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

Theorem: Exact imaginary part

Common-regulator evaluation · lines 1521–1552 · proof included or linked

IX-D-ST@1#thm:imag, IX-D-ST@1#eq:imag

For $t, s > 0$ and $R > 0$ ,

Im K (x, y) = \frac{t}{16 π R} [δ (τ - R) - δ (τ + R)] + \frac{1}{32 π R} [Θ (R - ∣ τ ∣) - Θ (R - u)] .

Away from the two cone boundaries it equals

Im K = ⎩ ⎨ ⎧ 0, (32 π R)^{- 1}, 0, R < ∣ t - s ∣, ∣ t - s ∣ < R < t + s, R > t + s .

The reflected cone is a jump locus, not an additional delta distribution.

Proof. Substitute Lemma lem:radial into (eq:imagdecomp), using $t + s > ∣ t - s ∣$ . The direct delta terms are retained separately.

Displayed formulas in this passage

IX-D-ST@1#eq:imag · line 1524
$Im K (x, y) = \frac{t}{16 π R} [δ (τ - R) - δ (τ + R)] + \frac{1}{32 π R} [Θ (R - ∣ τ ∣) - Θ (R - u)] .$
unlabeled · line 1536
$Im K = ⎩ ⎨ ⎧ 0, (32 π R)^{- 1}, 0, R < ∣ t - s ∣, ∣ t - s ∣ < R < t + s, R > t + s .$