 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/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

\int ∣ f ∣^{2} d x = (2 π)^{- 1} \int ∣ f ∣^{2} d k

. For smooth rapidly decaying functions its bilinear form follows by inserting the inverse transform of

g

into

\int f \overline{g} d x

and interchanging the integrals:

\int f (x) \overline{g (x)} d x = \frac{1}{2 π} \int f (k) \overline{g (k)} d k .

Take

g = f

. Approximation by such functions extends the identity to L2 space, which includes the compact triangular function here.

For that triangle,

\int ∣ f ∣^{2} d x = 2 \int_{0}^{1} (1 - x)^{2} d x = 2/3

, while

f (k) = [sin (k /2) / (k /2)]^{2}

. Thus

\frac{2}{3} = \frac{1}{2 π} \int_{- \infty}^{\infty} (\frac{s i n ( k /2 )}{k /2})^{4} d k = \frac{2}{π} \int_{0}^{\infty} (\frac{s i n u}{u})^{4} d u,

\int_{0}^{\infty} \frac{sin ^{4} u}{u ^{4}} d u = \frac{π}{3} .

The continuous limiting integrand at zero is one. This is the sinc fourth-power integral.

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