Ytukyg

Let $$phi_k(x)=sum_{1le n le x \(n,x)=1} n^k$$
What's the asymptotic behavior of
$$sum_{n=1}^xphi_k(n)?$$

According to the wikipedia $sum^x_{n=1} phi_0 (n) approx frac{3}{pi^2}x^2 $. It also appears in page $69$ and $70$ which are $30$ and $31$ of this pdf.

The possible routes

Route 1 (For someone who wants some practice with Abel Summations): There should be an approach which is an analog to the techniques shown here: sum of the divisor functions and I think that $sum_{n=1}^x frac{phi_k(n)}{n^{k+1}}$ is always on the order of a linear function. So that might be the place to start.

If no one takes this route I will almost certainly post my own answer in 2 or 3 weeks and ask this community for help verifying my proof. This is the most obvious route for me to take to make progress on this.

Route 2: Also it would be particularly interesting to see an argument which isn't an analog of the linked post and which exploits what we already know about the asymptotic behavior of $sum sigma_k(n)$ to make claims about $sum phi_k(n)$. I am not sure this possible but it may be a route forward.

This is a work in progress. For now I have copied an argument for the case $k=0$ and started to make progress on modifying for other $k$.

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

Before we prove this we will need two lemmata: Lemma (2.14) of this and the other lemma.

Lemma 2.14
$$sum_{n le x}sum_{d|n} g(d) fbigg(frac{n}{d} bigg)=sum_{d le x} g(d)Fbigg(frac{x}{d} bigg) text{ where } F(x)=sum_{nle x} f(n)
$$

This lemma is essentially what we get when apply the partial summation formula to a Dirichlet convolution which we let $*$ denote.

Lemma 2.17
The last equality of the claim's proof will follow through another lemma
which says $sum frac{mu(d)}{d^2} to frac{6}{pi^2}$

This lemma I still need to flush out. And it will likely need to be modified to make the broader argument I want to below.

Proof of the claim

Let $f(x)=ximplies F(x)=sum_{n le x} n = frac{1}{2} lfloor xrfloor(lfloor xrfloor +1) in frac{1}{2} x^2+O(x)$

Next we exploit (2.14) of the linked document.

$id=phi * 1 implies phi= id* mu$
We will take $f=id, g= mu$ and applying (2.14). All of these symbols are defined on the wiki.

$$
begin{align}
& sum_{nle x} phi(n) \
& =sum_{dle x} mu(d) F bigg(frac{x}{d} bigg) \
& in sum_{dle x} bigg( mu(d) bigg( frac{1}{2} bigg(frac{x}{d} bigg)^2+Obigg(frac{x}{d} bigg) bigg) bigg) \
& text{In this next line it doesn't look to me like $mu$ distributes. Why is this ok? }\
&=frac{1}{2}x^2 sum_{d le x} frac{mu (d)}{d^2}+Obigg(x sum_{d le x} frac{1}{d} bigg) \
&= frac{3}{pi^2}x^2+O(x log x)
end{align}
$$
As desired. $square$

Now. Let $k>1$ be given (I am sure we can find a broader context that this is true... but baby steps). What we want to do is use Reuns comment which is that $phi_k=f*g$ where $f(x)=sigma_k(x)$ and $g(x)= mu(x) x^k$

First we note that $F(x) in frac{zeta(k+1)}{k+1}x^{k+1}+O(x^k)$ which we get from here.
Then we will echo the argument above:
$$
begin{align}
& sum_{nle x} phi_k(n) \
& =sum_{dle x} mu(d)d^k F bigg(frac{x}{d} bigg) \
& in sum_{dle x} bigg( mu(d)d^k times bigg( frac{zeta(k+1)}{k+1} bigg(frac{x}{d} bigg)^{k+1}+Obigg(frac{x}{d} bigg)^k bigg) bigg) \
&=frac{zeta(k+1)}{k+1} x^{k+1}
sum_{dle x} frac{mu(d)}{d} + Obigg(x^k sum_{d le x} frac{1}{d^k} bigg)
end{align}
$$
So that's not the worst place to leave it for now. I need to next get bounds on $sum_{dle x} mu(d) / d approx frac{1}{ln(x)+ gamma}$

So that would mean that
$sum_{n le x} phi_k(n) approx frac{zeta(k+1)x^{k+1}}{(k+1)(ln(x)+gamma)}$

I think there must be an error in my work somewhere. Things just are not lining up numerically.

Not an answer but maybe a useful update:

We can get $ sum_{n=1}^xphi_{1}(n)=frac{1}{2}sum_{n=1}^x nphi_0(n)$ according to this. So we can apply Abel's Summation formula to this and we get

$$frac{1}{2}sum^x_{n=1} nphi_0(n) \ approx frac{1}{2}bigg ( x^3 bigg( frac{3}{pi^2} bigg)-int_1^x {frac{3}{pi^2}t^2}dt bigg) \
approx frac{1}{ pi^2} x^3 $$

This looks close to correct. Maybe this can help to generalize.

Reuns asks: What is $$sum_{d|m} mu(d)sum_{ndle m} (dn)^k$$

I dunno. Maybe:

$$sum_{d|m} mu(d)d^ksum_{ndle m} n^k$$

$$sum_{d|m} mu(d)d^kf_k(lfloor m/d rfloor)$$

Where $f_k(x)=sum_{n=1}^x n^k $ is a Faulhaber sum. This may not be what is being sought though...