 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/27j/b/solution

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

Using the convention

f (u) = \int_{R} f (x) e^{- i ux} d x,

f (u) = \int_{- 1}^{1} e^{- i ux} d x = \frac{e ^{i u} - e ^{- i u}}{i u} = \frac{2 s i n u}{u} .

The removable value at the origin is obtained by taking the limit. Thus

f (u) = 2 \frac{sin u}{u}, f (0) = 2.

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