Proof of ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}

Consider the function f(t):=t2  (πtπ)f(t):=t^2\ \ (-\pi\leq t\leq \pi), extended to all of R{\mathbb R} periodically with period 2π2\pi. Developping ff into a Fourier series we get t2=π23+k=14(1)kk2cos(kt)(πtπ).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:=πt:=\pi here we easily find ζ(2)=π26\zeta(2)={\pi^2\over6}. For ζ(4)\zeta(4) we use Parseval's formula f2=k=ck2\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2 . Here f2=12πππt4dt=π45\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5} and the ckc_k are the complex Fourier coefficients of ff. Therefore c0=π23c_0={\pi^2\over3} and c±k2=14ak2=4k4|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}  (k1)\ (k\geq1). Putting it all together gives ζ(4)=π490\zeta(4)={\pi^4\over 90}.

here here

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

Note by Rajdeep Dhingra
4 years, 6 months ago

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This was an easy proof , but nicely done :)

A Brilliant Member - 4 years, 6 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 - 4 years, 6 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 ζ(4)=π490\zeta{(4)} = \dfrac{{\pi}^{4}}{90} .

Ronak Agarwal - 4 years, 6 months ago

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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 ?

Rajdeep Dhingra - 4 years, 6 months ago

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

Ronak Agarwal - 4 years, 6 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 ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}. Can We ?

Rajdeep Dhingra - 4 years, 6 months ago

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

Ronak Agarwal - 4 years, 6 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 - 4 years, 6 months ago

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@Rajdeep Dhingra Using what integral.

Ronak Agarwal - 4 years, 6 months ago

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@Ronak Agarwal 0π/2x2ln(cos(x))\int_{0}^{\pi/2}{x^2 \ln(\cos(x))}

Rajdeep Dhingra - 4 years, 6 months ago

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@Rajdeep Dhingra You mean I have to prove the result ζ(4)=π490 \zeta(4) = \dfrac{{\pi}^{4}}{90} from this integral without using :

cos(x)=eix+eix2 cos(x) = \dfrac{e^{ix}+e^{-ix}}{2}

Ronak Agarwal - 4 years, 6 months ago

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@Ronak Agarwal NO, I meant how to prove it.

Rajdeep Dhingra - 4 years, 6 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 - 4 years, 6 months ago

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@Ronak Agarwal Fine , But when I solved the question by putting ζ(4)=π490\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 ζ(4)=π490\zeta(4) = \frac{\pi^4}{90} then only are proof for it will be valid.

Rajdeep Dhingra - 4 years, 6 months ago

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