Manuscript IX-D-ST@1 · IX-D-ST.tex · proposition · b1dc631a

Proposition: Exact direct-plus-reflected subtraction

Common-regulator evaluation · lines 1438–1468 · proof included or linked

IX-D-ST@1#prop:subtractionkernel, IX-D-ST@1#eq:Kdifference, IX-D-ST@1#eq:imagdecomp

One has

K - K = \frac{i t}{2} D (τ, z) + \frac{i}{4} (H (τ, z) - H (u, z)),

and therefore

Im K = \frac{t}{4} D (τ, z) + \frac{1}{8} (H (τ, z) - H (u, z)) .

Proof. Conjugate at fixed $x, y$ and replace $p$ by $- p$ . For $q = ∣ p ∣$ , the direct difference is

- \frac{t}{4 q} (e^{- i q τ} - e^{i q τ}) + \frac{i}{8 q ^{2}} (e^{- i q τ} + e^{i q τ}) = \frac{i t}{2 q} sin (q τ) + \frac{i}{4 q ^{2}} cos (q τ) .

The reflected difference is $- i cos (q u) / (4 q^{2})$ . The identities hold with the same Abel factor, and Lemma lem:Fourier permits its distributional removal.

Displayed formulas in this passage

IX-D-ST@1#eq:Kdifference · line 1441
$K - K = \frac{i t}{2} D (τ, z) + \frac{i}{4} (H (τ, z) - H (u, z)),$
IX-D-ST@1#eq:imagdecomp · line 1448
$Im K = \frac{t}{4} D (τ, z) + \frac{1}{8} (H (τ, z) - H (u, z)) .$
unlabeled · line 1459
$- \frac{t}{4 q} (e^{- i q τ} - e^{i q τ}) + \frac{i}{8 q ^{2}} (e^{- i q τ} + e^{i q τ}) = \frac{i t}{2 q} sin (q τ) + \frac{i}{4 q ^{2}} cos (q τ) .$

Proposition: Exact direct-plus-reflected subtraction

Displayed formulas in this passage

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