Periodic sine with an endpoint-correcting linear term ($b_n=2(-1{)}^n\sin L/\{n\pi [1-(n\pi /L{)}^2]\}$)

For $L>0$, extend $f_L(x)=\sin x-x\sin L/L$ from $[-L,L]$ periodically. The linear term makes both endpoint values zero and the endpoint first derivatives match. The function is odd, so its Fourier series has only sine terms, with the displayed Fourier coefficients. Indeed, for $k_n=n\pi /L\ne 1$,subtracting and multiplying by $2/L$ proves the formula. The coefficients decay as $O(n^{-3})$. At $L=m\pi$, the single coefficient $b_m$ has removable limit $1$, with all others zero. Multiplying $b_n$ by $e^{-\kappa k_n^{2}t}$ solves the corresponding heat equation on the circle.