Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V / 4

Write $a=2\pi m+r$, where $m$ is an integer and $0\le r<2\pi$. The change of variables $x=y+2\pi m$ and the fact that $f$ is a periodic function giveSplit at $2\pi$, then substitute $y=z+2\pi$ in the second piece:ThereforeThis proves translation invariance of an integral over one period without assuming that $f$ is differentiable.

First let $b>0$. Substitute $x=-y$ in the negative half of the integral. Since $g$ is an odd function,The two halves cancel, soFor $b=0$ both limits coincide. For $b<0$, reverse the limits and apply the result with $-b>0$. Thus the identity holds for every real $b$.

The real Fourier series of $f$ is writtenwhere the Fourier coefficients areBy part (a)(i), any other interval of length $2\pi$ gives the same Fourier coefficients, because each integrand is a periodic function with period $2\pi$. The notation $\sim$ associates $f$ with its Fourier series; existence of the integrals alone does not assert pointwise convergence.

On $(-\pi ,\pi )$ the function is odd, so its mean and all its cosine Fourier coefficients vanish:The assigned endpoint value does not affect these integrals. The product of the two odd functions $x|x|$ and $\sin nx$ is even, hence the sine Fourier coefficients areTwo applications of integration by parts giveThereforeThe periodic extension has a jump discontinuity at every odd multiple of $\pi$: its left and right limits are ${\pi }^2$ and $-{\pi }^2$. Its assigned value at each such point is $-{\pi }^2$. Thus it is odd away from these points, but it is not a globally odd function pointwise. The Fourier series will take the midpoint value zero at these jumps.

By Dirichlet's theorem on Fourier series, a piecewise continuously differentiable periodic function has a Fourier series converging at each point to the average of its one-sided limits. In particular it converges to the function at every point of continuity. Our periodic extension satisfies these hypotheses, and $x=\pi /2$ is a point of continuity with $f(\pi /2)={\pi }^2/4$.

At this point $\sin (n\pi /2)=0$ for even $n$. For $n=2j-1$,Substituting in the Fourier series from part (c) provesThe sum is taken in its displayed order. Discarding its zero even-indexed terms is valid and does not rearrange the remaining terms.