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 $\sum_{n\le x}\frac{1}{\mathrm{rad}(n)}.$
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.