Monday, 17 November 2014

complex analysis - Sum over all inverse zeta nontrivial zeros



Starting from the Hadamard product for the Riemann Zeta Function (assuming the product is taken over matching pairs of zeros)
ζ(s)=e(log(2π)1γ/2)s2(s1)Γ(1+s/2)ρ(1sρ)es/ρ
can one derive the exact value of ρ1ρ
to be
ρ1ρ=log(2π)+1+γ/2
What implications does this have?



Answer



Let's start with the logarithm of the Hadamard product and take the derivative relatively to s (remembering that ψ(z):=ddzlogΓ(z) with ψ the digamma function) :
logζ(s)=(log(2π)1γ/2)slog(2(s1)Γ(1+s/2))+ρlog(1sρ)+sρζ(s)ζ(s)=log(2π)1γ/2(1s1+12ψ(s2+1))+ρ1sρ+1ρζ(s)ζ(s)+1s1=log(2π)1γ/212ψ(s2+1)+ρ1sρ+1ρ
taking the limit as s1 and remembering that ζ(s)1s1=γ+O(s1) and ψ(32)=ψ(12)+2=γlog(4)+2 we get :
γ=log(2π)1γ/2+(γ/2+log(2)1)+ρ11ρ+1ρρ11ρ+1ρ=log(4π)+2+γ
With the answer the double of your series (as it should since if ρ is a root then 1ρ will also be one).



I considered here only a formal derivation (the series is only conditionally convergent : the roots must be sorted with increasing absolute imaginary parts, the limit inside the series should be justified) ; for detailed proofs see for example Edwards' book p.67.



For the sum of integer powers of the roots see Mathworld starting at (5).



Other references were proposed in this answer as this MO thread that may be more useful for your question concerning 'implications' (see the answer and comment to Micah Milinovich).


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