 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/4/b/solution

k \neq = 0

g (k) = \int_{- 1}^{1} e^{- ik x} d x = \frac{e ^{ik} - e ^{- ik}}{ik} = \frac{2 sin k}{k} .

k = 0

the defining integral is the interval length, so

g (0) = 2

. This also equals the continuous limit of the displayed formula.

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