= Solution
{c}
On $(-\pi,\pi)$ the <function> is <odd>, so its mean and all its <cosine> <Fourier coefficients> vanish:
$$
a_0=0,\qquad a_n=0\quad(n\geq1).
$$
The assigned endpoint value does not affect these <integrals>. The product of the two <odd functions> $x|x|$ and $\sin nx$ is <even>, hence the <sine> <Fourier coefficients> are
$$
b_n=\frac2\pi\int_0^\pi x^2\sin nx\,dx.
$$
Two applications of <integration by parts> give
$$
\begin{aligned}
\int_0^\pi x^2\sin nx\,dx
&=\left[-\frac{x^2\cos nx}{n}\right]_0^\pi+\frac2n\int_0^\pi x\cos nx\,dx\\
&=-\frac{\pi^2(-1)^n}{n}+\frac2n\left[\frac{x\sin nx}{n}+\frac{\cos nx}{n^2}\right]_0^\pi\\
&=-\frac{\pi^2(-1)^n}{n}+\frac{2[(-1)^n-1]}{n^3}.
\end{aligned}
$$
Therefore
$$
\boxed{f(x)\sim\sum_{n=1}^{\infty}\left[\frac{2\pi(-1)^{n+1}}n+\frac{4[(-1)^n-1]}{\pi n^3}\right]\sin nx.}
$$
The <periodic extension> has a <jump discontinuity> at every odd multiple of $\pi$: its left and right limits are $\pi^2$ and $-\pi^2$. Its assigned value at each such point is $-\pi^2$. Thus it is <odd> away from these points, but it is not a globally <odd function> pointwise. The <Fourier series> will take the midpoint value zero at these jumps.
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