## Estimating the number of squarefree integers up to $X$

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.