Series involving the zeta function, e and pi
We show the proof of this beutiful series posted by @infseriesbot. exp(ζ(2)2−ζ(2)3+ζ(4)4−ζ(4)5+ζ(6)6−...)=√2πe
To prove this result we use two concepts:
- The Taylor series expansion of the log-Gamma function lnΓ(x)
- The Euler's reflection formula for the Gamma function Γ(x)
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(z−1)+∞∑k=2(−1)kζ(k)k(z−1)k|z−1|<1
where γ is the Euler-Mascheroni constant.
If we let z=1+t and then z=1−t in the expansion series we obtain two equations:
ln(Γ(t+1))=γt+∞∑k=2(−1)kζ(k)ktkln(Γ(1−t))=−γt+∞∑k=2ζ(k)ktk
Adding both equations:
ln(Γ(t+1))+ln(Γ(1−t))=∞∑k=1ζ(2k)2kt2k
Integrating both sides:
∫10(ln(Γ(t+1))+ln(Γ(1−t)))dt=∞∑k=1ζ(2k)k∫10t2kdt⟹12∫10(ln(Γ(t+1))+ln(Γ(1−t)))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(1−t)=πsin(πt)t∉Z
Therefore
12∫10(ln(Γ(t+1))+ln(Γ(1−t)))dt=12∫10(ln(Γ(t+1))(Γ(1−t)))dt=12∫10(ln(t⋅Γ(t))(Γ(1−t)))dt=12∫10ln(t⋅πsin(πt))dt=12ln(√π)−12−12∫10ln(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=2∫120ln(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=12−t)=ln(2)+2∫10ln(sin(πt))dt+2∫10ln(sin(πw))dw
⟹∫10ln(sin(tπ))dt=ln(2)+∫10ln(sin(tπ))dt+∫10ln(sin(tπ))dt⟹∫10ln(sin(πt))dt=−ln(2)
We conclude from results (1),(2) and (3) our claim:
ζ(2)2−ζ(2)3+ζ(4)4−ζ(4)5+ζ(6)6−...=12ln(√π)−12−∫10ln(sin(tπ))dt=12ln(√π)−12+12ln(2)
exp(ζ(2)2−ζ(2)3+ζ(4)4−ζ(4)5+ζ(6)6−...)=√2πe
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