I recently wrote an answer to a question on MSE about estimating the number of squarefree integers up to $X$. Although the result is known and not too hard, I very much like the proof and my approach. So I write it down here.

First, let’s see if we can understand why this “should” be true from an analytic perspective.

We know that

$$ \sum_{n \geq 1} \frac{\mu(n)^2}{n^s} = \frac{\zeta(s)}{\zeta(2s)},$$

and a general way of extracting information from Dirichlet series is to perform a cutoff integral transform (or a type of Mellin transform). In this case, we get that

$$ \sum_{n \leq X} \mu(n)^2 = \frac{1}{2\pi i} \int_{(2)} \frac{\zeta(s)}{\zeta(2s)} X^s \frac{ds}{s},$$

where the contour is the vertical line $\text{Re }s = 2$. By Cauchy’s theorem, we shift the line of integration left and poles contribute terms or large order. The pole of $\zeta(s)$ at $s = 1$ has residue

$$ \frac{X}{\zeta(2)},$$

so we expect this to be the leading order. Naively, since we know that there are no zeroes of $\zeta(2s)$ on the line $\text{Re } s = \frac{1}{2}$, we might expect to push our line to exactly there, leading to an error of $O(\sqrt X)$. But in fact, we know more. We know the zero-free region, which allows us to extend the line of integration ever so slightly inwards, leading to a $o(\sqrt X)$ result (or more specifically, something along the lines of $O(\sqrt X e^{-c (\log X)^\alpha})$ where $\alpha$ and $c$ come from the strength of our known zero-free region.

In this heuristic analysis, I have omitted bounding the top, bottom, and left boundaries of the rectangles of integration. But proceeding in a similar way as in the proof of the analytic prime number theorem, you could proceed here. So we expect the answer to look like

$$ \frac{X}{\zeta(2)} + O(\sqrt X e^{-c (\log X)^\alpha})$$

using no more than the zero-free region that goes into the prime number theorem.

We will now prove this result, but in an entirely elementary way (except that I will refer to a result from the prime number theorem). This is below the fold.

We do this in a series of steps.

Lemma$$\sum_{d^2 \mid n} \mu(d) = \begin{cases} 1 & \text{if } n \text{ is squarefree} \

0 & \text{else} \end{cases}$$

*Proof.* This comes almost immediately upon noticing that this is a multiplicative function, and it’s trivial to prove it for prime powers. $\spadesuit$

So to sum up the squarefree numbers up to $X$, we look at

$$ \sum_{n \leq X} \sum_{d^2 \mid n} \mu(d) = \sum_{d^2e \leq X} \mu(d)= \sum_{d^2 \leq X} \mu(d) \left\lfloor \frac{X}{d^2} \right\rfloor.$$

This last expression is written in one of the links in Marty’s answer, and they prove it with inclusion-exclusion. I happen to find this derivation more intuitive, but it’s our launching point forwards.

We approximate the floored bit. Notice that

$$ \left \lfloor \frac{X}{d^2} \right \rfloor = \frac{X}{d^2} + E(X,d)$$

with $\lvert E(x,d) \rvert \leq 1$ (we think of it as the Error of the approximation). So the number of squarefree numbers up to $X$ is

$$ \sum_{d^2 \leq X} \mu(d) \frac{X}{d^2} + \sum_{d^2 \leq X}\mu(d) E(x,d).$$

We look at the two terms separately.

### The first term

The first term can be approximated by the infinite series plus an error term.

$$ \sum_{d^2 \leq X} \frac{\mu(d)}{d^2} = X\sum_{d \geq 1} \frac{\mu(d)}{d^2} – X\sum_{d > \sqrt X} \frac{\mu(d)}{d^2} = \frac{X}{\zeta(2)} – X\sum_{d > \sqrt X} \frac{\mu(d)}{d^2}.$$

We must now be a bit careful. If we perform the naive bound, by bounding $\mu(n) \leq 1$, then this last sum is of size $O(\sqrt X)$. That’s too big!

So instead, we integrate by parts (Riemann-Stieltjes integration) or equivalently we perform summation by parts to see that

$$ \sum_{d > \sqrt X} \frac{\mu(d)}{d^2} = O\left( \int_{\sqrt X}^\infty \frac{M(t)}{t^3} dt \right)$$

where

$$ M(t) = \sum_{n \leq t} \mu(n).$$

By the prime number theorem (and as you mention in your question), we know that $M(t) = o(t)$. (In fact, the analytic prime number theorem in one of the easy forms is that $M(X) = O(X e^{-c (\log X)^{1/9}})$, which we might use here). This means that this last term is bounded by

$$ o(\sqrt X)$$

if we just use that $M(t) = o(t)$, or

$$ O(\sqrt X e^{-c (\log X)^{1/9}})$$

if we use more. This completes the first term $\spadesuit$

### Second term

This is easier now. We again use integration by parts. Notice that

$$ \begin{align}

\sum_{d \leq \sqrt X} \mu(d) E(X,d) &= \sum_{d \leq \sqrt X} (M(d) – M(d – 1))E(X,d) \\

&= M(\lfloor \sqrt X \rfloor) E(X, \lfloor \sqrt X \rfloor) + \sum_{d \leq \sqrt X – 1} M(d) (E(X, d) – E(X, d+1)).

\end{align}$$

By using that $E(X,d) \leq 1$ and $M(X) = o(\sqrt X)$ (or the stronger version), we match the results from the first part. $\spadesuit$

Putting these two results together, we have proven that the number of squarefree integers up to $X$ is

$$\frac{X}{\zeta(2)} + o(\sqrt X)$$

using only that $M(X) = o(\sqrt X)$, and alternately

$$ \frac{X}{\zeta(2)} + O(\sqrt X e^{-c (\log X)^{1/9}})$$

using a bit of the zero-free region. This completes the proof. $\diamondsuit$