I'm trying to get a closed-form expression for a particular type of sum, or at least a good way to approximate such sums numerically. I've tried using Mathematica, Maxima, etc, to no avail so far.
The simplest example of the type of sum I'm looking at is this:
$$ \sum_{m,n\in\Bbb Z^2} \left( \frac{1}{e^{|a\cdot m|}} \cdot \frac{1}{e^{|b\cdot n|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot n|}} \right)$$
where I'm assuming that $a, b \in \Bbb R > 0$, so that the entire sum converges absolutely. The summation is taken from $-\infty$ to $\infty$ on both variables.
If that last term weren't there, we'd have
$$ \sum_{m,n\in\Bbb Z^2} \left( \frac{1}{e^{|a\cdot m|}} \cdot \frac{1}{e^{|b\cdot n|}} \right) = \left( \sum_{m\in\Bbb Z} \frac{1}{e^{|a\cdot m|}} \right)\cdot \left( \sum_{n\in\Bbb Z} \frac{1}{e^{|b\cdot n|}} \right) = \frac{(e^{|a|}+1)(e^{|b|}+1)}{(e^{|a|}-1)(e^{|b|}-1)}$$
and we know this converges whenever $a,b>0$. However, the presence of the last term makes it difficult to do it this way.
Does there exist any basic method for evaluating sums of this form?
The general version of the sums I'm looking at are basically the same as above, but with $n$ indices, typically for small $n$ no larger than 6-8 or so, which can almost be split into a product of sums, except for the presence of one or more extra "non-separable" terms of the type mentioned above where multiple indices are summed in a single exponent. The example I gave is the simplest nontrivial one on two indices. For three indices, an example would be
$$ \sum_{m,n,o\in\Bbb Z^3} \left( \frac{1}{e^{|a\cdot m|}} \cdot \frac{1}{e^{|b\cdot n|}} \cdot \frac{1}{e^{|c\cdot o|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot n|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot o|}} \cdot \frac{1}{e^{|a\cdot n + b\cdot o|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot n + c\cdot o|}} \right)$$
This is in some sense the "maximal" sum on three indices, where there are four of these non-separable terms. However, I am also looking at such sums where only some of those terms appear, for instance
$$ \sum_{m,n,o\in\Bbb Z^3} \left( \frac{1}{e^{|a\cdot m|}} \cdot \frac{1}{e^{|b\cdot n|}} \cdot \frac{1}{e^{|c\cdot o|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot n|}} \cdot \frac{1}{e^{|a\cdot m + b\cdot n + c\cdot o|}} \right)$$
I would be happy even for an approximate answer or way to speed up the numeric calculation, which can be slow to carry out to a reasonable number of terms even when the number of indices is only about 6 or so.
EDIT - as an addendum, I realized I missed one important variation of the original sums I was looking at, and that's where the index in the exponent isn't just a sum of $a \cdot m$ and $b \cdot n$, but where the multiplicands on $m$ and $n$ can be different. I'll rewrite the original as a sum on indices $n_1, n_2, ..., n_d$:
$$ \sum_{n_1,n_2\in\Bbb Z^2} \left( \frac{1}{e^{|c_1\cdot n_1|}} \cdot \frac{1}{e^{|c_2\cdot n_2|}} \cdot \frac{1}{e^{|c_3\cdot n_1 + c_4 \cdot n_2|}} \right)$$
and the "maximal" 3-term sum would be:
$${\small \sum_{n_1,n_2,n_3\in\Bbb Z^3} \left( \frac{1}{e^{|c_1\cdot n_1|}} \cdot \frac{1}{e^{|c_2\cdot n_2|}} \cdot \frac{1}{e^{|c_3\cdot n_3|}} \cdot \frac{1}{e^{|c_4 \cdot n_1 + c_5 \cdot n_2|}} \cdot \frac{1}{e^{|c_6 \cdot n_1 + c_7 \cdot c_3|}} \cdot \frac{1}{e^{|c_8 \cdot n_2 + c_9 \cdot n_3|}} \cdot \frac{1}{e^{|c_{10} \cdot n_1 + c_{11} \cdot n_2 + c_{12}\cdot n_3|}} \right)}$$
so you can get a different term on each exponent.
It would be useful to at least get a beginning start on the more basic version, though.
Not an answer but some thoughts (yet gonna try it numerically).
The most straightforward idea is to split the range of summation into pieces where arguments of the absolute values retain the same sign each, and compute the resulting sums separately. Say, one of such sums for your "simplest example" would be $$\sum_{\substack{m,n\geqslant 0\\ am-bn\geqslant 0}}e^{-am-bn-(am-bn)}=\sum_{m=0}^{\infty}\left(\Big\lfloor\frac{am}{b}\Big\rfloor+1\right)e^{-2am}$$ which can be computed rather efficiently (and, for rational $a/b$, even analytically). I still need a look at the general setting in somewhat more systematic way, but this may well be the way to go.
Yet another idea I should probably follow sharper is to use $$e^{-|t|}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{e^{itx}\ dx}{1+x^2},\qquad\sum_{n\in\Bbb{Z}}e^{inx}=2\pi\sum_{n\in\Bbb{Z}}\delta(x-2n\pi)$$ (the latter is in the generalized sense). In fact, using the first of these, we get $$\sum_{m,n\in\mathbb{Z}}e^{-a|m|-b|n|-|cm+dn|}=\frac1\pi\int_{-\infty}^\infty f(a,c,x)f(b,d,x)\frac{dx}{1+x^2},\\f(\alpha,\beta,x):=\sum_{n\in\mathbb{Z}}e^{-\alpha|n|+i\beta nx}=\frac{\sinh\alpha}{\cosh\alpha-\cos\beta x}\qquad(\alpha>0)$$
But the integral remains. In the general case it even becomes multiple. I wonder what it gives from the computational point of view (are there variants of double-exponential methods applicable?).