On the integral of Hardy's function

If $Z(t) = \chi^{-1/2}(1/2+it)\zeta(1/2+it)$ denotes Hardy's function, where $\zeta(s) = \chi(s)\zeta(1-s)$ is the functional equation of the Riemann zeta-function, then it is proved that $$ \int_0^T Z(t)\d t = O_\e(T^{1/4+\e}). $$