Squarefree totient sum 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!
 A: This answer gives a Tauberian argument:
As the Greg Martin suggests in the comment section, the Dirichlet series
$$
F(s)=\sum_{n=1}^\infty{\mu^2(n)\varphi(n)\over n^s}=s\int_1^\infty{A(x)\over x^{s+1}}\mathrm dx\tag1
$$
has a simple pole at $s=2$ with residue
$$
K=\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$
which implies
$$
G(s)={F(s)\over s}-{K/2\over s-2}
$$
is analytic in a neighborhood containing $s=2$. Plugging in (1), we have
$$
\begin{aligned}
G(s)
&=\int_1^\infty{A(x)-Kx^2/2\over x^{s+1}}\mathrm dx \\
&=\int_1^\infty{A(\sqrt t)-Kt/2\over2t^{s/2+1}}\mathrm dt
\end{aligned}
$$
Now, Wiener-Ikehara Tauberian theorem guarantees that the RHS converges at $s=2$, so we know that
$$
\int_1^\infty{2A(\sqrt t)/K-t\over t^2}\mathrm dt
$$
converges. Now, for convenience, we set $B(x)=2A(\sqrt x)/K$, so the above formula becomes
$$
\int_1^\infty{B(t)-t\over t^2}\mathrm dt\tag2
$$
from this, we will show that $B(t)\sim t$ therefore $A(x)\sim Kx^2/2$:
If $B(t)\nsim t$ then there exists a constant $\lambda>1$ such that there exists a strictly increasing sequence $t_n$ such that $B(t_n)\ge\lambda t_n$ for all $n$, meaning
$$
\begin{aligned}
\int_{t_n}^{\lambda t_n}{B(t)-t\over t^2}\mathrm dt
&\ge\int_{t_n}^{\lambda t_n}{\lambda t_n-t\over t^2}\mathrm dt
=\int_{t_n}^{\lambda t_n}{\lambda-t/t_n\over(t/t_n)^2}\mathrm d(t/t_n) \\
&=\int_1^\lambda{\lambda-u\over u^2}\mathrm du>0
\end{aligned}
$$
which violates Cauchy's criterion for convergence of (2), so we have
$$
\limsup_{t\to\infty}{B(t)\over t}\le1
$$
Similar argument can be established for $\liminf_{t\to\infty}B(t)/t\ge1$. Consequently, we see that $B(t)\sim t$, indicating
$$
\lim_{x\to\infty}{A(x)\over x^2}=\frac K2=\frac12\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right)
$$
A: $$\sum_{n\ge 1} \mu(n)^2 \phi(n) n^{-s}=\prod_p (1+(p-1)p^{-s})=
\prod_p \frac{1-p^{-s}+(p-p^2) p^{-2s}}{1-p^{1-s}}
$$ $$=(\sum_{d\ge 1} d^{1-s})(\sum_{m\ge 1} a(m) m^{-s})$$ 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.
