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](https://math.stackexchange.com/questions/806095/double-harmonic-summation) and [this](https://math.stackexchange.com/questions/596826/asymptotic-expansion-for-harmonic-sum-in-two-variables) other questions previously posted on MSE, for which no conclusive answer was provided despite multiple bounties. 



Asymptotic estimate of double summation
Using the decomposition
$$
H(x) := \sum\_{a \leq x} \frac{1}{a} = \log(x) + \gamma - \frac{\psi(x)}{x} + \int\_{x}^{+\infty} \frac{\psi(t) d t}{t^2},
$$
where $\psi(t) = \{ t \} - \frac{1}{2}$, one gets
$$
k(j) = - \sum\_{n \geq 1} \frac{\psi(nj)}{n^2j } + \tilde{k}(j),
$$
where $\tilde{k}$ is an explicit Lipschitz function (see below). In particular :


* if $j$ is irrational, then $k$ is continuous at $j$.
* if $j = \frac{a}{q}$ is rational (with $a$ and $q$ coprime), then $k$ has a jump discontinuity, with $k(j^+) = k(j)$, and $k(j^+) - k(j^-) = \frac{\zeta(2)}{aq}$.


Moreover, $\tilde{k}$ has the form
$$
\tilde{k}(j)= -\frac{1}{2} \log(j)^2 + \gamma \log(j) + \frac{1}{j} \int\_{1}^{+ \infty} \frac{\psi(jt) H(t) d t}{t^2} + c,
$$
for some constant $c$.


