Finding expression for $\sum\limits_{n\leq x}\frac{\varphi(n)}{n^\alpha}$.

Question

In Apostol's book on number theory, there is a problem related to average order of arithmetic functions:

For $x\geq 2 $ show that, $\sum\limits_{n\leq x}\frac{\varphi(n)}{n^\alpha}=\begin{cases}\frac{x^{2-\alpha}}{2-\alpha}\frac{1}{\zeta(2)}+\frac{\zeta(\alpha-1)}{\zeta(\alpha)}+O(x^{1-\alpha}\log x),\text{ if } \alpha>1,\alpha\neq 2,\\ \frac{x^{2-\alpha}}{2-\alpha}\frac{1}{\zeta(2)}+O(x^{1-\alpha}\log x),\text{ if }\alpha\leq1.\end{cases}$

I am unable to prove this. Please somebody help me to do this problem.

Answer 1

By the identity $\varphi = \mu \ast \text{id},$ we have \begin{align*} \sum_{n\leq x} \frac{\varphi(n)}{n^\alpha}&= \sum_{n\leq x}\frac{1}{n^\alpha} \sum_{d|n} \mu(d)\frac{n}{d} \\ &= \sum_{\substack{n\leq x \\ n=dk}} \frac{\mu(d)}{d^\alpha k^\alpha}\frac{kd}{d}\\ &= \sum_{d\leq x} \frac{\mu(d)}{d^\alpha}\left(\sum_{k \leq x/d} \frac{1}{k^{\alpha-1}}\right). \end{align*} Now split in two cases, the formulas below can be found as an application of Euler's summation formula.

• For $\alpha >1, \alpha \neq 2,$ you can use $$ \sum_{k \leq x/d} \frac{1}{k^{\alpha-1}} = \frac{(x/d)^{2-\alpha}}{2-\alpha} + \zeta(\alpha-1)+ \mathcal{O}\left(\left(\frac{x}{d}\right)^{1-\alpha}\right). $$
• For $ \alpha \leq 1$ use $$ \sum_{k \leq x/d} \frac{1}{k^{\alpha-1}} = \frac{(x/d)^{2-\alpha}}{2-\alpha}+ \mathcal{O}\left(\left(\frac{x}{d}\right)^{1-\alpha}\right). $$

Then you should be able to proceed. (Later on after substituting the results, you might also want to complete the resulting sum into a series and subtract the tail, I can update this answer if you wish).