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<math display="block"> \int_{-\infty}^{\infty} \exp\left( - \frac 1 2 a x^2 + Jx\right ) dx </math>
This integral can be performed by [[completing the square]]:
<math display="block"> \left( -{1 \over 2} a x^2 + Jx\right ) = -{1 \over 2} a \left ( x^2 - { 2 Jx \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 } </math>
Line 277:
==== Modified Coulomb potential with mass ====
<math display="block">\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]</math>
where the hat indicates a [[unit vector]] in three dimensional space. The derivation of this result is as follows:
<math display="block">\begin{align}
&\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]
Line 327:
<math display="block">\int_0^{\infty} {k\; dk \over k^2 +m^2} J_0 \left( kr \right)=K_0 (mr). </math>
See [[Abramowitz and Stegun]].<ref name="AbramowitzStegun">{{cite book| author1=M. Abramowitz | author2 = I. Stegun| title=Handbook of Mathematical Functions| publisher=Dover| year=1965| isbn=0486-61272-4| url-access=registration| url=https://archive.org/details/handbookofmathe000abra}}</ref>{{rp|at=§11.4.44}}
For <math> mr \ll 1 </math>, we have<ref name="Jackson"/>{{rp|p=116}}
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