Estimate on the number of integers in an interval coprime to a given integer

Question

Let $A,B$ be two positive numbers, I wonder how to show the following estimate

$$\sum_{A\le p \le A+B, (p,q)=1} 1 = \frac{B\varphi(q)}{q}+\ O(q^{\epsilon})$$

for any $\epsilon > 0$ and positive integer $q$.

The formula looks intuitively quite true, since $\frac{\varphi(q)}{q}$ is like the "frequency" of number of integers coprime to $q$ and $B$ is just the length of the interval. But I have no idea how to give the error term. $O(q^{\epsilon})$ may or may not be optimal, I don't know.

If the proof is well-known but too long, please feel free to let me know the reference. (The problem might be just in thr book of Apostol but I can't find it)

Answer 1

The standard trick of writing the indicator function of $(p,q)=1$ as $e((p,q)) = \sum_{d\mid (p,q)} \mu(d)$ and then switching the order of summation works here, as already covered in this answer: https://math.stackexchange.com/a/3158036/30402

The error term is bounded by $d(q)$, which is well-known to be $O_\epsilon(q^\epsilon)$ for every $\epsilon>0$ (some references here). More precisely, it is bounded above by something of size $\exp((\ln 2 + o(1)) \log q / \log \log q)$. Not to say that this is the best possible error term, just the best one afforded by this method.

Answer 2

This is essentially found in the Fundamental Lemma of Sieve Theory, which is a deep and interesting result. As its name suggests, it is quite important in the area of sieve theory. Granville and Soundarajan have some good notes on this, which you can find here. Their Theorem 1.4.2 (which can be found on page 37 of the notes I linked) is as follows. I've changed their notation to reflect the notation in your question.

Theorem (The Fundamental Lemma of Sieve Theory). Let $q$ be an integer with $(q,l)=1$, such that every prime factor of $q$ is $\leq (B/l)^{1/u}$ for some given $u = \frac{\log A}{\log B}\geq 1$. Then, uniformly [in the variables $A,B,q,l,a$], we have $$ \sum_{\substack{A \leq n \leq A+B\\ (n,q)=1 \\n\equiv a\ (\text{mod}\ l)}} 1 = \frac{B}{l} \frac{\phi(q)}{q}\left(1 + O(u^{-u/2}) \right) + O\left(\left(\frac{B}{l}\right)^{3/4+o(1)}\right). $$ Taking $a=l=1$, we get that $$ \sum_{\substack{A \leq n \leq A+B\\ (n,q)=1 }} 1 = \frac{B\phi(q)}{q}\left(1 + O(u^{-u/2}) \right) + O\left(B^{3/4+o(1)}\right). $$ The proof of the Fundamental Lemma is a bit too long to be written here. Whether or not this estimate constitutes and asymptotic or just and upper bound depends on how $B$ and $q$ grow with $A$.

It should be noted that the error term you have written is certainly false. In general, you should always expect your error term to depend on the length of your sum in some way. Except in very specific cases (such as the Polya-Vonogradov inequality), this is universally true of problems in analytic number theory.