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So I am trying to show the following: $$\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}=O(1) $$ so I tried partial summation as following:

Let $A(x)=\sum_{n \leq x} \frac{\mu(n)}{n}$, then we have $$\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}= \int_1^{x} \frac{A(t)\mathrm dt}{t},$$ and $A(t)$ is clearly very small, or $o(1)$ for all $t \in [1,x]$. My question is how to go from here to conclude that the error term is $O(1)$?

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    $\begingroup$ For what it's worth, I showed $\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}=O(1)$, using generalized Möbius inversion, as part of my answer to a previous question of yours. $\endgroup$
    – Mike Spivey
    Sep 29, 2011 at 3:13
  • $\begingroup$ If you go the partial summation route, you will need an estimate of $A(t)$ that's more precise than $o(1)$ to conclude the $O(1)$ result you want at the end. For example, if $A(t) = \frac{1}{1+ \log t}$, then $$\int_1^{x} \frac{A(t) dt}{t} = \left.\log (1+\log t)\right|_1^x =\log (1+\log x),$$ which is not $O(1)$. $\endgroup$
    – Mike Spivey
    Sep 29, 2011 at 3:30
  • $\begingroup$ So I think in your answer, there is a error:$$\sum_{n \leq x} \frac{1}{n}=\log{x}+\gamma+O(\frac{1}{x})$$, thus when use Generalized Mobius inversion, we have $$\sum_{n \leq x} \frac{\mu(n)}{n} \log{\frac{x}{n}}=-\sum_{n \leq x} \frac{\mu(n)}{n}\gamma-\sum_{n \leq x} \mu(n) O(\frac{1}{x})+1=-o(1)-\sum_{n \leq x} \mu(n) O(\frac{1}{x})+1$$, However, the middle part doesn't seems to be O(1). $\endgroup$
    – Rob
    Sep 29, 2011 at 5:56
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    $\begingroup$ I've corrected the mistake in my other answer. $\endgroup$
    – Mike Spivey
    Sep 29, 2011 at 13:13
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    $\begingroup$ Sure. Anything that is $o(1)$ is also $O(1)$, and so you have $O(1) + O(1) + O(1) = O(1)$. $\endgroup$
    – Mike Spivey
    Sep 30, 2011 at 0:16

1 Answer 1

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As you mentioned, we have $$\sum_{n\leq x}\frac{\mu(n)}{n} \log\frac{x}{n}=\int_1^x \left(\sum_{m\leq t}\frac{\mu(m)}{m}\right) \frac{dt}{t}.$$

You need a particular bound on the inner sum. For instance, since $\sum_{m\leq t}\frac{\mu(m)}{m}\ll (\log t)^{-2}$, the result follows.

Here is cool trick however. Notice that $$\sum_{nd\leq x}\frac{\mu(n)}{nd}=\sum_{k\leq x}\frac{1}{k}\sum_{d|k}\mu(d)=1.$$ Now, write the first sum as $$\sum_{n\leq x}\frac{\mu(n)}{n}\sum_{k\leq\frac{x}{n}}\frac{1}{k}=\sum_{n\leq x}\frac{\mu(n)}{n}\left(\log\left(\frac{x}{n}\right)+\gamma+O\left(\frac{n}{x}\right)\right).$$

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