Manuscript Em01-E@2 · Em01-E-v2.tex · lemma · 54f8f8c2

Lemma: Exact rescaled complete integral

Evaluated complete degenerate specialization › Exact integral and density · lines 1778–1825 · proof included or linked

Em01-E@2#lem:appintegral, Em01-E@2#eq:appintegral

For all sufficiently small scales,

t_{δ} (Φ) = - \frac{i}{2} j = 0 \sum^{2} δ^{j - 2} \int d τ \int_{R} d S \int_{0}^{\infty} d T p (S) p^{(2 - j)} (S - T) \times \frac{1}{2 π} \int_{0}^{2 π} F_{j} (τ; \frac{δ T}{2}, \frac{δ S}{2}, 2 B δ T e_{θ}) d θ,

where $e_{θ} = (cos θ, sin θ)$ . The tangential integration is over the fixed coordinate range of the patch; zero extension of the tests makes its relevant support compact.

Proof. Integrate $z$ against the incidence. The endpoint densities are

d^{4} b = 2 d B d u d^{2} X, d^{4} y = 2 d A d v d^{2} Y .

On the patch the future retarded kernel is

\frac{1}{2 π} δ_{D} (4 B u - ∣ X ∣^{2}) .

Its $u$ -integration contributes $1/ (4 B)$ . The combined factor is therefore

4 \cdot \frac{1}{2 π} \cdot \frac{1}{4 B} = \frac{1}{2 π B} .

With angular averaging and $Q_{X} = ∣ X ∣^{2}$ , $d^{2} X = π d Q_{X}$ . Set $t = Q_{X} / (2 B)$ and $s = 2 v$ . The resulting measure factor is exactly $d s d t /2$ . Finally $s = δ S, t = δ T$ contributes $δ^{2}$ . Substitution in (eq:appexact) proves the identity.

Displayed formulas in this passage

Em01-E@2#eq:appintegral · line 1781
$t_{δ} (Φ) = - \frac{i}{2} j = 0 \sum^{2} δ^{j - 2} \int d τ \int_{R} d S \int_{0}^{\infty} d T p (S) p^{(2 - j)} (S - T) \times \frac{1}{2 π} \int_{0}^{2 π} F_{j} (τ; \frac{δ T}{2}, \frac{δ S}{2}, 2 B δ T e_{θ}) d θ,$
unlabeled · line 1803
$d^{4} b = 2 d B d u d^{2} X, d^{4} y = 2 d A d v d^{2} Y .$
unlabeled · line 1808
$\frac{1}{2 π} δ_{D} (4 B u - ∣ X ∣^{2}) .$
unlabeled · line 1813
$4 \cdot \frac{1}{2 π} \cdot \frac{1}{4 B} = \frac{1}{2 π B} .$

Lemma: Exact rescaled complete integral

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