 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/4/c/solution

The real odd function has a purely imaginary odd Fourier transform. Setting

A = π /2

and using a product-to-sum identity,

f (k) = - 2 i \int_{0}^{A} sin x sin (k x) d x = - i [\frac{sin [( k - 1 ) A ]}{k - 1} - \frac{sin [( k + 1 ) A ]}{k + 1}] = \frac{2 ik cos ( πk /2 )}{k ^{2} - 1}, k \neq = \pm 1.

The apparent singularities are removable. Direct evaluation or limits give

f (1) = - \frac{iπ}{2}, f (- 1) = \frac{iπ}{2}, f (0) = 0.

In particular, the transform stays finite where its numerator and denominator both vanish.

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