On the integral of Hardy’s function

Abstract.

If $Z(t) = \chi^{-1/2}(\frac{1}{2} + it) \zeta ( \frac{1}{2} + it)$ denotes Hardy’s function, where $\zeta(s) = \chi(s)\zeta(1 - s)$ , then it is proved that

$\int_0^T Z(t)dt = O_{\varepsilon}(T^{1/4+\varepsilon})$.