On the integral of Hardy's function

Aleksandar Ivić

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^…