$\sum_{k=1}^\infty\frac{1}{k^4} = \frac{\pi^4}{90}$ 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

Does this answer your question? Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$?; the answer by Bach uses exactly Parseval's identity. — Dietrich Burde, Nov 18 '21 at 17:52
Thanks, It was helpful but we have not had this "zeta stuff" and our notation differs a bit. — Gunners, Nov 18 '21 at 17:59
Yes, but then the integral is wrong. You need to integrate $(x-1/2)^4$. You will then get the right result — Andrei, Nov 18 '21 at 18:13
@Andrei Nice! Thanks for catching the mistake. I really thought I had some knowledge gap regarding this Parseval's Theorem. — Gunners, Nov 18 '21 at 19:07

score 2 · Accepted Answer · answered Nov 18 '21 at 18:18

2

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 $$\frac1{40}-\frac1{72}=\frac{9-5}{40\cdot 9}=\frac1{90}$$

answered Nov 18 '21 at 18:18

Andrei

37,370

1

Thanks a lot for the work! :) – Gunners Nov 18 '21 at 19:08

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

1 Answers1