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.


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 '19 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 '19 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 '19 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.