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 \begin{equation} -\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})\}. \end{equation}

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.


1 Answer 1


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) $$B_k =\frac{k}{\phi(k)}$$

  • $\begingroup$ I am sorry but I still have some doubts. In the second equation, I think maybe $f(x/d)$ is not equal to $\sum_{n \leq x, d|n}\frac{\mu(n)}{n}$? $\endgroup$
    – Ke Wang
    Sep 1, 2019 at 11:52
  • $\begingroup$ @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 $\sum_{n\le x\gcd(n,k)=1}\frac{\nu(n)}{n}=\sum_{d| n} \mu(d)\frac{\nu(d)}{d}\sum_{n\le x/d,\gcd(n,k)=1}\frac{\nu(n)}{n}$ $\endgroup$
    – reuns
    Sep 1, 2019 at 20:53
  • $\begingroup$ Otherwise you can look at $\nu(n)=(-1)^{\Omega(n)}$ the completely multiplicative version of $\mu$ this time we do have $\sum_{n\le x\gcd(n,k)=1}\frac{\nu(n)}{n}=\sum_{d| k} \mu(d)\frac{\nu(d)}{d}\sum_{n\le x/d}\frac{\nu(n)}{n}= \sum_{d | k} \frac1d O((\log x/d)^{-A})=O(\frac{k}{\phi(k)} (\log x/k)^{-A})$ $\endgroup$
    – reuns
    Sep 1, 2019 at 20:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .