Skip to main content
13 events
when toggle format what by license comment
Jul 5, 2017 at 17:54 comment added Anixx By the way, when $k=0$ the integral should be simply $\pi\delta(0)$. Is it the case where notation $\cal P$ plays role?
Jul 5, 2017 at 16:04 comment added Carlo Beenakker the Wikipedia article has several applications of the Sokhotski–Plemelj integral relation, and a Google search gives many more pointers; a Fourier-transform formulation is here (appendix).
Jul 5, 2017 at 15:56 comment added Anixx Oh, possibly I understood now, in this notation $\cal P$ is not a factor, so we can simply disregard it. But does this method allow to integrate an arbitrary function (or at least, could we obtain values better than in Fourier transform?)
Jul 5, 2017 at 15:36 comment added Anixx Purely logically the integral should be positive because sine function is initially positive. It is analogious to the series 1-1+1-1+... which sums up to 1/2, while if the first term is -1, the sum would be -1/2
Jul 5, 2017 at 13:42 history edited Carlo Beenakker CC BY-SA 3.0
added 56 characters in body
Jul 5, 2017 at 13:25 comment added Carlo Beenakker indeed, my mistake, corrected, thanks; $g(x)$ is an arbitrary function.
Jul 5, 2017 at 13:22 history edited Carlo Beenakker CC BY-SA 3.0
deleted 10 characters in body
Jul 5, 2017 at 13:11 comment added Anixx If you get -1, you definitely made some mistake, I think. It is still unclear though how the first and second formulas are related (what is g(x) for instance).
Jul 5, 2017 at 13:08 history edited Carlo Beenakker CC BY-SA 3.0
added 113 characters in body
Jul 5, 2017 at 12:55 history edited Carlo Beenakker CC BY-SA 3.0
added 106 characters in body
Jul 5, 2017 at 12:03 comment added Anixx Still not clear, principal value of what integral?...
Jul 5, 2017 at 12:01 history edited Carlo Beenakker CC BY-SA 3.0
added 89 characters in body
Jul 5, 2017 at 11:59 history answered Carlo Beenakker CC BY-SA 3.0