= Solution
Take the odd extension $H(x)=x(\pi-|x|)$ on $[-\pi,\pi]$ and then extend with period $2\pi$. It agrees with the required arch for $0\leq x\leq\pi$ and is continuous at its joins. Its <Fourier sine series> has
$$
b_n=\frac2\pi\int_0^\pi x(\pi-x)\sin nx\,dx.
$$
For clarity, let $h=x(\pi-x)$, so $h'=\pi-2x$ and $h''=-2$. Two integrations by parts, with the first endpoint term zero, give
$$
\begin{aligned}
\int_0^\pi h\sin nx\,dx
&=\frac1n\int_0^\pi h'\cos nx\,dx\\
&=-\frac1{n^2}\int_0^\pi h''\sin nx\,dx
=\frac{2[1-(-1)^n]}{n^3}.
\end{aligned}
$$
Consequently $b_n=4[1-(-1)^n]/(\pi n^3)$: the even coefficients vanish and the odd ones are $8/(\pi n^3)$. This proves
$$
\boxed{h(x)=\frac8\pi\sum_{n=0}^{\infty}\frac{\sin((2n+1)x)}{(2n+1)^3},\qquad0\leq x\leq\pi.}
$$
The <Fourier sine series of an odd quadratic arch> converges to this continuous extension; its coefficients are also absolutely summable. Finally evaluate at $x=\pi/2$. The left side is $\pi^2/4$ and the <sine> factors are $(-1)^n$, so
$$
\boxed{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}.}
$$
The evaluation is at a <continuity> point, so it does not invoke a jump midpoint.
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