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

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