  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 $\Delta^*(x)
=
  -\Delta(x) + 2\Delta(2x) - {1\over2}\Delta(4x)$ and we set
  $\int_0^T E^*(t) dt = 3\pi T/4 + R(T)$, then we obtain $$ R(T) =
O_\epsilon(T^{593/912+\epsilon}), \int_0^TR^4(t) dt \ll_\epsilon
T^{3+\epsilon}, $$ and $$ \int_0^TR^2(t) dt = T^2P_3(\log T) +
O_\epsilon(T^{11/6+\epsilon}), $$ where $P_3(y)$ is a cubic polynomial in $y$
with positive leading coefficient.
