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

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