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G. [47] and the web site of Odlyzko [48]) but a mathematical proof is still absent. 2 Trace Formula for the Riemann Zeros Let us fix a test function h(r) exactly as it was done for the Selberg trace formula in Sect. e. • h(r) is a function analytical in the region |Im r| ≤ 1/2 + δ, • h(−r) = h(r), • |h(r)| ≤ A(1 + |r|)−2−δ . 44 Eugene Bogomolny Denote as in that Section g(u) = +∞ 1 2π and define −∞ +∞ H(s) = −∞ h(r)e−iru dr g(u)e(s−1/2)u du . Now let us compute the integral 1 2πi dsH(s) ζ (s) ζ(s) where the contour of integration is taken over the rectangle −η ≤ Re s ≤ 1+η and −T ≤ Im s ≤ T with 0 < η < δ and T → +∞.
55]) x . N (p < x) = ln x As ln p ≡ Tp this expression has the form similar to number of periodic orbits of chaotic systems with h = 1 N (Tp < T ) = eT . T Due to these similarities number-theoretical zeta functions play the role of simple (but by far non-trivial) models of quantum chaos. Notice that the overall signs of the oscillating part of trace formulas for the Riemann zeta function and dynamical systems are different. According to Connes [30] it may be interpreted as Riemann zeros belong not to a spectrum of a certain self-adjoint operator but to an ’absorption’ spectrum.
1 Functional Equation The possibility of this continuation is connected with the fact that the Riemann zeta function satisfies the important functional equation ζ(s) = ϕ(s)ζ(1 − s) where (37) πs Γ (1 − s) . g. [55]). g. [32], Vol. 1, Sect. 1). Therefore if Re s > 1 ∞ Γ (s/2)ζ(s) = xs/2−1 Ψ (x)dx π s/2 0 where Ψ (x) is given by the following series ∞ Ψ (x) = e−πn 2 x . n=1 Using the Poisson summation formula (5) one obtains ∞ e−πn 2 n=−∞ x ∞ 2 1 =√ e−πn /x x n=−∞ which leads to the identity 1 2Ψ (x) + 1 = √ x 1 2Ψ ( ) + 1 x .