Consider the function f(t):=t2 (−π≤t≤π), extended to all of R periodically with period 2π. Developping f into a Fourier series we get
t2=3π2+k=1∑∞k24(−1)kcos(kt)(−π≤t≤π).
If we put t:=π here we easily find ζ(2)=6π2. For ζ(4) we use Parseval's formula
∥f∥2=k=−∞∑∞∣ck∣2 . Here
∥f∥2=2π1∫−ππt4dt=5π4
and the ck are the complex Fourier coefficients of f. Therefore c0=3π2 and ∣c±k∣2=41ak2=k44 (k≥1). Putting it all together gives ζ(4)=90π4.
here
Source - http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90
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Comments
This was an easy proof , but nicely done :)
I think he forgot to mention the source . :D
http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90
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)=90π4.
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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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ζ(4)=90π4. Can We ?
I meant without proving the result of that integral in an another way we can't proveLog in to reply
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∫0π/2x2ln(cos(x))
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ζ(4)=90π4 from this integral without using :
You mean I have to prove the resultcos(x)=2eix+e−ix
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ζ(4)=90π4 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)=90π4 then only are proof for it will be valid.
Fine , But when I solved the question by putting