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On the mean square of the zeta-function and the divisor problem

Abstract

Let Δ(x)\Delta(x) denote the error term in the Dirichlet divisor problem, and E(T)E(T) the error term in the asymptotic formula for the mean square of ζ(1/2+it)|\zeta(1/2+it)|. If E(t)=E(t)2πΔ(t/2π)E^*(t) = E(t) - 2\pi\Delta^*(t/2\pi) with Δ(x)=Δ(x)+2Δ(2x)12Δ(4x)\Delta^*(x) = -\Delta(x) + 2\Delta(2x) - {1\over2}\Delta(4x), then we obtain the asymptotic formula 0T(E(t))2dt=T4/3P3(logT)+Oϵ(T7/6+ϵ), \int_0^T (E^*(t))^2 {\rm d} t = T^{4/3}P_3(\log T) + O_\epsilon(T^{7/6+\epsilon}), where P3P_3 is a polynomial of degree three in logT\log T with positive leading coefficient. The exponent 7/6 in the error term is the limit of the method.Comment: 10 page

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