= Solution
For an <odd> integrable <function>, both $f(x)$ and $f(x)\cos nx$ are <odd>. Their <integrals> over $[-\pi,\pi]$ therefore vanish, so the constant and <cosine> <Fourier coefficients> are zero. For $f(x)=\sin ax$, the remaining <Fourier sine coefficients> are
$$
b_n=\frac2\pi\int_0^\pi\sin ax\sin nx\,dx=\frac1\pi\left[\frac{\sin\pi(a-n)}{a-n}-\frac{\sin\pi(a+n)}{a+n}\right]=\frac{2(-1)^{n+1}n\sin\pi a}{\pi(n^2-a^2)}.
$$
The product-to-sum identity in the first step reduces the <integral> to two <cosines>. Here $a$ is real and noninteger, so no denominator vanishes. Consequently the <Fourier series of a noninteger-frequency sine> is
$$
\boxed{\sin ax=\sum_{n=1}^\infty\frac{2(-1)^{n+1}n\sin\pi a}{\pi(n^2-a^2)}\sin nx,\qquad -\pi<x<\pi.}
$$
At $x=\pm\pi$ every term is zero. The <Fourier series> converges there to the average of the two limits of the <periodic extension>, namely zero, rather than to the nonzero endpoint values $\pm\sin\pi a$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2-noninteger-sine-fourier.png]
{title=Fourier approximation to a noninteger-frequency sine and its periodic jump}
{description=Original illustration with $a=1/2$. Increasing the number of modes improves the interior approximation while the periodic endpoint jumps retain their Gibbs overshoot. The series has value zero at each endpoint.}
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