Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 15 / c

Again the function is even, so all sine coefficients vanish. Its constant term is $a_0/2=\sinh 1$. With $k=n\pi$, an antiderivative isThus $a_n=2(-1{)}^n\sinh 1/(1+n^2{\pi }^2)$. In the requested factored notation,This is the period-two specialization of the Fourier series of periodic hyperbolic sine and cosine. The function values match at the endpoints, although its derivative has a jump in the periodic extension.