$\sum_{k=1}^\infty\frac{1}{k^4} = \frac{\pi^4}{90}$ using Parseval’s Theorem and Fourier series

Question

Prove $$\sum_{k=1}^\infty\frac{1}{k^4} = \frac{\pi^4}{90}$$ using Parseval's Theorem and Fourier Series of $$f(x)=(x-\frac{1}{2})^2$$ which is $$\frac{1}{12}+\sum_{k\in \mathbb{N}}\frac{1}{\pi^2 k^2}$$
and Parseval for our case is $$ 2 \int_{0}^1 |f(x)|^2 dx = 2|a_o|^2 + \sum_{k=1}^\infty (|a_k|^2 + |b_k|^2)$$
I integrated first the left part of equality
$$2 \int_{0}^1 |f(x)|^2 dx = \frac{1}{6}$$
then I evaluated the right side of equality and that only $a_k$ is considered here
$$2 \frac{1}{144} + \sum_{k=1}^\infty \frac{1}{\pi^4 k^4}$$

$$\frac{1}{6} = \frac{1}{72} + \sum_{k=1}^\infty \frac{1}{\pi^4 k^4}$$
rearranging somehow doesn't bring me to the desired result. Am I missing something along the way.
Any hints or solution clarification is appreciated and Thanks

Answer 1

With the correct $f(x)$, you get $$\int_0^1(x-1/2)^4dx=\int_{-1/2}^{1/2}y^4dy=2\frac{(1/2)^5}5=\frac1{80}$$ So you get $$2\frac1{80}=\frac1{72}+ \sum_{k=1}^\infty \frac{1}{\pi^4 k^4}$$ Then $$\frac1{40}-\frac1{72}=\frac{9-5}{40\cdot 9}=\frac1{90}$$