= Fourier sine series of an odd quadratic arch
{c}
{title2=$x(L-|x|)=\frac{8L^2}{\pi^3}\sum_{m\geq0}\frac{\sin((2m+1)\pi x/L)}{(2m+1)^3}$}
Extend $F_L(x)=x(L-|x|)$ from $[-L,L]$ to a <periodic function> of period $2L$, with $L>0$. It is odd and continuous, so its <Fourier series> contains only sine terms. Its <Fourier coefficients> are
$$
b_n=\frac2L\int_0^Lx(L-x)\sin(n\pi x/L)\,dx
=\frac{4L^2(1-(-1)^n)}{n^3\pi^3}.
$$
Two applications of <integration by parts> prove the coefficient formula. The first has no endpoint contribution because $x(L-x)$ vanishes at both endpoints; the second leaves a multiple of $1-\cos(n\pi)$. Thus only odd frequencies survive, giving the displayed <Fourier sine series>. Its absolutely summable coefficients give uniform convergence, and <Fourier orthogonality> identifies its sum with $F_L$.
Taking $L=1$, $\int_{-1}^1F_1(x)^2dx=2\int_0^1x^2(1-x)^2dx=1/15$. <Parseval's identity> therefore gives
$$
\sum_{m=0}^\infty\frac1{(2m+1)^6}=\frac{\pi^6}{960}.
$$
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