 Normalized angle integration on the circle

On the circle group $\mathbb{T}$, normalized Haar measure is the pushforward measure of uniform angle on $[0,2\pi )$. For a continuous function $f:\mathbb{T}\to \mathbb{C}$, defineThe factor $1/(2\pi )$ makes the integral of the constant function $1$ equal to $1$. For $g=e^{i\varphi }$, a change of angle givesThe last equality uses the $2\pi$-periodic function $u\mapsto f(e^{iu})$. Since the circle group is an abelian group, this also verifies right invariance. An arc of angular length $L$ has normalized Haar measure $L/(2\pi )$. In particular, the arc $A=\{e^{i\theta }:\pi /6\le \theta \le \pi /2\}$ and its quarter-turn $iA$ both have measure $(\pi /2-\pi /6)/(2\pi )=1/6$.

The continuous unitary characters ${\chi }_n(e^{i\theta })=e^{in\theta }$, indexed by integers, give a fully computed family of integrals:Here $n$ must be an integer, so $z^n$ is well defined and $e^{2\pi in}=1$. Every nonconstant mode cancels exactly over a full turn. For example, the integral of $f(e^{i\theta })=3+2\cos \theta -\sin (2\theta )$ is $3$, and the integral of ${\cos }^2\theta =(1+\cos (2\theta ))/2$ is $1/2$. More generally,This is the orthogonality of complex exponentials underlying the Fourier basis and Fourier series: invariant averaging separates distinct frequencies and retains the constant component.