Example 3.7 . [04ZM] Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.
Source coverage notes Source fidelity gap: the original arXiv HTML omits an author TeX footnote attached to equation e:def-omega, including its reference to Remark r:error-function. The retained HTML is preserved as published; the original author TeX remains available. TeX correspondence is not complete. Complete original source context · Original author HTML
Example 3.7 .
The above normalization constant is chosen such that in the case Q ≡ ℝ 3 Q\equiv\mathbb{R}^{3} P ≡ 0 3 ∈ ℝ 3 P\equiv 0^{3}\in\mathbb{R}^{3}
(3.43) G P = 1 2 | y | d y 1 ∧ d y 2 ∧ d y 3 G_{P}=\frac{1}{2|y|}dy_{1}\wedge dy_{2}\wedge dy_{3}
solves the current equation Δ 0 G P = 2 π ⋅ δ P \Delta_{0}G_{P}=2\pi\cdot\delta_{P} Δ 0 \Delta_{0} ℝ 3 \mathbb{R}^{3}