Fourier sine series of a squared sine (${\sin }^2(\pi x/L)=\frac{8}{\pi }{\sum }_{m\ge 0}\frac{\sin ((2m+1)\pi x/L)}{(2m+1)[4-(2m+1{)}^2]}$)

On $0\le x\le L$, the Fourier sine series of ${\sin }^2(\pi x/L)$ has zero even-index coefficients and odd-index coefficients $b_n=8/[\pi n(4-n^2)]$. Substitute $u=\pi x/L$ in the coefficient integral and use ${\sin }^{2}u\sin (nu)=\frac{1}{2}\sin (nu)-\frac{1}{4}\sin ((n+2)u)-\frac{1}{4}\sin ((n-2)u)$. For odd $n$, integrating gives $b_n=(2/\pi )[1/n-\frac{1}{2}(1/(n+2)+1/(n-2))]$; reflection about the midpoint makes the even coefficients zero. The $O(n^{-3})$ decay gives absolute and uniform convergence, and Fourier orthogonality identifies the sum. In the heat equation with zero endpoint temperatures, multiply each coefficient by $\exp [-\kappa (n\pi /L{)}^{2}t]$. With insulated endpoints, the same initial profile is instead just the constant and second cosine modes of a Fourier cosine series.