The answer to the MO question here refers to a result from Montgomery and Vaughan, stating that $$\lim\sup \frac{R(x)}{x\sqrt{\log\log(x)}}>0,$$ where $R(x)$ is the approximation error $$R(x)=\sum_{n\leq x}\phi(n)-\frac{3x^2}{\pi^2}.$$

By Mellin inversion and Cauchy's theorem, this approximation error is related to the zeros of $\zeta(s)$ by $$R(x)=\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{\zeta(s-1)x^sds}{\zeta(s)s},$$ where $1<\sigma<2$.

I had initially thought it was a typo, that $x$ should be under the root, that the result is conditional and appears to be the law of the iterated logarithm.

Thus, is it a typo or, if not, why does it not suggest that $\zeta(s)$ has zeros arbitrarily close to $\sigma=1$?

I am assuming it does not suggest that because no corollaries were mentioned.

  • $\begingroup$ I don't know, but I can give you the Montgomery citation: Fluctuations in the mean of Euler's phi function, Proc. Indian Acad. Sci. Math. Sci. 97 (1987), no. 1-3, 239–245 (1988), MR0983617 (90e:11138). $\endgroup$ Feb 6, 2013 at 22:34
  • $\begingroup$ Perhaps I am totally confused, but AFAIK, the result is actually Omega plus/minus result and since the summands can be of size x and by continuity a sublinear (in x) lower boudn for the error would not really be a result. So the x should not be under the root. I do not oversee the rest of your question at the moment. $\endgroup$
    – user9072
    Feb 6, 2013 at 22:46
  • $\begingroup$ Sorry the comment before does not quite make sense, I still leave it for the moment. In any case when summing things of size x a sqrt lower bound the size of x would seem strange to me. $\endgroup$
    – user9072
    Feb 6, 2013 at 22:49
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    $\begingroup$ I don't think the integral you have written down has very proper convergence properties, on the vertical lines. If you smooth the counting function, with a Mellin transform that decays enough as $t\rightarrow\infty$, then I think I agree that the zeros are more closely linked. $\endgroup$
    – Junkie
    Feb 6, 2013 at 22:55
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    $\begingroup$ ... that should read $R(n)-R(n-1)$ instead of $R(n+1)-R(n)$ of course. In particular, when $p$ is prime we have $R(p)-R(p-1) = (1-6/\pi^2) p + O(1)$ and so this already rules out any universal bound of the form $R(x)=o(x)$. $\endgroup$
    – Terry Tao
    Feb 6, 2013 at 23:47

1 Answer 1


It's not a typo - I've read the proof before (the proof, by the way, doesn't invoke complex analysis at any point; it's very hands-on).

We're used to using Perron's formula / Mellin transforms to write the summatory function of an arithmetic function (such as $\Lambda(n)$ or $\phi(n)$) as a integral over a vertical contour, then moving that contour to the left and saying that the asymptotic behavior of the summatory function is governed by the residues of the poles we move the contour over. But that can only be rigorously established when the remaining part of the contour can be suitably estimated. I think if you try to apply this method here, you'll find that the integrand simply cannot be estimated nicely enough to make the argument work.

  • $\begingroup$ Okay, thank you for this explanation Greg Martin. So, is this scenario therefore similar to, say, the inverse Mellin transform of $\zeta(s)/-1/(s-1)$, where an $O(1)$ contribution persists regardless of where one integrates? $\endgroup$ Feb 7, 2013 at 7:23
  • $\begingroup$ That is, assuming RH, is it the case that the line of integration still cannot be moved to anywhere to the right of that line, even though there would be no poles there? That is, is this a limitation of Cauchy's theorem? $\endgroup$ Feb 7, 2013 at 7:32
  • $\begingroup$ @Kevin: It is not a limitation of Cauchy's theorem, but integrals being large. $\endgroup$
    – GH from MO
    Feb 7, 2013 at 17:58

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