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Feb 25, 2022 at 1:45 comment added Ruvi Lecamwasam Oh I see, yes that makes sense, thanks!
Feb 24, 2022 at 17:36 comment added AfterShave You misunderstand by answer, I am taking the surface term and integrating it over a test function in $x$. The idea is that these expressions are distributions and only their effect on test functions under the integral sign matters.
Feb 24, 2022 at 13:30 comment added Ruvi Lecamwasam That was my first instinct. Unfortunately the integral is over k, not x, so it is k that's going to infinity.
Feb 24, 2022 at 10:17 history answered AfterShave CC BY-SA 4.0