# Proof of $\zeta(4) = \frac{\pi^4}{90}$

Consider the function $f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of ${\mathbb R}$ periodically with period $2\pi$. Developping $f$ into a Fourier series we get $t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).$ If we put $t:=\pi$ here we easily find $\zeta(2)={\pi^2\over6}$. For $\zeta(4)$ we use Parseval's formula $\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2$ . Here $\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}$ and the $c_k$ are the complex Fourier coefficients of $f$. Therefore $c_0={\pi^2\over3}$ and $|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}$ $\ (k\geq1)$. Putting it all together gives $\zeta(4)={\pi^4\over 90}$.

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Source - http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90

Note by Rajdeep Dhingra
4 years, 8 months ago

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This was an easy proof , but nicely done :)

- 4 years, 8 months ago

I think he forgot to mention the source . :D

http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90

- 4 years, 8 months ago

Remember You just recently posted a solution to my question. I don't know whether you realize or not that it contained a proof that $\zeta{(4)} = \dfrac{{\pi}^{4}}{90}$.

- 4 years, 8 months ago

Yes , You are right . But without showing the result of that integral using a different way we can't prove it. Can we ?

- 4 years, 8 months ago

I can't understand what you are trying to say here kindly clarify your statement.

- 4 years, 8 months ago

I meant without proving the result of that integral in an another way we can't prove $\zeta(4) = \frac{\pi^4}{90}$. Can We ?

- 4 years, 8 months ago

You have posted a proof that doesn't use that method.

- 4 years, 8 months ago

I am asking just to gain knowledge not check whether my way is right. Could you prove it using that integral.

- 4 years, 8 months ago

Using what integral.

- 4 years, 8 months ago

$\int_{0}^{\pi/2}{x^2 \ln(\cos(x))}$

- 4 years, 8 months ago

You mean I have to prove the result $\zeta(4) = \dfrac{{\pi}^{4}}{90}$ from this integral without using :

$cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}$

- 4 years, 8 months ago

NO, I meant how to prove it.

- 4 years, 8 months ago

You have posted a solution for solving that integral, in your solution observe that the imaginary part of the integral is zero and you will observe that the result gets proved.

- 4 years, 8 months ago

Fine , But when I solved the question by putting $\zeta(4) = \frac{\pi^4}{90}$ then I can't possibly use the answer to prove this.First , we need to solve that question without using the fact that $\zeta(4) = \frac{\pi^4}{90}$ then only are proof for it will be valid.

- 4 years, 8 months ago