TITLE: Squarefree totient sum
QUESTION [8 upvotes]: Does anybody have a reference/proof for the asymptotic growth rate of
$$A(x) = \!\!\!\!\!\!\sum_{\substack{n \leqslant x \\ n \ \text{squarefree}}} \!\!\!\!\!\! \varphi(n)$$
as $x \to \infty$? Here $\varphi(n)$ denotes Euler's totient function. Numerical evidence suggests it grows like $cx^2$ where $c \approx 0.21$.
I tried using the identity
$$ \varphi(n) = n \sum_{d \mid n} \frac{\mu(d)}{d}$$
along with the fact that $\mu(n)^2 = 1$ iff $n$ is squarefree, and I got the expression
$$A(x) = \sum_{d \le x} \mu(d) \sum_{m \le x/d} m \mu(md)^2.$$
I realize that $md$ is squarefree iff $m$ and $d$ are both squarefree and coprime, but it feels like that just makes the problem harder. Any help would be appreciated!

REPLY [2 votes]: $$\sum_{n\ge 1} \mu(n)^2 \phi(n) n^{-s}=\prod_p (1+(p-1)p^{-s})=
\prod_p \frac{1-p^{-s}+(p-p^2) p^{-2s}}{1-p^{1-s}}
$$ $$=(\sum_{d\ge 1} d^{1-s})(\sum_{m\ge 1} a(m) m^{-s})$$ where $\sum_{m\ge 1} \frac{a(m)}{m^s}$ converges absolutely for $s> 3/2$, whence $\sum_{m\le x} |a(m)|=O(x^{5/3})$ and $$\sum_{n\le x} \mu(n)^2 \phi(n)=\sum_{n\le x} \sum_{dm=n} a(m) d=\sum_{m\le x} a(m) \sum_{d\le x/m} d$$ $$=\frac12\sum_{m\le x} a(m) (\lfloor x/m\rfloor^2+\lfloor x/m\rfloor)$$
$$=\frac12\sum_{m\le x/\log x} a(m) \frac{x^2}{m^2} (1+o(1))+
\sum_{x/\log x<m\le x} a(m) O(\log^2 x)$$
$$ = \frac{x^2}2\sum_{m\ge 1} \frac{a_m}{m^2}\ \ + o(x^2)$$
$\sum_{m\ge 1} \frac{a(m)}{m^2}=\prod_p (1-2p^{-2}+p^{-3})$ doesn't have a closed form.