Calculating $\sum_{k = 1}^{\infty} k^{-8}$ using Parseval´s identity I calculated the Fourier series of $x^2$ getting
$$x^2=\frac{\pi^2}{3}+4\sum_{k=1}^\infty \frac{(-1)^k}{k^2} \cos(kx)$$
Then, integrating this equation two times i got
$$\frac{x^4-2\pi^2x^2}{12}=4\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^4} \cos(kx)$$
Using the Parseval's identity $\sum c_k^2=\int f^2$ i obtained that
$$\sum_{k=1}^\infty \frac{1}{k^8}=\frac{107}{362880}\pi^8 $$
which, according to Wolfram Alpha should be $\sum_{k=1}^\infty \frac{1} {k^8}=\frac{\pi^8} {9450}$
I can't find where my error is. Is my proceeding wrong or am I just missing something?
 A: Starting with
$$ \frac{\theta^2}{4} - \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \, \cos(n \theta) $$
then, by integration, the series
$$ \frac{\theta^3}{12} - \frac{\zeta(2) \, \theta}{2} + c_{1} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \, \sin(n \theta) $$
is obtained. The constant of integration, $c_{1}$ is found to be zero by evaluating the series at $\theta = 0$ or $\theta = \pi$. Now integrating once again the resulting series is
$$ \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + c_{2} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} \, \cos(n \theta). $$
Evaluating this series at $\theta = 0$, or $\theta = \pi$, the constant $c_{2}$ can be found and leads to the series
$$ \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^4} \, \cos(n \theta). $$
Now, by Pasreval's theorem for Fourier Series, it can be seen that
\begin{align}
\sum_{n=1}^{\infty} \left( \frac{(-1)^{n-1}}{n^4} \right)^2 &= \frac{1}{\pi} \, \int_{-\pi}^{\pi} \left( \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} \right)^2 \, d\theta \\
\sum_{n=1}^{\infty} \frac{1}{n^8} &= \frac{2}{\pi} \, \int_{0}^{\pi} \left( \frac{\theta^4}{48} - \frac{\zeta(2) \, \theta^2}{4} + \frac{7 \, \zeta(4)}{8} \right)^2 \, d\theta \\
\zeta(8) &= \frac{2}{\pi} \, \left[ \frac{\theta^9}{20736} - \frac{\zeta(2) \, \theta^7}{672} + \frac{37 \, \zeta^{2}(2) \, \theta^5}{2400} - \frac{7 \, \zeta^{3}(2) \, \theta^3}{120} + \frac{49 \, \zeta^{4}(2) \, \theta}{400} \right]_{0}^{\pi} \\
&= \frac{\pi^8}{9450}.
\end{align}
Comparing this result to that from MathWorld Zeta Function then it is determined that
$$ \zeta(8) = \frac{\pi^8}{9450} $$
is the correct result.
