Common integrals in quantum field theory: Difference between revisions

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===Gaussian integral===

The first integral, with broad application outside of quantum field theory, is the Gaussian 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:

:<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>

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

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

:<math display="block"> \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>.

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

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

:<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>

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.

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===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>

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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}}.

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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>

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

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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>

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]].

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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>

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

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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^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.

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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>

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>

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

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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^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.

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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>

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

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

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

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

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

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

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

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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>

From the characteristic equation we know

:<math display="block"> {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>

The eigenvectors can be written as:

:<math display="block">\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>

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>

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

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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}}.

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If we define

:<math display="block"> \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>

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

:<math display="block">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^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 display="block">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>

The eigenvectors are

:<math display="block">{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 }\\ 1 \end{bmatrix}</math>

where

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

Then

:<math display="block">\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}

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The diagonal matrix becomes

:<math display="block">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====

With the diagonalization the integral can be written

:<math display="block">\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^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"> dy^2 = dx^2 </math>

The integrations can now be performed.

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

\int \exp\left( - \frac{1}{2} x^\mathsf{T} A x \right) d^2x &
={}& \int \exp\left( - \frac 1 2 \sum_{j=1}^2 \lambda_{j} y_j^2 \right) d^2y \\

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

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

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

:<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>

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

:<math display="block">\int \exp\left(-\frac{1}{2} x \cdot A \cdot x +iJ \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>

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

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

where

:<math display="block">\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]].

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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>

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>

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.

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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'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]].

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

:<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)^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.

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==

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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>

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]].

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The [[Dirac delta distribution]] in [[spacetime]] can be written as a [[Fourier transform]]<ref>Zee, p. 23.</ref>

:<math display="block"> \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===

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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]].

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

:<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>Zee, p. 26, 29.</ref>

:<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>

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

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

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

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

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

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

&={}& \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}

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

&={}& 2 \int_0^{\infty} \frac{k^2 dk}{(2\pi)^2} \frac{1}{k^2 + m^2} \left\{\frac{1}{kr} \sin(kr) + \frac{2}{(kr)^2} \cos(kr)- \frac{2}{(kr)^3} \sin(kr) \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>

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====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} &\\
={}& \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 } \\

&={}&\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] \\

&={}&{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\} \\

&={}&{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 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

:<math display="block">{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 display="block">\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]].

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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=978-0-12-384933-5 |lccn=2014010276 <!-- |url=https://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>{{cite book|author=Jackson, John D.|title=Classical Electrodynamics (3rd ed.)| publisher=Wiley| year=1998| isbn=0-471-30932-X}} p. 113</ref>

:<math display="block">\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>

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]].

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====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>{{cite book| author=M. Abramowitz and I. Stegun| title=Handbook of Mathematical Functions| publisher=Dover| year=1965| isbn=0486-61272-4| url-access=registration| url=https://archive.org/details/handbookofmathe000abra}} Section 11.4.44</ref>

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For <math> mr \ll 1 </math>, we have<ref>Jackson, p. 116</ref>

:<math display="block">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]].

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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>

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>

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>

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]].

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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>

===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>

:<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]].