On the integral of Hardy's function

TL;DR

This paper proves that the integral of Hardy's function over [0,T] grows at most on the order of T^{1/4+ε}, providing a new bound related to the Riemann zeta-function's properties.

Contribution

It establishes a new upper bound for the integral of Hardy's function, advancing understanding of its growth behavior.

Findings

Integral of Hardy's function is O(T^{1/4+ε})

Provides bounds related to the Riemann zeta-function

Enhances understanding of Hardy's function growth

Abstract

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