Given vectors $m_{j}\in\mathbb{Z}^{n},M=(m_{1} \ldots m_{n}),\det(M)\not =0$. Is it possible to find a closed form of: $$S=\frac{1}{{\pi}^n}\sum_{k\in \mathbb{Z}^n}\prod_{j=1}^n\frac{1}{1+(m_{j}^{T}k)^2}$$ (I'm just asking the possibility of finding closed form, but it would be great if one can find it explicitly)
I observed that if $|\det(M)|=1$, then $S=\coth^{n}(\pi)$ because $M$ acts like an automorphism of $\mathbb{Z}^n$, thus we can make "substitution" and seperate $S$ into product of sums. However, it is not clear how to find $S$ if $|\det(M)|\not=1$. For instance:
$$S_2=\frac{1}{\pi^{2}}\sum_{(a,b)\in\mathbb{Z}^2}\frac{1}{(1+(a+2b)^2)(1+(3a+2b)^2)}=\frac{\cosh^{2}(\pi)+\cosh(\pi)+2}{4\cosh^{2}(\pi)-4}$$
I used residue theorem to calculate $S_2$, the intermidiate form looks like this:
$$S_2=\frac{2}{\pi}\sum_{b\in\mathbb{Z}}\left(\frac{3i\cot\left(\frac{1}{3}(2\pi b+i\pi)\right)}{16(2b^{2}-ib+1)}+\frac{\coth(\pi-i2\pi b)}{16(2b^2+i3b-1)}\right)$$
From here I could use residue theorem again but I can't do it by hand so I let my free WolframAlpha do it for me, the calculator had hard time calculating it as well (but I got the result after all).
Is this the only way to compute the sum? I haven't checked higher dimensions so I doubt that using residue theorem repeatedly works. Can we somehow ultilize $n$ complex variables simultaneously to solve the problem?-It might answer this related question as well.
Also, in this problem, can we exploit the structure of the lattice formed by matrix $M$ to compute the sum faster?
