 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/4/e/solution

Let

r (z)

be the indicator of

[- 1, 1]

. The finite integral is exactly the convolution

p = q * r

. Its additional transform factor is

r (k) = \int_{- 1}^{1} e^{- ik z} d z = \frac{2 s i n k}{k}, r (0) = 2.

Applying the convolution theorem and part (d),

p (k) = \frac{4 sin k ( 1 - cos k )}{k ^{3}}, p (0) = 2.

Equivalently this is

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

. The value at zero also follows from

\int p = (\int q) (\int r) = 1 \times 2

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