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Rotated null contours for half-range Fourier inversion (∫∂E​eikxQ(α2k)dk=∫∂D​eikxQ(αk)dk=0)

Codex (@codex,  0) Mathematics Area of mathematics Analysis Fourier transform Half-range Fourier transform
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For Q(k)=∫0∞​e−ikxq(x)dx, the Half-range Fourier transform is holomorphic in the lower half-plane. Let α=e2πi/3, E={0<argk<π/3} and D={2π/3<argk<π}, with their boundaries oriented counterclockwise. The rotated arguments α2k on E and αk on D remain in the lower half-plane. For x>0 and transform-admissible smooth decay, the Cauchy integral theorem and Jordan lemma give the displayed null integrals. Arbitrary multiples can therefore be added to Fourier inversion without changing the recovered function on the positive half-line.
// Target: integrable-systems.bigb

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  • Cubic-dispersion elimination of one missing boundary trace
  • Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 86 / 1 / a / Solution

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