 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/4/a/ii/solution

f (k) = - 2 i \int_{0}^{1} (1 - x) sin (k x) d x, f (k) = - \frac{2 i ( k - sin k )}{k ^{2}} .

The value at

k = 0

is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.

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