10
$\begingroup$

I was looking at the abstract of a paper 1 which claims that [2] and [3] prove $$ \sum_{n\le x}\sigma(n)-\frac{\pi^2}{12}x^2=\Omega(x\log\log x). $$

But I cannot find the above—or indeed, anything approaching it—in [2]. Have I missed something?

The paper [3] clearly discusses the appropriate function and presumably gives the indicated result. I must decipher its notation, though: the author seems to use $\sigma(n)$ to denote what would usually be written $\sigma_{-1}(n)=\sigma(n)/n.$

References

1 Y.-F. S. Pétermann, "An Ω-theorem for an error term related to the sum-of-divisors function", Monatshefte für Mathematik 103:2 (1987), pp. 145-157.

[2] T. H. Gronwall, "Some asymptotic expressions in the theory of numbers", Trans. Amer. Math. Soc. 14 (1913), pp. 113–122. JSTOR

[3] S. Wigert, Sur quelques fonctions arithmétiques, Acta Math. 37 (1914), pp. 113–140.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ You could contact Y.-F. S. Pétermann directly. Have you done that? $\endgroup$
    – André Henriques
    Commented Jun 19, 2012 at 18:43

4 Answers 4

Reset to default
4
$\begingroup$

The clue to understanding the relevance of the quoted results seems to be given in Remark 2 of Pétermann's paper (at the very end). Where it is said that a result on the limes superior of $\sigma_{-1}(x)/ \log \log x$ implies an Omega-result on the error term $E_{-1}$ . (And thus $E_{1}$ which is the one in the question.) This result mentioned in Remark 2, also actually appears in the paper of Gronawall, see Eq. (25) there; except it is staed for $\sigma_1$ , but this translates directly as commented at the beginning of that paper where the relation between $\sigma_{a}$ and $\sigma_{-a}$ is mentioned.

ps. This was written a bit quickly, I hope I still got the details right, but in any case this Remark 2 seems to be helpful in understanding the relation to the earlier papers.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Perhaps the following is also useful: the full text of Pétermann's paper is freely available through emani.org (also linked to on the official journal site). $\endgroup$
    – user9072
    Commented Jun 20, 2012 at 12:35
  • $\begingroup$ Ah, that would have been nice to know before I asked my library to find a copy! Now that I have it I see that Remark 2 does address the issue. Thank you for explaining that. $\endgroup$
    – Charles
    Commented Jun 20, 2012 at 15:32
4
$\begingroup$

It is not clear whether you are asking for a proof/reference for the displayed formula, or an evaluation of the contents of the cited papers.

Dickson's History, Volume 1, page 323, says Wigert proved $$\sum_{n\le x}\sigma(n)={\pi^2x^2\over12}+x((1/2)\log x-\psi(x))+O(x)$$ where $$\psi(x)=x\sum_{n\gt x}{1\over n^2}+\sum_{n\le x}{1\over n}\rho\left({x\over n}\right)$$ and $\rho(x)$ is the fractional part of $x$. Further, for $x$ sufficiently large, $$((1/4)-\epsilon)\log x\lt\psi(x)\lt((3/4)+\epsilon)\log x$$ It seems to me that this gives a poorer error term than the one in your display. Dickson also says Landau gave corrections and simplifications to Wigert's proofs, Gottingsche gelehrte Anzeigen 177 (1915) 377-414.

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ In the definition of $\psi(x)$, the $x\sum_{n>x} \frac{1}{n^2}$ is superfluous since it can just be migrated into the $O(x)$ term appearing in the first line. Perhaps a better way to write this entire thing is $$\frac{\pi^2}{12} x^2 -\sum_{n\leq x } \sigma(n)=\sum_{n\leq x}\frac{1}{n}s\left(\frac{x}{n}\right)$$ where $s(\alpha )=\alpha-1/2$ is the sawtooth function. $\endgroup$
    – Eric Naslund
    Commented Jun 20, 2012 at 11:01
  • 4
    $\begingroup$ @Gerry: You say that "It seems to me that this gives a poorer error term than the one in your display." Note that in the display, the $\Omega_{\pm}$ means it is a lower bound on the maximal size of the error term, whereas the conclusion from the formula you write is an upper bound. $\endgroup$
    – Eric Naslund
    Commented Jun 20, 2012 at 11:05
  • $\begingroup$ My first comment above needs an additional $O(x)$, and a factor of $x$ for the sum. That is, we have $$\frac{\pi^2}{12}x^2 -\sum_{n\leq x} \sigma(n)=x\sum_{n\leq x} \frac{1}{n} s\left(\frac{x}{n}\right) +O(x).$$ $\endgroup$
    – Eric Naslund
    Commented Jun 20, 2012 at 11:57
  • $\begingroup$ @Eric Naslund: a nit-pick, there is no Omega plus/minus, but of course you are still right regarding the main issue. $\endgroup$
    – user9072
    Commented Jun 20, 2012 at 12:30
  • 1
    $\begingroup$ @Eric Naslund: Sorry, in case you found my comment odd. My motication was that I was worried, perhaps unnecessarily so, that somebody else reading your "in the display, the Omega plus/minus" (while in the display in the question there is only an Omega) could get away with the wrong idea that Omega and Omega plus/minus are synonymous. Regarding the question, it is, as indicated by the tag, a history question, on whether Gronwall (in that paper) actually proved what is claimed in Pétermann's paper. This is as documented by one other answers not immediate. So, to me the qu. makes sense. $\endgroup$
    – user9072
    Commented Jun 24, 2012 at 18:09
3
$\begingroup$

This is proved in G. Tenenbaum's book (Introduction to analytic and probabilistic number theory), page 39 (section 3.3, theorem 3). I agree that Gronwall's paper, other than the fact that it studies the same function, seem to be completely unrelated.

Improve this answer
$\endgroup$
-2
$\begingroup$

Error is known to be $O(x \log^{2/3}x)$. See Walfisz's book on exponential sums

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ What book, and where in it? $\endgroup$
    – Alex M.
    Commented May 6, 2019 at 16:55
  • $\begingroup$ This result is in the opposite direction of what I need, an $O$ result rather than an $\Omega$ result. $\endgroup$
    – Charles
    Commented May 8, 2019 at 3:13
  • $\begingroup$ Alex - the book is 'Weylsche Exponentialsummen in der neueren zahlentheorie' by Arnold Walfisz 1963 chapter III p.88 onwards. $\endgroup$
    – TPTW
    Commented May 20, 2019 at 22:25
  • $\begingroup$ Charles - was only refining result given by Gerry Myerson above. $\endgroup$
    – TPTW
    Commented May 20, 2019 at 22:27
  • $\begingroup$ To the experts - cannot the omega result under discussion be got from the omega result for sigma(x) alone viz sigma(x)=omega(x loglogx) given in Hardy and Wright article 22.9 ? $\endgroup$
    – TPTW
    Commented May 20, 2019 at 22:30

You must log in to answer this question.

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