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

Let $r(z)$ be the indicator of $[-1,1]$. The finite integral is exactly the convolution $p=q\ast r$. Its additional transform factor isApplying the convolution theorem and part (d),Equivalently this is $2(\sin k/k)[\sin (k/2)/(k/2){]}^2$. The value at zero also follows from $\int p=(\int q)(\int r)=1\times 2$.