On the behaviour of $\sum_{\substack{n\leq x\\(n,k)=1}}\frac{1}{\sqrt[\alpha]{n}}$ For integers $a,b\geq 1$ we denote with $(a,b)$ their greatest common divisor, and the $\alpha$th root with $ \sqrt[\alpha]{x} =x^{1/\alpha}$, where $\alpha\geq 2$. Here with $\mu(n)$ we denote the Möbius function. Then I did calculations following the ideas of [1], for a different example, to get
$$\sum_{\substack{n\leq x\\(n,k)=1}}\frac{1}{\sqrt[\alpha]{n}}=\frac{1}{\alpha} \left( \sum_{d\mid k} \frac{\mu(d)}{\sqrt[\alpha]{d}} \right)\log x+O(1),$$ where $k$ denotes a positive number. But I don't know how do calculations for 
$$\sum_{d\mid k} \frac{\mu(d)}{\sqrt[\alpha]{d}}$$ with the purpose to obtain a more eleaborated and concise result.

Question For $\alpha\geq 2$, what's your approach to get the asymptotic behaviour of
  $$\sum_{\substack{n\leq x\\(n,k)=1}}\frac{1}{\sqrt[\alpha]{n}},$$
  as $x$ is large?. Many thanks.

My thoughts were about Möbius inversion, but I didn't get nothing useful.

[1] Murty, Problems in Analytic Number Theory, Second Edition, GTM 206 Springer (2008). I am saying that the ideas of my calculations were from exercises likes to Exercise 1.5.1 or Exercise 1.5.8.
 A: When $f$ is a smooth function, one method to obtain the asymptotic behaviour of
$$F_k(x) := \sum_{\substack{n \leqslant x \\ (n,k) = 1}} f(n)$$
is using Abel's sum formula. We know pretty well how many $n \leqslant x$ that are coprime to $k$ we have:
$$N_k(x) = \sum_{\substack{n \leqslant x \\ (n,k) = 1}} 1 = \sum_{n \leqslant x} \chi_0(n) = \frac{\varphi(k)}{k}\cdot x + O_k(1),$$
where $\chi_0$ is the principal character modulo $k$ and $\varphi$ is Euler's totient function. The bound on the error term depends on $k$ as indicated, but on nothing else. Thus Abel's formula yields
$$F_k(x) = N_k(x)\cdot f(x) - \int_1^x N_k(t)\cdot f'(t)\,dt\,.$$
With $f(t) = t^{-s}$ and $s \neq 1$, this gives
\begin{align}
\sum_{\substack{n \leqslant x \\ (n,k) = 1}} \frac{1}{n^s}
&= \frac{N_k(x)}{x^s} + s \int_1^x \frac{N_k(t)}{t^{s+1}}\,dt \\
&= \frac{\varphi(k)}{k}\Biggl(x^{1-s} + s\int_1^x \frac{t}{t^{s+1}}\,dt\Biggr)
+ \frac{O_k(1)}{x^s} + s\int_1^x \frac{O_k(1)}{t^{s+1}}\,dt \\
&= \frac{\varphi(k)}{k}\cdot \Biggl(\frac{x^{1-s}}{1-s} - \frac{s}{1-s}\Biggr)
+ \frac{O_k(1)}{x^s} + s\int_1^x \frac{O_k(1)}{t^{s+1}}\,dt\,.
\end{align}
When $\sigma = \operatorname{Re} s$ is non-positive this yields
$$\frac{\varphi(k)}{k}\cdot \frac{x^{1-s}}{1-s} + O_k(x^{-\sigma})$$
and when $\sigma > 0$ the integral converges as $x \to \infty$, and writing it as $\int_1^{\infty} - \int_x^{\infty}$ we obtain
$$\sum_{\substack{n \leqslant x \\ (n,k) = 1}} \frac{1}{n^s} = \frac{\varphi(k)}{k}\cdot \frac{x^{1-s}}{1-s} + C + O_k(x^{-\sigma})\,.$$
For $s = 1$ we obtain
\begin{align}
\sum_{\substack{ n \leqslant x \\ (n,k) = 1}} \frac{1}{n}
&= \frac{N_k(x)}{x} + \int_1^x \frac{N_k(t)}{t^2}\,dt \\
&= \frac{\varphi(k)}{k}\Biggl(1 + \int_1^x \frac{t}{t^2}\,dt\Biggr) + O_k(x^{-1}) + \int_1^x \frac{O_k(1)}{t^2}\,dt \\
&= \frac{\varphi(k)}{k}\log x + C + O_k(x^{-1})\,.
\end{align}
This method yields good asymptotics with little effort, but the downside is that one gets no useful formula for the constant term $C$. And its applicability is restricted by the requirement that $f$ be differentiable. If $f$ were a non-smooth arithmetical function like the divisor function we couldn't use this at all.
Another method is to inject the Möbius function, using
$$\sum_{d \mid m} \mu(d) = \begin{cases} 1 &\text{if } m = 1, \\ 0 &\text{if } m > 1.\end{cases}$$
Then we can write
\begin{align}
\sum_{\substack{n \leqslant x \\ (n,k) = 1}} f(n)
&= \sum_{n \leqslant x} \Biggl(\sum_{d \mid (n,k)} \mu(d)\Biggr)f(n) \\
&= \sum_{d \mid k} \mu(d) \Biggl(\sum_{m \leqslant x/d} f(dm)\Biggr)
\end{align}
changing the order of summation. This is of course only helpful if we can compute
$$\sum_{m \leqslant x/d} f(dm)$$
sufficiently accurately. If $f$ is a completely multiplicative function, like in the cases we're looking at here, this is usually easier than for general $f$, since we can pull $f(d)$ out of the sum. For $f(n) = 1/n$ we have
$$\sum_{n \leqslant y} \frac{1}{n} = \log y + \gamma + O(y^{-1})$$
and thus
\begin{align}
\sum_{\substack{n \leqslant x \\ (n,k) = 1}} \frac{1}{n}
&= \sum_{d \mid k} \frac{\mu(d)}{d}\sum_{m \leqslant x/d} \frac{1}{m} \\
&= \sum_{d \mid k} \frac{\mu(d)}{d}\bigl(\log x + \gamma - \log d + O(d/x)\bigr) \\
&= \Biggl(\sum_{d \mid k} \frac{\mu(d)}{d}\Biggr)\log x + \gamma \sum_{d \mid k} \frac{\mu(d)}{d} - \sum_{d \mid k} \frac{\mu(d)\log d}{d} + O_k(x^{-1})\,.
\end{align}
Now we note that for a multiplicative (it need not be completely multiplicative) function $g$ we have
$$\sum_{d \mid k} \mu(d)\cdot g(d) = \prod_{p \mid k}\bigl(1 - g(p)\bigr),$$
so
$$\sum_{d \mid k} \frac{\mu(d)}{d} = \prod_{p \mid k} \biggl(1 - \frac{1}{p}\biggr) = \frac{\varphi(k)}{k},$$
as we also know from the first part. We thus have gained an explicit formula for the constant term (which is still infeasible to compute for large $k$ in general, since we need the factorisation of $k$).
For $f(n) = 1/n^s$ with $s\neq 1$ we find
$$\sum_{n \leqslant y} \frac{1}{n^s} = \frac{y^{1-s}}{1-s} + \zeta(s) + O(y^{-\sigma})$$
(where $\sigma = \operatorname{Re} s$) by Abel's sum formula, Euler's sum formula or the Euler-Maclaurin formula. (For $\sigma \leqslant 0$, the appearance of the constant term $\zeta(s)$ is spurious, this should be absorbed in the error term then, or one should use the Euler-Maclaurin formula to get more terms in the asymptotic expansion.) Plugging that in, we obtain
\begin{align}
\sum_{\substack{n \leqslant x \\ (n,k) = 1}} \frac{1}{n^s}
&= \sum_{d \mid k} \frac{\mu(d)}{d^s} \sum_{m \leqslant x/d} \frac{1}{m^s} \\
&= \sum_{d \mid k} \frac{\mu(d)}{d^s} \biggl(\frac{x^{1-s}}{1-s}d^{s-1} + \zeta(s) + O(d^{\sigma}x^{-\sigma})\biggr) \\
&= \Biggl(\sum_{d\mid k} \frac{\mu(d)}{d}\Biggr)\frac{x^{1-s}}{1-s} + \zeta(s)\sum_{d \mid k} \frac{\mu(d)}{d^s} + O_k(x^{-\sigma})\,.
\end{align}
Again we find an expression for the constant term (which however is only relevant for $\sigma > 0$, unless we take more terms of the asymptotic expansion into account), namely
$$\zeta(s) \sum_{d \mid k} \frac{\mu(d)}{d^s} = \zeta(s)\prod_{p \mid k}\biggl(1 - \frac{1}{p^s}\biggr) = L(s,\chi_0)\,.$$
