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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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    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](http://math.stackexchange.com/questions/66820/how-do-i-prove-sum-n-leq-x-frac-mu-nn-log2-fracxn-2-logxo/67025#67025) to a previous question of yours.2011-09-29
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    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 $O(1)$.2011-09-29
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    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).2011-09-29
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    The middle part is actually $o(1)$; according to Apostol's *Introduction to Analytic Number Theory* (Theorems 4.14 and 4.15) this is equivalent to the prime number theorem. (But you're right that I made an error in my other answer. Thanks for bringing this to my attention; I'll go back and fix it in the morning.)2011-09-29
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    I've corrected the mistake in my other answer.2011-09-29
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    So you are saying that $-o(1)-o(1)+1=O(1)?$2011-09-29
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    Sure. Anything that is $o(1)$ is also $O(1)$, and so you have $O(1) + O(1) + O(1) = O(1)$.2011-09-30
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    Related post on MathOverflow: [$\sum_{d\leq x} (\mu(d)/d) \log x/d$: elementary estimates?](http://mathoverflow.net/q/260235)2017-01-22

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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 $$\sum_{n\leq x}\frac{\mu(n)}{n}\sum_{k\leq\frac{x}{n}}\frac{1}{k}=\sum_{n\leq x}\frac{\mu(n)}{n}\left(\log\left(\frac{x}{n}\right)+\gamma+O\left(\frac{n}{x}\right)\right).$$