Common integrals in quantum field theory

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Common integrals in quantum field theory are all variations and generalizations of Gaussian integrals to the complex plane and to multiple dimensions.^{[1] : 13–15} Other integrals can be approximated by versions of the Gaussian integral. Fourier integrals are also considered.

The first integral, with broad application outside of quantum field theory, is the Gaussian integral. $${\displaystyle G\equiv \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}},円dx}$${\displaystyle G\equiv \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}},円dx}

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

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

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

$${\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円dx={\sqrt {2\pi \over a}}}$${\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円dx={\sqrt {2\pi \over a}}} where we have scaled $${\displaystyle x\to {x \over {\sqrt {a}}}.}$${\displaystyle x\to {x \over {\sqrt {a}}}.}

$${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-{1 \over 2}ax^{2}},円dx=-2{d \over da}\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円dx=-2{d \over da}\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{1 \over a}}$${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-{1 \over 2}ax^{2}},円dx=-2{d \over da}\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円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 $${\displaystyle \int _{-\infty }^{\infty }x^{4}e^{-{1 \over 2}ax^{2}},円dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\left({2\pi \over a}\right)^{1 \over 2}=\left({2\pi \over a}\right)^{1 \over 2}{3 \over a^{2}}}$${\displaystyle \int _{-\infty }^{\infty }x^{4}e^{-{1 \over 2}ax^{2}},円dx=\left(-2{d \over da}\right)\left(-2{d \over da}\right)\int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}},円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 $${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-{1 \over 2}ax^{2}},円dx=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)\left(2n-3\right)\cdots 5\cdot 3\cdot 1=\left({2\pi \over a}\right)^{1 \over {2}}{1 \over a^{n}}\left(2n-1\right)!!}$${\displaystyle \int _{-\infty }^{\infty }x^{2n}e^{-{1 \over 2}ax^{2}},円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.

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

This integral can be performed by completing the square: $${\displaystyle \left(-{1 \over 2}ax^{2}+Jx\right)=-{1 \over 2}a\left(x^{2}-{2Jx \over a}+{J^{2} \over a^{2}}-{J^{2} \over a^{2}}\right)=-{1 \over 2}a\left(x-{J \over a}\right)^{2}+{J^{2} \over 2a}}$${\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: $${\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+Jx\right),円dx&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left[-{1 \over 2}a\left(x-{J \over a}\right)^{2}\right],円dx\\[8pt]&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}aw^{2}\right),円dw\\[8pt]&=\left({2\pi \over a}\right)^{1 \over 2}\exp \left({J^{2} \over 2a}\right)\end{aligned}}}$${\displaystyle {\begin{aligned}\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+Jx\right),円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],円dx\\[8pt]&=\exp \left({J^{2} \over 2a}\right)\int _{-\infty }^{\infty }\exp \left(-{1 \over 2}aw^{2}\right),円dw\\[8pt]&=\left({2\pi \over a}\right)^{1 \over 2}\exp \left({J^{2} \over 2a}\right)\end{aligned}}}

The integral $${\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over 2}ax^{2}+iJx\right)dx=\left({2\pi \over a}\right)^{1 \over 2}\exp \left(-{J^{2} \over 2a}\right)}$${\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) $${\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx.}$${\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 $${\displaystyle \left({1 \over 2}iax^{2}+iJx\right)={1 \over 2}ia\left(x^{2}+{2Jx \over a}+\left({J \over a}\right)^{2}-\left({J \over a}\right)^{2}\right)=-{1 \over 2}{a \over i}\left(x+{J \over a}\right)^{2}-{iJ^{2} \over 2a}.}$${\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 $${\displaystyle \int _{-\infty }^{\infty }\exp \left({1 \over 2}iax^{2}+iJx\right)dx=\left({2\pi i \over a}\right)^{1 \over 2}\exp \left({-iJ^{2} \over 2a}\right).}$${\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 one-dimensional integrals can be generalized to multiple dimensions.^[2] $${\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+J\cdot x\right)d^{n}x={\sqrt {\frac {(2\pi )^{n}}{\det A}}}\exp \left({1 \over 2}J\cdot A^{-1}\cdot J\right)}$${\displaystyle \int \exp \left(-{\frac {1}{2}}x\cdot A\cdot x+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)}

Here A is a real positive definite symmetric matrix.

This integral is performed by diagonalization of A with an orthogonal transformation $${\displaystyle D=O^{-1}AO=O^{\text{T}}AO}$${\displaystyle 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.

The Gaussian integral in two dimensions is $${\displaystyle \int \exp \left(-{\frac {1}{2}}A_{ij}x^{i}x^{j}\right)d^{2}x={\sqrt {\frac {(2\pi )^{2}}{\det A}}}}$${\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 $${\displaystyle A={\begin{bmatrix}a&c\\c&b\end{bmatrix}}}$${\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 $${\displaystyle A_{ij}x^{i}x^{j}\equiv x^{\text{T}}Ax=x^{\text{T}}\left(OO^{\text{T}}\right)A\left(OO^{\text{T}}\right)x=\left(x^{\text{T}}O\right)\left(O^{\text{T}}AO\right)\left(O^{\text{T}}x\right)}$${\displaystyle 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)} 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 $${\displaystyle {\begin{bmatrix}a&c\\c&b\end{bmatrix}}{\begin{bmatrix}u\\v\end{bmatrix}}=\lambda {\begin{bmatrix}u\\v\end{bmatrix}}.}$${\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 (a-\lambda )(b-\lambda )-c^{2}=0} $${\displaystyle \lambda ^{2}-\lambda (a+b)+ab-c^{2}=0,}$${\displaystyle \lambda ^{2}-\lambda (a+b)+ab-c^{2}=0,} which are found using the quadratic equation: $${\displaystyle {\begin{aligned}\lambda _{\pm }&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a+b)^{2}-4(ab-c^{2})}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {a^{2}+2ab+b^{2}-4ab+4c^{2}}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a-b)^{2}+4c^{2}}}.\end{aligned}}}$${\displaystyle {\begin{aligned}\lambda _{\pm }&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a+b)^{2}-4(ab-c^{2})}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {a^{2}+2ab+b^{2}-4ab+4c^{2}}}.\\&={\tfrac {1}{2}}(a+b)\pm {\tfrac {1}{2}}{\sqrt {(a-b)^{2}+4c^{2}}}.\end{aligned}}}

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

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

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

$${\displaystyle O={\begin{bmatrix}{\frac {1}{\eta }}&-{\frac {b-\lambda _{+}}{c\eta }}\\-{\frac {a-\lambda _{-}}{c\eta }}&{\frac {1}{\eta }}\end{bmatrix}}.}$${\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 $${\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}}$${\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}} then the orthogonal matrix can be written $${\displaystyle O={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}}$${\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: $${\displaystyle O^{-1}=O^{\text{T}}={\begin{bmatrix}\cos(\theta )&\sin(\theta )\\-\sin(\theta )&\cos(\theta )\end{bmatrix}}.}$${\displaystyle O^{-1}=O^{\text{T}}={\begin{bmatrix}\cos(\theta )&\sin(\theta )\\-\sin(\theta )&\cos(\theta )\end{bmatrix}}.}

The diagonal matrix becomes $${\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\[1ex]0&\lambda _{+}\end{bmatrix}}}$${\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\[1ex]0&\lambda _{+}\end{bmatrix}}} with eigenvectors $${\displaystyle {\begin{bmatrix}1\0円\end{bmatrix}},\qquad {\begin{bmatrix}0\1円\end{bmatrix}}}$${\displaystyle {\begin{bmatrix}1\0円\end{bmatrix}},\qquad {\begin{bmatrix}0\1円\end{bmatrix}}}

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

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

The eigenvectors are $${\displaystyle {1 \over \eta }{\begin{bmatrix}1\\[1ex]-{1 \over 2}-{{\sqrt {5}} \over 2}\end{bmatrix}},\qquad {1 \over \eta }{\begin{bmatrix}{1 \over 2}+{{\sqrt {5}} \over 2}\\[1ex]1\end{bmatrix}}}$${\displaystyle {1 \over \eta }{\begin{bmatrix}1\\[1ex]-{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}}} where $${\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.}$${\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.}

Then $${\displaystyle {\begin{aligned}O&={\begin{bmatrix}{\frac {1}{\eta }}&{\frac {1}{\eta }}\left({1 \over 2}+{{\sqrt {5}} \over 2}\right)\\{\frac {1}{\eta }}\left(-{1 \over 2}-{{\sqrt {5}} \over 2}\right)&{1 \over \eta }\end{bmatrix}}\\O^{-1}&={\begin{bmatrix}{\frac {1}{\eta }}&{\frac {1}{\eta }}\left(-{1 \over 2}-{{\sqrt {5}} \over 2}\right)\\{\frac {1}{\eta }}\left({1 \over 2}+{{\sqrt {5}} \over 2}\right)&{\frac {1}{\eta }}\end{bmatrix}}\end{aligned}}}$${\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 $${\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\0円&\lambda _{+}\end{bmatrix}}={\begin{bmatrix}{3 \over 2}-{{\sqrt {5}} \over 2}&0\0円&{3 \over 2}+{{\sqrt {5}} \over 2}\end{bmatrix}}}$${\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\0円&\lambda _{+}\end{bmatrix}}={\begin{bmatrix}{3 \over 2}-{{\sqrt {5}} \over 2}&0\0円&{3 \over 2}+{{\sqrt {5}} \over 2}\end{bmatrix}}} with eigenvectors $${\displaystyle {\begin{bmatrix}1\0円\end{bmatrix}},\qquad {\begin{bmatrix}0\1円\end{bmatrix}}}$${\displaystyle {\begin{bmatrix}1\0円\end{bmatrix}},\qquad {\begin{bmatrix}0\1円\end{bmatrix}}}

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

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

The integrations can now be performed: $${\displaystyle {\begin{aligned}\int \exp \left(-{\frac {1}{2}}x^{\mathsf {T}}Ax\right)d^{2}x={}&\int \exp \left(-{\frac {1}{2}}\sum _{j=1}^{2}\lambda _{j}y_{j}^{2}\right)d^{2}y\\[1ex]={}&\prod _{j=1}^{2}\left({2\pi \over \lambda _{j}}\right)^{1/2}\\={}&\left({(2\pi )^{2} \over \prod _{j=1}^{2}\lambda _{j}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(O^{-1}AO\right)}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(A\right)}}\right)^{1/2}\end{aligned}}}$${\displaystyle {\begin{aligned}\int \exp \left(-{\frac {1}{2}}x^{\mathsf {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\\[1ex]={}&\prod _{j=1}^{2}\left({2\pi \over \lambda _{j}}\right)^{1/2}\\={}&\left({(2\pi )^{2} \over \prod _{j=1}^{2}\lambda _{j}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(O^{-1}AO\right)}}\right)^{1/2}\\[1ex]={}&\left({(2\pi )^{2} \over \det {\left(A\right)}}\right)^{1/2}\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.

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

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

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

As an example consider the integral^{[1] : 21‒22} $${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi }$${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi } where ${\displaystyle {\hat {A}}}${\displaystyle {\hat {A}}} is a differential operator with ${\displaystyle \varphi }${\displaystyle \varphi } and J functions of spacetime, and ${\displaystyle D\varphi }${\displaystyle D\varphi } indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is $${\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi \;\propto \;\exp \left({1 \over 2}\int d^{4}x\;d^{4}yJ(x)D(x-y)J(y)\right)}$${\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 $${\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)}$${\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)} and D(x − y), called the propagator, is the inverse of ${\displaystyle {\hat {A}}}${\displaystyle {\hat {A}}}, and ${\displaystyle \delta ^{4}(x-y)}${\displaystyle \delta ^{4}(x-y)} is the Dirac delta function.

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

In quantum field theory n-dimensional integrals of the form $${\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over \hbar }f(q)\right)d^{n}q}$${\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over \hbar }f(q)\right)d^{n}q} appear often. Here ${\displaystyle \hbar }${\displaystyle \hbar } is the reduced Planck constant and f is a function with a positive minimum at ${\displaystyle q=q_{0}}${\displaystyle 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 $${\displaystyle \int _{-\infty }^{\infty }\exp \left[-{1 \over \hbar }\left(f\left(q_{0}\right)+{1 \over 2}\left(q-q_{0}\right)^{2}f^{\prime \prime }\left(q-q_{0}\right)+\cdots \right)\right]d^{n}q.}$${\displaystyle \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^{n}q.}Here ${\displaystyle f^{\prime \prime }}${\displaystyle 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. $${\displaystyle \int _{-\infty }^{\infty }\exp \left[-{1 \over \hbar }(f(q))\right]d^{n}q\approx \exp \left[-{1 \over \hbar }\left(f\left(q_{0}\right)\right)\right]{\sqrt {(2\pi \hbar )^{n} \over \det f^{\prime \prime }}}.}$${\displaystyle \int _{-\infty }^{\infty }\exp \left[-{1 \over \hbar }(f(q))\right]d^{n}q\approx \exp \left[-{1 \over \hbar }\left(f\left(q_{0}\right)\right)\right]{\sqrt {(2\pi \hbar )^{n} \over \det f^{\prime \prime }}}.}

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

The Dirac delta distribution in spacetime can be written as a Fourier transform ^{[1] : 23} $${\displaystyle \int {\frac {d^{4}k}{(2\pi )^{4}}}\exp(ik(x-y))=\delta ^{4}(x-y).}$${\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 N} $${\displaystyle \int {\frac {d^{N}k}{(2\pi )^{N}}}\exp(ik(x-y))=\delta ^{N}(x-y).}$${\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 $${\displaystyle -{1 \over 4\pi }\nabla ^{2}\left({1 \over r}\right)=\delta \left(\mathbf {r} \right)}$${\displaystyle -{1 \over 4\pi }\nabla ^{2}\left({1 \over r}\right)=\delta \left(\mathbf {r} \right)}where$${\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} }$${\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 $${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}}={1 \over 4\pi r}.}$${\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 ^{[1] : 26, 29} $${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}+m^{2}}={e^{-mr} \over 4\pi r}}$${\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 $${\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} ,\qquad k^{2}=\mathbf {k} \cdot \mathbf {k} .}$${\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: $${\displaystyle {\begin{aligned}\int {\frac {d^{3}k}{(2\pi )^{3}}}{\frac {\exp \left(i\mathbf {k} \cdot \mathbf {r} \right)}{k^{2}+m^{2}}}={}&\int _{0}^{\infty }{\frac {k^{2}dk}{(2\pi )^{2}}}\int _{-1}^{1}du{e^{ikru} \over k^{2}+m^{2}}\\[1ex]={}&{2 \over r}\int _{0}^{\infty }{\frac {kdk}{(2\pi )^{2}}}{\sin(kr) \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over (k+im)(k-im)}\\[1ex]={}&{1 \over ir}{\frac {2\pi i}{(2\pi )^{2}}}{\frac {im}{2im}}e^{-mr}\\[1ex]={}&{\frac {1}{4\pi r}}e^{-mr}\end{aligned}}}$${\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}}\\[1ex]={}&{2 \over r}\int _{0}^{\infty }{\frac {kdk}{(2\pi )^{2}}}{\sin(kr) \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over k^{2}+m^{2}}\\[1ex]={}&{1 \over ir}\int _{-\infty }^{\infty }{\frac {kdk}{(2\pi )^{2}}}{e^{ikr} \over (k+im)(k-im)}\\[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{aligned}}}

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

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

$${\displaystyle \int {\frac {d^{3}k}{(2\pi )^{3}}}\left[\mathbf {1} -\mathbf {\hat {k}} \mathbf {\hat {k}} \right]{\exp \left(i\mathbf {k} \cdot \mathbf {r} \right) \over k^{2}+m^{2}}={1 \over 2}{e^{-mr} \over 4\pi r}\left\{{2 \over (mr)^{2}}\left(e^{mr}-1\right)-{2 \over mr}\right\}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]}$${\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 $${\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}$${\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 $${\displaystyle {\frac {1}{4\pi m^{2}r^{3}}}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}$${\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^{[4] [5] : 113} $${\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\exp \left(ip\cos(\varphi )\right)=J_{0}(p)}$${\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\exp \left(ip\cos(\varphi )\right)=J_{0}(p)} and $${\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\cos(\varphi )\exp \left(ip\cos(\varphi )\right)=iJ_{1}(p).}$${\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.

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

See Abramowitz and Stegun.^{[6] : §11.4.44}

For ${\displaystyle mr\ll 1}${\displaystyle mr\ll 1}, we have^{[5] : 116} $${\displaystyle K_{0}(mr)\to -\ln \left({mr \over 2}\right)+0.5772.}$${\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^[4] $${\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)=I_{1}(mr)K_{1}(mr).}$${\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 $${\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left[1-{1 \over 8}(mr)^{2}\right].}$${\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 $${\displaystyle \int _{o}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{1}^{2}(kr)\to {1 \over 2}\left({1 \over mr}\right).}$${\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, $${\displaystyle \int _{0}^{\infty }{k\;dk \over k^{2}+m^{2}}J_{\nu }^{2}(kr)=I_{\nu }(mr)K_{\nu }(mr)\qquad \Re (\nu )>-1.}$${\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^{[6] : §11.4.28} $${\displaystyle {2a^{2n+2} \over n!}\int _{0}^{\infty }{dr}\;r^{2n+1}\exp \left(-a^{2}r^{2}\right)J_{0}(kr)=M\left(n+1,1,-{k^{2} \over 4a^{2}}\right).}$${\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.

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

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