<s>[INST] Using Parseval' s theorem to evaluate a sum.. On the function $f(x)$ = $x^3$ on $(-1,1)$ find Fourier coefficients for this function and then use Parseval's Theorem to evaluate:
$$\sum_{n=1}^{\infty}\frac{1}{n^6}$$
Current work:
I have used Mathematica to find the sum, which results in $\frac{\pi^6}{945}$
Since $a_0$ and $a_n = 0$ as it is an odd function, my problem is calculating $b_n$.
I am using Mathematica and resulting in: 
$$\int_a^b x^3 \sin{(n x \pi)} \,dx= \frac17$$ in which $a$ is $-1$ and $b$ is $1$, the right hand side is calculated from the average value formula, by squaring Integral$[f(x)dx]$.
This is the problem, as my result for the $b_n$ coefficients, is resulting in a nasty rational: 
$$-2(-6+n^2\pi^2)cos(n\pi) / n^3\pi^3$$
where I can't seem to factor out a $\frac{1}{n^6}$ term to evaluate the sum using the theorem. 
 [/INST] When you integrate (using integration by parts or a table)
\begin{equation*}
  b_n=\int_{-1}^{1}x^3\sin (n \pi x) \, dx = 2\int_{0}^{1}x^3 \sin (n \pi x) \, dx
\end{equation*}
you get
\begin{equation*}
  -\frac{2(-1)^n}{n\pi}\left( 1-\frac{6}{n^2\pi^2} \right).
\end{equation*}
Squaring this, we obtain
\begin{equation*}
  \frac{4}{n^2\pi^2}-\frac{48}{n^4\pi^4}+\frac{144}{n^6\pi^6}.
\end{equation*}
If we use the fact that
\begin{equation*}
  \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6},  \quad \sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}
\end{equation*}
together with Parseval's identity
\begin{equation*}
  \frac{1}{1}\int_{-1}^{1}|x^3|^2 \, dx = \sum_{n=1}^{\infty}|b_n|^2\quad (\text{since } a_n=0)
\end{equation*}
the result follows.
</s>