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Explanation - everything real-valued: $$ (x,y) = \mbox{cartesian coordinates} \\ \phi = \mbox{angle, in polar coordinates} \\ H(x) = \begin{cases} 0 & \mbox{for} & x < 0 \\ 1 & \mbox{for} & x > 0 \end{cases} \quad \mbox{: Heaviside function}\\ \delta(y) = \begin{cases} 0 & \mbox{for} & y \ne 0 \\ \infty & \mbox{for} & y = 0 \end{cases} \quad \mbox{and} \quad \int_{-\infty}^{+\infty} \delta(y) \,dy = 1 \quad \mbox{: Dirac delta} $$ The meaning of the formula is a topological one:

Crossing number = Winding number

Here is a visualization. WLOG the origin $(0,0)$ is taken at the black dots inside/outside a domain. The integrations are carried out over the boundary of that domain.

A proof of the formula is found at this place.
I want to know if this formula has been found by others (< 2005); any references would be quite welcome.

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  • $\begingroup$ if we attend just to the notation we will have that $$\oint H(x)\delta(y) d(y)=H(x)\oint \delta(y)d(y)=i H(x)\int_{-\pi}^{\pi}e^{is}\delta(e^{i s})ds=0$$ $\endgroup$ – Masacroso Oct 8 at 16:24
  • $\begingroup$ @Masacroso: The infinitesimal $dy$ is not meant to be an arc-length increment. I've updated the question accordingly. Sorry if I've caused any confusion. $\endgroup$ – Han de Bruijn Oct 9 at 12:08
  • $\begingroup$ I think I have seen this in computational geometry for solid (CAD) operations: take a point, shoot a ray out to infinity, count the number of intersections, odd number of intersections means the point is on the inside of the domain. I think you may find it in books and papers on "constructive solid geometry", perhaps under "ray shooting". $\endgroup$ – Jap88 Nov 21 at 1:55
  • $\begingroup$ @Jap88: Hmm, I've always thought that CSG (Computational Solid Geometry) is 3-D. But my statement is 2-D (though it is "ray shooting" indeed). How does the 3-D compare with the 2-D? $\endgroup$ – Han de Bruijn Nov 21 at 14:46
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    $\begingroup$ From a ray shooting perspective 2D and 3D is more or less the same thing. Formulating it as integrals will be a bit different I think in 3D and probably quite a bit more difficult and technical. $\endgroup$ – Jap88 Nov 21 at 15:20

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