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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 givesSince both numerator sines equal $(-1{)}^n\sin (\pi \lambda )$, the Fourier series of a noninteger-frequency sine isIt 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 $(1/\pi ){\int }_{-\pi }^{\pi }{f}^2={\sum }_{n\ge 1}{b}_n^2$. Hence

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 areThe 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 isAt $x=\pm \pi$ 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 $\pm \sin \pi a$.