Manuscript IX-E-2@1 · IX-E-2.tex · theorem · e70d9077

Theorem: Exact imaginary part

Evaluated imaginary part and its support · lines 837–877 · proof included or linked

IX-E-2@1#eq:imag-evaluated, IX-E-2@1#eq:imag-regions

On $t, s > 0$ 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 direct and reflected cone boundaries,

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

Proof. Substitute (eq:H-evaluated) and (eq:D-evaluated) into (eq:imag-decomposition). Since $t, s > 0$ ,

u = t + s > ∣ τ ∣.

The Heaviside difference is therefore one precisely in the open region $∣ τ ∣ < R < u$ , and zero in the other two open regions. The direct-cone delta contribution is kept separately.

Displayed formulas in this passage

IX-E-2@1#eq:imag-evaluated · line 839
$Im K (x, y) = \frac{t}{16 π R} [δ (τ - R) - δ (τ + R)] + \frac{1}{32 π R} [Θ (R - ∣ τ ∣) - Θ (R - u)] .$
IX-E-2@1#eq:imag-regions · line 856
$Im K = ⎩ ⎨ ⎧ 0, \frac{1}{32 π R}, 0, R < ∣ t - s ∣, ∣ t - s ∣ < R < t + s, R > t + s .$
unlabeled · line 871
$u = t + s > ∣ τ ∣.$

Theorem: Exact imaginary part

Displayed formulas in this passage

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