On the Integral of Hardy’s Function

(1) Z(t) = χ−1/2( 1 2 + it)ζ( 1 2 + it), χ(s) = 2sπs−1 sin( 1 2 πs)Γ(1− s), so that the functional equation for ζ(s) has the form ζ(s) = χ(s)ζ(1 − s). Since χ(s)χ(1− s) = 1, it follows that |Z(t)| = |ζ( 1 2 + it)|, and that Z(t) is a real-valued function of t. The function Z(t) plays an important rôle in the theory of the distribution of zeros of ζ(s) on the “critical line” Re s = 1 2 (see e.g., [1]–[3] and [5]-[6]). The result on the integral of Z(t) is contained in the following