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 \(\zeta(4)={\pi^4\over 90}\).

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

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## Comments

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TopNewestI think he forgot to mention the source . :D

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

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Comment deleted Feb 25, 2015

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Please answer the reply I made to your comment.

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This was an easy proof , but nicely done :)

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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 \(\zeta{(4)} = \dfrac{{\pi}^{4}}{90} \).

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

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

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

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