# An estimation from Iwaniec-Kowalski book

I am reading the book 'Analytic Number Theory' by Iwaniec and Kowalski and I am confused with an estimation from Chapter 19.

On page 446, they say

If $$cd \leq z^{1/2}$$ we derive from the Prime Number Theorem that $$$$-\sum_{\substack{m \leq z/cd \\ (m,cd)=1}} \frac{\mu(m)}{m} \log cdm = \frac{cd}{\varphi(cd)} \{1+O(\tau(cd)(\log z)^{-A})\}.$$$$

The notations in the equation are explained as follows: Here $$c,d$$ are positive square-free integers, $$\mu$$ is the Möbius function, $$\varphi$$ is Euler's totient function and $$\tau$$ is the divisor function. Also, $$z$$ is a large parameter and $$A>0$$ is arbitrarily large.

I cannot understand how to use PNT to derive this estimation.

Thanks for any help.

I am sorry for mistakenly posting this question on MathOverflow before.

All the $$O$$-constants depend only on $$A$$.

• The PNT says that for any fixed $$A$$ as $$x \to \infty$$ $$\sum_{n \le x} \frac{\mu(n)}{n} =O( (\log x)^{-A})$$

• To estimate the $$\sum_{n\le x,\gcd(n,k)=1}$$ sum we'll show by induction, starting with $$k=1$$ to obtain it for $$k=p,pq,pqr,\ldots$$,

that if $$\sum_{n \le x,\gcd(n,k)=1} \frac{\mu(n)}{n}=O(C_{A,k} (\log x)^{-A}),\qquad C_{A,k}=\prod_{q | k}\frac{A+1}{(1-q^{-1})^2}$$ Then for $$p \nmid k$$ and $$x \ge p$$ $$\qquad\qquad\scriptstyle( \text{since }\sum_{n,\gcd(n,kp)=1} \mu(n)n^{-s}= \frac1{1-p^{-s}}\sum_{n,\gcd(n,k)=1} \mu(n)n^{-s})$$ $$\sum_{n \le x,\gcd(n,kp)=1} \frac{\mu(n)}{n}= \sum_{l=0}^{\log_p(x)}\frac1{p^l}\sum_{n \le x/p^l,\gcd(n,k)=1} \frac{\mu(n)}{n}$$ $$= O(C_{A,k}\sum_{l= 0}^{\log_p(x)}\frac1{p^l} (\log x/p^l)^{-A}) = O(C_{A,k}\sum_{l= 0}^{\log_p(x)}\frac1{p^l} (\log x)^{-A} (1-\frac{\log p^l}{\log x})^{-A})$$ $$= O(C_{A,k}\sum_{l\ge 0}\frac1{p^l} (\log x)^{-A} (1+A\frac{l\log p}{\log x})) =O(C_{A,kp} (\log x)^{-A})$$

• From partial summation of $$\sum_{n > x, \gcd(n,k)=1}\frac{\mu(n)}{n}\log n$$ we obtain that

$$-\sum_{n \le x, \gcd(n,k)=1}\frac{\mu(n)}{n}\log n = B_k+O(C_{A,k}(\log x)^{1-A})$$ Note $$C_{A,k} \approx (A+1)^{\omega(k)}$$ while Iwaniec expects $$\tau(k) \approx 2^{\omega(k)}$$

• To find $$B_k$$ let $$F_k(s)=\sum_{n,\gcd(n,k)=1}\!\! \mu(n) n^{-s} = \frac{1}{\zeta(s)\prod_{p | k} (1-p^{-s})}, \ F_k'(s)=-\sum_{n,\gcd(n,k)=1}\!\!\! \mu(n) n^{-s}\log n$$ Then $$\lim_{s \to 1^+} F_k'(s)= \frac1{\prod_{p | k} (1-p^{-1})} = \frac{k}{\phi(k)}$$ so that (from partial summation of the Dirichlet series) $$B_k =\frac{k}{\phi(k)}$$

• I am sorry but I still have some doubts. In the second equation, I think maybe $f(x/d)$ is not equal to $\sum_{n \leq x, d|n}\frac{\mu(n)}{n}$? – Ke Wang Sep 1 '19 at 11:52
• @KeWang You are completely right. I wrote an elementary estimate, unfortunately I obtained an additional $(A+1)^{\omega(k)}$ term. If you really need $O(2^{\omega(k)} (\log x)^{-A})$ not only $O(2^{\omega(k)} (\log x)^{-A}))$ I think the next step is to find what result gives the Tauberian theorem with $\int_{(2)} \frac{1}{\zeta(s)\prod_{p | k}(1-p^{-s})} \frac{x^s}{s}ds$. Otherwise you can look at $\nu(n)=(-1)^{\Omega(n)}$ the completely multiplicative version of $\mu$ then $\sum_{n\le x\gcd(n,k)=1}\frac{\nu(n)}{n}=\sum_{d| n} \mu(d)\frac{\nu(d)}{d}\sum_{n\le x/d,\gcd(n,k)=1}\frac{\nu(n)}{n}$ – reuns Sep 1 '19 at 20:53
• Otherwise you can look at $\nu(n)=(-1)^{\Omega(n)}$ the completely multiplicative version of $\mu$ this time we do have $\sum_{n\le x\gcd(n,k)=1}\frac{\nu(n)}{n}=\sum_{d| k} \mu(d)\frac{\nu(d)}{d}\sum_{n\le x/d}\frac{\nu(n)}{n}= \sum_{d | k} \frac1d O((\log x/d)^{-A})=O(\frac{k}{\phi(k)} (\log x/k)^{-A})$ – reuns Sep 1 '19 at 20:57