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:  
 $$S_2=\frac{1}{\pi^{2}}\sum_{(a,b)\in\mathbb{Z}^2}\frac{1}{(1+(a+2b)^2)(1+(3a+2b)^2)}=\frac{\cosh^{2}(\pi)+\cosh(\pi)+2}{4\cosh(\pi)^2-4}$$
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).

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1) 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][1] as well.

2) Also, in this problem, can we exploit the structure of the lattice formed by matrix $M$ to compute the sum faster?
  
3) Not a question, just my guess, it seems to me that using $\cosh(\pi)$ and integers is enough to represent closed form of $S$ (in case it exists).


  [1]: https://mathoverflow.net/questions/483456/closed-form-of-frac1-pin-int-mathbbrndv-prod-i-1l-frac11