= Normalized angle integration on the circle
{title2=$d\mu=d\theta/(2\pi)$}
On the <circle group> $\mathbb T$, normalized <Haar measure> is the <pushforward measure> of uniform angle on $[0,2\pi)$. For a <continuous function> $f:\mathbb T\to\mathbb C$, define
$$
\int_{\mathbb T}f(z)\,d\mu(z)=\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta.
$$
The factor $1/(2\pi)$ makes the <integral> of the constant function $1$ equal to $1$. For $g=e^{i\phi}$, a change of angle gives
$$
\int_{\mathbb T}f(gz)\,d\mu(z)
=\frac1{2\pi}\int_\phi^{2\pi+\phi}f(e^{iu})\,du
=\frac1{2\pi}\int_0^{2\pi}f(e^{iu})\,du.
$$
The 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 $A=\{e^{i\theta}:\pi/6\leq\theta\leq\pi/2\}$ and its quarter-turn $iA$ both have <measure> $(\pi/2-\pi/6)/(2\pi)=1/6$.
\Image[/circle-group-rotation-invariance.png]
{title=A quarter-turn changes the position of an arc but preserves its normalized Haar measure}
{description=Multiplication by $i=e^{i\pi/2}$ carries the blue arc onto the orange arc. Both span $\pi/3$ and have <Haar measure> $1/6$. The diagram is generated by \a[circle-group-rotation-invariance.py][circle-group-rotation-invariance.py].}
The <continuous unitary characters> $\chi_n(e^{i\theta})=e^{in\theta}$, indexed by <integers>, give a fully computed family of <integrals>:
$$
\int_{\mathbb T}z^n\,d\mu(z)
=\frac1{2\pi}\int_0^{2\pi}e^{in\theta}\,d\theta
=\begin{cases}
1,&n=0,\\
\displaystyle\frac{e^{2\pi in}-1}{2\pi in}=0,&n\ne0.
\end{cases}
$$
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,
$$
\int_{\mathbb T}\chi_n(z)\overline{\chi_m(z)}\,d\mu(z)
=\begin{cases}1,&n=m,\\0,&n\ne m.\end{cases}
$$
This is the <orthogonality of complex exponentials> underlying the <Fourier basis> and <Fourier series>: invariant averaging separates distinct frequencies and retains the constant component.
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