<s>[INST] Fourier expansion of $4x(1-x)$ Im triyng expand $4x(1-x)$ in a Fourier sine series in the interval $(-1\leq x \leq 1)$
where:
$f(x) = 4x(1-x), (0\leq x \leq 1)$
$f(x) = 4x(1+x), (-1\leq x \leq 0)$
But on my calculations the coeficient $b_n$ is going to zero, where is supose to be $b_n=\frac{32}{n^3\pi^3}$ if n is odd and $b_n=0$ if n is even. I'm not seeing where am i wrong.
so far what i got is:
$b_n = \int_{-1}^{0} (4x^2+4x)(\sin(nx\pi))dx + \int_{0}^{1} (4x^2-4x)(\sin(nx\pi))dx = \frac {(4\pi\sin(n\pi)+8\cos(n\pi)-8}{n^3\pi^3} - \frac {4\pi\sin(n\pi)+8\cos(n\pi)-8}{n^3\pi^3}$
Which is zero
 [/INST] It would take forever to typeset all the IBP, but we have
\begin{align}
b_n &= \int_{-1}^0 (4x+4x^2)\sin(n\pi x)dx+\int_{0}^1 (4x-4x^2)\sin(n\pi x)dx\\
&= \int_{-1}^0 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx+\int_{0}^1 4x\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&= \int_{-1}^1 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&= 2\int_{0}^1 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&=\frac{8(\sin(\pi n)-\pi n cos(\pi n))}{\pi^2 n^2}+\frac{4((\pi^2 n^2-2)\cos(\pi n)-2\pi n  \sin(\pi n)+2)}{\pi^3 n^3}+\frac{4(\pi^2 n^2-2)\cos(n \pi)-8 \pi n \sin(\pi n)+8}{\pi^3 n^3},
\end{align}
where we've used that $4x \sin(n \pi x)$ is even. You can check that each integral gives each of the terms listed in the last equality. This yields, as you should check,
$$b_n=
\begin{cases}
0 & \text{$n$ even} \\
\frac{32}{\pi^3 n^3} & \text{$n$ odd}
\end{cases}$$
</s>