For $m \geq 1$ an integer and $k>0$ a real number, define $$F_{k,m}(x) = \sum\limits_{a_1\cdots a_m \leq x} \dfrac{1}{(a_1 \cdots a_m)^k}$$ In particular, we have $$F_{k,3}(x) = \sum\limits_{abc \leq x} \dfrac{1}{(abc)^k}$$ (Here the product is over the $m$-tuples $(a_1, ..., a_m)$ of integers greater than or equal to $1$ such that the product $a_1 \cdots a_m$ is smaller than or equal to $x$).
My question: what is the asymptotics of $F_{k,m}(x)$ as $x \to \infty$? At least are there some good upper bounds?
Ideas: I tried for $m=3$, when $k \neq 1$: $$F_{k,3}(x) \ll \sum_{a \leq x} \sum_{b \leq \frac x {a}} \dfrac{(x/ab)^{-k+1}}{-k+1} \ll x^{-k+1} \sum_{a \leq x} \frac 1 a \sum_{b \leq \frac x {a}} \frac 1 b \ll x^{-k+1} \sum_{a \leq x} \frac 1 a \ln(x/a) \leq x^{-k+1} \ln(x) \sum_{a \leq x} \frac 1 a \ll x^{-k+1} \ln(x)^2$$
When $k=1$, I got $F_{1,3}(x) \ll \ln(x)^3$ Apparently we have $F_{1,3}(x) = \dfrac{\ln(x)^3}{6} + O(\ln(x)^2)$. How can I find it? What about the general case for $k>0$ (and for $m>3$)?
Thank you.
With $|m|_\times = \prod_{i=1}^l m_i$ $$\zeta(s+k)^l = \sum_{m \in \mathbb{Z}_{\ge 1}^l} |m|_\times^{-s-k} = s \int_1^\infty (\sum_{m \in \mathbb{Z}_{\ge 1}^l, |m|_\times < x} |m|_\times^{-k}) x^{-s-1}dx$$ By inverse Mellin transform and the residue theorem $$\begin{eqnarray}\sum_{m \in \mathbb{Z}_{\ge 1}^l, |m|_\times < x} |m|_\times^{-k} &=& \frac{1}{2i\pi}\int_{2-i\infty}^{2+i\infty} \zeta(s+k)^l\frac{x^s}{s}ds \\ &=& Res(\zeta(s+k)^l\frac{x^s}{s},0)+Res(\zeta(s+k)^l\frac{x^s}{s},1-k)+\frac{1}{2i\pi}\int_{\frac{l-1}{l}-k+\epsilon-i\infty}^{\frac{l-1}{l}-k+\epsilon+i\infty} \zeta(s+k)^l\frac{x^s}{s}ds \\ &=& \zeta(k)^l-\frac{(l-1)!}{k-1}x^{1-k} (\log x)^{l-1}+\sum_{j=0}^{l-2} a_{l,k,j}x^{1-k} (\log x)^j+\mathcal{O}(x^{\frac{l-1}{l}-k+\epsilon})\end{eqnarray}$$ Where you'll compute the coefficients $a_{l,k,j}$ in term of $\zeta(s) = \frac{1}{s-1}+\sum_{n=0}^\infty \gamma_n (s-1)^n$, or for a fixed $l,k$ using WA