Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 4 / c

The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.