Manuscript XVI-B@1 · XVI-B.tex · lemma · 6a0abffd

Lemma: A smooth-amplitude integral identity

Integral and scalar diagnostics › Integration by parts can introduce a second switch derivative · lines 994–1028 · proof included or linked

XVI-B@1#lem:ibp, XVI-B@1#eq:ibp

Let $a$ be smooth on a neighborhood of $[0, ε]$ , independent of $λ$ , and let $λ > 0$ . Define the illustrative integral

I_{toy}^{μν} (λ) = - i λ B^{μν} \int_{0}^{ε} a (t) σ^{'} (t) e^{2 i λ t} d t .

Then

I_{toy}^{μν} (λ) = \frac{B ^{μν}}{2} \int_{0}^{ε} (a^{'} σ^{'} + a σ^{''}) e^{2 i λ t} d t .

Proof. Set $b = a σ^{'}$ . By (eq:flat), $b (0) = b (ε) = 0$ . One integration by parts gives

\int_{0}^{ε} b e^{2 i λ t} d t = - \frac{1}{2 i λ} \int_{0}^{ε} b^{'} e^{2 i λ t} d t .

Since $b^{'} = a^{'} σ^{'} + a σ^{''}$ , multiplication by $- i λ B^{μν}$ proves (eq:ibp).

Displayed formulas in this passage

XVI-B@1#eq:ibp · line 1006
$I_{toy}^{μν} (λ) = \frac{B ^{μν}}{2} \int_{0}^{ε} (a^{'} σ^{'} + a σ^{''}) e^{2 i λ t} d t .$
unlabeled · line 999
$I_{toy}^{μν} (λ) = - i λ B^{μν} \int_{0}^{ε} a (t) σ^{'} (t) e^{2 i λ t} d t .$
unlabeled · line 1020
$\int_{0}^{ε} b e^{2 i λ t} d t = - \frac{1}{2 i λ} \int_{0}^{ε} b^{'} e^{2 i λ t} d t .$