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Let $d_{sf}(n)$ be the number of square-free divisors of $n$, and let $D_{sf}(k)=\sum_{n=1}^{k} d_{sf}(n)$ denote the corresponding summation function. Mertens showed that the asymptotic expansion of $D_{sf}(k)$ is

$$\displaystyle D_{sf}(k)=\frac{1}{\zeta(2)}k \log(k)+\ (\frac{2\gamma-1}{\zeta(2)}-\frac{2\zeta'(2)}{\zeta^2(2)})k+O(k^{\frac{1}{2}} \log(k))$$

where the error term was recently reported to be, under the Riemann hypothesis, $O(k^{\frac{1}{4}+\epsilon})$.

It is also known that, for $Re(s)>1$, the following asymptotic expansion holds:

$$\sum_{n=1}^\infty \frac{d_{sf}(n)}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)}$$

I would like to determine an asymptotic expansion for the weigthed summation corresponding to the case $s=1$, i.e.

$$\sum_{n=1}^x \frac{d_{sf}(n)}{n}$$

which expresses the average number of square free divisors for $n\leq x$, when $x$ tends to $\infty$. After some calculations, I suppose that this asymptotic expansion may be given by $$\frac{1}{2\zeta(2)} \log^2(x)+O(\log(x))$$ but I would like to get a confirmation of this. Based on the growth rate of the summation, I also hypothesize that a more accurate estimate could be written as $$\frac{1}{2\zeta(2)} \log^2(x)+a \log(x)+b+O(\frac{1}{\sqrt{x}})$$ but again I would like to get a formal proof.

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This is a very basic question that follows trivially via summation by parts. We have that \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = \frac{1}{x} \sum_{n \leq x} d_{sf}(n) + \int_{1}^{x} \frac{1}{t^2} \sum_{n \leq t} d_{sf}(n) \, dt. \] Using the expression \[\sum_{n \leq x} d_{sf}(n) = A x \log x + B x + E(x), \] where the error term $E(x)$ satisfies $E(x) = O(\sqrt{x} \log(x + 2))$, we see that \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = A \log x + B + O\left(\frac{\log x}{\sqrt{x}}\right) + \int_{1}^{x} \left(\frac{A \log t}{t} + \frac{B}{t} + \frac{E(t)}{t^2}\right) \, dt. \] The integral is equal to \[\frac{A (\log x)^2}{2} + B \log x + \int_{1}^{\infty} \frac{E(t)}{t^2} \, dt + O\left(\int_{x}^{\infty} \frac{\log t}{t^{3/2}} \, dt\right), \] and this last term is \[O\left(\frac{\log x}{\sqrt{x}}\right). \] So if we write \[C = \frac{A}{2}, \qquad D = A + B, \qquad E = B + \int_{1}^{\infty} \frac{E(t)}{t^2} \, dt, \] then we have the asymptotic \[\sum_{n \leq x} \frac{d_{sf}(n)}{n} = C (\log x)^2 + D \log x + E + O\left(\frac{\log x}{\sqrt{x}}\right). \]

As an aside, you write that the error term was recently reduced to $O(k^{1/2 + \varepsilon})$, but this is a weaker error term than $O(k^{1/2} \log k)$.

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  • $\begingroup$ Thank you for your answer. Regarding the error, I meant that it has been recently hypothesized to be $O(x^{\frac{1}{4}+\epsilon})$, which is stronger than the bound given by Mertens. I corrected the typo in the OP. What about the convergence of the integral in the constant term? $\endgroup$ – Anatoly Nov 30 '15 at 7:59
  • $\begingroup$ The integral in the constant term clearly converges by the direct comparison test with the function $\log x/x^{3/2}$. As for your statement about the error, the fact that it ought to be $O(x^{1/4 + \varepsilon})$ is an extremely straightforward consequence of the Riemann hypothesis, in the same way as bounds for the error term in the prime number theorem. It is very easy to show this assuming RH and not at all surprising. $\endgroup$ – Peter Humphries Nov 30 '15 at 15:48
  • $\begingroup$ Yes, the reduced bound directly comes from the assumption of the RH. Regarding the integral, my question was not on whether it converges (it is straightforward to show it) but on its convergence limit, which seems to be approximately $-0.3$. $\endgroup$ – Anatoly Dec 1 '15 at 12:40
  • $\begingroup$ Oh, determining the value of the integral is not at all straightforward. Perhaps a better approach would be to calculate the sum using Perron's inversion formula instead of summation by parts, in which case it might be more clear what the constant term is. $\endgroup$ – Peter Humphries Dec 1 '15 at 15:35

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