Estimate related to the Möbius function I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less or equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
\mathcal{J}_T(x,k) = \{ n \in \mathbb{N} : n \le x, \Omega(n)=k , n \mbox{ is square free } \}.$$
Other way to describe this cardinal, using the Möbius function, is 
$$ |\mathcal{J}_T(x,k) | = \sum_{\Omega(n) = k, n \le x  } |\mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using 
$$
\sum_{\Omega(n) = k, n \le x  } |\mu(n)| \le \sum_{n \le x  } |\mu(n)| \le \frac{6x}{\pi^2} + O(\sqrt{x}) \ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
 A: You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method. 
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$\sum_\text{$n$ square-free}z^{\omega(n)}n^{-s}=\prod_p\left(1+\frac{z}{p^s}\right).$$
Then you will need to "factor out" $\zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
A: It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
\begin{align*}
\#\{ n\le x\colon \omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!} \\
\#\{ n\le x\colon \Omega(n) = k \} &\sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}.
\end{align*}
(Here $\omega$ and $\Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $\Omega(n) = k$ satisfies $\omega(n) = j$ for some $1\le j\le k-1$. Therefore the bounds
\begin{align*}
\#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\le \#\{ n\le x\colon \Omega(n) = k \} \\
\#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} &\ge \#\{ n\le x\colon \Omega(n) = k \} - \sum_{j=1}^{k-1} \#\{ n\le x\colon \omega(n) = j \},
\end{align*}
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
\#\{ n\le x\colon \Omega(n) = k,\, n\text{ is squarefree} \} \sim \frac x{\log x} \frac{(\log\log x)^{k-1}}{(k-1)!}.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(\log\log x)$ is what is needed).
