How to show $\sum_{d>x,P^+(d)\le y} \mu(d)/d = O (\log{y}\cdot e^{-\log{x}/2\log{y}})$?

Question

I can't see why $$\sum_{d>x,P^+(d)\le y} \mu(d)/d = O (\log{y}\cdot e^{-\log{x}/2\log{y}})\;,$$

where $P^+(d)$ is the greatest prime factor of $d$.

Can anyone give a hint? Thanks.

Answer 1

Hint: Let $S(x,y)=\left\{ n\leq x:\ P(n)\leq y\right\}$, and let $\psi(x,y)=|S(x,y)|$ be the size of this set. Then, since $|\mu(d)|\leq1$, we may bound your sum above by $\sum_{d>x,\ d\in S(x,y)}\frac{1}{d}.$ Writing this as a Riemann Stieltjes integral, you are looking

$$\sum_{d>x,\ d\in S(x,y)}\frac{1}{d}=\int_{x}^{\infty}\frac{1}{t}d\left(\psi(t,y)\right)=-\frac{\psi(x,y)}{x}+\int_{x}^{\infty}\frac{\psi(t,y)}{t^{2}}dt.$$ Now, the result you desire will follow if you can show that

$$\psi(t,y)\ll te^{-\log t/\log y}.$$

Remarks: Usually when talking about the friable/smooth integers, we let $u=\frac{\log x}{\log y}$. You are trying to prove that $$\psi(x,y)\ll x e^{-u/2}.$$ In fact there are much stronger results than this. We expect that $\psi(x,y)\asymp x u^{-u}$. See the Dickmann De Bruijn rho function, or Hildebrand and Tenenbaum's survey article for more details.