I am trying to compute the Fourier series of function $f(x)=x(\pi-x)$ on the interval $(0,\pi)$
The final answer here is $f(x)=\frac{8}{\pi} \sum_{k odd\geq 1} \frac{sin(kx)}{k^3}$
I compute the Fourier coefficients as follows.
$f(n)=\frac{1}{\pi} \int_{0}^{\pi} x(\pi-x)e^{-2inx}dx$
=$\frac{1}{\pi}[\frac{(-1)x(pi-x)e^{-2inx}}{2in}]\Big|_0^\pi + \frac{1}{\pi} \int_{0}^{\pi} \frac{e^{-2inx}(\pi-2x)}{2in}dx$
=$0+\frac{1}{\pi} \int_{0}^{\pi} \frac{e^{-2inx}(\pi-2x)}{2in}dx$
$\frac{1}{\pi}[\frac{e^{-2inx}(\pi-2x)}{4n^2}] \Big|_0^\pi - \frac{1}{\pi} \int_{0}^{\pi} \frac{e^{-2inx}(-2)}{4n^2}dx$
=$\frac{-1}{2n^2}+[\frac{e^{-2inx}}{8in^3}] \Big|_0^\pi$
=$\frac{-1}{2n^2}$
Now, I have been working on this problem for a while and know that this cannot be correct. I must have made an error setting it up or evaluating because the $n^3$ term in the denominator must survive!
Thank you for your time.
Your calculation is correct and the coefficients are indeed given by $c_n= - \frac{1}{2n^2}$ for $n \in \mathbb{Z} \setminus \{0\}$ . Since $c_0 = \frac{\pi^2}{6}$ , the desired Fourier series with period $\pi$ is $$ f(x) = x (\pi - x) = \sum \limits_{n \in \mathbb{Z}} c_n \mathrm{e}^{2 \mathrm{i} n x} = \frac{\pi^2}{6} - \sum \limits_{n=1}^\infty \frac{\cos(2nx)}{n^2} \, , \, x \in [0,\pi) \, .$$ The series you wanted to find, $$ \frac{8}{\pi} \sum \limits_{k=0}^\infty \frac{\sin[(2k+1)x]}{(2k+1)^3} \, ,$$ is the Fourier series of $$ g(x) = \begin{cases} x (\pi - x) \, , \, 0 \leq x < \pi \\ x(\pi+x) \, , - \pi \leq x < 0\end{cases} $$ with period $2 \pi$ .