Integers Free of Small Prime Factors I am trying to understand (a version of) the elementary proof of the Prime Number Theorem. I've been following Tenenbaum and Mendès France's book The Prime Numbers and Their Distributions. My goal is to understand each step as fully as possible, since many of the details have been left out by the authors.
Although I've managed to get through most of their account, I seem to be stuck in one of the crucial parts of the argument. One wishes to understand the behaviour of $\frac{M(x)}{x}$ as x tends to infinity, where $M(x) = \sum_{d \leq x} \mu (d)$ is Mertens function. To this end one might find useful defining the sets
$$
S(x,y) := \{n \leq x \,|\, P^{+}(n) \leq y \} \\
T(x,y) := \{n \leq x \,|\, P^{-}(n) > y \}
$$
for non-negative numbers x and y. $P^{+}(n)$ denotes the greatest prime factor of the integer n and $P^{-}(n)$ its smallest.
Setting
$$
\Psi (x,y) := |S(x,y)| \,, \\
\Phi (x,y) := |T(x,y)| \,,\, \mathrm{and} \\
u := \frac{\log(x)}{\log(y)}
$$
the idea is to stablish the following:
$$
\Psi (x,y) \ll xe^{-u/2} \, (x \geq 1, \, y \geq 2) \\
\Phi (x,y)  = x \prod_{p \leq y}(1-\frac{1}{p}) + O(xe^{-u/2}\log(y))
$$
and, finally,
$$
\limsup_{x \, \to \, +\infty} \frac{|M(x)|}{x} \leq \prod_{p \leq y}(1-\frac{1}{p}) \int ^{\infty} _{1} \frac{|M(x,y)|}{x^{2}} \mathrm{d}x
$$
($M(x,y) := \sum_{n \in S(x,y)} \mu (n)$).

I had no major trouble with the first of the three results.
Regarding the second one, after some manipulations, the problem is reduced to justifying the bound

$$
\sum _{d > x \, , \, P^{+}(d) \leq y} \frac{\mu (d)}{d} \ll e^{-u/2}\log(y)
$$
The authors suggest partial integration. However, all I could deduce was that
$$
\sum _{d > x \, , \, P^{+}(d) \leq y} \frac{\mu (d)}{d} \ll \int ^{\infty} _{x} \frac{|M(t,y)|}{t^{2}} \mathrm{d}t
$$
would be enough, but I was not able to find a reason why that last bound should hold.
I'd appreciate if someone gave me a hand with this. Thanks.
 A: I must admit that, with some help, I was finally able to 'solve' this.
The answer is the following: first, one needs to rearrange the sumation.
$$
\sum_{d>x \, ,\, P^{+}(d) \leq y} \frac{\mu (d)}{d} = \sum _{d>x} \mu (d) \, \mathbb{I} _{\{P^{+} \leq y\}} (d) \frac{1}{d}
$$
Setting $a_{d} = \mu (d) \, \mathbb{I} _{\{P^{+} \leq y\}} (d)$, and $f(t) = \frac{1}{t}$, and using Abel's summation formula, the last expression is equal to
$$
\frac{-1}{x}\,(\sum _{d \leq x} a_{d}) + \int _{x} ^{+ \infty} (\sum _{d \leq t} a_{d}) \, f'(t) \, \mathrm{d}t = \\
= \frac{-1}{x}\,(\sum _{d \leq x} \mu (d) \, \mathbb{I} _{\{P^{+} \leq y\}} (d)) + \int _{x} ^{+ \infty} (\sum _{d \leq t} \mu (d) \, \mathbb{I} _{\{P^{+} \leq y\}} (d))\frac{\mathrm{d}t}{t^{2}}
$$
which is asymptotically bounded by
$$
|S(x,y)| \frac{1}{x} + \int _{x} ^{+ \infty} \frac{|M(t,y)|}{t^{2}} \mathrm{d}t 
$$
Since $e^{-u/2} \log (y)$ is a bound on the integral, and, making use of the bound on $|S(x,y)|$, $|S(x,y)| \frac{1}{x} \ll e^{-u/2}$, the desired result is obtained.
