= Fourier series of a noninteger-frequency cosine
{c}
{title2=$\cos(\mu x),\quad\mu\in\mathbb R\setminus\mathbb Z$}
Extend $\cos(\mu x)$ from $[-\pi,\pi]$ with period $2\pi$. It is an <even function>, so all sine <Fourier coefficients> vanish. The product-to-sum identity gives
$$
a_0=\frac{2\sin(\pi\mu)}{\pi\mu},\qquad
a_n=\frac{2\mu(-1)^n\sin(\pi\mu)}{\pi(\mu^2-n^2)}\quad(n\geq1).
$$
Indeed, integrate $\cos(\mu x)\cos(nx)$ on $[0,\pi]$ and use $\sin(\pi(\mu\pm n))=(-1)^n\sin(\pi\mu)$. Thus
$$
f(x)=\frac{\sin(\pi\mu)}{\pi\mu}
+\frac{2\mu\sin(\pi\mu)}\pi\sum_{n=1}^\infty\frac{(-1)^n\cos(nx)}{\mu^2-n^2}.
$$
The coefficients are $O(n^{-2})$, so the series converges absolutely and uniformly. The <Dirichlet-Jordan convergence theorem> identifies it with the periodic function everywhere, including the endpoints because the two endpoint values agree. Substituting a noninteger rational frequency and a convenient $x$ yields explicit <series> sums. Integer frequencies require taking removable limits rather than substituting into a zero denominator.
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