Closed form of $\frac{1}{\pi^{n}}\int_{\mathbb{R}^n}dV \prod_{i=1}^{l}\frac{1}{1+(v_{i}^{T}x)^2}$

Question

Is it possible to find closed form of $$I=\frac{1}{\pi^{n}}\int_{\mathbb{R}^n}dV \prod_{i=1}^{l}\frac{1}{1+(v_{i}^{T}x)^2}$$ in terms of vectors $v_i$?

Where $x=(x_1,\ldots,x_{n}),\ dV=dx_1\wedge\ldots\wedge dx_{n},\ l \geq n,\ \operatorname{span}(\{v_1,\ldots,v_l\})=\mathbb{R}^n$
Whenever I input the entries of $v_{i}$ as integers, I always get $I$ as a rational number, that's why I believe there is a closed form of $I$. Below is one of the cases I found (though I did not see any clear pattern so that I can guess the closed form of $I$)
$$\frac{1}{\pi^{2}}\int_{\mathbb{R}^2}dV\prod_{i=1}^{3}\frac{1}{1+(v_{i}^{T}x)^2}=\frac{1}{|\det(v_1\ v_2)|+|\det(v_2\ v_3)|+|\det(v_3\ v_1)|}$$ So far, I have only used residue theorem to evaluate these integrals and I have to apply it repeatedly each integral, which prevented me from using the "symmetry" of variables, also, the amount of calculations escalated very quickly.

I'm not only interested in the closed form of $I$, I'm also looking for a technique to deal with this kind of problem more efficiently (something like multivariate residue theorem combined with linear algebra, I imagine).

Answer 1

Here my reformulated comment as answer:

We first consider the case $l=n$ and start with the integral \begin{align} \tag{1a}\label{eq:1a} I_0 &= \frac{1}{\pi^{n}}\int_{\mathbb{R}^n} \mathrm d^n \xi \prod_{i=1}^{n}\frac{1}{1+(e_{i}^{T}\xi)^2} \\ \tag{1b}\label{eq:1b} &= \frac{1}{\pi^{n}}\prod_{i=1}^{n}\int_{-\infty}^\infty \frac{\mathrm d \xi_i}{1+\xi_i^2} \\ \tag{1c}\label{eq:1c} &=1, \end{align} with base vectors $e_i$, e.g., $e_1=(1,0,\ldots,0)$.

We now define the matrix \begin{align} \tag{2}\label{eq:2} V = \{v_1,\ldots,v_n\} \quad\Leftrightarrow\quad e_i^T\,V=v_i^T \end{align} such that \begin{align} \tag{3a}\label{eq:3a} I_0=1 &= \frac{1}{\pi^{n}}\int_{\mathbb{R}^n} \mathrm d^n \xi \prod_{i=1}^{n}\frac{1}{1+(e_{i}^{T} V V^{-1}\xi)^2} \\ \tag{3b}\label{eq:3b} &= \frac{1}{\pi^{n}}\int_{\mathbb{R}^n} \mathrm d^n \xi \prod_{i=1}^{n}\frac{1}{1+(v_{i}^{T} V^{-1}\xi)^2} \\ \tag{3c}\label{eq:3c} &= \frac{1}{|\det V^{-1}|}\frac{1}{\pi^{n}}\int_{\mathbb{R}^n} \mathrm d^n x \prod_{i=1}^{n}\frac{1}{1+(v_{i}^{T} x)^2} \\ \tag{3d}\label{eq:3d} &= |\det V|\,I, \end{align} with $x=V^{-1}\xi$, where we introduced the Jacobi determinant $|\det V|$ in the change of variables.

I did not work out the case $l>n$. I guess that one can extend the integration from $\mathbb{R}^n$ to $\mathbb{R}^l$ and let the additional components of $v_{i>n}$ go to infinity using a certain protocol, such that the corresponding (normalized) Lorentzians become Dirac delta distributions and drop out again. This might lead to a sum of Jacobi determinants in \eqref{eq:3d}. However, this is only a guess.

Answer 2

Partial results

I just realized the problem is a lot harder than I thought due to heavy algebraic expressions. I suspect that we might never find out the most general solution even if it exists. Therefore, I think predicting patterns is worth trying here.

The list below is a collection of found cases:

1. Case $n=1$, not simplified (trivial, somewhat unuseful) $$\frac{1}{\pi}\int_{\mathbb{R}}dx\prod_{i=1}^{l}\frac{1}{1+(a_{i}x)^2}=\sum_{i=1}^{l} \frac{a_{i}^{2l-3}}{\prod_{j\not =i}(a_{i}^{2}-a_{j}^{2})}$$

2. Case $n=1,\ l=3$, simplified (trivial)

$$\frac{1}{\pi}\int_{\mathbb{R}} dx\prod_{i=1}^3\frac{1}{1+(a_{i}x)^2}=\frac{a_{1}a_2+a_{2}a_{3}+a_{3}a_{1}}{(a_{1}+a_{2})(a_{2}+a_{3})(a_3+a_1)}$$

3. Case $n=1,\ l=4$, simplified (non-trivial, but provable)
   $$\frac{1}{\pi}\int_{\mathbb{R}}dx\prod_{i=1}^{4}\frac{1}{1+(a_{i}x)^2}=\frac{\sum_{\sigma\in S_4}a_{\sigma(1)}^2 a_{\sigma(2)}^2 a_{\sigma(3)}^1 a_{\sigma(4)}^0+2\sum_{\sigma\in S_4}a_{\sigma(1)}^2 a_{\sigma(2)}^1 a_{\sigma(3)}^1 a_{\sigma(4)}^1}{\prod_{i\not= j}(a_{i}+a_{j})}$$ note: case 1 to case 3, wlog, assumed $a_i\geq 0$
4. Case $l=n$ (trivial, see FredHucht's answer) $$\frac{1}{\pi^n}\int_{\mathbb{R}^n}d^{n}x\prod_{i=1}^{n}\frac{1}{1+(v_{i}^T x)^2}=\frac{1}{|\det(v_{1}\ldots v_{n})|}$$
5. Case $n=2, l=3$ (UNSOLVED) $$\frac{1}{\pi^{2}}\int_{\mathbb{R}^2}d^{2}x\prod_{i=1}^{3}\frac{1}{1+(v_{i}^T x)^2}=\frac{1}{D_{12}+D_{23}+D_{31}}$$
6. Case $n=2, l=4$ (UNSOLVED) $$\frac{1}{\pi^{2}}\int_{\mathbb{R}^2}d^{2}x\prod_{i=1}^{4}\frac{1}{1+(v_{i}^T x)^2}=\frac{\sum D_{aa'}D_{bb'}D_{cc'}-\sum_{i\in [4]} D_{i\mu_1}D_{i\mu_2}D_{i\mu_3}}{\prod_{\{i,j,k\}\subset [4]}\left(D_{ij}+D_{jk}+D_{ki}\right)}$$ $D_{ij}=|\det(v_{i}\ v_{j})|$. In the case 6 right above, the left sum sums over all triples of distinct determinants.

Honestly, I have made a lot of guesses and computation to find formulas but I couldn't continue when $l-n$ grows. I expect that the case $l=n+1$ and the case $l=n+2$ can be solved but I'm not sure about $l\geq n+3$ .
I hope these partial results are helpful.