Squarefree totient sum

Question

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!

Answer 1

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

$$ \lim_{x\to\infty}{A(x)\over x^2}=\frac K2=\frac12\prod_p\left(1+{p-1\over p^2}\right)\left(1-\frac1p\right) $$

Answer 2

$$\sum_{n\ge 1} \mu(n)^2 \phi(n) n^{-s}=\prod_p (1+(p-1)p^{-s})= \prod_p \frac{1-p^{-s}+(p-p^2) p^{-2s}}{1-p^{1-s}} $$ $$=(\sum_{d\ge 1} d^{1-s})(\sum_{m\ge 1} a(m) m^{-s})$$ 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.