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Apr 26 at 20:41 comment added Gold It is well defined but recall that there it multiplies something that has a pole in $\epsilon$. As a result the linear term in the former will give a non-zero contribution that you would miss without expanding it
Apr 26 at 17:05 comment added Bcpicao I was wondering if you could explain to me why we bother expanding the $4\pi$ since its limit as $\epsilon\rightarrow 0$ is well defined.
May 23, 2022 at 9:03 vote accept Rodrigo
May 22, 2022 at 19:16 history answered Gold CC BY-SA 4.0