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



Estimate related to the Möbius function
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).


