 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/d/solution

The squared norm of the original function is

\int_{R} ∣ f (x) ∣^{2} d x = \int_{- π /2}^{π /2} sin^{2} x d x = \frac{π}{2} .

Applying the Parseval identity to part (c), using the evenness of the squared transform, gives

\frac{π}{2} = \frac{1}{2 π} \int_{R} \frac{4 k ^{2} c o s ^{2} ( πk /2 )}{( k ^{2} - 1 ) ^{2}} d k = \frac{4}{π} \int_{0}^{\infty} \frac{k ^{2} c o s ^{2} ( πk /2 )}{( k ^{2} - 1 ) ^{2}} d k .

Thus the last integral equals

π^{2} /8

. Now set

u = πk /2

. Since

\frac{k ^{2} d k}{( k ^{2} - 1 ) ^{2}} = \frac{π}{2} \frac{u ^{2} d u}{( u ^{2} - π ^{2} /4 ) ^{2}},

we obtain

\int_{0}^{\infty} \frac{u ^{2} cos ^{2} u}{( u ^{2} - π ^{2} /4 ) ^{2}} d u = \frac{2}{π} \frac{π ^{2}}{8} = \frac{π}{4} .

The apparent singularity at

u = π /2

is removable: the continuous value of the integrand there is

1/4

. No principal-value interpretation is needed.

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