How to show $\sum_{d\mid k}\frac{\mu (d)}{d}\left(\log\left(\frac{x}{d}\right)+O(1)\right)=\left(\sum_{d\mid k}\frac{\mu (d)}{d}\right)\log x+O(1)$

Question

How to show this is true.

$$\sum_{d\mid k}\frac{\mu (d)}{d}\left(\log\left(\frac{x}{d}\right)+O(1)\right)=\left(\sum_{d\mid k}\frac{\mu (d)}{d}\right)\log x+O(1)$$

I'm studying the book which is called Problems in Analytic Number Theory. I saw this but didn't quite understand. Isn't a term missing on the right hand side? Like this one?

$$\left(\sum_{d\mid k}\frac{\mu (d)}{d}\log d\right)+O(1)$$

Where did it go?

Answer 1

The "missing term" contributes only to the remainder, since: $$\begin{eqnarray*}\sum_{d|k}\mu(d)\frac{\log d}{d}=\lim_{s\to 1^+}\frac{d}{ds}\sum_{d|k}\frac{\mu(d)}{d^s}&=&\lim_{s\to 1^+}\frac{d}{ds}\prod_{p|k}\left(1-\frac{1}{p^s}\right)\\&=&\frac{\phi(k)}{k}\sum_{p|k}\frac{\log p}{p-1}\\&\ll&\log(k)\log\log(k).\end{eqnarray*}$$