Again a question about $\mathrm{rad}(n).$

Let $\mathrm{rad}(n)$ denote the radical of an integer $n$, which is the product of the distinct prime numbers dividing $n$. Or equivalently, $$\mathrm{rad}(n)=\prod_{\scriptstyle p\mid n\atop p\text{ prime}}p.$$ Assume $\mathrm{rad}(1)=1$, so that $\mathrm{rad}(n)$ is multiplicative.

I was trying to obtain asymptotics for the sum $$\sum_{n\le x}\frac{1}{\mathrm{rad}(n)}.$$

If we define the Dirichlet series of $\frac{1}{\mathrm{rad}(n)}$ to be $$R(s)=\sum_{n\ge 1}\frac{1}{n^s\mathrm{rad}(n)},$$ using the multiplicity of $\mathrm{rad}(n)$, we can derive that \begin{align} \begin{split} R(s)&=\prod_{p}\left(1+\frac{p^{-1}}{p^s}+\frac{p^{-1}}{p^{2s}}+\frac{p^{-1}}{p^{3s}}+\cdots\right)\\ &=\prod_{p}\left(1+\frac{p^{-1}}{p^s}\frac{1}{1-\frac{1}{p^s}}\right). \end{split} \end{align}

At first, I want to use Perron's formula here, so I have to find analytic continuation for $R(s)$ and then use residue theorem to evaluate the integral in Perron's formula. However, the Dirichlet series $R(s)$ is quite quirky.

Both $$ \sum_{n\ge 1}|\frac{p^{-1}}{p^s}\frac{1}{1-\frac{1}{p^s}}|^2 \le\sum_{n\ge 1}\frac{1}{p^2}\frac{1}{|p^\sigma -1|^2} $$ and $$ \sum_{n\ge 1}|\frac{p^{-1}}{p^s}\frac{1}{1-\frac{1}{p^s}}| \le\sum_{n\ge 1}\frac{1}{p}\frac{1}{|p^\sigma -1|}. $$ converge for all $\Re(s)=\sigma>0$ while the second sum diverges at $s=0,$ so the Euler product for $R(s)$ converges for all $\Re(s)>0$ and the abssica of absolute convergence for $R(s)$ is $\sigma_{a}=0.$ It seems $R(s)$ has a pole at $s=0.$

I tried to "extract" Rieman zeta function out of $R(s).$ $$R(s)=\prod_{p}\frac{1-p^{-s}+p^{-(s+1)}}{1-p^{-s}}=\zeta(s)\prod_{p}\left(1-p^{-s}+p^{-(s+1)}\right) $$ It seems no good, so I tried it out with $$ R(s)=\frac{R(s)}{\zeta(s+1)}\zeta(s+1)=\zeta(s+1)\prod_{p}\left( 1-\frac{1}{p^{2s+2}}+\frac{1}{p^{2s+1}}\frac{1-p^{-(s+1)}}{1-p^{-s}} \right). $$ The product still explodes at $s=0.$ I also experimented with extracting out $[\zeta(s+1)]^2$, $[\zeta(s+1)]^3$ and $\zeta(2s+1),$ still geting exploded products at $s=0.$

I can't handle $R(s),$ so I tried with elementary methods. Since $$ \frac{1}{\mathrm{rad}}*\mu(n)=\frac{\mu(n)\varphi(n)}{n}, $$ $$ \sum_{n\le x}\frac{1}{\mathrm{rad}(n)}=\sum_{n\le x}\sum_{d|n}\frac{\mu(d)\varphi(d)}{d}=\sum_{n\le x}\frac{\mu(n)\varphi(n)}{n}\left[\frac{x}{n}\right] =x\sum_{n\le x}\frac{\mu(n)\varphi(n)}{n^2}+\mathcal{O}\left(\sum_{n\le x}\frac{|\mu(n)|\varphi(n)}{n}\right). $$ However, $\sum_{n\ge 1}\frac{\mu(n)\varphi(n)}{n^2}$ still diverges.

Thanks for any advice regarding asymptotics for $\sum_{n\le x}\frac{1}{\mathrm{rad}(n)}.$

  • 1
    $\begingroup$ This ends up being tricky because most of the mass of the sum comes from numbers with very few small prime factors, but large exponents. By consider all possible combinations of the first few primes in a way that is similar to this problem concerning Ramanujan's letter on $3$-smooth numbers, we can show that it grows faster than $C_A(\log x)^A$ for any $A>0$ where $C_A$ is a positive constant depending on $A$. $\endgroup$ – Eric Naslund Nov 9 '15 at 18:03
  • $\begingroup$ @EricNaslund! Thanks a lot! So it's not possible to obtain asymptotics using elementary functions? $\endgroup$ – Zhenhua Liu Nov 10 '15 at 0:04
  • $\begingroup$ Can its order be something like $\frac{Cx}{(\log x)^A}$? $\endgroup$ – Zhenhua Liu Nov 10 '15 at 7:54
  • $\begingroup$ Hmmm, I don't think it will be like that. It should be possible to obtain the asymptotic, it 's just a bit trickier $\endgroup$ – Eric Naslund Nov 10 '15 at 8:03
  • $\begingroup$ @EricNaslund. Thanks. Can you give me some hints or advice on what method to use to obtain the aymptotic? $\endgroup$ – Zhenhua Liu Nov 11 '15 at 0:57

After some search, I found out that this problem had already been solved.

Let $$K(x)=\sum_{n\le x}\frac{1}{\text{rad}(n)}.$$

In 1962, de Bruijn published the result $$K(x) = \exp\bigg(\{1+o(1)\} \sqrt{ \frac{8\log x}{ \log_2 x}}\bigg),$$ in the article, On the number of integers x whose prime factors divide n, Illinois J. Math.

Recently, better asymptotic formula has been obtained by Robert and Tenenbaum in theorem 4.3 of the article Sur la r´epartition du noyau d’un entier, Indag. Math. Their result is $$ K(x) = \frac{1 }{2} e^\gamma F(\log x)(\log_2 x) \bigg\{ 1 + \sum_{1\le j\le N} \frac{R_j(\log_3x)}{(\log_2x)^j} + \mathcal{O}_N\bigg( \big(\frac{\log_3 x}{ \log_2 x}\big)^{ N+1}\bigg)\bigg\}. $$ Please refer to their article for full proof.

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  • $\begingroup$ $\log_2x$ means $\log\log x,$ and $\log_3x$ means $\log\log\log x.$ $\endgroup$ – Zhenhua Liu Oct 29 '16 at 17:09

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