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
Variations on a simple Gaussian integral
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Gaussian integral
edit
The first integral, with broad application outside of quantum field theory, is the Gaussian integral.
G
≡
∫
−
∞
∞
e
−
1
2
x
2
d
x
{\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:
G
2
=
(
∫
−
∞
∞
e
−
1
2
x
2
d
x
)
⋅
(
∫
−
∞
∞
e
−
1
2
y
2
d
y
)
=
2
π
∫
0
∞
r
e
−
1
2
r
2
d
r
=
2
π
∫
0
∞
e
−
w
d
w
=
2
π
.
{\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
∫
−
∞
∞
e
−
1
2
x
2
d
x
=
2
π
.
{\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}x^{2}}\,dx={\sqrt {2\pi }}.}
Slight generalization of the Gaussian integral
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∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
2
π
a
{\displaystyle \int _{-\infty }^{\infty }e^{-{1 \over 2}ax^{2}}\,dx={\sqrt {2\pi \over a}}}
x
→
x
a
.
{\displaystyle x\to {x \over {\sqrt {a}}}.}
Integrals of exponents and even powers of x
edit
∫
−
∞
∞
x
2
e
−
1
2
a
x
2
d
x
=
−
2
d
d
a
∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
−
2
d
d
a
(
2
π
a
)
1
2
=
(
2
π
a
)
1
2
1
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}}
∫
−
∞
∞
x
4
e
−
1
2
a
x
2
d
x
=
(
−
2
d
d
a
)
(
−
2
d
d
a
)
∫
−
∞
∞
e
−
1
2
a
x
2
d
x
=
(
−
2
d
d
a
)
(
−
2
d
d
a
)
(
2
π
a
)
1
2
=
(
2
π
a
)
1
2
3
a
2
{\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
∫
−
∞
∞
x
2
n
e
−
1
2
a
x
2
d
x
=
(
2
π
a
)
1
2
1
a
n
(
2
n
−
1
)
(
2
n
−
3
)
⋯
5
⋅
3
⋅
1
=
(
2
π
a
)
1
2
1
a
n
(
2
n
−
1
)
!
!
{\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.
Integrals with a linear term in the argument of the exponent
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∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
J
x
)
d
x
{\displaystyle \int _{-\infty }^{\infty }\exp \left(-{\frac {1}{2}}ax^{2}+Jx\right)dx}
This integral can be performed by completing the square:
(
−
1
2
a
x
2
+
J
x
)
=
−
1
2
a
(
x
2
−
2
J
x
a
+
J
2
a
2
−
J
2
a
2
)
=
−
1
2
a
(
x
−
J
a
)
2
+
J
2
2
a
{\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:
∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
J
x
)
d
x
=
exp
(
J
2
2
a
)
∫
−
∞
∞
exp
[
−
1
2
a
(
x
−
J
a
)
2
]
d
x
=
exp
(
J
2
2
a
)
∫
−
∞
∞
exp
(
−
1
2
a
w
2
)
d
w
=
(
2
π
a
)
1
2
exp
(
J
2
2
a
)
{\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}}}
Integrals with an imaginary linear term in the argument of the exponent
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The integral
∫
−
∞
∞
exp
(
−
1
2
a
x
2
+
i
J
x
)
d
x
=
(
2
π
a
)
1
2
exp
(
−
J
2
2
a
)
{\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)}
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
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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 )
∫
−
∞
∞
exp
(
1
2
i
a
x
2
+
i
J
x
)
d
x
.
{\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
(
1
2
i
a
x
2
+
i
J
x
)
=
1
2
i
a
(
x
2
+
2
J
x
a
+
(
J
a
)
2
−
(
J
a
)
2
)
=
−
1
2
a
i
(
x
+
J
a
)
2
−
i
J
2
2
a
.
{\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
∫
−
∞
∞
exp
(
1
2
i
a
x
2
+
i
J
x
)
d
x
=
(
2
π
i
a
)
1
2
exp
(
−
i
J
2
2
a
)
.
{\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 .
Gaussian integrals in higher dimensions
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The one-dimensional integrals can be generalized to multiple dimensions.[2]
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
1
2
J
⋅
A
−
1
⋅
J
)
{\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
D
=
O
−
1
A
O
=
O
T
A
O
{\displaystyle D=O^{-1}AO=O^{\text{T}}AO}
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
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The Gaussian integral in two dimensions is
∫
exp
(
−
1
2
A
i
j
x
i
x
j
)
d
2
x
=
(
2
π
)
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}}}}
A is a two-dimensional symmetric matrix with components specified as
A
=
[
a
c
c
b
]
{\displaystyle A={\begin{bmatrix}a&c\\c&b\end{bmatrix}}}
Einstein summation convention .
Diagonalize the matrix
edit
The first step is to diagonalize the matrix.[3]
A
i
j
x
i
x
j
≡
x
T
A
x
=
x
T
(
O
O
T
)
A
(
O
O
T
)
x
=
(
x
T
O
)
(
O
T
A
O
)
(
O
T
x
)
{\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)}
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 ≡ OT AO Eigenvalues of A
edit
To find the eigenvectors of A one first finds the eigenvalues λ of A given by
[
a
c
c
b
]
[
u
v
]
=
λ
[
u
v
]
.
{\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
(
a
−
λ
)
(
b
−
λ
)
−
c
2
=
0
{\displaystyle (a-\lambda )(b-\lambda )-c^{2}=0}
λ
2
−
λ
(
a
+
b
)
+
a
b
−
c
2
=
0
,
{\displaystyle \lambda ^{2}-\lambda (a+b)+ab-c^{2}=0,}
quadratic equation :
λ
±
=
1
2
(
a
+
b
)
±
1
2
(
a
+
b
)
2
−
4
(
a
b
−
c
2
)
.
=
1
2
(
a
+
b
)
±
1
2
a
2
+
2
a
b
+
b
2
−
4
a
b
+
4
c
2
.
=
1
2
(
a
+
b
)
±
1
2
(
a
−
b
)
2
+
4
c
2
.
{\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}}}
Eigenvectors of A
edit
Substitution of the eigenvalues back into the eigenvector equation yields
v
=
−
(
a
−
λ
±
)
u
c
,
v
=
−
c
u
(
b
−
λ
±
)
.
{\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
a
−
λ
±
c
=
c
b
−
λ
±
.
{\displaystyle {a-\lambda _{\pm } \over c}={c \over b-\lambda _{\pm }}.}
Also note
a
−
λ
±
c
=
−
b
−
λ
∓
c
.
{\displaystyle {a-\lambda _{\pm } \over c}=-{b-\lambda _{\mp } \over c}.}
The eigenvectors can be written as:
[
1
η
−
a
−
λ
−
c
η
]
,
[
−
b
−
λ
+
c
η
1
η
]
{\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}}}
η is a normalizing factor given by,
η
=
1
+
(
a
−
λ
−
c
)
2
=
1
+
(
b
−
λ
+
c
)
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.
Construction of the orthogonal matrix
edit
The orthogonal matrix is constructed by assigning the normalized eigenvectors as columns in the orthogonal matrix
O
=
[
1
η
−
b
−
λ
+
c
η
−
a
−
λ
−
c
η
1
η
]
.
{\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
sin
(
θ
)
=
−
a
−
λ
−
c
η
{\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}}
O
=
[
cos
(
θ
)
−
sin
(
θ
)
sin
(
θ
)
cos
(
θ
)
]
{\displaystyle O={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}}
O
−
1
=
O
T
=
[
cos
(
θ
)
sin
(
θ
)
−
sin
(
θ
)
cos
(
θ
)
]
.
{\displaystyle O^{-1}=O^{\text{T}}={\begin{bmatrix}\cos(\theta )&\sin(\theta )\\-\sin(\theta )&\cos(\theta )\end{bmatrix}}.}
Diagonal matrix
edit
The diagonal matrix becomes
D
=
O
T
A
O
=
[
λ
−
0
0
λ
+
]
{\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\[1ex]0&\lambda _{+}\end{bmatrix}}}
[
1
0
]
,
[
0
1
]
{\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}
Numerical example
edit
A
=
[
2
1
1
1
]
{\displaystyle A={\begin{bmatrix}2&1\\1&1\end{bmatrix}}}
The eigenvalues are
λ
±
=
3
2
±
5
2
.
{\displaystyle \lambda _{\pm }={3 \over 2}\pm {{\sqrt {5}} \over 2}.}
The eigenvectors are
1
η
[
1
−
1
2
−
5
2
]
,
1
η
[
1
2
+
5
2
1
]
{\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}}}
η
=
5
2
+
5
2
.
{\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.}
Then
O
=
[
1
η
1
η
(
1
2
+
5
2
)
1
η
(
−
1
2
−
5
2
)
1
η
]
O
−
1
=
[
1
η
1
η
(
−
1
2
−
5
2
)
1
η
(
1
2
+
5
2
)
1
η
]
{\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
D
=
O
T
A
O
=
[
λ
−
0
0
λ
+
]
=
[
3
2
−
5
2
0
0
3
2
+
5
2
]
{\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\0&\lambda _{+}\end{bmatrix}}={\begin{bmatrix}{3 \over 2}-{{\sqrt {5}} \over 2}&0\\0&{3 \over 2}+{{\sqrt {5}} \over 2}\end{bmatrix}}}
[
1
0
]
,
[
0
1
]
{\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}}
Rescale the variables and integrate
edit
With the diagonalization the integral can be written
∫
exp
(
−
1
2
x
T
A
x
)
d
2
x
=
∫
exp
(
−
1
2
∑
j
=
1
2
λ
j
y
j
2
)
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)\,d^{2}y}
y
=
O
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
d
2
y
=
d
2
x
{\displaystyle d^{2}y=d^{2}x}
The integrations can now be performed:
∫
exp
(
−
1
2
x
T
A
x
)
d
2
x
=
∫
exp
(
−
1
2
∑
j
=
1
2
λ
j
y
j
2
)
d
2
y
=
∏
j
=
1
2
(
2
π
λ
j
)
1
/
2
=
(
(
2
π
)
2
∏
j
=
1
2
λ
j
)
1
/
2
=
(
(
2
π
)
2
det
(
O
−
1
A
O
)
)
1
/
2
=
(
(
2
π
)
2
det
(
A
)
)
1
/
2
{\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}}}
Integrals with complex and linear terms in multiple dimensions
edit
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
edit
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
1
2
J
⋅
A
−
1
⋅
J
)
{\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)}
Integrals with an imaginary linear term
edit
∫
exp
(
−
1
2
x
⋅
A
⋅
x
+
i
J
⋅
x
)
d
n
x
=
(
2
π
)
n
det
A
exp
(
−
1
2
J
⋅
A
−
1
⋅
J
)
{\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)}
Integrals with a complex quadratic term
edit
∫
exp
(
i
2
x
⋅
A
⋅
x
+
i
J
⋅
x
)
d
n
x
=
(
2
π
i
)
n
det
A
exp
(
−
i
2
J
⋅
A
−
1
⋅
J
)
{\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)}
Integrals with differential operators in the argument
edit
As an example consider the integral[1] : 21‒22
∫
exp
[
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
J
φ
)
]
D
φ
{\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi }
A
^
{\displaystyle {\hat {A}}}
φ
{\displaystyle \varphi }
J functions of spacetime , and
D
φ
{\displaystyle D\varphi }
∫
exp
[
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
J
φ
)
]
D
φ
∝
exp
(
1
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
{\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)}
A
^
D
(
x
−
y
)
=
δ
4
(
x
−
y
)
{\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)}
D (x − y )propagator , is the inverse of
A
^
{\displaystyle {\hat {A}}}
δ
4
(
x
−
y
)
{\displaystyle \delta ^{4}(x-y)}
Dirac delta function .
Similar arguments yield
∫
exp
[
∫
d
4
x
(
−
1
2
φ
A
^
φ
+
i
J
φ
)
]
D
φ
∝
exp
(
−
1
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
,
{\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),}
∫
exp
[
i
∫
d
4
x
(
1
2
φ
A
^
φ
+
J
φ
)
]
D
φ
∝
exp
(
−
i
2
∫
d
4
x
d
4
y
J
(
x
)
D
(
x
−
y
)
J
(
y
)
)
.
{\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.
Integrals that can be approximated by the method of steepest descent
edit
Integrals that can be approximated by the method of stationary phase
edit
Fourier integrals
edit
Dirac delta distribution
edit
The Dirac delta distribution in spacetime can be written as a Fourier transform [1] : 23
∫
d
4
k
(
2
π
)
4
exp
(
i
k
(
x
−
y
)
)
=
δ
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
N
{\displaystyle N}
∫
d
N
k
(
2
π
)
N
exp
(
i
k
(
x
−
y
)
)
=
δ
N
(
x
−
y
)
.
{\displaystyle \int {\frac {d^{N}k}{(2\pi )^{N}}}\exp(ik(x-y))=\delta ^{N}(x-y).}
Fourier integrals of forms of the Coulomb potential
edit
Laplacian of 1/r
edit
While not an integral, the identity in three-dimensional Euclidean space
−
1
4
π
∇
2
(
1
r
)
=
δ
(
r
)
{\displaystyle -{1 \over 4\pi }\nabla ^{2}\left({1 \over r}\right)=\delta \left(\mathbf {r} \right)}
r
2
=
r
⋅
r
{\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} }
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
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
=
1
4
π
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}.}
Yukawa potential: the Coulomb potential with mass
edit
The Yukawa potential in three dimensions can be represented as an integral over a Fourier transform [1] : 26, 29
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
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}}
r
2
=
r
⋅
r
,
k
2
=
k
⋅
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:
∫
d
3
k
(
2
π
)
3
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
e
i
k
r
u
k
2
+
m
2
=
2
r
∫
0
∞
k
d
k
(
2
π
)
2
sin
(
k
r
)
k
2
+
m
2
=
1
i
r
∫
−
∞
∞
k
d
k
(
2
π
)
2
e
i
k
r
k
2
+
m
2
=
1
i
r
∫
−
∞
∞
k
d
k
(
2
π
)
2
e
i
k
r
(
k
+
i
m
)
(
k
−
i
m
)
=
1
i
r
2
π
i
(
2
π
)
2
i
m
2
i
m
e
−
m
r
=
1
4
π
r
e
−
m
r
{\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}}}
Modified Coulomb potential with mass
edit
∫
d
3
k
(
2
π
)
3
(
k
^
⋅
r
^
)
2
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
r
[
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
]
{\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]}
∫
d
3
k
(
2
π
)
3
(
k
^
⋅
r
^
)
2
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
u
2
e
i
k
r
u
k
2
+
m
2
=
2
∫
0
∞
k
2
d
k
(
2
π
)
2
1
k
2
+
m
2
[
1
k
r
sin
(
k
r
)
+
2
(
k
r
)
2
cos
(
k
r
)
−
2
(
k
r
)
3
sin
(
k
r
)
]
=
e
−
m
r
4
π
r
[
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
]
{\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 .
Longitudinal potential with mass
edit
∫
d
3
k
(
2
π
)
3
k
^
k
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
1
2
e
−
m
r
4
π
r
(
[
1
−
r
^
r
^
]
+
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
[
1
+
r
^
r
^
]
)
{\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)}
∫
d
3
k
(
2
π
)
3
k
^
k
^
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
∫
d
3
k
(
2
π
)
3
[
(
k
^
⋅
r
^
)
2
r
^
r
^
+
(
k
^
⋅
θ
^
)
2
θ
^
θ
^
+
(
k
^
⋅
ϕ
^
)
2
ϕ
^
ϕ
^
]
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
e
−
m
r
4
π
r
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
{
1
−
1
2
[
1
−
r
^
r
^
]
}
+
∫
0
∞
k
2
d
k
(
2
π
)
2
∫
−
1
1
d
u
e
i
k
r
u
k
2
+
m
2
1
2
[
1
−
r
^
r
^
]
=
1
2
e
−
m
r
4
π
r
[
1
−
r
^
r
^
]
+
e
−
m
r
4
π
r
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
{
1
2
[
1
+
r
^
r
^
]
}
=
1
2
e
−
m
r
4
π
r
(
[
1
−
r
^
r
^
]
+
{
1
+
2
m
r
−
2
(
m
r
)
2
(
e
m
r
−
1
)
}
[
1
+
r
^
r
^
]
)
{\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
1
2
1
4
π
r
[
1
−
r
^
r
^
]
.
{\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}
Transverse potential with mass
edit
∫
d
3
k
(
2
π
)
3
[
1
−
k
^
k
^
]
exp
(
i
k
⋅
r
)
k
2
+
m
2
=
1
2
e
−
m
r
4
π
r
{
2
(
m
r
)
2
(
e
m
r
−
1
)
−
2
m
r
}
[
1
+
r
^
r
^
]
{\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
1
2
1
4
π
r
[
1
+
r
^
r
^
]
.
{\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
1
4
π
m
2
r
3
[
1
+
r
^
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 .
Angular integration in cylindrical coordinates
edit
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
∫
0
2
π
d
φ
2
π
exp
(
i
p
cos
(
φ
)
)
=
J
0
(
p
)
{\displaystyle \int _{0}^{2\pi }{d\varphi \over 2\pi }\exp \left(ip\cos(\varphi )\right)=J_{0}(p)}
∫
0
2
π
d
φ
2
π
cos
(
φ
)
exp
(
i
p
cos
(
φ
)
)
=
i
J
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 .
Bessel functions
edit
Integration of the cylindrical propagator with mass
edit
First power of a Bessel function
edit
∫
0
∞
k
d
k
k
2
+
m
2
J
0
(
k
r
)
=
K
0
(
m
r
)
.
{\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
m
r
≪
1
{\displaystyle mr\ll 1}
[5] : 116
K
0
(
m
r
)
→
−
ln
(
m
r
2
)
+
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 .
Squares of Bessel functions
edit
The integration of the propagator in cylindrical coordinates is[4]
∫
0
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
=
I
1
(
m
r
)
K
1
(
m
r
)
.
{\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
∫
o
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
→
1
2
[
1
−
1
8
(
m
r
)
2
]
.
{\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
∫
o
∞
k
d
k
k
2
+
m
2
J
1
2
(
k
r
)
→
1
2
(
1
m
r
)
.
{\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,
∫
0
∞
k
d
k
k
2
+
m
2
J
ν
2
(
k
r
)
=
I
ν
(
m
r
)
K
ν
(
m
r
)
ℜ
(
ν
)
>
−
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.}
Integration over a magnetic wave function
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The two-dimensional integral over a magnetic wave function is[6] : §11.4.28
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{\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 .
See also
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References
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^ a b c d A. Zee (2003). Quantum Field Theory in a Nutshell . Princeton University. ISBN 0-691-01019-6 . ^ Frederick W. Byron and Robert W. Fuller (1969). Mathematics of Classical and Quantum Physics . Addison-Wesley. ISBN 0-201-00746-0 . ^ Herbert S. Wilf (1978). Mathematics for the Physical Sciences ISBN 0-486-63635-6 . ^ a b Gradshteyn, Izrail Solomonovich ; Ryzhik, Iosif Moiseevich ; Geronimus, Yuri Veniaminovich ; Tseytlin, Michail Yulyevich ; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo (eds.). Table of Integrals, Series, and Products Academic Press, Inc. ISBN 978-0-12-384933-5 . LCCN 2014010276.^ a b Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X . ^ a b M. Abramowitz; I. Stegun (1965). Handbook of Mathematical Functions ISBN 0486-61272-4 .
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70339555c7cfe2e9f4a72d4ce69f1ef4453b3618)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe97f77101c09c71f165e75e77a7d426127537cd)
![{\displaystyle D=O^{\text{T}}AO={\begin{bmatrix}\lambda _{-}&0\\[1ex]0&\lambda _{+}\end{bmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69ab2aeeb9a40d61a2ad78fa4c34ed8a2dca135f)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ff8bf406c2f47aaf23a78525b3fa339f78127a5)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f8f687a33248a374fdf7f3b4f76432efe1f6131)
![{\displaystyle \int \exp \left[\int d^{4}x\left(-{\frac {1}{2}}\varphi {\hat {A}}\varphi +J\varphi \right)\right]D\varphi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/460de24ce7d3cc04ca135dd19131951d5ba69162)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19e34ab56be6e9dddf0a9aa4f311c53f4c51eed7)
![{\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),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb42be62a36ea9f64e3dbad84d40fcb32207ff82)
![{\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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72eb4772e5501f24af95a79c52b739bce4c17a82)
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e85ad2161a23775ac29a098700f4a4de5a75aff)
![{\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 }}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8eba633ce9f0389c5ca6af92228a18beab782b93)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a96177e909f4d4ef8f07b5e761487b368c0df869)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2b4cd6a378e4674ea1e152606038a5b76aa4d2)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b10eac72cc589461947a3f54b3eadd207c0986d0)
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/269613b0badd1f2f8d1803e244f8f4f0d2476f93)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52e53c36f2a0b590917c2d25f7bfd6df0fcd758e)
![{\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} -\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c38e982b576ea3398166a359d432c892aac37fd)
![{\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39e9912d48f12733a4a2c14c65be61b35ed4d023)
![{\displaystyle {1 \over 2}{1 \over 4\pi r}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9adfb28ff636f84346674533a4261f61fd8cdf47)
![{\displaystyle {\frac {1}{4\pi m^{2}r^{3}}}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8eaf7ee08f4624fab70dac29c3fe1ae9295687dc)
![{\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].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12094b15be99cdf437b2e9615b23ae2a9636cd09)
