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)−1−γ/2)s2(s−1)Γ(1+s/2)∏ρ(1−sρ)es/ρ
can one derive the exact value of ∑ρ1ρ
to be
∑ρ1ρ=−log(2√π)+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)−1−γ/2)s−log(2(s−1)Γ(1+s/2))+∑ρlog(1−sρ)+sρζ′(s)ζ(s)=log(2π)−1−γ/2−(1s−1+12ψ(s2+1))+∑ρ1s−ρ+1ρζ′(s)ζ(s)+1s−1=log(2π)−1−γ/2−12ψ(s2+1)+∑ρ1s−ρ+1ρ
taking the limit as s→1 and remembering that ζ(s)−1s−1=γ+O(s−1) and ψ(32)=ψ(12)+2=−γ−log(4)+2 we get :
γ=log(2π)−1−γ/2+(γ/2+log(2)−1)+∑ρ11−ρ+1ρ∑ρ11−ρ+1ρ=−log(4π)+2+γ
With the answer the double of your series (as it should since if ρ is a root then 1−ρ 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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