I want to ask on the estimates of the sum $$ \sum_{n=1}^{\infty} \mu(n)M\Big(\frac{x}{n}\Big)=\frac{1}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{x^s}{s\zeta(s)^2}ds.$$ It seems that the estimate of the sum be $x\exp(-c\sqrt{\log x})$ for some positive constant $c$, but it is little known to the sign-changes of the Mertens function. Also, the sum can be represented as $$ \sum_{1 \leq n \leq x}\bigg( \sum_{n=ab}\mu(a)\mu(b) \bigg). $$ It is easy to find the condition that $(the \ inner \ sum)=0$ if every degree of the prime factors of $n$ is bigger than $1$, where $k(n)$ is the square-free kernel of an integer. But also the representation has some difficulties.

So I thought that it is more good to deal with the integral, then there have to be some informations on the integrations with the integrand $\zeta(s)^{-2}$. I want to ask the estimates of the sum or the informations on the integrals which involve the integrand $\zeta(s)^{-2}$.

Additionally, the exact expression of the complex integral above with the zeros of the Riemann zeta function is given by $$ 4+2\sum_{|\text{Re } \rho| <1}\frac{x^{\rho}}{\zeta'(\rho)\rho}\phi_{(0,\rho)}+\sum_{n=1}^{\infty}\text{Res}\bigg[\frac{x^s}{s\zeta(s)^2},-2n\bigg],$$ where $$\phi_{(0,\rho)}=\sum_{k=1}^{\infty}\bigg\{ \frac{\mu(k)}{k^{\rho}}-\frac{1}{\zeta'(\rho)}\log\Big(\frac{k+1}{k}\Big) \bigg\} .$$ (Ref: N. A. Carella, "Simple Zeros Of The Zeta Function", page 3, http://arxiv.org/ftp/arxiv/papers/1306/1306.0458.pdf)

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    $\begingroup$ By the way, I would always be very dubious of Carella's calculations, as he has an ungodly number of incorrect results on the arXiv. $\endgroup$ Nov 3, 2015 at 20:19
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    $\begingroup$ I would say that it is good to have a secondary source to verify the calculations. N. A. Carella has at least one incorrect result posted on ArXiv, and others which appear unverified. Gerhard "Also Has ArXiv Unverified Results" Paseman, 2015.11.03. $\endgroup$ Nov 3, 2015 at 20:54

1 Answer 1


One can treat this sum with the Dirichlet hyperbola method: $$\begin{align} \sum_{ab\leq x}\mu(a)\mu(b) &= \sum_{\substack{a\leq\sqrt x\\b\leq x/a}}\mu(a)\mu(b)+\sum_{\substack{b\leq\sqrt x\\a\leq x/b}}\mu(a)\mu(b)-\sum_{a,b\leq\sqrt x}\mu(a)\mu(b)\\ &= 2\sum_{n\leq\sqrt x}\mu(n)M\left(\frac{x}{n}\right)-M(\sqrt x)^2\\ &\ll\sum_{n\leq\sqrt{x}}M\left(\frac{x}{n}\right)+M(\sqrt x)^2\\ &\ll\sum_{n\leq\sqrt{x}}\frac{x}{n}\exp\left(-c_1\sqrt{\log\frac{x}{n}}\right)+x\exp\left(-2c_1\sqrt{\log\sqrt{x}}\right)\\ &\ll\sum_{n\leq\sqrt{x}}\frac{x}{n}\exp\left(-c_2\sqrt{\log x}\right)+x\exp\left(-c_2\sqrt{\log x}\right)\\ &\ll x\left(\log x\right)\exp\left(-c_2\sqrt{\log x}\right)\\ &\ll x\exp\left(-c_3\sqrt{\log x}\right). \end{align}$$

Alternately, one can start from your Mellin transform representation and apply the usual technique as in the proof of the prime number theorem (truncation, contour shift, zero free region and lower bounds for $\zeta(s)$) to get the same result. In fact one can go the other way, i.e. upper bounds for your sums can be turned into a zero free region and a lower bound for $\zeta(s)$, so the above bound cannot be improved significantly. Specifically, one can replace $\sqrt{\log x}$ by something close to $(\log x)^{3/5}$, but that is the limit of current knowledge.

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    $\begingroup$ Oh I missed the fact that it suffices to deal with $\sqrt x$. Always thanks to your answers GH. $\endgroup$
    – B . O
    Nov 4, 2015 at 5:46

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