Common integrals in quantum field theory

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

${\displaystyle G\equiv \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx}$

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

Note:

${\displaystyle G^{2}=\left(\int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx\right)\cdot \left(\int _{-\infty }^{\infty }e^{-{1 \over 2}y^{2}}\,dy\right)=2\pi \int _{0}^{\infty }re^{-{1 \over 2}r^{2}}\,dr=2\pi \int _{0}^{\infty }e^{-w}\,dw=2\pi .}$

Thus we obtain

${\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx={\sqrt {2\pi }}.}$

${\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx={\sqrt {2\pi \over a}}}$

where we have scaled

${\displaystyle x\to {x \over {\sqrt {a}}}}$.

${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-{1 \over 2}ax^{2}}\,dx=-2{d \over da}\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx=-2{d \over da}\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{1 \over a}}$

and

${\displaystyle \int _{-\infty }^{\infty }x^{4}e^{-{1 \over 2}ax^{2}}\,dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{3 \over a^{2}}}$

In general

${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-{1 \over 2}ax^{2}}\,dx=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)\left(2n-3\right)\cdots 5\cdot 3\cdot 1=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)!!}$

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

${\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+Jx\right)dx}$

This integral can be performed by completing the square:

${\displaystyle \left(-{1 \over 2}ax^{2}+Jx\right)=-{1 \over 2}a\left(x^{2}-{2Jx \over a}+{J^{2} \over a^{2}}-{J^{2} \over a^{2}}\right)=-{1 \over 2}a\left(x-{J \over a}\right)^{2}+{J^{2} \over 2a}}$

Therefore:

${\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+Jx\right)\,dx&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left[-{1 \over 2}a\left(x-{J \over a}\right)^{2}\right]\,dx\\[8pt]&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}aw^{2}\right)\,dw\\[8pt]&=\left({2\pi \over a}\right)^{1 \over 2}\exp \left({J^{2} \over 2a}\right)\end{aligned}}}$

The integral

${\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+iJx\right)dx=\left({2\pi \over a}\right)^{1 \over 2}\exp \left(-{J^{2} \over 2a}\right)}$

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.

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)

${\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx.}$

We now assume that a and J may be complex.

Completing the square

${\displaystyle \left({1 \over 2}iax^{2}+iJx\right)={1 \over 2}ia\left(x^{2}+{2Jx \over a}+\left({J \over a}\right)^{2}-\left({J \over a}\right)^{2}\right)=-{1 \over 2}{a \over i}\left(x+{J \over a}\right)^{2}-{iJ^{2} \over 2a}.}$

By analogy with the previous integrals

${\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx=\left({2\pi i \over a}\right)^{1 \over 2}\exp \left({-iJ^{2} \over 2a}\right).}$

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.

The Gaussian integral in two dimensions is

${\displaystyle \int \exp \left(-{\frac {1}{2}}A_{ij}x^{i}x^{j}\right)d^{2}x={\sqrt {\frac {(2\pi )^{2}}{\det A}}}}$

where A is a two-dimensional symmetric matrix with components specified as

${\displaystyle A={\begin{bmatrix}a&c\\c&b\end{bmatrix}}}$

and we have used the Einstein summation convention.

The first step is to diagonalize the matrix.[3] Note that

${\displaystyle A_{ij}x^{i}x^{j}\equiv x^{T}Ax=x^{T}\left(OO^{T}\right)A\left(OO^{T}\right)x=\left(x^{T}O\right)\left(O^{T}AO\right)\left(O^{T}x\right)}$

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.

To find the eigenvectors of A one first finds the eigenvalues λ of A given by

${\displaystyle {\begin{bmatrix}a&c\\c&b\end{bmatrix}}{\begin{bmatrix}u\\v\end{bmatrix}}=\lambda {\begin{bmatrix}u\\v\end{bmatrix}}.}$

The eigenvalues are solutions of the characteristic polynomial

${\displaystyle (a-\lambda )(b-\lambda )-c^{2}=0}$

${\displaystyle \lambda ^{2}-\lambda (a+b)+ab-c^{2}=0}$

which are found using the quadratic equation:

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

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

${\displaystyle \lambda _{\pm }={1 \over 2}(a+b)\pm {1 \over 2}{\sqrt {(a-b)^{2}+4c^{2}}}.}$

Substitution of the eigenvalues back into the eigenvector equation yields

${\displaystyle v=-{\left(a-\lambda _{\pm }\right)u \over c},\qquad v=-{cu \over \left(b-\lambda _{\pm }\right)}.}$

From the characteristic equation we know

${\displaystyle {a-\lambda _{\pm } \over c}={c \over b-\lambda _{\pm }}.}$

Also note

${\displaystyle {a-\lambda _{\pm } \over c}=-{b-\lambda _{\mp } \over c}.}$

The eigenvectors can be written as:

${\displaystyle {\begin{bmatrix}{\frac {1}{\eta }}\\-{\frac {a-\lambda _{-}}{c\eta }}\end{bmatrix}},\qquad {\begin{bmatrix}-{\frac {b-\lambda _{+}}{c\eta }}\\{\frac {1}{\eta }}\end{bmatrix}}}$

for the two eigenvectors. Here η is a normalizing factor given by

${\displaystyle \eta ={\sqrt {1+\left({\frac {a-\lambda _{-}}{c}}\right)^{2}}}={\sqrt {1+\left({\frac {b-\lambda _{+}}{c}}\right)^{2}}}.}$

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

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

${\displaystyle O={\begin{bmatrix}{\frac {1}{\eta }}&-{\frac {b-\lambda _{+}}{c\eta }}\\-{\frac {a-\lambda _{-}}{c\eta }}&{\frac {1}{\eta }}\end{bmatrix}}.}$

Note that det(O) = 1.

If we define

${\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}}$

then the orthogonal matrix can be written

${\displaystyle O={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}}$

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

${\displaystyle O^{-1}=O^{T}={\begin{bmatrix}\cos(\theta )&\sin(\theta )\\-\sin(\theta )&\cos(\theta )\end{bmatrix}}.}$

The diagonal matrix becomes

${\displaystyle D=O^{T}AO={\begin{bmatrix}\lambda _{-}&0\\0&\lambda _{+}\end{bmatrix}}}$

with eigenvectors

${\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}$

${\displaystyle A={\begin{bmatrix}2&1\\1&1\end{bmatrix}}}$

The eigenvalues are

${\displaystyle \lambda _{\pm }={3 \over 2}\pm {{\sqrt {5}} \over 2}.}$

The eigenvectors are

${\displaystyle {1 \over \eta }{\begin{bmatrix}1\\-{1 \over 2}-{{\sqrt {5}} \over 2}\end{bmatrix}},\qquad {1 \over \eta }{\begin{bmatrix}{1 \over 2}+{{\sqrt {5}} \over 2}\\1\end{bmatrix}}}$

where

${\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.}$

Then

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

The diagonal matrix becomes

${\displaystyle D=O^{T}AO={\begin{bmatrix}\lambda _{-}&0\\0&\lambda _{+}\end{bmatrix}}={\begin{bmatrix}{3 \over 2}-{{\sqrt {5}} \over 2}&0\\0&{3 \over 2}+{{\sqrt {5}} \over 2}\end{bmatrix}}}$

with eigenvectors

${\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}$

With the diagonalization the integral can be written

${\displaystyle \int \exp \left(-{\frac {1}{2}}x^{T}Ax\right)d^{2}x=\int \exp \left(-{\frac {1}{2}}\sum _{j=1}^{2}\lambda _{j}y_{j}^{2}\right)\,d^{2}y}$

where

${\displaystyle y=O^{T}x.}$

Since the coordinate transformation is simply a rotation of coordinates the Jacobian determinant of the transformation is one yielding

${\displaystyle dy^{2}=dx^{2}}$

The integrations can now be performed.

${\displaystyle {\begin{aligned}\int \exp \left(-{\frac {1}{2}}x^{T}Ax\right)d^{2}x&=\int \exp \left(-{\frac {1}{2}}\sum _{j=1}^{2}\lambda _{j}y_{j}^{2}\right)d^{2}y\\&=\prod _{j=1}^{2}\left({2\pi \over \lambda _{j}}\right)^{1 \over 2}\\&=\left({(2\pi )^{2} \over \prod _{j=1}^{2}\lambda _{j}}\right)^{1 \over 2}\\&=\left({(2\pi )^{2} \over \det {\left(O^{-1}AO\right)}}\right)^{1 \over 2}\\&=\left({(2\pi )^{2} \over \det {\left(A\right)}}\right)^{1 \over 2}\end{aligned}}}$

which is the advertised solution.

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

${\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+J\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left({1 \over 2}J\cdot A^{-1}\cdot J\right)}$

${\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+iJ\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left(-{1 \over 2}J\cdot A^{-1}\cdot J\right)}$

${\displaystyle \int \exp \left({\frac {i}{2}}x\cdot A\cdot x+iJ\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi i)^{n}}{\det A}}}\exp \left(-{i \over 2}J\cdot A^{-1}\cdot J\right)}$

As an example consider the integral[4]

${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi }$

where ${\displaystyle {\hat {A}}}$ is a differential operator with ${\displaystyle \varphi }$ and J functions of spacetime, and ${\displaystyle D\varphi }$ indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is

${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi \;\propto \;\exp \left({1 \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right)}$

where

${\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)}$

and D(x − y), called the propagator, is the inverse of ${\displaystyle {\hat {A}}}$, and ${\displaystyle \delta ^{4}(x-y)}$ is the Dirac delta function.

Similar arguments yield

${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +iJ\varphi \right)\right]D\varphi \;\propto \;\exp \left(-{1 \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right),}$

and

${\displaystyle \int \exp \left[i\int d^{4}x\left({\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi \;\propto \;\exp \left(-{i \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right).}$

See Path-integral formulation of virtual-particle exchange for an application of this integral.

The Dirac delta distribution in spacetime can be written as a Fourier transform[5]

${\displaystyle \int {\frac {d^{4}k}{(2\pi )^{4}}}\exp(ik(x-y))=\delta ^{4}(x-y).}$

In general, for any dimension ${\displaystyle N}$

${\displaystyle \int {\frac {d^{N}k}{(2\pi )^{N}}}\exp(ik(x-y))=\delta ^{N}(x-y).}$

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

${\displaystyle -{1 \over 4\pi }\nabla ^{2}\left({1 \over r}\right)=\delta \left(\mathbf {r} \right)}$

where

${\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} }$

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

${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}}={1 \over 4\pi r}.}$

The Yukawa potential in three dimensions can be represented as an integral over a Fourier transform[6]

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

where

${\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} ,\qquad k^{2}=\mathbf {k} \cdot \mathbf {k} .}$

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:

${\displaystyle {\begin{aligned}\int {\frac {d^{3}k}{(2\pi )^{3}}}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}&=\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du{e^{ikru} \over k^{2}+m^{2}}\\&={2 \over r}\int _{0}^{\infty }{\frac {kdk}{(2\pi )^{2}}}{\sin(kr) \over k^{2}+m^{2}}\\&={1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over k^{2}+m^{2}}\\&={1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over (k+im)(k-im)}\\&={1 \over ir}{\frac {2\pi i}{(2\pi )^{2}}}{\frac {im}{2im}}e^{-mr}\\&={\frac {1}{4\pi r}}e^{-mr}\end{aligned}}}$

${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={\frac {e^{-mr}}{4\pi r}}\left\{1+{\frac {2}{mr}}-{\frac {2}{(mr)^{2}}}\left(e^{mr}-1\right)\right\}}$

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

${\displaystyle {\begin{aligned}\int {\frac {d^{3}k}{(2\pi )^{3}}}\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}&=\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}duu^{2}{\frac {e^{ikru}}{k^{2}+m^{2}}}\\&=2\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}{\frac {1}{k^{2}+m^{2}}}\left\{{\frac {1}{kr}}\sin(kr)+{\frac {2}{(kr)^{2}}}\cos(kr)-{\frac {2}{(kr)^{3}}}\sin(kr)\right\}\\&={\frac {e^{-mr}}{4\pi r}}\left\{1+{\frac {2}{mr}}-{\frac {2}{(mr)^{2}}}\left(e^{mr}-1\right)\right\}\end{aligned}}}$

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.

${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\mathbf {\hat {k}} \mathbf {\hat {k}} {\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left(\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right)}$

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

${\displaystyle {\begin{aligned}\int {\frac {d^{3}k}{(2\pi )^{3}}}\mathbf {\hat {k}} \mathbf {\hat {k}} {\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}&=\int {\frac {d^{3}k}{(2\pi )^{3}}}\left[\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {r}} \right)^{2}\mathbf {\hat {r}} \mathbf {\hat {r}} +\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {\theta }} \right)^{2}\mathbf {\hat {\theta }} \mathbf {\hat {\theta }} +\left(\mathbf {\hat {k}} \cdot \mathbf {\hat {\phi }} \right)^{2}\mathbf {\hat {\phi }} \mathbf {\hat {\phi }} \right]{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}\\&={\frac {e^{-mr}}{4\pi r}}\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left\{\mathbf {1} -{1 \over 2}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right\}+\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du{\frac {e^{ikru}}{k^{2}+m^{2}}}{1 \over 2}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\\&={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+{e^{-mr} \over 4\pi r}\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left\{{1 \over 2}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right\}\\&={1 \over 2}{\frac {e^{-mr}}{4\pi r}}\left(\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right]+\left\{1+{\frac {2}{mr}}-{2 \over (mr)^{2}}\left(e^{mr}-1\right)\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\right)\end{aligned}}}$

Note that in the small m limit the integral reduces to

${\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}$

${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\left[\mathbf {1} -\mathbf {\hat {k}} \mathbf {\hat {k}} \right]{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}+m^{2}}={1 \over 2}{e^{-mr} \over 4\pi r}\left\{{2 \over (mr)^{2}}\left(e^{mr}-1\right)-{2 \over mr}\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]}$

In the small mr limit the integral goes to

${\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}$

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

${\displaystyle {\frac {1}{4\pi m^{2}r^{3}}}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}$

For applications of this integral see Darwin Lagrangian and Darwin interaction in a vacuum.

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[7][8]

${\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\exp \left(ip\cos(\varphi )\right)=J_{0}(p)}$

and

${\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\cos(\varphi )\exp \left(ip\cos(\varphi )\right)=iJ_{1}(p).}$

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

${\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{0}\left(kr\right)=K_{0}(mr).}$

See Abramowitz and Stegun.[9]

For ${\displaystyle mr\ll 1}$, we have[10]

${\displaystyle K_{0}(mr)\to -\ln \left({mr \over 2}\right)+0.5772.}$

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

The integration of the propagator in cylindrical coordinates is[7]

${\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)=I_{1}(mr)K_{1}(mr).}$

For small mr the integral becomes

${\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left[1-{1 \over 8}(mr)^{2}\right].}$

For large mr the integral becomes

${\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left({1 \over mr}\right).}$

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

In general

${\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{\nu }^{2}(kr)=I_{\nu }(mr)K_{\nu }(mr)\qquad \Re (\nu )>-1.}$

The two-dimensional integral over a magnetic wave function is[11]

${\displaystyle {2a^{2n+2} \over n!}\int _{0}^{\infty }{dr}\;r^{2n+1}\exp \left(-a^{2}r^{2}\right)J_{0}(kr)=M\left(n+1,1,-{k^{2} \over 4a^{2}}\right).}$

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