Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 27J / c / Solution

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With the standard Fourier convention, <Plancherel theorem> applied to $\mathbf1_{[-1,1]}$ gives
$$
\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^2dx=\pi,
\qquad
\int_0^\infty\left(\frac{\sin x}{x}\right)^2dx=\frac\pi2.
$$

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