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The identity contained in the last two displayed equations in the following passage (from page 110 in Ayoub's An Introduction to the Analytic Theory of Numbers, 1963) gives us right away a simple proof that the prime number theorem implies $M(x) = o(x)$. (Here $M(x)$ is the Mertens function $\sum_{n\leq x} \mu(n)$.) What is more, it's really useful in deriving explicit bounds on how fast $M(x)/x$ goes to $0$; until now, there was nothing better - the 'natural' analytic approach to bounding $M(x)$ is numerically hopeless. (More about that soon, but that's another matter.)

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I find it unlikely that the identity appeared in Ayoub's textbook for the first time: it's a textbook, and not a particularly early one. However, the first proof that PNT implies $M(x)=o(x)$ (in Landau's thesis; in fact, it's basically Landau's thesis, together with PNT implying $\sum_n \mu(n)/n = o(1)$) seems rather more roundabout, though maybe it's just prolixly written. Any idea of where the identity can first be found?

Rules: full credit for anything the same as the above with different notation; high partial credit for finding the first occurence of $$\sum_{n\leq x} \mu(n) \log n = - 1 - \sum_{m\leq x} \mu(m) \left(\psi\left(\frac{x}{m}\right) - \left\lfloor \frac{x}{m}\right\rfloor\right),$$ which is enough to prove $\text{PNT}\Rightarrow M(x) = 0$. (Reason: $(\log x) \sum_{n\leq x} \mu = \sum_{n\leq x}\mu(n) \log n + \sum_{n\leq x} \mu(n) \log \frac{x}{n}$, and $\sum_{n\leq x} \log \frac{x}{n}\leq x$.)

Note: found the citation to Ayoub in Schoenfeld's 1969 paper on explicit bounds on $M(x)$.

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  • $\begingroup$ OTOH Ayoub is careful to give references at the end of each chapter and discusses them at length, so maybe this identity is really his. $\endgroup$
    – H A Helfgott
    Commented Nov 15 at 23:58
  • $\begingroup$ Are you aware the second Chebyshev function $$\psi(x)=\sum\limits_{n=1}^x \Lambda(n)\tag{1}$$ can be evaluated as $$\psi(x)=-\sum\limits_{n=1}^x \mu(n)\, \log(n)\, \left\lfloor \frac{x}{n}\right\rfloor\tag{2}$$ and also as $$\psi(x)=\sum\limits_{n=1}^x \log(n)\, M\left(\frac{x}{n}\right)\tag{3}$$ where $$M(x)=\sum\limits_{n=1}^x \mu(n)\tag{4}$$ is Mertens function? $\endgroup$
    – Steven Clark
    Commented Nov 16 at 3:37
  • $\begingroup$ Also the prime-power counting function $$K(x)=\sum_{p^k\le x} 1=\sum\limits_{n=1}^{\log _2(x)} \pi \left(x^{1/n}\right)\tag{5}$$ with Moebius inversion $$\pi(x)=\sum_{p\le x} 1=\sum\limits_{n=1}^{\log_2(x)} \mu(n)\, K\left(x^{1/n}\right)\tag{6}$$ can be evaluated as $$K(x)=-\sum\limits_{n=1}^x \mu(n)\, \nu(n)\, \left\lfloor \frac{x}{n}\right\rfloor\tag{7}$$ where $\nu(n)$ is the number of distinct primes dividing $n$ and also as $$K(x)=\sum\limits_{n=1}^x \Omega(n)\, M\left(\frac{x}{n}\right)\tag{8}$$ where $\Omega(n)$ is the number of non-distinct primes dividing $n$. $\endgroup$
    – Steven Clark
    Commented Nov 16 at 4:10
  • $\begingroup$ I'm wondering if the formulas in your question are more closely related to formulas (1) to (4) in my first comment above or to the relationship $$\psi(x)=\sum\limits_{n=1}^x \mu(n)\, T\left(\frac{x}{n}\right)\tag{9}$$ where $$T(x)=\sum\limits_{n=1}^x \log(n)\tag{10}$$ is the log-step summatory function and $$T(x)=\sum\limits_{n=1}^x \psi \left(\frac{x}{n}\right)\tag{11}$$ is the Moebius inversion of formula (9) above. $\endgroup$
    – Steven Clark
    Commented Nov 16 at 4:39
  • $\begingroup$ @StevenClark Formulas (1)-(4) can't be compared to the identities I am asking about: if you plug in the trivial estimate for $M(x)$, you get a worse-than-trivial estimate for $\psi(x) = \sum_{n\leq x} \Lambda(n)$ (please, for all that's good, keep the notation $\psi(x)$ for this function). The identities in my question give you an estimate $M(x)=o(x)$ given any estimate of the form $\psi(x) = x + o(x)$. $\endgroup$
    – H A Helfgott
    Commented Nov 16 at 9:21

1 Answer 1

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I'm willing to give Landau half-marks for section 155 (chapter 41, part 11, vol II) in his Handbuch: he derives $M(x)=o(x)$ from PNT by noting, in effect, that $$\sum_{n\leq x} \mu(n) \log n = - \sum_{m\leq x} \mu(m) \left(\psi\left(\frac{x}{m}\right) - \frac{x}{m}\right) - x \sum_{m\leq x} \frac{\mu(m)}{m}.$$ He's already proved in section 153 that $|\sum_{m\leq x} \mu(m)/m|\leq 1$ using $\sum_{m\leq x} \mu(m) \lfloor x/m\rfloor = 1$ (the argument and the bound on $\sum_{m\leq x} \mu(m)/m$ are due to Gram, 1884, whom Landau properly cites). Landau actually separates the terms with $m$ higher than a certain $x/\xi$ (because readers needed things spelled out back then?) but does not attempt to get cancellation in $\mu$ there. (Neither does Ayoub.)

Landau says that section 155 is based on his own paper "Über den Zusammenhang einiger neueren Sätze in der analytischen Zahlentheorie" from 1906. It doesn't seem easy to find it online; I'll look it up in his Collected Works (which I don't have at home) when I can. (Of course maybe some Internet search hero will step up with a link first.)

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  • $\begingroup$ Typo fixed, thanks. $\endgroup$
    – H A Helfgott
    Commented Nov 16 at 21:22
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    $\begingroup$ You need to fix the same typo in your question above where you indicate $\text{PNT}\Rightarrow M(x) = 0$. $\endgroup$
    – Steven Clark
    Commented Nov 16 at 21:34
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    $\begingroup$ Tao cites the 1906 paper here so maybe he can furnish a copy. Gordon (1958) shows PNT implies the desired conclusion here by the very end of the paper: but he similarly mentions the Landau (1906) paper. At the least, 1958 < 1963, so this constitutes a slight lowering of the upper bound! $\endgroup$
    – Benjamin Dickman
    Commented Nov 16 at 22:06

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