I would like to determine an asymptotic expansion for the following double summation:

$$\sum_{a=1}^{N/\sqrt {j}} \sum_{b=a}^{ja} \frac{1}{ab}$$ where $j$ is a real number $\geq 1$ and $N$ tends to $\infty$. In practice, the summation includes all pairs of integers $a,b$ (with $a \leq b \leq ja $) such that the product $ab$ is $\leq N^2 $.

For $j=1$, the summation yields the infinite harmonic sum of squares, i.e. $\pi^2/6$. For $j>1$, the asymptotic expansion has the form $\log(N) \log(j) + O(1)$, where the $O(1)$ term converges to a value $k$ that depends on $j$. Interestingly, plotting $k$ vs $j$, the resulting function is discontinuous, with the most evident discontinuities occurring for $j$ integer. Is there any way to express this value explicitly? I am not necessarily searching a closed form, which may probably not exist. Rather, I would be interested to know potential alternative ways (e.g., using functions, series, and so on) to express the value of $k$, different from the trivial definition given by the difference between the double summation and the log term.

This question is a general formulation of a problem that is related to this and this other questions previously posted on MSE, for which no conclusive answer was provided despite multiple bounties.