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Sep 28, 2016 at 19:15 answer added user130529 timeline score: 1
Sep 26, 2016 at 3:23 comment added Sean E. Lake You can have a Dirac delta function that works with a rotation in complex coordinates, but it is decidedly non-analytic: $$\delta(x) = \lim_{\sigma\rightarrow 0} \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{x^*x}{2\sigma^2}\right). $$ It, also, would not spit out a factor of $i$.
Sep 26, 2016 at 1:10 answer added Sean E. Lake timeline score: 0
Sep 24, 2016 at 7:58 comment added flippiefanus Part of the problem could be the Dirac delta with an imaginary argument. What happens if one expresses Dirac delta in terms of its Fourier transform and then changes the variable to an imaginary number? The result does not seem to give a Dirac delta anymore.
Sep 24, 2016 at 6:43 comment added an offer can't refuse @flippiefanus It is, but my confusion is about the details about the analytical continuation. The listed questions and the concerns. For example, are the second attempt correct? why not equal to the first one. Why $\sqrt{-1}=i$ and $\sqrt{-1}=-1$ give different result, which one to choose, etc...
Sep 24, 2016 at 5:00 comment added flippiefanus Have you considered the Wick rotation? The analytic continuation in attempt 1 looks very similar to the Wick rotation to me.
Sep 22, 2016 at 12:38 history tweeted twitter.com/StackPhysics/status/778936574109188097
S Sep 22, 2016 at 4:53 history bounty started an offer can't refuse
S Sep 22, 2016 at 4:53 history notice added an offer can't refuse Canonical answer required
Sep 19, 2016 at 12:19 history edited Qmechanic CC BY-SA 3.0
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Sep 19, 2016 at 10:56 history edited an offer can't refuse CC BY-SA 3.0
layout edit and try to make the question clearer.
Sep 19, 2016 at 9:53 history asked an offer can't refuse CC BY-SA 3.0