Common integrals in quantum field theory

Template:SHORTDESC: Common integrals in quantum field theory are all variations and generalizations of Gaussian integrals to the complex plane and to multiple dimensions. Other integrals can be approximated by versions of the Gaussian integral. Fourier integrals are also considered.

Variations on a simple Gaussian integral

Gaussian integral

The first integral, with broad application outside of quantum field theory, is the Gaussian integral. $$G\equiv \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}{x}^2}dx$$

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

Note that, if we let $r=\sqrt{x^2+y^2}$ be the radius, then we can use the usual polar coordinate change of variables (which in particular renders $dxdy=rdrd\theta$) to get $$G^2=\left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}{x}^2}dx\right)\cdot \left(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}{y}^2}dy\right)=2\pi \underset{0}{\overset{\mathrm{\infty }}{\int }}r{e}^{-\frac{1}{2}{r}^2}dr=2\pi \underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-w}dw=2\pi .$$

Thus we obtain $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}{x}^2}dx=\sqrt{2\pi }.$$

Slight generalization of the Gaussian integral

$$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}a{x}^2}dx=\sqrt{\frac{2\pi }{a}}$$ where we have scaled $$x\to \frac{x}{\sqrt{a}}.$$

Integrals of exponents and even powers of x

$$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-\frac{1}{2}a{x}^2}dx=-2\frac{d}{da}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}a{x}^2}dx=-2\frac{d}{da}{\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\frac{1}{a}$$ and $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^4{e}^{-\frac{1}{2}a{x}^2}dx=(-2\frac{d}{da})(-2\frac{d}{da})\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{1}{2}a{x}^2}dx=(-2\frac{d}{da})(-2\frac{d}{da}){\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\frac{3}{a^2}$$

In general $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2n}{e}^{-\frac{1}{2}a{x}^2}dx={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\frac{1}{a^n}(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\frac{1}{a^n}(2n-1)!!$$

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

Integrals with a linear term in the argument of the exponent

$$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}a{x}^2+Jx)dx$$

This integral can be performed by completing the square: $$(-\frac{1}{2}a{x}^2+Jx)=-\frac{1}{2}a(x^2-\frac{2Jx}{a}+\frac{J^2}{a^2}-\frac{J^2}{a^2})=-\frac{1}{2}a{(x-\frac{J}{a})}^2+\frac{J^2}{2a}$$

Therefore: $$\begin{array}{cc}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}a{x}^2+Jx)dx& =\exp \left(\frac{J^2}{2a}\right)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp [-\frac{1}{2}a{(x-\frac{J}{a})}^2]dx\\ & =\exp \left(\frac{J^2}{2a}\right)\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}a{w}^2)dw\\ & ={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\exp \left(\frac{J^2}{2a}\right)\end{array}$$

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

The integral $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{2}a{x}^2+iJx)dx={\left(\frac{2\pi }{a}\right)}^{\frac{1}{2}}\exp (-\frac{J^2}{2a})$$ is proportional to the Fourier transform of the Gaussian where J is the conjugate variable of 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 a is, the narrower the Gaussian in x and the wider the Gaussian in 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) $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (\frac{1}{2}ia{x}^2+iJx)dx.$$

We now assume that a and J may be complex.

Completing the square $$(\frac{1}{2}ia{x}^2+iJx)=\frac{1}{2}ia(x^2+\frac{2Jx}{a}+{\left(\frac{J}{a}\right)}^2-{\left(\frac{J}{a}\right)}^2)=-\frac{1}{2}\frac{a}{i}{(x+\frac{J}{a})}^2-\frac{i{J}^2}{2a}.$$

By analogy with the previous integrals $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (\frac{1}{2}ia{x}^2+iJx)dx={\left(\frac{2\pi i}{a}\right)}^{\frac{1}{2}}\exp \left(\frac{-i{J}^2}{2a}\right).$$

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

Gaussian integrals in higher dimensions

The one-dimensional integrals can be generalized to multiple dimensions. $$\int \exp (-\frac{1}{2}x\cdot A\cdot x+J\cdot x)d^{n}x=\sqrt{\frac{(2\pi {)}^n}{\det A}}\exp (\frac{1}{2}J\cdot A^{-1}\cdot J)$$

Here A is a real positive definite symmetric matrix.

This integral is performed by diagonalization of A with an orthogonal transformation $$D=O^{-1}AO=O^{\text{T}}AO$$ where D is a diagonal matrix and O is an orthogonal matrix. This decouples the variables and allows the integration to be performed as 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 $$\int \exp (-\frac{1}{2}{A}_{ij}{x}^i{x}^j)d^{2}x=\sqrt{\frac{(2\pi {)}^2}{\det A}}$$ where A is a two-dimensional symmetric matrix with components specified as $$A=\left[\begin{array}{cc}a& c\\ c& b\end{array}\right]$$ and we have used the Einstein summation convention.

Diagonalize the matrix

The first step is to diagonalize the matrix. Note that $$A_{ij}{x}^i{x}^j\equiv x^{\text{T}}Ax=x^{\text{T}}\left(O{O}^{\text{T}}\right)A\left(O{O}^{\text{T}}\right)x=\left(x^{\text{T}}O\right)\left(O^{\text{T}}AO\right)\left(O^{\text{T}}x\right)$$ where, since A is a real symmetric matrix, we can choose O to be orthogonal, and hence also a unitary matrix. O can be obtained from the eigenvectors of A. We choose O such that: D ≡ O^TAO is diagonal.

Eigenvalues of A

To find the eigenvectors of A one first finds the eigenvalues λ of A given by $$\left[\begin{array}{cc}a& c\\ c& b\end{array}\right]\left[\begin{array}{c}u\\ v\end{array}\right]=\lambda \left[\begin{array}{c}u\\ v\end{array}\right].$$

The eigenvalues are solutions of the characteristic polynomial $$(a-\lambda )(b-\lambda )-c^2=0$$ $${\lambda }^2-\lambda (a+b)+ab-c^2=0,$$ which are found using the quadratic equation: $$\begin{array}{cc}{\lambda }_{\pm }& =\frac{1}{2}(a+b)\pm \frac{1}{2}\sqrt{(a+b{)}^2-4(ab-c^2)}.\\ & =\frac{1}{2}(a+b)\pm \frac{1}{2}\sqrt{a^2+2ab+b^2-4ab+4c^2}.\\ & =\frac{1}{2}(a+b)\pm \frac{1}{2}\sqrt{(a-b{)}^2+4c^2}.\end{array}$$

Eigenvectors of A

Substitution of the eigenvalues back into the eigenvector equation yields $$v=-\frac{(a-{\lambda }_{\pm })u}{c},v=-\frac{cu}{(b-{\lambda }_{\pm })}.$$

From the characteristic equation we know $$\frac{a-{\lambda }_{\pm }}{c}=\frac{c}{b-{\lambda }_{\pm }}.$$

Also note $$\frac{a-{\lambda }_{\pm }}{c}=-\frac{b-{\lambda }_{\mp }}{c}.$$

The eigenvectors can be written as: $$\left[\begin{array}{c}\frac{1}{\eta }\\ -\frac{a-{\lambda }_{-}}{c\eta }\end{array}\right],\left[\begin{array}{c}-\frac{b-{\lambda }_{+}}{c\eta }\\ \frac{1}{\eta }\end{array}\right]$$ for the two eigenvectors. Here η is a normalizing factor given by, $$\eta =\sqrt{1+{\left(\frac{a-{\lambda }_{-}}{c}\right)}^2}=\sqrt{1+{\left(\frac{b-{\lambda }_{+}}{c}\right)}^2}.$$

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

$$O=\left[\begin{array}{cc}\frac{1}{\eta }& -\frac{b-{\lambda }_{+}}{c\eta }\\ -\frac{a-{\lambda }_{-}}{c\eta }& \frac{1}{\eta }\end{array}\right].$$

Note that det(O) = 1.

If we define $$\sin (\theta )=-\frac{a-{\lambda }_{-}}{c\eta }$$ then the orthogonal matrix can be written $$O=\left[\begin{array}{cc}\cos (\theta )& -\sin (\theta )\\ \sin (\theta )& \cos (\theta )\end{array}\right]$$ which is simply a rotation of the eigenvectors with the inverse: $$O^{-1}=O^{\text{T}}=\left[\begin{array}{cc}\cos (\theta )& \sin (\theta )\\ -\sin (\theta )& \cos (\theta )\end{array}\right].$$

Diagonal matrix

The diagonal matrix becomes $$D=O^{\text{T}}AO=\left[\begin{array}{cc}{\lambda }_{-}& 0\\ 0& {\lambda }_{+}\end{array}\right]$$ with eigenvectors $$\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Numerical example

$$A=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$$

The eigenvalues are $${\lambda }_{\pm }=\frac{3}{2}\pm \frac{\sqrt{5}}{2}.$$

The eigenvectors are $$\frac{1}{\eta }\left[\begin{array}{c}1\\ -\frac{1}{2}-\frac{\sqrt{5}}{2}\end{array}\right],\frac{1}{\eta }\left[\begin{array}{c}\frac{1}{2}+\frac{\sqrt{5}}{2}\\ 1\end{array}\right]$$ where $$\eta =\sqrt{\frac{5}{2}+\frac{\sqrt{5}}{2}}.$$

Then $$\begin{array}{cc}O& =\left[\begin{array}{cc}\frac{1}{\eta }& \frac{1}{\eta }(\frac{1}{2}+\frac{\sqrt{5}}{2})\\ \frac{1}{\eta }(-\frac{1}{2}-\frac{\sqrt{5}}{2})& \frac{1}{\eta }\end{array}\right]\\ O^{-1}& =\left[\begin{array}{cc}\frac{1}{\eta }& \frac{1}{\eta }(-\frac{1}{2}-\frac{\sqrt{5}}{2})\\ \frac{1}{\eta }(\frac{1}{2}+\frac{\sqrt{5}}{2})& \frac{1}{\eta }\end{array}\right]\end{array}$$

The diagonal matrix becomes $$D=O^{\text{T}}AO=\left[\begin{array}{cc}{\lambda }_{-}& 0\\ 0& {\lambda }_{+}\end{array}\right]=\left[\begin{array}{cc}\frac{3}{2}-\frac{\sqrt{5}}{2}& 0\\ 0& \frac{3}{2}+\frac{\sqrt{5}}{2}\end{array}\right]$$ with eigenvectors $$\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]$$

Rescale the variables and integrate

With the diagonalization the integral can be written $$\int \exp (-\frac{1}{2}{x}^{\text{T}}Ax)d^{2}x=\int \exp (-\frac{1}{2}\sum _{j=1}^2{\lambda }_j{y}_j^2)d^{2}y$$ where $$y=O^{\text{T}}x.$$

Since the coordinate transformation is simply a rotation of coordinates the Jacobian determinant of the transformation is one yielding $$d^{2}y=d^{2}x$$

The integrations can now be performed: $$\begin{array}{cc}\int \exp (-\frac{1}{2}{x}^{𝖳}Ax)d^{2}x=& \int \exp (-\frac{1}{2}\sum _{j=1}^2{\lambda }_j{y}_j^2)d^{2}y\\ =& \prod _{j=1}^2{\left(\frac{2\pi }{{\lambda }_j}\right)}^{1/2}\\ =& {\left(\frac{(2\pi {)}^2}{\prod _{j=1}^2{\lambda }_j}\right)}^{1/2}\\ =& {\left(\frac{(2\pi {)}^2}{\det \left(O^{-1}AO\right)}\right)}^{1/2}\\ =& {\left(\frac{(2\pi {)}^2}{\det \left(A\right)}\right)}^{1/2}\end{array}$$ 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

$$\int \exp (-\frac{1}{2}{x}^T\cdot A\cdot x+J^T\cdot x)dx=\sqrt{\frac{(2\pi {)}^n}{\det A}}\exp (\frac{1}{2}{J}^T\cdot A^{-1}\cdot J)$$

Integrals with an imaginary linear term

$$\int \exp (-\frac{1}{2}{x}^T\cdot A\cdot x+i{J}^T\cdot x)dx=\sqrt{\frac{(2\pi {)}^n}{\det A}}\exp (-\frac{1}{2}{J}^T\cdot A^{-1}\cdot J)$$

Integrals with a complex quadratic term

$$\int \exp (\frac{i}{2}{x}^T\cdot A\cdot x+i{J}^T\cdot x)dx=\sqrt{\frac{(2\pi i{)}^n}{\det A}}\exp (-\frac{i}{2}{J}^T\cdot A^{-1}\cdot J)$$

Integrals with differential operators in the argument

As an example consider the integral $$\int \exp [\int d^{4}x(-\frac{1}{2}\phi \hat{A}\phi +J\phi )]D\phi$$ where $\hat{A}$ is a differential operator with $\phi$ and J functions of spacetime, and $D\phi$ indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is $$\int \exp [\int d^{4}x(-\frac{1}{2}\phi \hat{A}\phi +J\phi )]D\phi \propto \exp (\frac{1}{2}\int d^{4}x{d}^{4}yJ(x)D(x-y)J(y))$$ where $$\hat{A}D(x-y)={\delta }^4(x-y)$$ and D(x − y), called the propagator, is the inverse of $\hat{A}$, and ${\delta }^4(x-y)$ is the Dirac delta function.

Similar arguments yield $$\int \exp [\int d^{4}x(-\frac{1}{2}\phi \hat{A}\phi +iJ\phi )]D\phi \propto \exp (-\frac{1}{2}\int d^{4}x{d}^{4}yJ(x)D(x-y)J(y)),$$ and $$\int \exp [i\int d^{4}x(\frac{1}{2}\phi \hat{A}\phi +J\phi )]D\phi \propto \exp (-\frac{i}{2}\int d^{4}x{d}^{4}yJ(x)D(x-y)J(y)).$$

See 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 $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp (-\frac{1}{\hslash }f(q))d^{n}q$$ appear often. Here $\hslash$ is the reduced Planck constant and f is a function with a positive minimum at $q=q_0$. 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 $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp [-\frac{1}{\hslash }(f\left(q_0\right)+\frac{1}{2}{(q-q_0)}^2{f}^{\prime \prime }(q-q_0)+\cdots )]d^{n}q.$$Here $f^{\prime \prime }$ 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. $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\exp [-\frac{1}{\hslash }(f(q))]d^{n}q\approx \exp [-\frac{1}{\hslash }\left(f\left(q_0\right)\right)]\sqrt{\frac{(2\pi \hslash {)}^n}{\det f^{\prime \prime }}}.$$

Integrals that can be approximated by the method of stationary phase

A common integral is a path integral of the form $$\int \exp \left(\frac{i}{\hslash }S(q,\dot{q})\right)Dq$$ where $S(q,\dot{q})$ is the classical action and the integral is over all possible paths that a particle may take. In the limit of small $\hslash$ 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 mechanics.

Fourier integrals

Dirac delta distribution

The Dirac delta distribution in spacetime can be written as a Fourier transform $$\int \frac{d^{4}k}{(2\pi {)}^4}\exp (ik(x-y))={\delta }^4(x-y).$$

In general, for any dimension $N$ $$\int \frac{d^{N}k}{(2\pi {)}^N}\exp (ik(x-y))={\delta }^N(x-y).$$

Fourier integrals of forms of the Coulomb potential

Laplacian of 1/r

While not an integral, the identity in three-dimensional Euclidean space $$-\frac{1}{4\pi }{\mathrm{\nabla }}^2\left(\frac{1}{r}\right)=\delta \left(𝐫\right)$$where$$r^2=𝐫\cdot 𝐫$$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 $$\int \frac{d^{3}k}{(2\pi {)}^3}\frac{\exp (i𝐤\cdot 𝐫)}{k^2}=\frac{1}{4\pi r}.$$

Yukawa potential: the Coulomb potential with mass

The Yukawa potential in three dimensions can be represented as an integral over a Fourier transform $$\int \frac{d^{3}k}{(2\pi {)}^3}\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}=\frac{e^{-mr}}{4\pi r}$$ where $$r^2=𝐫\cdot 𝐫,k^2=𝐤\cdot 𝐤.$$

See Static forces and virtual-particle exchange for an application of this integral.

In the small m limit the integral reduces to ⁠1/4πr⁠.

To derive this result note: $$\begin{array}{cc}\int \frac{d^{3}k}{(2\pi {)}^3}\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}=& \underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{k^{2}dk}{(2\pi {)}^2}\underset{-1}{\overset{1}{\int }}du\frac{e^{ikru}}{k^2+m^2}\\ =& \frac{2}{r}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{(2\pi {)}^2}\frac{\sin (kr)}{k^2+m^2}\\ =& \frac{1}{ir}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{(2\pi {)}^2}\frac{e^{ikr}}{k^2+m^2}\\ =& \frac{1}{ir}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{(2\pi {)}^2}\frac{e^{ikr}}{(k+im)(k-im)}\\ =& \frac{1}{ir}\frac{2\pi i}{(2\pi {)}^2}\frac{im}{2im}{e}^{-mr}\\ =& \frac{1}{4\pi r}{e}^{-mr}\end{array}$$

Modified Coulomb potential with mass

$$\int \frac{d^{3}k}{(2\pi {)}^3}{(\widehat{𝐤}\cdot \widehat{𝐫})}^2\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}=\frac{e^{-mr}}{4\pi r}[1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)]$$ where the hat indicates a unit vector in three dimensional space. The derivation of this result is as follows: $$\begin{array}{cc}& \int \frac{d^{3}k}{(2\pi {)}^3}{(\widehat{𝐤}\cdot \widehat{𝐫})}^2\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}\\ & =\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{k^{2}dk}{(2\pi {)}^2}\underset{-1}{\overset{1}{\int }}du{u}^2\frac{e^{ikru}}{k^2+m^2}\\ & =2\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{k^{2}dk}{(2\pi {)}^2}\frac{1}{k^2+m^2}[\frac{1}{kr}\sin (kr)+\frac{2}{(kr{)}^2}\cos (kr)-\frac{2}{(kr{)}^3}\sin (kr)]\\ & =\frac{e^{-mr}}{4\pi r}[1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)]\end{array}$$

Note that in the small m limit the integral goes to the result for the Coulomb potential since the term in the brackets goes to 1.

Longitudinal potential with mass

$$\int \frac{d^{3}k}{(2\pi {)}^3}\widehat{𝐤}\widehat{𝐤}\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}=\frac{1}{2}\frac{e^{-mr}}{4\pi r}([\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}]+\{1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)\}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}])$$ where the hat indicates a unit vector in three dimensional space. The derivation for this result is as follows: $$\begin{array}{cc}& \int \frac{d^{3}k}{(2\pi {)}^3}\widehat{𝐤}\widehat{𝐤}\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}\\ & =\int \frac{d^{3}k}{(2\pi {)}^3}[{(\widehat{𝐤}\cdot \widehat{𝐫})}^2\widehat{𝐫}\widehat{𝐫}+{(\widehat{𝐤}\cdot \widehat{𝜽})}^2\widehat{𝜽}\widehat{𝜽}+{(\widehat{𝐤}\cdot \widehat{𝝓})}^2\widehat{𝝓}\widehat{𝝓}]\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}\\ & =\frac{e^{-mr}}{4\pi r}\{1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)\}\{\mathrm{𝟏}-\frac{1}{2}[\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}]\}+\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{k^{2}dk}{(2\pi {)}^2}\underset{-1}{\overset{1}{\int }}du\frac{e^{ikru}}{k^2+m^2}\frac{1}{2}[\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}]\\ & =\frac{1}{2}\frac{e^{-mr}}{4\pi r}[\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}]+\frac{e^{-mr}}{4\pi r}\{1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)\}\left\{\frac{1}{2}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}]\right\}\\ & =\frac{1}{2}\frac{e^{-mr}}{4\pi r}([\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}]+\{1+\frac{2}{mr}-\frac{2}{(mr{)}^2}(e^{mr}-1)\}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}])\end{array}$$

Note that in the small m limit the integral reduces to $$\frac{1}{2}\frac{1}{4\pi r}[\mathrm{𝟏}-\widehat{𝐫}\widehat{𝐫}].$$

Transverse potential with mass

$$\int \frac{d^{3}k}{(2\pi {)}^3}[\mathrm{𝟏}-\widehat{𝐤}\widehat{𝐤}]\frac{\exp (i𝐤\cdot 𝐫)}{k^2+m^2}=\frac{1}{2}\frac{e^{-mr}}{4\pi r}\{\frac{2}{(mr{)}^2}(e^{mr}-1)-\frac{2}{mr}\}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}]$$

In the small mr limit the integral goes to $$\frac{1}{2}\frac{1}{4\pi r}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}].$$

For large distance, the integral falls off as the inverse cube of r $$\frac{1}{4\pi m^2{r}^3}[\mathrm{𝟏}+\widehat{𝐫}\widehat{𝐫}].$$

For applications of this integral see Darwin Lagrangian and 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 $$\underset{0}{\overset{2\pi }{\int }}\frac{d\phi }{2\pi }\exp (ip\cos (\phi ))=J_0(p)$$ and $$\underset{0}{\overset{2\pi }{\int }}\frac{d\phi }{2\pi }\cos (\phi )\exp (ip\cos (\phi ))=i{J}_1(p).$$

For applications of these integrals see 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

$$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{k^2+m^2}{J}_0\left(kr\right)=K_0(mr).$$

See Abramowitz and Stegun.

For $mr\ll 1$, we have $$K_0(mr)\to -\ln \left(\frac{mr}{2}\right)+\mathrm{0.5772.}$$

For an application of this integral see Two line charges embedded in a plasma or electron gas.

Squares of Bessel functions

The integration of the propagator in cylindrical coordinates is $$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{k^2+m^2}{J}_1^2(kr)=I_1(mr)K_1(mr).$$

For small mr the integral becomes $$\underset{o}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{k^2+m^2}{J}_1^2(kr)\to \frac{1}{2}[1-\frac{1}{8}(mr{)}^2].$$

For large mr the integral becomes $$\underset{o}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{k^2+m^2}{J}_1^2(kr)\to \frac{1}{2}\left(\frac{1}{mr}\right).$$

For applications of this integral see Magnetic interaction between current loops in a simple plasma or electron gas.

In general, $$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{kdk}{k^2+m^2}{J}_{\nu }^2(kr)=I_{\nu }(mr)K_{\nu }(mr)\mathrm{\Re }(\nu )>-1.$$

Integration over a magnetic wave function

The two-dimensional integral over a magnetic wave function is $$\frac{2a^{2n+2}}{n!}\underset{0}{\overset{\mathrm{\infty }}{\int }}dr{r}^{2n+1}\exp (-a^2{r}^2)J_0(kr)=M(n+1,1,-\frac{k^2}{4a^2}).$$

Here, M is a confluent hypergeometric function. For an application of this integral see Charge density spread over a wave function.

See also

• Relation between Schrödinger's equation and the path integral formulation of quantum mechanics