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3

Apologies for not producing a most general answer for arbitrary Lie groups, (which you might tease with great effort out of WP ), but only a trail-map for your particular (charmed!) problem. I call it charmed because it should remind you of the Lorentz group, with a,b,c parameterizing Kx,Ky,Kz boosts and d,e,f the three J rotation angles. Decent treatments ...


0

So I actually integrated this a while ago, I thought I should share the result. It turned out rather simple with the right substitution. By the way I'm talking about both potential and charge disrtibution as a Debye-like. Point charge is straight forward... First we put the center of the charge density on the z-axis as mentioned in my question. I get $$ ...


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You will get infinity because in addition to $kQ_1Q_2/d$, it also includes the self-energy of the two point charges, which is infinity.


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This is just complementary to Alex's answer. For the second integral the book provides an analysis in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$ \frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}} $$ At the limit $r\rightarrow \infty$, the effect ...



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