Common integrals in quantum field theory: Difference between revisions

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

Note that, if we let <math>r=\sqrt{x^2+y^2}</math> be the radius, then we can use the usual polar coordinate change of variables (which in particular renders <math>dx\,dy=r\,dr\,d\theta</math>) to get

Note:

<math display="block"> 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} r e^{-{1 \over 2} r^2}\,dr = 2\pi \int_{0}^{\infty} e^{- w}\,dw = 2 \pi.</math>

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==== Integrals with a linear term in the argument ====

<math display="block">\int \exp\left(-\frac{1}{2} x^{T} \cdot A \cdot x + J^{T} \cdot x \right) d^nxdx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( {1\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

==== Integrals with an imaginary linear term ====

<math display="block">\int \exp\left(-\frac{1}{2} x^{T} \cdot A \cdot x +iJi J^{T} \cdot x \right) d^nxdx = \sqrt{\frac{(2\pi)^n}{\det A}} \exp \left( -{1\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

==== Integrals with a complex quadratic term ====

<math display="block">\int \exp\left(\frac{i}{2} x^{T} \cdot A \cdot x +iJi J^{T} \cdot x \right) d^nxdx =\sqrt{\frac{(2\pi i)^n}{\det A}} \exp \left( -{i\over 2} J^{T} \cdot A^{-1} \cdot J \right)</math>

=== Integrals with differential operators in the argument ===