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On the distribution of values of Hardy’s Z-functions in short intervals, II : The q-aspect

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An asymptotic formula for T/2TZ2(t)Z(t+U)dt(0<U=U(T)T1/2ε)\displaystyle{ \int _{T/2}^{T}Z^{2}(t)Z(t + U)dt\qquad (0 <U = U(T)\leqslant T^{1/2-\varepsilon }) } is derived, where Z(t):=ζ(12+it)(χ(12+it))1/2(tR),ζ(s)=χ(s)ζ(1s)\displaystyle{ Z(t):= \zeta \left ({1 \over 2} + it\right )\big(\chi \left ({1 \over 2} + it\right )\big)^{-1/2}\quad (t \in \mathbb{R}),\quad \zeta (s) =\chi (s)\zeta (1 - s) } is Hardy’s function. The cubic moment of Z(t) is also discussed, and a mean value result is presented which supports the author’s conjecture that \displaystyle{ \int _{1}^{T}Z^{3}(t)dt\; =\; O_{\varepsilon }(T^{3/4+\varepsilon }). }
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Some problems involving the classical Hardy function Z\left( t \right) = \zeta \left( {\frac{1} {2} + it} \right)\left( {\chi \left( {\frac{1} {2} + it} \right)} \right)^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}} , \zeta \left( s \right) = \chi \left( s \right) \zeta \left( {1 - s} \right) , are discussed. In particular we discuss the odd moments of Z(t) and the distribution of its positive and negative values.
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If Z(t) = c-1/2(\frac12 + it) z( \frac12 + it) Z(t) = \chi^{-1/2}(\frac{1}{2} + it) \zeta ( \frac{1}{2} + it) denotes Hardyz(s) = c(s)z(1 - s) \zeta(s) = \chi(s)\zeta(1 - s) , then it is proved that ò0T Z(t)dt = Oe(T1/4+e) \int_0^T Z(t)dt = O_{\varepsilon}(T^{1/4+\varepsilon}) .
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We investigate the distribution of positive and negative values of Hardy's function Z(t):=ζ(1/2+it)χ(1/2+it)1/2,ζ(s)=χ(s)ζ(1s). Z(t) := \zeta(1/2+it){\chi(1/2+it)}^{-1/2}, \quad \zeta(s) = \chi(s)\zeta(1-s). In particular we prove that μ(I+(T,T))  T  andμ(I(T,T))    T, \mu\bigl(I_{+}(T,T)\bigr) \;\gg T\; \qquad \hbox{and}\qquad \mu\bigl(I_{-}(T, T)\bigr) \; \gg \; T, where μ()\mu(\cdot) denotes the Lebesgue measure and \begin{align*} { I}_+(T,H) &\;=\; \bigl\{T< t\le T+H\,:\, Z(t)>0\bigr\}, { I}_-(T,H) &\;=\; \bigl\{T< t\le T+H\,:\, Z(t)<0\bigr\}. \end{align*}
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An analogue of Atkinson's formula is proved for the integral function F(T)=∫0TZ(t)dt of Hardy's function Z(t). As an application of this formula, we analyze the behavior of the function F(T) showing that it can be approximated by a simple step-function. It follows that F(T)=O(T1/4) and F(T)=Ω±(T1/4); these results were recently obtained by M.A. Korolev using an alternative method.
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