Let $f(x)=|x|$ I wish to compute corresponding complex Fourier coefficients $c_k$ given by
$$
c_k=\frac{1}{2T}\int_{-T}^Tf(x)e^{-ikx} dx
$$
for $T=\pi$. And then use the Parsevalle's equality, which says:
$$
\sum_{k=-\infty}^\infty |c_k|^2=\frac{1}{2T}\int_{-T}^T|f(x)|^2dx
$$
to obtain a result of the sum $\sum_{k=1}^\infty \frac{1}{(2k-1)^4}$.
Here is my go:
First, I compute the integral for $c_k$, I split by the absolute value around $0$
$$I:=\int_{-\pi}^\pi |x|e^{-ikx}dx = -\int_{-\pi}^0 xe^{-ikx}dx+\int_0^\pi xe^{-ikx}dx$$
Now,
$$
\int xe^{-ikx}dx = \frac{xi}{k}e^{-ikx}+\frac{e^{-ikx}}{k^2}+C
$$
plugging in the bounds I got
$$
I=\frac{2}{k^2}\bigg((-1)^k-1\bigg)
$$
which basically simplifies to: $\frac{-4}{k^2}$ for $k$ odd and $0$ for $k$ even.
So $c_k = \frac{-2}{\pi k^2}$ for $k$ odd and $c_k = 0$ for $k$ even. Oh yes, and $c_0=0$ (we didn't compute that by the integral above). Now, Parsevalle's equality gives:
$$
\sum_{k=-\infty}^\infty |c_k|^2=\sum_{k=-\infty}^\infty|c_{2k-1}|^2=2\sum_{k=1}^\infty \frac{4}{(2k-1)^4\pi^2} = \frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx = \frac{\pi^2}{3}
$$
which then yields
$$
\sum_{k=1}^\infty\frac{1}{(2k-1)^4} = \frac{\pi^4}{24}
$$
but it should be $\frac{\pi^4}{96}$, which is just a $1/4$ off. Can someone spot my mistake?
Best Answer
Your mistake is in suggestion that $c_0=0$. With the correct value $c_0=\frac\pi2$ you will obtain the desired equality.