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Suppose I have a rational function of $8$ coordinates $a,b,c,d,e,f,g,h$ that I want to integrate over $\mathbb{H}^4$:

(256*b*d*f*(a^5 + b^4*(c - 2*e) - a^4*(3*c + 2*e) - c*(c^2 + d^2)*(e^2 + f^2) + a^3*(2*b^2 + 3*c^2 + d^2 + 6*c*e + e^2 + f^2) + 
    b^2*(c^3 - 2*c^2*e + 2*d^2*e + c*(d^2 - e^2 - f^2)) + a*(b^4 + 2*c^3*e + 2*c*d^2*e + 3*c^2*(e^2 + f^2) + d^2*(e^2 + f^2) + 
      b^2*(-c^2 - 3*d^2 + 6*c*e + e^2 + f^2)) - a^2*(c^3 + 6*c^2*e + 2*d^2*e + 2*b^2*(c + 2*e) + c*(d^2 + 3*(e^2 + f^2))))*(e - g)*h*
   (c^5 + d^4*(e - 2*g) - c^4*(3*e + 2*g) - e*(e^2 + f^2)*(g^2 + h^2) + c^3*(2*d^2 + 3*e^2 + f^2 + 6*e*g + g^2 + h^2) + 
    d^2*(e^3 - 2*e^2*g + 2*f^2*g + e*(f^2 - g^2 - h^2)) + c*(d^4 + 2*e^3*g + 2*e*f^2*g + 3*e^2*(g^2 + h^2) + f^2*(g^2 + h^2) + 
      d^2*(-e^2 - 3*f^2 + 6*e*g + g^2 + h^2)) - c^2*(e^3 + 6*e^2*g + 2*f^2*g + 2*d^2*(e + 2*g) + e*(f^2 + 3*(g^2 + h^2)))))/
  ((a^2 + b^2)*((-1 + c)^2 + d^2)*((b + d)^2 + (a - c)^2)*((b - d)^2 + (a - c)^2)*((b + f)^2 + (a - e)^2)*((b - f)^2 + (a - e)^2)*((d + f)^2 + (c - e)^2)*((d - f)^2 + (c - e)^2)*((d + h)^2 + (c - g)^2)*((d - h)^2 + (c - g)^2)*((f + h)^2 + (e - g)^2)*((f - h)^2 + (e - g)^2))

Here $\mathbb{H}^4$ is the Cartesian product of four copies of the upper half plane $\mathbb{H} = \mathbb{R} \times \mathbb{R}_{>0}$. That is, the integration bounds are given by $a \in (-\infty,\infty), b \in (0, \infty), \ldots, g\in (-\infty,\infty), h\in(0,\infty)$.

I'd like to know the exact value of the integral, but accurate numerical approximations would also be appreciated. Since I have 9 more integrals of this type, general advice is also welcome.

I tried some of Mathematica's methods, but I didn't succeed.


The denominator factors as a product of $12$ quadratics of the form $x^2+y^2$:

$$(a^2 + b^2)((c - 1)^2 + d^2)((b + d)^2 + (a - c)^2)((b - d)^2 + (a - c)^2)((b + f)^2 + (a - e)^2)((b - f)^2 + (a - e)^2)((d + f)^2 + (c - e)^2)((d - f)^2 + (c - e)^2)((d + h)^2 + (c - g)^2)((d - h)^2 + (c - g)^2)((f + h)^2 + (e - g)^2)((f - h)^2 + (e - g)^2)$$

If $x^2+y^2 = 0$ then $x=y=0$, so the set of singularities is a union of (restrictions of) strict linear subspaces. Also, we have that $x$ involves only $a,c,e,g \in (-\infty,\infty)$ and $y$ involves only $b,d,f,h \in (0, \infty)$.


The numerator factors as $256bdfh(e-g)$ times $$a^5 + 2 a^3 b^2 + a b^4 - 3 a^4 c - 2 a^2 b^2 c + b^4 c + 3 a^3 c^2 - a b^2 c^2 - a^2 c^3 + b^2 c^3 + a^3 d^2 - 3 a b^2 d^2 - a^2 c d^2 + b^2 c d^2 - 2 a^4 e - 4 a^2 b^2 e - 2 b^4 e + 6 a^3 c e + 6 a b^2 c e - 6 a^2 c^2 e - 2 b^2 c^2 e + 2 a c^3 e - 2 a^2 d^2 e + 2 b^2 d^2 e + 2 a c d^2 e + a^3 e^2 + a b^2 e^2 - 3 a^2 c e^2 - b^2 c e^2 + 3 a c^2 e^2 - c^3 e^2 + a d^2 e^2 - c d^2 e^2 + a^3 f^2 + a b^2 f^2 - 3 a^2 c f^2 - b^2 c f^2 + 3 a c^2 f^2 - c^3 f^2 + a d^2 f^2 - c d^2 f^2$$

times $$c^5 + 2 c^3 d^2 + c d^4 - 3 c^4 e - 2 c^2 d^2 e + d^4 e + 3 c^3 e^2 - c d^2 e^2 - c^2 e^3 + d^2 e^3 + c^3 f^2 - 3 c d^2 f^2 - c^2 e f^2 + d^2 e f^2 - 2 c^4 g - 4 c^2 d^2 g - 2 d^4 g + 6 c^3 e g + 6 c d^2 e g - 6 c^2 e^2 g - 2 d^2 e^2 g + 2 c e^3 g - 2 c^2 f^2 g + 2 d^2 f^2 g + 2 c e f^2 g + c^3 g^2 + c d^2 g^2 - 3 c^2 e g^2 - d^2 e g^2 + 3 c e^2 g^2 - e^3 g^2 + c f^2 g^2 - e f^2 g^2 + c^3 h^2 + c d^2 h^2 - 3 c^2 e h^2 - d^2 e h^2 + 3 c e^2 h^2 - e^3 h^2 + c f^2 h^2 - e f^2 h^2,$$

and the latter two factors are both irreducible over $\mathbb{Q}$ according to Mathematica.

The two factors have the same shape: the first is mapped to the second by $$a \mapsto c, \quad b \mapsto d,\quad c \mapsto e,\quad d \mapsto f,\quad e \mapsto g,\quad f \mapsto h.$$

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  • 4
    $\begingroup$ Out of curiosity, why do you want to integrate this function? $\endgroup$ – carmichael561 May 6 '16 at 4:51
  • $\begingroup$ Do those expressions have some symmetries? And does he integral converge? $\endgroup$ – Andrew May 7 '16 at 17:02
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    $\begingroup$ Why is this so poorly formatted? $\endgroup$ – Chill2Macht May 9 '16 at 2:03
  • $\begingroup$ Since the denominator have a lot of factors, try some partial fractions decomposition. $\endgroup$ – Chen Wang May 19 '17 at 11:43

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