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H A Helfgott
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Counting square-free numbers smoothly

Let $f:\mathbb{R}^+\to \mathbb{R}^+$ be given by

$f(t) = 1$ if $0\leq t\leq 1$

$f(t) = 1/t$ if $t>1$.

Question:

What is the sum $\sum_{n\; \text{squarefree}} f(n/x)$, for $x$ large?

Just by the simplest inclusion-exclusion, one can get an estimate of the form $x/\zeta(2) + C \sqrt{x}$, where $C$ is a largish constant, in part by using highly optimized estimates for $\sum_{n\leq x: n\; \text{squarefree}} 1$ in the literature (e.g. Cohen and Dress, MR0952866). However, I'd imagine the smoothing inherent in the problem allows one to do better than that. How much better? (Experiments suggest a very small error term.)

H A Helfgott
  • 20.2k
  • 3
  • 43
  • 126