Number of squareful integers less than x

[Boorglar]

I'm doing the exercises from Introduction to Analytic Number Theory by A.J. Hildebrand (online pdf lecture notes) from Chapter 2: Arithmetic Functions II - Asymptotic Estimates, and some of them leave me stumped...

1. Homework Statement

Problem 2.14:
Obtain an asymptotic estimate with error term O(x^{\frac{1}{3}}) for the number of squarefull integers ≤ x, i.e., for the quantity S(x) = \left\{n ≤ x : p | n => p^{2} | n\right\}.

2. Homework Equations
The text describes a method known as the "convolution method" to evaluate sums of arithmetic functions asymptotically. In our case, the arithmetic function would be the characteristic function of the squarefull integers, a(n) = 1 if n is squarefull and 0 otherwise.

If a = f * g (Dirichlet convolution), then \sum_{n ≤ x} a(n) = \sum_{d ≤ x} g(d) F\left(\frac{x}{d}\right) where F(x) is the summatory function of f.

The Attempt at a Solution

I am not even sure how to start. I tried expressing a(n) = 1 * (\mu * a) thus using f = 1 and g = \mu * a. But I can't find a way to estimate the sum \sum_{d ≤ x} g(d) \left\lfloor{\frac{x}{d}}\right\rfloor Here \mu means the Mobius function.

 

[Geofleur]

Maybe it would be a good idea to investigate S(x) for different values of x. You could plot the number of squarefull integers for x up to some cutoff, for example. It couldn't hurt! Maybe it will reveal some pattern that will give you an idea of how to solve the problem.

 

[Boorglar]

So I finally solved the problem, in a totally different (and way simpler) method -_-

I used an earlier result from another problem, saying that squarefull integers can be written uniquely in the form a^2b^3.
So to count the squarefull integers \leq x you can just do \sum_{a^2b^3 \leq x} 1 = \sum_{b \leq x^{1/3}} \left\lfloor{\frac{\sqrt x}{b^{3/2}}}\right\rfloor which eventually simplifies to \zeta(3/2)\sqrt x + O(x^{1/3})

UPDATE:
Ah, turns out I was wrong... The representation a^2b^3 is not unique. For example, p^8 = (p^4)^2 * 1^3 = p^2 * (p^2)^3. I think we need the additional condition that b is squarefree...
So the sum is actually \sum_{a^2b^3 \leq x} µ^2(b) = \sum_{b \leq x^{1/3}}µ^2(b) \sum_{a \leq \frac{x^{1/2}}{b^{3/2}}}1 = \sqrt x \sum_{b \leq x^{1/3}}\frac{µ^2(b)}{b^{3/2}} + O(x^{1/3})

which still becomes something of the form C \sqrt x + O(x^{1/3}) but C = \sum_{b = 1}^{\infty} \frac{µ^2(b)}{b^{3/2}} which I can't really simplify...

After playing with this expression and using Dirichlet series of µ(n) and (µ^2 * µ)(n), I found that C = \frac{\zeta(3/2)}{\zeta(3)}.