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Thursday, June 17, 2021

Riemann zeta function

zeta series

Series involving the zeta function, e and pi

We show the proof of this beutiful series posted by @infseriesbot. exp(ζ(2)2ζ(2)3+ζ(4)4ζ(4)5+ζ(6)6...)=2πe
To prove this result we use two concepts:
  1. The Taylor series expansion of the log-Gamma function lnΓ(x)
  2. The Euler's reflection formula for the Gamma function Γ(x)
Lets start with the left hand side ζ(2)2ζ(2)3+ζ(4)4ζ(4)5+ζ(6)6...=n=1(12n12n+1)ζ(2n)=n=1ζ(2n)(2n)(2n+1)
What is the series of the right hand side? This series is related to the expansion series of the log-Gamma function: ln(Γ(z))=γ(z1)+k=2(1)kζ(k)k(z1)k|z1|<1
where γ is the Euler-Mascheroni constant. If we let z=1+t and then z=1t in the expansion series we obtain two equations: ln(Γ(t+1))=γt+k=2(1)kζ(k)ktkln(Γ(1t))=γt+k=2ζ(k)ktk
Adding both equations: ln(Γ(t+1))+ln(Γ(1t))=k=1ζ(2k)2kt2k
Integrating both sides: 10(ln(Γ(t+1))+ln(Γ(1t)))dt=k=1ζ(2k)k10t2kdt1210(ln(Γ(t+1))+ln(Γ(1t)))dt=k=1ζ(2k)2k(2k+1)
This is the series we were looking for in terms of a definite integral. Hopefully we can transform it to a maneuverable one. Here is where the Euler's reflection formula for the Gamma function is useful: Γ(t)Γ(1t)=πsin(πt)tZ
Therefore 1210(ln(Γ(t+1))+ln(Γ(1t)))dt=1210(ln(Γ(t+1))(Γ(1t)))dt=1210(ln(tΓ(t))(Γ(1t)))dt=1210ln(tπsin(πt))dt=12ln(π)121210ln(sin(tπ))dt
We just have to prove that 10ln(sin(tπ))dt=ln(2) 10ln(sin(tπ))dt=120ln(sin(2πt))dt=ln(2)+120(sin(πt))dt+120ln(cos(πt))dt(sin(2t)=2sin(t)cos(t))
We also know that 10ln(sin(tπ))dt=2120ln(sin(πt))dt(using sin(θ)=sin(θ))=120ln(sin(2πt))dt
Therefore 10ln(sin(tπ))dtln(2)+120ln(sin(πt))dt+120ln(cos(πt))dt=ln(2)+120ln(sin(πt))dt+120ln(sin(πw))dw(w=12t)=ln(2)+210ln(sin(πt))dt+210ln(sin(πw))dw
10ln(sin(tπ))dt=ln(2)+10ln(sin(tπ))dt+10ln(sin(tπ))dt10ln(sin(πt))dt=ln(2)
We conclude from results (1),(2) and (3) our claim: ζ(2)2ζ(2)3+ζ(4)4ζ(4)5+ζ(6)6...=12ln(π)1210ln(sin(tπ))dt=12ln(π)12+12ln(2)
exp(ζ(2)2ζ(2)3+ζ(4)4ζ(4)5+ζ(6)6...)=2πe

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