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Common integrals in quantum field theory
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<h2 id="variations-on-a-simple-gaussian-integral">Variations on a simple Gaussian integral</h2> <h3>Gaussian integral</h3> <p>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} </p><p>In physics the factor of 1/2 in the argument of the exponential is common. </p><p>Note that, if we let r = 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 d x d y = r d r d θ {\displaystyle dx\,dy=r\,dr\,d\theta } ) to get 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 .} </p><p>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 }}.} </p> <h3>Slight generalization of the Gaussian integral</h3> <p> ∫ − ∞ ∞ 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}}} where we have scaled x → x a . {\displaystyle x\to {x \over {\sqrt {a}}}.} </p> <h3>Integrals of exponents and even powers of <i>x</i></h3> <p> ∫ − ∞ ∞ 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}} and ∫ − ∞ ∞ 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}}} </p><p>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)!!} </p><p>Note that the integrals of exponents and odd powers of x are 0, due to <a href="/facts/Odd_function/tGHWJqAa">odd</a> symmetry. </p> <h3>Integrals with a linear term in the argument of the exponent</h3> <p> ∫ − ∞ ∞ exp ( − 1 2 a x 2 + J x ) d x {\displaystyle \int _{-\infty }^{\infty }\exp \left(-{\frac {1}{2}}ax^{2}+Jx\right)dx} </p><p>This integral can be performed by <a href="/facts/Completing_the_square/SN6xA8sc">completing the square</a>: ( − 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}} </p><p>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}}} </p> <h3>Integrals with an imaginary linear term in the argument of the exponent</h3> <p>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)} is proportional to the <a href="/facts/Fourier_transform/XXhKJtc8">Fourier transform</a> of the Gaussian where J is the <a href="/facts/Conjugate_variables/NQca4AKf">conjugate variable</a> of x. </p><p>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 <a href="/facts/Uncertainty_principle/UUSoSp4T">uncertainty principle</a>. </p><p>This integral is also known as the <a href="/facts/Hubbard%25E2%2580%2593Stratonovich_transformation/J0VlwXVd">Hubbard–Stratonovich transformation</a> used in field theory. </p> <h3>Integrals with a complex argument of the exponent</h3> <p>The integral of interest is (for an example of an application see <a href="/facts/Relation_between_Schr%25C3%25B6dinger%2527s_equation_and_the_path_integral_formulation_of_quantum_mechanics/ea9ns7Fs">Relation between Schrödinger's equation and the path integral formulation of quantum mechanics</a>) ∫ − ∞ ∞ 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.} </p><p>We now assume that a and J may be complex. </p><p>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}.} </p><p>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).} </p><p>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 <a href="/facts/Fresnel_integral/5W7z760j">Fresnel integral</a>. </p> <h2 id="gaussian-integrals-in-higher-dimensions">Gaussian integrals in higher dimensions</h2> <p>The one-dimensional integrals can be generalized to multiple dimensions.<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> ∫ 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)} </p><p>Here A is a real positive definite <a href="/facts/Symmetric_matrix/08CGwjNl">symmetric matrix</a>. </p><p>This integral is performed by <a href="/facts/Diagonalizable_matrix/y3MkyTCy">diagonalization</a> of A with an <a href="/facts/Orthogonal_matrix/WDXBXYB5">orthogonal transformation</a> D = O − 1 A O = O T A O {\displaystyle D=O^{-1}AO=O^{\text{T}}AO} where D is a <a href="/facts/Diagonal_matrix/tjIAHrE6">diagonal matrix</a> and O is an <a href="/facts/Orthogonal_matrix/WDXBXYB5">orthogonal matrix</a>. This decouples the variables and allows the integration to be performed as n one-dimensional integrations. </p><p>This is best illustrated with a two-dimensional example. </p> <h3>Example: Simple Gaussian integration in two dimensions</h3> <p>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}}}} where 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}}} and we have used the <a href="/facts/Einstein_summation_convention/UXwS43wU">Einstein summation convention</a>. </p> <h4>Diagonalize the matrix</h4> <p>The first step is to <a href="/facts/Diagonalizable_matrix/y3MkyTCy">diagonalize</a> the matrix.<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> Note that 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)} where, since A is a real <a href="/facts/Symmetric_matrix/08CGwjNl">symmetric matrix</a>, we can choose O to be <a href="/facts/Orthogonal_matrix/WDXBXYB5">orthogonal</a>, and hence also a <a href="/facts/Unitary_matrix/8fsxv3KD">unitary matrix</a>. O can be obtained from the <a href="/facts/Eigenvectors/8TjEoT8u">eigenvectors</a> of A. We choose O such that: <i>D</i> ≡ <i>OTAO</i> is diagonal. </p> <h5>Eigenvalues of <i>A</i></h5> <p>To find the eigenvectors of A one first finds the <a href="/facts/Eigenvalues/8TjEoT8u">eigenvalues</a> 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}}.} </p><p>The eigenvalues are solutions of the <a href="/facts/Characteristic_polynomial/7m8roDqo">characteristic polynomial</a> ( 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,} which are found using the <a href="/facts/Quadratic_equation/9IMnkJAT">quadratic equation</a>: 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}}} </p> <h5>Eigenvectors of <i>A</i></h5> <p>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)}.} </p><p>From the characteristic equation we know a − λ ± c = c b − λ ± . {\displaystyle {a-\lambda _{\pm } \over c}={c \over b-\lambda _{\pm }}.} </p><p>Also note a − λ ± c = − b − λ ∓ c . {\displaystyle {a-\lambda _{\pm } \over c}=-{b-\lambda _{\mp } \over c}.} </p><p>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}}} for the two eigenvectors. Here 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}}}.} </p><p>It is easily verified that the two eigenvectors are orthogonal to each other. </p> <h5>Construction of the orthogonal matrix</h5> <p>The orthogonal matrix is constructed by assigning the normalized eigenvectors as columns in the orthogonal matrix </p><p> 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}}.} </p><p>Note that det(<i>O</i>) = 1. </p><p>If we define sin ( θ ) = − a − λ − c η {\displaystyle \sin(\theta )=-{\frac {a-\lambda _{-}}{c\eta }}} then the orthogonal matrix can be written O = [ cos ( θ ) − sin ( θ ) sin ( θ ) cos ( θ ) ] {\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: 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}}.} </p> <h5>Diagonal matrix</h5> <p>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}}} with eigenvectors [ 1 0 ] , [ 0 1 ] {\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}} </p> <h5>Numerical example</h5> <p> A = [ 2 1 1 1 ] {\displaystyle A={\begin{bmatrix}2&1\\1&1\end{bmatrix}}} </p><p>The eigenvalues are λ ± = 3 2 ± 5 2 . {\displaystyle \lambda _{\pm }={3 \over 2}\pm {{\sqrt {5}} \over 2}.} </p><p>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}}} where η = 5 2 + 5 2 . {\displaystyle \eta ={\sqrt {{5 \over 2}+{{\sqrt {5}} \over 2}}}.} </p><p>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}}} </p><p>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}}} with eigenvectors [ 1 0 ] , [ 0 1 ] {\displaystyle {\begin{bmatrix}1\\0\end{bmatrix}},\qquad {\begin{bmatrix}0\\1\end{bmatrix}}} </p> <h4>Rescale the variables and integrate</h4> <p>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} where y = O T x . {\displaystyle y=O^{\text{T}}x.} </p><p>Since the coordinate transformation is simply a rotation of coordinates the <a href="/facts/Jacobian_matrix_and_determinant/0tNv6o8j">Jacobian</a> determinant of the transformation is one yielding d 2 y = d 2 x {\displaystyle d^{2}y=d^{2}x} </p><p>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}}} which is the advertised solution. </p> <h3>Integrals with complex and linear terms in multiple dimensions</h3> <p>With the two-dimensional example it is now easy to see the generalization to the complex plane and to multiple dimensions. </p> <h4>Integrals with a linear term in the argument</h4> <p> ∫ exp ( − 1 2 x T ⋅ A ⋅ x + J T ⋅ x ) d x = ( 2 π ) n det A exp ( 1 2 J T ⋅ A − 1 ⋅ J ) {\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)} </p> <h4>Integrals with an imaginary linear term</h4> <p> ∫ exp ( − 1 2 x T ⋅ A ⋅ x + i J T ⋅ x ) d x = ( 2 π ) n det A exp ( − 1 2 J T ⋅ A − 1 ⋅ J ) {\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)} </p> <h4>Integrals with a complex quadratic term</h4> <p> ∫ exp ( i 2 x T ⋅ A ⋅ x + i J T ⋅ x ) d x = ( 2 π i ) n det A exp ( − i 2 J T ⋅ A − 1 ⋅ J ) {\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)} </p> <h3>Integrals with differential operators in the argument</h3> <p>As an example consider the integral<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a>: 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 } where A ^ {\displaystyle {\hat {A}}} is a differential operator with φ {\displaystyle \varphi } and J functions of <a href="/facts/Spacetime/uRIN1x4b">spacetime</a>, and D φ {\displaystyle D\varphi } indicates integration over all possible paths. In analogy with the matrix version of this integral the solution is ∫ 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)} where A ^ D ( x − y ) = δ 4 ( x − y ) {\displaystyle {\hat {A}}D(x-y)=\delta ^{4}(x-y)} and <i>D</i>(<i>x</i> − <i>y</i>), called the <a href="/facts/Propagator/FER8tNKP">propagator</a>, is the inverse of A ^ {\displaystyle {\hat {A}}} , and δ 4 ( x − y ) {\displaystyle \delta ^{4}(x-y)} is the <a href="/facts/Dirac_delta_function/V8PjRP8T">Dirac delta function</a>. </p><p>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),} and ∫ 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).} </p><p>See <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Path-integral formulation of virtual-particle exchange</a> for an application of this integral. </p> <h2 id="integrals-that-can-be-approximated-by-the-method-of-steepest-descent">Integrals that can be approximated by the method of steepest descent</h2> <p>In quantum field theory n-dimensional integrals of the form ∫ − ∞ ∞ exp ( − 1 ℏ f ( q ) ) d n q {\displaystyle \int _{-\infty }^{\infty }\exp \left(-{1 \over \hbar }f(q)\right)d^{n}q} appear often. Here ℏ {\displaystyle \hbar } is the <a href="/facts/Reduced_Planck_constant/qGHAe3qq">reduced Planck constant</a> and <i>f</i> is a function with a positive minimum at q = q 0 {\displaystyle q=q_{0}} . These integrals can be approximated by the <a href="/facts/Method_of_steepest_descent/C6E59y70">method of steepest descent</a>. </p><p>For small values of the Planck constant, <i>f</i> can be expanded about its minimum ∫ − ∞ ∞ exp [ − 1 ℏ ( f ( q 0 ) + 1 2 ( q − q 0 ) 2 f ′ ′ ( q − q 0 ) + ⋯ ) ] 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 f ′ ′ {\displaystyle f^{\prime \prime }} is the n by n matrix of second derivatives evaluated at the minimum of the function. </p><p>If we neglect higher order terms this integral can be integrated explicitly. ∫ − ∞ ∞ exp [ − 1 ℏ ( f ( q ) ) ] d n q ≈ exp [ − 1 ℏ ( f ( q 0 ) ) ] ( 2 π ℏ ) n det f ′ ′ . {\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 }}}.} </p> <h2 id="integrals-that-can-be-approximated-by-the-method-of-stationary-phase">Integrals that can be approximated by the method of stationary phase</h2> <p>A common integral is a path integral of the form ∫ exp ( i ℏ S ( q , q ˙ ) ) D q {\displaystyle \int \exp \left({i \over \hbar }S\left(q,{\dot {q}}\right)\right)Dq} where S ( q , q ˙ ) {\displaystyle S\left(q,{\dot {q}}\right)} is the classical <a href="/facts/Action_(physics)/NhWS2Mts">action</a> and the integral is over all possible paths that a particle may take. In the limit of small ℏ {\displaystyle \hbar } the integral can be evaluated in the <a href="/facts/Stationary_phase_approximation/MYnwqtU9">stationary phase approximation</a>. In this approximation the integral is over the path in which the action is a minimum. Therefore, this approximation recovers the <a href="/facts/Classical_limit/Yu7aHtr6">classical limit</a> of <a href="/facts/Classical_mechanics/ykG8upsN">mechanics</a>. </p> <h2 id="fourier-integrals">Fourier integrals</h2> <h3>Dirac delta distribution</h3> <p>The <a href="/facts/Dirac_delta_distribution/V8PjRP8T">Dirac delta distribution</a> in <a href="/facts/Spacetime/uRIN1x4b">spacetime</a> can be written as a <a href="/facts/Fourier_transform/XXhKJtc8">Fourier transform</a><a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a>: 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).} </p><p>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).} </p> <h3>Fourier integrals of forms of the Coulomb potential</h3> <h4>Laplacian of 1/<i>r</i></h4> <p>While not an integral, the identity in three-dimensional <a href="/facts/Euclidean_space/R2UbzmzM">Euclidean space</a> − 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)} where r 2 = r ⋅ r {\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} } is a consequence of <a href="/facts/Gauss%2527s_theorem/pUGKF1N2">Gauss's theorem</a> and can be used to derive integral identities. For an example see <a href="/facts/Longitudinal_and_transverse_vector_fields/c6DIwyWn">Longitudinal and transverse vector fields</a>. </p><p>This identity implies that the <a href="/facts/Fourier_integral/XXhKJtc8">Fourier integral</a> representation of 1/<i>r</i> 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}.} </p> <h4>Yukawa potential: the Coulomb potential with mass</h4> <p>The <a href="/facts/Yukawa_potential/TmPzXYpx">Yukawa potential</a> in three dimensions can be represented as an integral over a <a href="/facts/Fourier_transform/XXhKJtc8">Fourier transform</a><a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a>: 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}} where r 2 = r ⋅ r , k 2 = k ⋅ k . {\displaystyle r^{2}=\mathbf {r} \cdot \mathbf {r} ,\qquad k^{2}=\mathbf {k} \cdot \mathbf {k} .} </p><p>See <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Static forces and virtual-particle exchange</a> for an application of this integral. </p><p>In the small m limit the integral reduces to 1/4<i>πr</i>. </p><p>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}}} </p> <h4>Modified Coulomb potential with mass</h4> <p> ∫ 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]} where the hat indicates a <a href="/facts/Unit_vector/GyaXneDn">unit vector</a> in three dimensional space. The derivation of this result is as follows: ∫ 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}}} </p><p>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. </p> <h4>Longitudinal potential with mass</h4> <p> ∫ 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)} where the hat indicates a unit vector in three dimensional space. The derivation for this result is as follows: ∫ 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}}} </p><p>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].} </p> <h4>Transverse potential with mass</h4> <p> ∫ 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]} </p><p>In the small <i>mr</i> 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].} </p><p>For large distance, the integral falls off as the inverse cube of <i>r</i> 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].} </p><p>For applications of this integral see <a href="/facts/Darwin_Lagrangian/AqCt6LlF">Darwin Lagrangian</a> and <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Darwin interaction in a vacuum</a>. </p> <h4>Angular integration in cylindrical coordinates</h4> <p>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<a class="footnote-ref" id="fnref:7" href="#fn:7"><sup>7</sup></a><a class="footnote-ref" id="fnref:8" href="#fn:8"><sup>8</sup></a>: 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)} and ∫ 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).} </p><p>For applications of these integrals see <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Magnetic interaction between current loops in a simple plasma or electron gas</a>. </p> <h2 id="bessel-functions">Bessel functions</h2> <h3>Integration of the cylindrical propagator with mass</h3> <h4>First power of a Bessel function</h4> <p> ∫ 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).} </p><p>See <a href="/facts/Abramowitz_and_Stegun/CNGSV3Ed">Abramowitz and Stegun</a>.<a class="footnote-ref" id="fnref:9" href="#fn:9"><sup>9</sup></a>: §11.4.44 </p><p>For m r ≪ 1 {\displaystyle mr\ll 1} , we have<a class="footnote-ref" id="fnref:10" href="#fn:10"><sup>10</sup></a>: 116 K 0 ( m r ) → − ln ( m r 2 ) + 0.5772. {\displaystyle K_{0}(mr)\to -\ln \left({mr \over 2}\right)+0.5772.} </p><p>For an application of this integral see <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Two line charges embedded in a plasma or electron gas</a>. </p> <h4>Squares of Bessel functions</h4> <p>The integration of the propagator in cylindrical coordinates is<a class="footnote-ref" id="fnref:11" href="#fn:11"><sup>11</sup></a> ∫ 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).} </p><p>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].} </p><p>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).} </p><p>For applications of this integral see <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Magnetic interaction between current loops in a simple plasma or electron gas</a>. </p><p>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.} </p> <h3>Integration over a magnetic wave function</h3> <p>The two-dimensional integral over a magnetic wave function is<a class="footnote-ref" id="fnref:12" href="#fn:12"><sup>12</sup></a>: §11.4.28 2 a 2 n + 2 n ! ∫ 0 ∞ d r r 2 n + 1 exp ( − a 2 r 2 ) J 0 ( k r ) = M ( n + 1 , 1 , − k 2 4 a 2 ) . {\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).} </p><p>Here, <i>M</i> is a <a href="/facts/Confluent_hypergeometric_function/D0FxgWUa">confluent hypergeometric function</a>. For an application of this integral see <a href="/facts/Static_forces_and_virtual-particle_exchange/jgH4Np4C">Charge density spread over a wave function</a>. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Relation_between_Schr%25C3%25B6dinger%2527s_equation_and_the_path_integral_formulation_of_quantum_mechanics/ea9ns7Fs">Relation between Schrödinger's equation and the path integral formulation of quantum mechanics</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>A. Zee (2003). Quantum Field Theory in a Nutshell. Princeton University. ISBN 0-691-01019-6. <a href="0-691-01019-6" target="_blank">0-691-01019-6</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Frederick W. Byron and Robert W. Fuller (1969). Mathematics of Classical and Quantum Physics. Addison-Wesley. ISBN 0-201-00746-0. <a href="0-201-00746-0" target="_blank">0-201-00746-0</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>Herbert S. Wilf (1978). Mathematics for the Physical Sciences. Dover. ISBN 0-486-63635-6. <a href="0-486-63635-6" target="_blank">0-486-63635-6</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>A. Zee (2003). Quantum Field Theory in a Nutshell. Princeton University. ISBN 0-691-01019-6. <a href="0-691-01019-6" target="_blank">0-691-01019-6</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>A. Zee (2003). Quantum Field Theory in a Nutshell. Princeton University. ISBN 0-691-01019-6. <a href="0-691-01019-6" target="_blank">0-691-01019-6</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>A. Zee (2003). Quantum Field Theory in a Nutshell. Princeton University. ISBN 0-691-01019-6. <a href="0-691-01019-6" target="_blank">0-691-01019-6</a> <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> <li id="fn:7"><p>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. Translated by Scripta Technica, Inc. (8 ed.). Academic Press, Inc. ISBN 978-0-12-384933-5. LCCN 2014010276. <a href="978-0-12-384933-5" target="_blank">978-0-12-384933-5</a> <a href="#fnref:7" class="footnote-back-ref">↩</a></p></li> <li id="fn:8"><p>Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X. <a href="0-471-30932-X" target="_blank">0-471-30932-X</a> <a href="#fnref:8" class="footnote-back-ref">↩</a></p></li> <li id="fn:9"><p>M. Abramowitz; I. Stegun (1965). Handbook of Mathematical Functions. Dover. 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