 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-2/7a/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

φ

and nonzero

p

, substitution

u = pt

with the orientation handled by

∣ p ∣

gives

\int φ (t) δ (pt) d t = \frac{1}{∣ p ∣} φ (0) .

Hence the Dirac delta scaling identity is $δ (pt) = δ (t) /∣ p ∣$ .

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