4
$\begingroup$

Let $$F(x) = \sum_{d\leq x} \frac{\mu(d)}{d} \log \frac{x}{d}.$$

s it possible/feasible to give an elementary proof of the fact that $F(x)= 1 + o(1)$ (and, ideally, $1+O(1/\log x)$, or better)? By "elementary" I mean "using the properties of $\zeta(s)$ only for $\Re(s)\geq 1$, and of preference only for $s$ real". (Call work with $s$ complex, $\Re(s)\geq 1$, "semi-elementary" if you wish.) I'd also need for it to be possible to make the bounds nicely explicit.

[Note: I am well aware of Ramaré's and Balazard's work, which relies on estimates on $\sum_{m\leq x} \mu(m)$ (derived in turn from estimates on $\sum_{m\leq x} \Lambda(m)$). I am looking for (semi-)elementary estimates, in part because I would like something that can be easily adapted to analogous sums.]

Improve this question
$\endgroup$
3
  • $\begingroup$ In case you (ore other users) will find there something useful (in the answer or in the comments), I'll add a link to math.SE post Showing that $\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n} = O(1)$.. Found using Approach0. $\endgroup$
    – Martin Sleziak
    Commented Jan 22, 2017 at 11:59
  • $\begingroup$ Probably a very naive question of mine, but would it help in any way to write the considered sum as $ \sum_{d\leqslant x}\frac{\mu(d)}{d}\log x-\sum_{d\leqslant x}\mu(d)(\log' d.\log d) $? $\endgroup$
    – Sylvain JULIEN
    Commented Jan 22, 2017 at 12:55
  • $\begingroup$ according to Sylvain julien , Try to check this Erdos bound , equation 3 and 4 :terrytao.wordpress.com/tag/divisor-function, in the sense of conditionally convergent series. Assuming one can justify this (which, ultimately, requires one to exclude zeroes of the Riemann zeta function on the line Re(s)=1, you can obtain your estimation $\endgroup$
    – zeraoulia rafik
    Commented Jan 22, 2017 at 23:33

1 Answer 1

Reset to default
1
$\begingroup$

In chapter 4, p. 15-18 of the book

H. Iwaniec, Lectures on the Riemann Zeta Function, American Mathematical Society, University Lecture Series nº 62, 2014

there is an elementary proof of the prime number theorem with a rest $\psi(x)=x +O(x(\log x)^{-A})$ with arbitrary A. In page 18 he propose as exercise to proof $$ \sum_{m\le x} \frac{\mu(m)}{m}\ll (\log x)^{-A},\qquad \sum_{m\le x} \frac{\mu(m)}{m}\log m=-1+O((\log x)^{-A}.$$ From this your assertion follows. So it is definitely feasible.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Er - I'd like to see it done? $\endgroup$
    – H A Helfgott
    Commented Jan 25, 2017 at 18:54
  • $\begingroup$ He uses "semi-elementary methods" to prove $M(x)\ll x (\log x)^{-A}$. From this I think there is no much difficulty. Essentially this is done by Landau when he proves $\sum \mu(n) \log n/n=-1$. (Primzhalen p. 612-613) $\endgroup$
    – juan
    Commented Jan 25, 2017 at 19:49
  • $\begingroup$ Sure, I agree it's not difficult if you assume that. But what are "semi-elementary methods"? Do you have to work with $\zeta(s)$ to the left of the line $\Re(s)=1$? $\endgroup$
    – H A Helfgott
    Commented Jan 25, 2017 at 19:51
  • $\begingroup$ But Iwaniec do not use zeta at Re s=1 only for $\sigma>1$. Well it is true he uses the behavior near $\sigma=1$ but he appear to prove this in page 10 again without using $\sigma=1$. $\endgroup$
    – juan
    Commented Jan 25, 2017 at 19:56
  • $\begingroup$ Iwaniec treat the analytic continuation in Chapter 6 of his book. $\endgroup$
    – juan
    Commented Jan 25, 2017 at 20:02

You must log in to answer this question.

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