Sum of the sum-of-divisors function

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.

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You could contact Y.-F. S. Pétermann directly. Have you done that? –  André Henriques Jun 19 '12 at 18:43

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.

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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). –  quid Jun 20 '12 at 12:35
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. –  Charles Jun 20 '12 at 15:32

This is proved in G. Tenenaum'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.

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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.

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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. –  Eric Naslund Jun 20 '12 at 11:01
@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. –  Eric Naslund Jun 20 '12 at 11:05
My first comment above needs an additional $O(x)$, and a factor of $x$ for the sum. That is, we have $$\frac{\pi^2}{12}x^2 -\sum_{n\leq x} \sigma(n)=x\sum_{n\leq x} \frac{1}{n} s\left(\frac{x}{n}\right) +O(x).$$ –  Eric Naslund Jun 20 '12 at 11:57
@Eric Naslund: a nit-pick, there is no Omega plus/minus, but of course you are still right regarding the main issue. –  quid Jun 20 '12 at 12:30
@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. –  quid Jun 24 '12 at 18:09