Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 5 / i / Solution

The triangular function is real and even, so its Fourier transform reduces to a cosine integral:For $k\ne 0$, one integration by parts gives ${\int }_0^1(1-x)\cos (kx)dx=(1-\cos k)/k^2$. Therefore the Fourier transform of a triangular function isThe zero-frequency value is the removable limit and also the area of the triangle. The transform is a squared sinc function, which explains its nonnegativity despite the oscillating Fourier exponential.