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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
1 year, 11 months ago

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I think he forgot to mention the source . :D

http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90 Shivang Jindal · 1 year, 11 months ago

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Comment deleted Feb 25, 2015

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@Ronak Agarwal Please answer the reply I made to your comment. Rajdeep Dhingra · 1 year, 11 months ago

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This was an easy proof , but nicely done :) Azhaghu Roopesh M · 1 year, 11 months ago

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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 \(\zeta{(4)} = \dfrac{{\pi}^{4}}{90} \). Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal Yes , You are right . But without showing the result of that integral using a different way we can't prove it. Can we ? Rajdeep Dhingra · 1 year, 11 months ago

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@Rajdeep Dhingra I can't understand what you are trying to say here kindly clarify your statement. Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal 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 ? Rajdeep Dhingra · 1 year, 11 months ago

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@Rajdeep Dhingra You have posted a proof that doesn't use that method. Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal I am asking just to gain knowledge not check whether my way is right. Could you prove it using that integral. Rajdeep Dhingra · 1 year, 11 months ago

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@Rajdeep Dhingra Using what integral. Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal \(\int_{0}^{\pi/2}{x^2 \ln(\cos(x))}\) Rajdeep Dhingra · 1 year, 11 months ago

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

\( cos(x) = \dfrac{e^{ix}+e^{-ix}}{2} \) Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal NO, I meant how to prove it. Rajdeep Dhingra · 1 year, 11 months ago

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@Rajdeep Dhingra 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. Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal 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. Rajdeep Dhingra · 1 year, 11 months ago

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