= Solution
The function is odd, so its <Fourier series> has no constant or cosine terms. If $\lambda=m\in\mathbb Z$, it is already a Fourier basis function: for $m>0$ the only sine coefficient is $b_m=1$; for $m<0$ it is $b_{|m|}=-1$; for $m=0$ the function is zero.
For real noninteger $\lambda$, the product-to-sum identity gives
$$
b_n=\frac2\pi\int_0^\pi\sin(\lambda x)\sin(nx)\,dx
=\frac1\pi\left[\frac{\sin((\lambda-n)\pi)}{\lambda-n}-\frac{\sin((\lambda+n)\pi)}{\lambda+n}\right].
$$
Since both numerator sines equal $(-1)^n\sin(\pi\lambda)$, the <Fourier series of a noninteger-frequency sine> is
$$
\boxed{\sin(\lambda x)\sim\sum_{n\geq1}\frac{2(-1)^{n+1}n\sin(\pi\lambda)}{\pi(n^2-\lambda^2)}\sin(nx).}
$$
It equals the function for $-\pi<x<\pi$. At either periodic endpoint it equals the average of the two one-sided limits, namely zero; the noninteger-frequency function has a jump in its periodic extension there. Equality also holds in the square-integral sense on the full interval.
For $\lambda=1/2$, $b_n=8(-1)^{n+1}n/[\pi(4n^2-1)]$ and $\int_{-\pi}^\pi\sin^2(x/2)\,dx=\pi$. The <Parseval identity> for a real sine series is $(1/\pi)\int_{-\pi}^\pi f^2=\sum_{n\geq1}b_n^2$. Hence
$$
1=\frac{64}{\pi^2}\sum_{n\geq1}\frac{n^2}{(4n^2-1)^2},\qquad
\boxed{\sum_{n\geq1}\frac{n^2}{(4n^2-1)^2}=\frac{\pi^2}{64}.}
$$
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