a^2 } \right ) = -{1 \over 2} a \left ( x - { J \over a } \right )^2 + { J^2 \over 2a } </math>

Therefore:

<math display="block">\begin{align}

\int_{-\infty}^\infty \exp\left( -{1 \over 2} a x^2 + Jx\right) \, dx

:<math>\begin{align}

\int_{-\infty}^\infty \exp\left( -{1 \over 2} a x^2 + Jx\right) \, dx &= \exp\left( { J^2 \over 2a } \right ) \int_{-\infty}^\infty \exp \left [ -{1 \over 2} a \left ( x - { J \over a } \right )^2 \right ] \, dx \\[8pt]

&= \exp\left( { J^2 \over 2a } \right )\int_{-\infty}^\infty \exp\left( -{1 \over 2} a w^2 \right) \, dw \\[8pt]

&= \left ( {2\pi \over a } \right ) ^{1\over 2} \exp\left( { J^2 \over 2a }\right )

\end{align}</math>

=== Integrals with an imaginary linear term in the argument of the exponent ===

The integral

<math display="block"> \int_{-\infty}^{\infty} \exp\left( -{1 \over 2} a x^2 +iJx\right ) dx = \left ( {2\pi \over a } \right ) ^{1\over 2} \exp\left( -{ J^2 \over 2a }\right ) </math>

is proportional to the [[Fourier transform]] of the Gaussian where {{mvar|J}} is the [[conjugate variables|conjugate variable]] of {{mvar|x}}.

By again completing the square we see that the Fourier transform of a Gaussian is also a Gaussian, but in the conjugate variable. The larger {{mvar|a}} is, the narrower the Gaussian in {{mvar|x}} and the wider the Gaussian in {{mvar|J}}. This is a demonstration of the [[uncertainty principle]].

:<math> \int_{-\infty}^{\infty} \exp\left( -{1 \over 2} a x^2 + iJx\right ) dx = \left ( {2\pi \over a } \right ) ^{1\over 2} \exp\left( -{ J^2 \over 2a }\right ) </math>

This integral is also known as the [[Hubbard–Stratonovich transformation]] used in field theory.

is proportional to the [[Fourier transform]] of the gaussian where {{mvar|J}} is the [[conjugate variables|conjugate variable]] of {{mvar|x}}.

=== Integrals with a complex argument of the exponent ===

By again completing the square we see that the Fourier transform of a gaussian is also a gaussian, but in the conjugate variable. The larger {{mvar|a}} is, the narrower the gaussian in {{mvar|x}} and the wider the gaussian in {{mvar|J}}. This is a demonstration of the [[uncertainty principle]].

This integral is also known as the [[Hubbard-Stratonovich transformation]] used in field theory.

===Integrals with a complex argument of the exponent===

The integral of interest is (for an example of an application see [[Relation between Schrödinger's equation and the path integral formulation of quantum mechanics]])

<math display="block"> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx. </math>

:<math> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx. </math>

We now assume that {{mvar|a}} and {{mvar|J}} may be complex.

Completing the square

<math display="block"> \left( {1 \over 2} i a x^2 + iJx\right ) = {1\over 2} ia \left ( x^2 + {2Jx \over a} + \left ( { J \over a} \right )^2 - \left ( { J \over a} \right )^2 \right ) = -{1\over 2} {a \over i} \left ( x + {J\over a} \right )^2 - { iJ^2 \over 2a}. </math>

:<math> \left( {1 \over 2} i a x^2 + iJx\right ) = {1\over 2} ia \left ( x^2 + {2Jx \over a} + \left ( { J \over a} \right )^2 - \left ( { J \over a} \right )^2 \right ) = -{1\over 2} {a \over i} \left ( x + {J\over a} \right )^2 - { iJ^2 \over 2a}. </math>

By analogy with the previous integrals

<math display="block"> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx = \left ( {2\pi i \over a } \right ) ^{1\over 2} \exp\left( { -iJ^2 \over 2a }\right ). </math>

This result is valid as an integration in the complex plane as long as {{mvar|a}} is non-zero and has a semi-positive imaginary part. See [[Fresnel integral]].

:<math> \int_{-\infty}^{\infty} \exp\left( {1 \over 2} i a x^2 + iJx\right ) dx = \left ( {2\pi i \over a } \right ) ^{1\over 2} \exp\left( { -iJ^2 \over 2a }\right ). </math>

== Gaussian integrals in higher dimensions ==

This result is valid as an integration in the complex plane as long as {{mvar|a}} is non-zero and has a semi-positive imaginary part.

==Gaussian integrals in higher dimensions==

The one-dimensional integrals can be generalized to multiple dimensions.<ref>{{cite book | author=Frederick W. Byron and Robert W. Fuller| title=Mathematics of Classical and Quantum Physics | publisher= Addison-Wesley| year=1969 | isbn=0-201-00746-0}}</ref>

<math display="block">\int \exp\left( - \frac 1 2 x \cdot A \cdot x +J \cdot x \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J \cdot A^{-1} \cdot J \right)</math>

:<math>\int \exp\left( - \frac 1 2 x \cdot A \cdot x +J \cdot x \right) d^nx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J \cdot A^{-1} \cdot J \right)</math>

Here {{mvar|A}} is a real positive definite [[symmetric matrix]].

This integral is performed by [[Diagonalizable matrix|diagonalization]] of {{mvar|A}} with an [[orthogonal matrix|orthogonal transformation]]

<math display="block">D= O^{-1} A O = O^\text{T} A O</math>

:<math>D= O^{-1} A O = O^T A O</math>

where {{mvar|D}} is a [[diagonal matrix]] and {{mvar|O}} is an [[orthogonal matrix]]. This decouples the variables and allows the integration to be performed as {{mvar|n}} one-dimensional integrations.

This is best illustrated with a two-dimensional example.

=== Example: Simple Gaussian integration in two dimensions ===

The Gaussian integral in two dimensions is

<math display="block">\int \exp\left( - \frac 1 2 A_{ij} x^i x^j \right) d^2x = \sqrt{\frac{(2\pi)^2}{\det A}}</math>

:<math>\int \exp\left( - \frac 1 2 A_{ij} x^i x^j \right) d^2x = \sqrt{\frac{(2\pi)^2}{\det A}}</math>

where {{mvar|A}} is a two-dimensional symmetric matrix with components specified as

<math display="block"> A = \begin{bmatrix} a&c\\ c&b\end{bmatrix}</math>

:<math> A = \begin{bmatrix} a&c\\ c&b\end{bmatrix}</math>

and we have used the [[Einstein summation convention]].

==== Diagonalize the matrix ====

The first step is to [[Diagonalizable matrix|diagonalize]] the matrix.<ref>{{cite book | author=Herbert S. Wilf| title=Mathematics for the Physical Sciences | publisher= Dover| year=1978 | isbn=0-486-63635-6| url-access=registration| url=https://archive.org/details/mathematicsforph0000wilf_w9m6}}</ref> Note that

<math display="block">A_{ij} x^i x^j \equiv x^\text{T}Ax = x^\text{T} \left(OO^\text{T}\right) A \left(OO^\text{T}\right) x = \left(x^\text{T}O \right) \left(O^\text{T}AO \right) \left(O^\text{T}x \right) </math>

:<math>A_{ij} x^i x^j \equiv x^TAx = x^T \left(OO^T\right) A \left(OO^T\right) x = \left(x^TO \right) \left(O^TAO \right) \left(O^Tx \right) </math>

where, since {{mvar|A}} is a real [[symmetric matrix]], we can choose {{mvar|O}} to be [[orthogonal matrix|orthogonal]], and hence also a [[unitary matrix]]. {{mvar|O}} can be obtained from the [[eigenvectors]] of {{mvar|A}}. We choose {{mvar|O}} such that: {{math|''D'' ≡ ''O^TAO''}} is diagonal.

===== Eigenvalues of ''A'' =====

To find the eigenvectors of {{mvar|A}} one first finds the [[eigenvalues]] {{mvar|λ}} of {{mvar|A}} given by

<math display="block"> \begin{bmatrix}a&c\\ c&b\end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=\lambda \begin{bmatrix}u\\ v\end{bmatrix}.</math>

:<math> \begin{bmatrix}a&c\\ c&b\end{bmatrix} \begin{bmatrix} u\\ v \end{bmatrix}=\lambda \begin{bmatrix}u\\ v\end{bmatrix}.</math>

The eigenvalues are solutions of the [[characteristic polynomial]]

<math display="block">( a - \lambda)( b-\lambda) -c^2 = 0 </math>

:<math display="block">( a\lambda^2 - \lambda)( a+b-\lambda) + ab -c^2 = 0 ,</math>

:<math>\lambda^2 - \lambda(a+b) + ab -c^2 = 0 </math>

which are found using the [[quadratic equation]]:

<math display="block">\begin{align}

:<math> \lambda_{\pm} = {1\over 2} ( a+b) \pm {1\over 2}\sqrt{(a+b)^2-4(ab - c^2)}. </math>

:<math> \lambda_{\pm} &= \tfrac{1\over }{2} ( a+b) \pm \tfrac{1\over }{2} \sqrt{(a^2 +2ab + b)^2 -4ab4(ab +- 4cc^2)}. </math>\\

:<math> \lambda_{\pm} &= \tfrac{1\over }{2} ( a+b) \pm \tfrac{1\over }{2} \sqrt{(a-^2 +2ab + b)^2 -4ab + 4c^2}. </math>\\

&= \tfrac{1}{2} ( a+b) \pm \tfrac{1}{2} \sqrt{(a-b)^2+4c^2}.

\end{align}</math>

===== Eigenvectors of ''A'' =====

Substitution of the eigenvalues back into the eigenvector equation yields

<math display="block"> v = -{ \left( a - \lambda_{\pm} \right)u \over c }, \qquad v = -{cu \over \left( b - \lambda_{\pm} \right)}.</math>

:<math> v = -{ \left( a - \lambda_{\pm} \right)u \over c }, \qquad v = -{cu \over \left( b - \lambda_{\pm} \right)}.</math>

From the characteristic equation we know

<math display="block"> {a - \lambda_{\pm} \over c } = {c \over b - \lambda_{\pm}}.</math>

:<math> {a - \lambda_{\pm} \over c } = {c \over b - \lambda_{\pm}}.</math>

Also note

<math display="block"> { a - \lambda_{\pm} \over c } = -{ b - \lambda_{\mp} \over c}. </math>

:<math> { a - \lambda_{\pm} \over c } = -{ b - \lambda_{\mp} \over c}. </math>

The eigenvectors can be written as:

<math display="block">\begin{bmatrix} \frac{1}{\eta} \\[1ex] -\frac{a - \lambda_-}{c\eta} \end{bmatrix},

\qquad

:<math>\begin{bmatrix} \frac{1}{\eta}\\ -\frac{a - \lambda_-}{c\eta} \end{bmatrix}, \qquad \begin{bmatrix}-\frac{b - \lambda_+}{c\eta} \\ \frac{1}{\eta} \end{bmatrix} </math>

\begin{bmatrix} -\frac{b - \lambda_+}{c\eta} \\[1ex] \frac{1}{\eta} \end{bmatrix} </math>

for the two eigenvectors. Here {{mvar|η}} is a normalizing factor given by,

<math display="block">\eta = \sqrt{1 + \left(\frac{a -\lambda_{-}}{c} \right)^2 } = \sqrt{1 + \left(\frac{b - \lambda_{+}}{c} \right)^2}.</math>

:<math>\eta = \sqrt{1 + \left(\frac{a - \lambda_{-}}{c} \right)^2 } = \sqrt{1 + \left(\frac{b - \lambda_{+}}{c} \right)^2}.</math>

It is easily verified that the two eigenvectors are orthogonal to each other.

===== Construction of the orthogonal matrix =====

The orthogonal matrix is constructed by assigning the normalized eigenvectors as columns in the orthogonal matrix

:<math display="block">O = \begin{bmatrix} \frac{1}{\eta} & -\frac{b - \lambda_{+}}{c \eta} \\ -\frac{a - \lambda_{-}}{c \eta} &\frac{1}{\eta}\end{bmatrix}.</math>

Note that {{math|det(''O'') {{=}} 1}}.

If we define

<math display="block"> \sin(\theta) = -\frac{a - \lambda_{-}}{c \eta }</math>

:<math> \sin(\theta) = -\frac{a - \lambda_{-}}{c \eta }</math>

then the orthogonal matrix can be written

<math display="block">O = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix} </math>

:<math>O = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix} </math>

which is simply a rotation of the eigenvectors with the inverse:

<math display="block">O^{-1} = O^\text{T} = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}.</math>

===== Diagonal matrix =====

:<math>O^{-1} = O^T = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}.</math>

=====Diagonal matrix=====

The diagonal matrix becomes

<math display="block"> D = O^\text{T} A O = \begin{bmatrix}\lambda_{-}& 0 \\[1ex] 0 & \lambda_{+}\end{bmatrix}</math>

:<math> D = O^T A O = \begin{bmatrix}\lambda_{-}&0\\ 0 & \lambda_{+}\end{bmatrix}</math>

with eigenvectors

<math display="block">\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>

===== Numerical example =====

:<math>\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>

<math display="block">A = \begin{bmatrix} 2&1\\ 1 & 1\end{bmatrix}</math>

=====Numerical example=====

:<math>A = \begin{bmatrix} 2&1\\ 1 & 1\end{bmatrix}</math>

The eigenvalues are

<math display="block">\lambda_{\pm} = {3\over 2} \pm {\sqrt{ 5} \over 2}.</math>

:<math>\lambda_{\pm} = {3\over 2} \pm {\sqrt{ 5} \over 2}.</math>

The eigenvectors are

<math display="block">{1\over \eta}\begin{bmatrix} 1 \\[1ex] -{1\over 2} - {\sqrt{5} \over 2} \end{bmatrix}, \qquad

:<math>{1\over \eta}\begin{bmatrix} 1\\ -{1\over 2} - {\sqrt{5} \over 2}\end{bmatrix}, \qquad {1\over \eta} \begin{bmatrix} {1\over 2} + {\sqrt{5} \over 2 } \\ [1ex] 1 \end{bmatrix}</math>

where

<math display="block">\eta = \sqrt{{5\over 2} + {\sqrt{5}\over 2}}.</math>

:<math>\eta = \sqrt{{5\over 2} + {\sqrt{5}\over 2}}.</math>

Then

<math display="block">\begin{align}

:<math>\begin{align}

O &= \begin{bmatrix} \frac{1}{\eta} & \frac{1}{\eta} \left({1\over 2} + {\sqrt{5} \over 2}\right) \\ \frac{1}{\eta} \left(-{1\over 2} - {\sqrt{ 5} \over 2}\right) & {1\over \eta}\end{bmatrix} \\

O^{-1} &= \begin{bmatrix} \frac{1}{\eta} & \frac{1}{\eta} \left(-{1\over 2} - {\sqrt{5} \over 2}\right) \\ \frac{1}{\eta} \left({1\over 2} + {\sqrt{5} \over 2}\right) & \frac{1}{\eta} \end{bmatrix}

Line 204 ⟶ 165:

The diagonal matrix becomes

<math display="block">D = O^\text{T}AO = \begin{bmatrix} \lambda_- &0\\ 0 & \lambda_+ \end{bmatrix} = \begin{bmatrix} {3\over 2} - {\sqrt{5} \over 2}& 0\\ 0 & {3\over 2} + {\sqrt{5} \over 2}\end{bmatrix} </math>

:<math>D = O^TAO = \begin{bmatrix} \lambda_- &0\\ 0 & \lambda_+ \end{bmatrix} = \begin{bmatrix} {3\over 2} - {\sqrt{5} \over 2}& 0\\ 0 & {3\over 2} + {\sqrt{5} \over 2}\end{bmatrix} </math>

with eigenvectors

<math display="block">\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>

==== Rescale the variables and integrate ====

:<math>\begin{bmatrix} 1\\ 0\end{bmatrix}, \qquad \begin{bmatrix} 0\\ 1 \end{bmatrix}</math>

====Rescale the variables and integrate====

With the diagonalization the integral can be written

<math display="block">\int \exp\left( - \frac 1 2 x^\text{T} A x \right) d^2x = \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) \, d^2y</math>

:<math>\int \exp\left( - \frac 1 2 x^T A x \right) d^2x = \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) \, d^2y</math>

where

<math display="block">y = O^\text{T} x.</math>

:<math>y = O^T x.</math>

Since the coordinate transformation is simply a rotation of coordinates the [[Jacobian matrix and determinant|Jacobian]] determinant of the transformation is one yielding

<math display="block"> d^2y = d^2x </math>

The integrations can now be performed:

:<math> dy^2 = dx^2 </math>

<math display="block">\begin{align}

\int \exp\left( - \frac{1}{2} x^\mathsf{T} A x \right) d^2x

The integrations can now be performed.

={}& \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) d^2y \\[1ex]

={}& \prod_{j=1}^2 \left( { 2\pi \over \lambda_j } \right)^{1/2} \\

:<math>\begin{align}

\int={}& \exp\left( -{ \frac{1}{(2} x^T A x \rightpi) d^2x2 &= \intover \exp\left( - \frac 1 2 \sum_prod_{j=1}^2 \lambda_{jlambda_j } y_j^2 \right) d^2y{1/2} \\[1ex]

&= \prod_{j=1}^2& \left( { (2\pi)^2 \over \lambda_jdet{ \left( O^{-1}AO \right)} } \right)^{1\over /2} \\ [1ex]

&={}& \left( { (2\pi)^2 \over \prod_det{j=1}^2 \lambda_jleft( A \right)} } \right)^{1\over /2} \\

&= \left( { (2\pi)^2 \over \det{ \left( O^{-1}AO \right)} } \right)^{1\over 2} \\

&= \left( { (2\pi)^2 \over \det{ \left( A \right)} } \right)^{1\over 2}

\end{align}</math>

which is the advertised solution.

=== Integrals with complex and linear terms in multiple dimensions ===

With the two-dimensional example it is now easy to see the generalization to the complex plane and to multiple dimensions.

==== Integrals with a linear term in the argument ====

:<math display="block">\int \exp\left(-\frac{1}{2} x^{T} \cdot A \cdot x + J^{T} \cdot x \right) d^nxdx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

==== Integrals with an imaginary linear term ====

:<math display="block">\int \exp\left(-\frac{1}{2} x^{T} \cdot A \cdot x +iJi J^{T} \cdot x \right) d^nxdx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( -{1\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

==== Integrals with a complex quadratic term ====

:<math display="block">\int \exp\left(\frac{i}{2} x^{T} \cdot A \cdot x +iJi J^{T} \cdot x \right) d^nxdx =\sqrt{\frac{(2\pi i)^n}{\det A}} \exp \left( -{i\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

===Integrals with differential operators in the argument===

As an example consider the integral<ref>Zee, pp. 21-22.</ref>

:<math>\int \exp\left[ \int d^4x \left (-\frac{1}{2} \varphi \hat A \varphi + J \varphi \right) \right ] D\varphi</math>

=== Integrals with differential operators in the argument ===

As an example consider the integral<ref name="Zee"/>{{rp|pp=21‒22}}

<math display="block">\int \exp\left[ \int d^4x \left (-\frac{1}{2} \varphi \hat A \varphi + J \varphi \right) \right ] D\varphi</math>

where <math> \hat A </math> is a differential operator with <math> \varphi </math> and {{mvar|J}} functions of [[spacetime]], and <math> D\varphi </math> indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is

<math display="block">\int \exp\left[ \int d^4x \left (-\frac 1 2 \varphi \hat A \varphi +J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( {1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right)</math>

:<math>\int \exp\left[ \int d^4x \left (-\frac 1 2 \varphi \hat A \varphi +J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( {1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right)</math>

where

<math display="block">\hat A D( x - y) = \delta^4 ( x - y)</math>

:<math>\hat A D( x - y) = \delta^4 ( x - y)</math>

and {{math|''D''(''x'' − ''y'')}}, called the [[propagator]], is the inverse of <math> \hat A</math>, and <math> \delta^4( x - y)</math> is the [[Dirac delta function]].

Similar arguments yield

<math display="block">\int \exp\left[\int d^4x \left (-\frac 1 2 \varphi \hat A \varphi + i J \varphi \right) \right ] D\varphi \; \propto \; \exp \left( - { 1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right),</math>

:<math>\int \exp\left[\int d^4x \left (-\frac 1 2 \varphi \hat A \varphi + i J \varphi \right) \right ] D\varphi \; \propto \; \exp \left( - { 1\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right),</math>

and

<math display="block">\int \exp\left[ i \int d^4x \left ( \frac 1 2 \varphi \hat A \varphi + J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( -{ i\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right).</math>

:<math>\int \exp\left[ i \int d^4x \left ( \frac 1 2 \varphi \hat A \varphi + J\varphi \right) \right ] D\varphi \; \propto \; \exp \left( -{ i\over 2} \int d^4x \; d^4y J(x) D( x - y) J(y) \right).</math>

See [[Static forces and virtual-particle exchange#Path-integral formulation of virtual-particle exchange|Path-integral formulation of virtual-particle exchange]] for an application of this integral.

== Integrals that can be approximated by the method of steepest descent ==

In quantum field theory n-dimensional integrals of the form

<math display="block">\int_{-\infty}^{\infty} \exp\left( -{1 \over \hbar} f(q) \right ) d^nq</math>

appear often. Here <math>\hbar</math> is the [[reduced Planck constant]] and ''f'' is a function with a positive minimum at <math> q=q_0</math>. These integrals can be approximated by the [[method of steepest descent]].

For small values of the Planck constant, ''f'' can be expanded about its minimum

:<math>\int_{-\infty}^{\infty} \exp\left( -{1 \over \hbar} f(q) \right ) d^nq</math>

<math display="block">\int_{-\infty}^{\infty} \exp\left[ -{1 \over \hbar} \left( f\left( q_0 \right) + {1\over 2} \left( q-q_0\right)^2 f^{\prime \prime} \left( q-q_0\right) + \cdots \right ) \right] d^nq.</math>Here <math> f^{\prime \prime} </math> is the n by n matrix of second derivatives evaluated at the minimum of the function.

appear often. Here <math>\hbar</math> is the [[reduced Planck's constant]] and f is a function with a positive minimum at <math> q=q_0</math>. These integrals can be approximated by the [[method of steepest descent]].

For small values of Planck's constant, f can be expanded about its minimum

:<math>\int_{-\infty}^{\infty} \exp\left[ -{1 \over \hbar} \left( f\left( q_0 \right) + {1\over 2} \left( q-q_0\right)^2f^{\prime \prime} \left( q-q_0\right) + \cdots \right ) \right] d^nq</math>.

Here <math> f^{\prime \prime} </math> is the n by n matrix of second derivatives evaluated at the minimum of the function.

If we neglect higher order terms this integral can be integrated explicitly.

<math display="block">\int_{-\infty}^{\infty} \exp\left[ -{1 \over \hbar} (f(q)) \right] d^nq \approx \exp\left[ -{1 \over \hbar} \left( f\left( q_0 \right) \right ) \right] \sqrt{ (2 \pi \hbar )^n \over \det f^{\prime \prime} }.</math>

== Integrals that can be approximated by the method of stationary phase ==

:<math>\int_{-\infty}^{\infty} \exp\left[ -{1 \over \hbar} (f(q)) \right] d^nq \approx \exp\left[ -{1 \over \hbar} \left( f\left( q_0 \right) \right ) \right] \sqrt{ (2 \pi \hbar )^n \over \det f^{\prime \prime} }.</math>

==Integrals that can be approximated by the method of stationary phase==

A common integral is a path integral of the form

<math display="block"> \int \exp\left( {i \over \hbar} S\left( q, \dot q \right) \right ) Dq </math>

:<math> \int \exp\left( {i \over \hbar} S\left( q, \dot q \right) \right ) Dq </math>

where <math> S\left( q, \dot q \right) </math> is the classical [[Action (physics)|action]] and the integral is over all possible paths that a particle may take. In the limit of small <math> \hbar </math> the integral can be evaluated in the [[stationary phase approximation]]. In this approximation the integral is over the path in which the action is a minimum. Therefore, this approximation recovers the [[classical limit]] of [[classical mechanics|mechanics]].

== Fourier integrals ==

=== Dirac delta distribution ===

The [[Dirac delta distribution]] in [[spacetime]] can be written as a [[Fourier transform]]<ref name="Zee"/>Zee, {{rp|p. =23.</ref>}}

<math display="block"> \int \frac{d^4 k}{(2\pi)^4} \exp(ik ( x-y)) = \delta^4 ( x-y).</math>

:<math> \int \frac{d^4 k}{(2\pi)^4} \exp(ik ( x-y)) = \delta^4 ( x-y).</math>

In general, for any dimension <math> N </math>

<math display="block"> \int \frac{d^N k}{(2\pi)^N} \exp(ik ( x-y)) = \delta^N ( x-y).</math>

=== Fourier integrals of forms of the Coulomb potential ===

:<math> \int \frac{d^N k}{(2\pi)^N} \exp(ik ( x-y)) = \delta^N ( x-y).</math>

===Fourier integrals of forms of the Coulomb potential===

==== Laplacian of 1/''r'' ====

While not an integral, the identity in three-dimensional [[Euclidean space]]

<math display="block">-{1 \over 4\pi} \nabla^2 \left( {1 \over r} \right) = \delta \left( \mathbf r \right) </math>where<math display="block">r^2 = \mathbf r \cdot \mathbf r</math>is a consequence of [[Gauss's theorem]] and can be used to derive integral identities. For an example see [[Longitudinal and transverse vector fields]].

This identity implies that the [[Fourier integral]] representation of 1/''r'' is

:<math>-{1 \over 4\pi} \nabla^2 \left( {1 \over r} \right) = \delta \left( \mathbf r \right) </math>

<math display="block">\int \frac{d^3 k}{(2\pi)^3} { \exp \left ( i\mathbf k \cdot \mathbf r \right) \over k^2 } = {1 \over 4 \pi r }.</math>

==== Yukawa potential: the Coulomb potential with mass ====

The [[Yukawa potential]] in three dimensions can be represented as an integral over a [[Fourier transform]]<ref name="Zee"/>{{rp|p=26, 29}}

<math display="block">\int \frac{d^3 k}{(2\pi)^3} { \exp \left ( i\mathbf k \cdot \mathbf r \right) \over k^2 +m^2 } = {e^{-mr} \over 4 \pi r } </math>

where

<math display="block">r^2 = \mathbf{r} \cdot \mathbf r, \qquad k^2 = \mathbf k \cdot \mathbf k.</math>

:<math>r^2 = \mathbf r \cdot \mathbf r</math>

is a consequence of [[Gauss's theorem]] and can be used to derive integral identities. For an example see [[Longitudinal and transverse vector fields]].

This identity implies that the [[Fourier integral]] representation of 1/r is

:<math>\int \frac{d^3 k}{(2\pi)^3} { \exp \left ( i\mathbf k \cdot \mathbf r \right) \over k^2 } = {1 \over 4 \pi r }.</math>

====Yukawa Potential: The Coulomb potential with mass====

The [[Yukawa potential]] in three dimensions can be represented as an integral over a [[Fourier transform]]<ref>Zee, p. 26, 29.</ref>

:<math>\int \frac{d^3 k}{(2\pi)^3} { \exp \left ( i\mathbf k \cdot \mathbf r \right) \over k^2 +m^2 } = {e^{-mr} \over 4 \pi r } </math>

where

:<math>r^2 = \mathbf{r} \cdot \mathbf r, \qquad k^2 = \mathbf k \cdot \mathbf k.</math>

See [[Static forces and virtual-particle exchange#Selected examples|Static forces and virtual-particle exchange]] for an application of this integral.

Line 342 ⟶ 265:

To derive this result note:

<math display="block">\begin{align}

\int \frac{d^3 k}{(2\pi)^3} \frac{\exp \left (i \mathbf k \cdot \mathbf r\right)}{k^2 +m^2}

:<math>\begin{align}

\int \frac={d^3 k}{(2\pi)^3} \frac{\exp \left (i \mathbf k \cdot \mathbf r\right)}{k^2 +m^2} &= \int_0^{\infty} \frac{k^2 dk}{(2\pi)^2} \int_{-1}^1 du {e^{ikru}\over k^2 + m^2} \\[1ex]

&={}& {2\over r} \int_0^{\infty} \frac{k dk}{(2\pi)^2} {\sin(kr) \over k^2 + m^2} \\[1ex]

&={}& {1\over ir} \int_{-\infty}^{\infty} \frac{k dk}{(2\pi)^2} {e^{ikr} \over k^2 + m^2} \\[1ex]

&={}& {1\over ir} \int_{-\infty}^{\infty} \frac{k dk}{(2\pi)^2} {e^{ikr} \over (k + i m)(k - i m)} \\[1ex]

&={}& {1\over ir} \frac{2\pi i}{(2\pi)^2} \frac{im}{2im} e^{-mr} \\[1ex]

&={}& \frac{1}{4 \pi r} e^{-mr}

\end{align}</math>

==== Modified Coulomb potential with mass ====

:<math display="block">\int \frac{d^3 k}{(2\pi)^3} \left(\mathbf{\hat{k}}\cdot \mathbf{\hat{r}}\right)^2 \frac{\exp \left (i\mathbf{k} \cdot \mathbf{r} \right)}{k^2 +m^2} = \frac{e^{-mr}}{4 \pi r} \left\{[1 + \frac{2}{mr} - \frac{2}{(mr)^2} \left(e^{mr}-1 \right) \right \}]</math>

where the hat indicates a [[unit vector]] in three dimensional space. The derivation of this result is as follows:

<math display="block">\begin{align}

where the hat indicates a unit vector in three dimensional space. The derivation of this result is as follows:

&\int \frac{d^3 k}{(2\pi)^3} \left(\mathbf{\hat k}\cdot \mathbf{\hat r}\right)^2 \frac{\exp \left (i\mathbf{k}\cdot \mathbf{r}\right )}{k^2 +m^2} \\[1ex]

&= \int_0^{\infty} \frac{k^2 dk}{(2\pi)^2} \int_{-1}^{1} du \ u^2 \frac{e^{ikru}}{k^2 + m^2} \\[1ex]

:<math>\begin{align}

&= 2 \intint_0^{\infty} \frac{dk^32 kdk}{(2\pi)^32} \left(\mathbffrac{\hat k1}\cdot \mathbf{\hatk^2 r}\right)+ m^2 \frac{\exp} \left (i[\mathbffrac{k1}\cdot \mathbf{rkr} \right sin(kr)}{k^2 +m^2} &= \int_0^{\infty} \frac{k^2 dk}{(2\pikr)^2} \int_{cos(kr)-1}^{1} du u^2 \frac{e^{ikru}2}{k(kr)^23} +\sin(kr) m^2}\right] \\[1ex]

&= 2 \int_0frac{e^{\infty-mr} \frac{k^2 dk}{(24\pi)^2 r} \frac{left[1}{k^2 + m^2} \left\{\frac{12}{krmr} \sin(kr) +- \frac{2}{(krmr)^2} \cosleft(kr)- \frace^{2}{(kr)^3mr}-1 \sin(krright) \right \} \\]

&= \frac{e^{-mr}}{4\pi r} \left\{1 + \frac{2}{mr} - \frac{2}{(mr)^2} \left(e^{mr}-1 \right) \right \}

\end{align} </math>

Note that in the small {{mvar|m}} limit the integral goes to the result for the Coulomb potential since the term in the brackets goes to {{math|1}}.

==== Longitudinal potential with mass ====

:<math display="block">\int \frac{d^3 k}{(2\pi)^3} \mathbf{\hat{k}} \mathbf{\hat{k}} \frac{\exp \left ( i\mathbf{k} \cdot \mathbf{r} \right )}{k^2 +m^2 } = {1\over 2} \frac{e^{-mr}}{4\pi r} \left (\left[ \mathbf{1}- \mathbf{\hat{r}} \mathbf{\hat{r}} \right] + \left\{1 + \frac{2}{mr} - {2 \over (mr)^2} \left(e^{mr} -1 \right) \right \} \left[\mathbf{1}+ \mathbf{\hat{r}} \mathbf{\hat{r}}\right] \right ) </math>

where the hat indicates a unit vector in three dimensional space. The derivation for this result is as follows:

<math display="block">\begin{align}

& \int \frac{d^3 k}{(2\pi)^3} \mathbf{\hat k} \mathbf{\hat k} \frac{\exp \left (i\mathbf k \cdot \mathbf r \right)}{k^2 +m^2} \\[1ex]

:<math>\begin{align}

\int \frac{d^3 k}{(2\pi)^3} \mathbf{\hat k} \mathbf{\hat k} \frac{\exp \left (i\mathbf k \cdot \mathbf r \right)}{k^2 +m^2} &= \int \frac{d^3 k}{(2\pi)^3} \left[ \left( \mathbf{\hat k}\cdot \mathbf{\hat r}\right)^2\mathbf{\hat r} \mathbf{\hat r} + \left( \mathbf{\hat k}\cdot \mathbf{\hat \theta}\right)^2\mathbf{\hat \theta} \mathbf{\hat \theta} + \left( \mathbf{\hat k}\cdot \mathbf{\hat \phi}\right)^2\mathbf{\hat \phi} \mathbf{\hat \phi} \right] \frac{\exp \left (i\mathbf k \cdot \mathbf r \right )}{k^2 +m^2 } \\[1ex]

&= \frac{e^{-mr}}{4 \pi r}\left\{ 1+ \frac{2}{mr}- {2\over (mr)^2 } \left( e^{mr} -1 \right) \right \} \left\{\mathbf 1 - {1\over 2} \left[\mathbf 1 - \mathbf{\hat r} \mathbf{\hat r}\right] \right\} + \int_0^{\infty} \frac{k^2 dk}{(2\pi)^2 } \int_{-1}^{1} du \frac{e^{ikru}}{k^2 + m^2} {1\over 2} \left[ \mathbf 1 - \mathbf{\hat r} \mathbf{\hat r} \right] \\[1ex]

&= {1\over 2} \frac{e^{-mr}}{4 \pi r} \left[ \mathbf 1 - \mathbf{\hat r} \mathbf{\hat r} \right]+ {e^{-mr} \over 4 \pi r } \left\{ 1+\frac{2}{mr} - {2\over (mr)^2} \left( e^{mr} -1 \right) \right \} \left\{ {1\over 2} \left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right] \right\} \\[1ex]

&= {1\over 2} \frac{e^{-mr}}{4\pi r} \left (\left[ \mathbf{1}- \mathbf{\hat{r}} \mathbf{\hat{r}} \right] + \left\{1 + \frac{2}{mr} - {2 \over (mr)^2} \left(e^{mr} -1 \right) \right \} \left[\mathbf{1}+ \mathbf{\hat{r}} \mathbf{\hat{r}}\right] \right )

\end{align}</math>

Note that in the small {{mvar|m}} limit the integral reduces to

<math display="block">{1\over 2} {1 \over 4 \pi r } \left[ \mathbf 1 - \mathbf{\hat r} \mathbf{\hat r} \right]. </math>

==== Transverse potential with mass ====

:<math>{1\over 2} {1 \over 4 \pi r } \left[ \mathbf 1 - \mathbf{\hat r} \mathbf{\hat r} \right]. </math>

<math display="block">\int \frac{d^3 k}{(2\pi)^3} \left[\mathbf{1} - \mathbf{\hat{k}} \mathbf{\hat{k}} \right] { \exp \left ( i \mathbf{k} \cdot \mathbf{r}\right ) \over k^2 +m^2 } = {1\over 2} {e^{-mr} \over 4 \pi r} \left\{ {2 \over (mr)^2 } \left( e^{mr} -1 \right) - {2\over mr} \right \} \left[\mathbf{1} + \mathbf{\hat{r}} \mathbf{\hat{r}}\right]</math>

In the small ''mr'' limit the integral goes to

====Transverse potential with mass====

:<math display="block">\int \frac{d^3 k}{(2\pi)^3} \left[\mathbf{1} - \mathbf{\hat{k}} \mathbf{\hat{k}} \right] { \exp \left ( i \mathbf{k} \cdot \mathbf{r}\right ) \over k^2 +m^2 } = {1\over 2} {e^{-mr} \over 4 \pi r} \left\{ {2 \over (mr)^2 } \left( e^{mr} -1 \right) - {2\over mr} \right \} \left[\mathbf{ 1} + \mathbf{\hat{ r}} \mathbf{\hat{ r}}\right].</math>

For large distance, the integral falls off as the inverse cube of ''r''

In the small mr limit the integral goes to

<math display="block">\frac{1}{4 \pi m^2r^3 }\left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right].</math>

:<math>{1\over 2} {1 \over 4 \pi r } \left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right].</math>

For large distance, the integral falls off as the inverse cube of r

:<math>\frac{1}{4 \pi m^2r^3 }\left[\mathbf 1 + \mathbf{\hat r} \mathbf{\hat r}\right].</math>

For applications of this integral see [[Darwin Lagrangian]] and [[Static forces and virtual-particle exchange#Darwin interaction in a vacuum|Darwin interaction in a vacuum]].

==== Angular integration in cylindrical coordinates ====

There are two important integrals. The angular integration of an exponential in cylindrical coordinates can be written in terms of Bessel functions of the first kind<ref name="Zwillinger_2014">{{cite book |author-first1=Izrail Solomonovich |author-last1=Gradshteyn |author-link1=Izrail Solomonovich Gradshteyn |author-first2=Iosif Moiseevich |author-last2=Ryzhik |author-link2=Iosif Moiseevich Ryzhik |author-first3=Yuri Veniaminovich |author-last3=Geronimus |author-link3=Yuri Veniaminovich Geronimus |author-first4=Michail Yulyevich |author-last4=Tseytlin |author-link4=Michail Yulyevich Tseytlin |author-first5=Alan |author-last5=Jeffrey |editor-first1=Daniel |editor-last1=Zwillinger |editor-first2=Victor Hugo |editor-last2=Moll |editor-link2=Victor Hugo Moll |translator=Scripta Technica, Inc. |title=Table of Integrals, Series, and Products |publisher=[[Academic Press, Inc.]] |date=2015 |orig-year=October 2014 |edition=8 |language=English |isbn=0-12-384933-0 |id={{ISBN|978-0-12-384933-5}} |lccn=2014010276 <!-- |url=httphttps://books.google.com/books?id=NjnLAwAAQBAJ |access-date=2016-02-21 -->|title-link=Gradshteyn and Ryzhik <!-- |pages= was on page 402 and 679 in 1965 edition, but page numbers probably changed meanwhile -->}}</ref><ref name="Jackson">{{cite book|author last = Jackson, | first = John D.|title=Classical Electrodynamics (| edition = 3rd ed.)| publisher=Wiley| year=1998| isbn=0-471-30932-X}} p. 113</ref>{{rp|p=113}}

<math display="block">\int_0^{2 \pi} {d\varphi \over 2 \pi} \exp\left( i p \cos( \varphi) \right)=J_0 (p)</math>

:<math>\int_0^{2 \pi} {d\varphi \over 2 \pi} \exp\left( i p \cos( \varphi) \right)=J_0 (p)</math>

and

<math display="block">\int_0^{2 \pi} {d\varphi \over 2 \pi} \cos( \varphi) \exp\left( i p \cos( \varphi) \right) = i J_1 (p). </math>

:<math>\int_0^{2 \pi} {d\varphi \over 2 \pi} \cos( \varphi) \exp\left( i p \cos( \varphi) \right) = i J_1 (p). </math>

For applications of these integrals see [[Static forces and virtual-particle exchange#Magnetic interaction between current loops in a simple plasma or electron gas|Magnetic interaction between current loops in a simple plasma or electron gas]].

== Bessel functions ==

=== Integration of the cylindrical propagator with mass ===

==== First power of a Bessel function ====

:<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_0 \left( kr \right)=K_0 (mr). </math>

See [[Abramowitz and Stegun]].<ref name="AbramowitzStegun">{{cite book | authorauthor1=M. Abramowitz and| author2 = I. Stegun| title=Handbook of Mathematical Functions| publisher= Dover| year=1965 | isbn=0486-61272-4}}| Sectionurl-access=registration| 11url=https://archive.4.44org/details/handbookofmathe000abra}}</ref>{{rp|at=§11.4.44}}

For <math> mr <<\ll 1 </math>, we have<ref name="Jackson"/>Jackson, {{rp|p. =116</ref>}}

<math display="block">K_0 (mr) \to -\ln \left( {mr \over 2}\right) + 0.5772.</math>

:<math>K_0 (mr) \to -\ln \left( {mr \over 2}\right) + 0.5772.</math>

For an application of this integral see [[Static forces and virtual-particle exchange#Two line charges embedded in a plasma or electron gas|Two line charges embedded in a plasma or electron gas]].

==== Squares of Bessel functions ====

The integration of the propagator in cylindrical coordinates is<ref name="Zwillinger_2014"/>

<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) =I_1 (mr)K_1 (mr).</math>

:<math>\int_0^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) =I_1 (mr)K_1 (mr).</math>

For small mr the integral becomes

<math display="block">\int_o^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) \to {1\over 2 }\left[ 1 - {1\over 8} (mr)^2 \right].</math>

:<math>\int_o^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) \to {1\over 2 }\left[ 1 - {1\over 8} (mr)^2 \right].</math>

For large mr the integral becomes

<math display="block">\int_o^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) \to {1\over 2} \left( {1\over mr}\right).</math>

:<math>\int_o^{\infty} {k\; dk \over k^2 +m^2} J_1^2 (kr) \to {1\over 2} \left( {1\over mr}\right).</math>

For applications of this integral see [[Static forces and virtual-particle exchange#Magnetic interaction between current loops in a simple plasma or electron gas|Magnetic interaction between current loops in a simple plasma or electron gas]].

In general,

<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_{\nu}^2 (kr) = I_{\nu} (mr)K_{\nu} (mr) \qquad \Re (\nu) > -1. </math>

:<math>\int_0^{\infty} {k\; dk \over k^2 +m^2} J_{\nu}^2 (kr) = I_{\nu} (mr)K_{\nu} (mr) \qquad \Re (\nu) > -1. </math>

===Integration over a magnetic wave function===

The two-dimensional integral over a magnetic wave function is<ref>Abramowitz and Stegun, Section 11.4.28</ref>

=== Integration over a magnetic wave function ===

:<math>{2 a^{2n+2}\over n!} \int_0^{\infty} { dr }\;r^{2n+1}\exp\left( -a^2 r^2\right) J_{0} (kr) = M\left( n+1, 1, -{k^2 \over 4a^2}\right).</math>

The two-dimensional integral over a magnetic wave function is<ref name="AbramowitzStegun"/>{{rp|at=§11.4.28}}

<math display="block">{2 a^{2n+2}\over n!} \int_0^{\infty} { dr }\;r^{2n+1}\exp\left( -a^2 r^2\right) J_{0} (kr) = M\left( n+1, 1, -{k^2 \over 4a^2}\right).</math>

Here, ''M'' is a [[confluent hypergeometric function]]. For an application of this integral see [[Static forces and virtual-particle exchange#Charge density spread over a wave function|Charge density spread over a wave function]].

== See also ==

* [[Relation between Schrödinger's equation and the path integral formulation of quantum mechanics]]

== References ==

{{reflist}}

{{Reflist}}<!--added under references heading by script-assisted edit-->

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[[Category:Quantum field theory|*]]