On the order of the error function of the square-full integers |
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Authors: | D Suryanarayana |
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Institution: | 1. Department of Mathematics, Andhra University, Waltair, India
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Abstract: | LetL(x) denote the number of square full integers ≤x. By a square-full integer, we mean a positive integer all of whose prime factors have multiplicity at least two. It is well known that $$\left. {L(x)} \right| \sim \frac{{\zeta ({3 \mathord{\left/ {\vphantom {3 2}} \right. \kern-\nulldelimiterspace} 2})}}{{\zeta (3)}}x^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-\nulldelimiterspace} 2}} + \frac{{\zeta ({2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-\nulldelimiterspace} 3})}}{{\zeta (2)}}x^{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-\nulldelimiterspace} 3}} ,$$ where ζ(s) denotes the Riemann Zeta function. Let Δ(x) denote the error function in the asymptotic formula forL(x). On the basis of the Riemann hypothesis (R.H.), it is known that \(\Delta (x) = O(x^{\tfrac{{13}}{{81}} + \varepsilon } )\) for every ε>0. In this paper, we prove the following results on the assumption of R.H.: (1) $$\frac{1}{x}\int\limits_1^x {\Delta (t)dt} = O(x^{\tfrac{1}{{12}} + \varepsilon } ),$$ (2) $$\int\limits_1^x {\frac{{\Delta (t)}}{t}\log } ^{v - 1} \left( {\frac{x}{t}} \right) = O(x^{\tfrac{1}{{12}} + \varepsilon } )$$ for any integer ν≥1. In fact, we prove some general results and deduce the above from them. On the basis of (1) and (2) above, we conjecture that \(\Delta (x) = O(x^{{1 \mathord{\left/ {\vphantom {1 {12}}} \right. \kern-0em} {12}} + \varepsilon } )\) under the assumption of R.H. |
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