Average order of $\phi(n)$

Question

Theorem 3.7 of the book Analytic Number Theory by Apostol states:

$$\sum_{n\le x} \phi(n)= \frac{3}{\pi^2} x^2 + O(x\log x)$$

and then it claims : Hence the average order of $\phi(n)$ is $\frac{3n}{\pi^2}$.

But I don't understand it because if we do this:

$$\frac{1}{x} \sum_{n\le x} \phi(n)= \frac{3x}{\pi^2} + O(\log x),$$ then it seems we get something else.

So, how is the average order calculated ? Why are we ignoring the term $O(\log x)$?

Any hint/help will be appreciated. Thanks.

Answer 1

I doubt that Apostol uses a different definition for average order from the more widely accepted one. According to Hardy & Wright's, Ténenbaum's, and Montgomery & Vaughan's textbooks, we say that $b_n$ is an average order of $a_n$ if

$$ \sum_{n\le x}a_n\sim\sum_{n\le x}b_n $$

Using this relatively more accepted definition, we have

$$ \sum_{n\le x}{6n\over\pi^2}\sim{3x^2\over\pi^2}\sim\sum_{n\le x}\varphi(n) $$

This means that the average order of $\varphi(n)$ is $6n/\pi^2$. However, if the definition of Apostol is used, we have

$$ \frac1x\sum_{n\le x}\varphi(n)\sim{3x\over\pi^2} $$

Consequently, Apostol's version of average order for $\varphi(n)$ is $3n/\pi^2$.