 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/4/c/solution

\int_{R} ∣ h (x) ∣^{2} d x = \frac{1}{2 π} \int_{R} ∣ h (k) ∣^{2} d k .

Apply it to

h = g

. Its squared norm is

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

, so

\frac{2}{3} = \frac{1}{2 π} \int_{R} \frac{16 s i n ^{4} ( k /2 )}{k ^{4}} d k .

Under

k = 2 x

d k

becomes

2 sin^{4} x d x / x^{4}

. Consequently

\int_{- \infty}^{\infty} \frac{sin ^{4} x}{x ^{4}} d x = \frac{2 π}{3} .

The integrand has continuous value

1

at zero and decays integrably at infinity, so this is an ordinary convergent improper integral.

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