Evaluate a certain two-dimensional integral involving an inverse hyperbolic tangent

Question

Evaluate the integral of

(1/(π (-1 + y^2))) 8 x ((2 + (-3 + u^2) x^2 - 2 y^2) Sqrt[1 - u^2 x^2 - y^2] (-1 + y^2) + 
     3 u^2 (-1 + u^2) x^4 Sqrt[1 - y^2] 
       ArcTanh[Sqrt[(-1 + u^2 x^2 + y^2)/(-1 + y^2)]])

over $x \in [0,1]$, $y \in [-1,1]$, subject to $u \geq 1$, thereby obtaining a (complex-valued) function of $u$. More specifically (as Anton Antonov requested in a comment),

Integrate[
  (1/(π (-1 + y^2))) 8 x 
    ((2 + (-3 + u^2) x^2 - 2 y^2) Sqrt[1 - u^2 x^2 - y^2] (-1 + y^2) + 
      3 u^2 (-1 + u^2) x^4 Sqrt[1 - y^2] 
        ArcTanh[Sqrt[(-1 + u^2 x^2 + y^2)/(-1 + y^2)]]), 
  {x, 0, 1}, 
  {y, -1, 1}, 
  Assumptions -> u > 1]

This problem started as a four-dimensional integration with (three-dimensional) integrand

(π^2 x (1 + (-1 + u^2) x^2 - y^2 - z^2) (-1 + u^2 x^2 + y^2 + z^2)) / 
  (2 (-1 + y^2 + z^2))

over

u > 1 && -(1/u) < x Sin[θ] < 1/u && 
-(Sqrt[1 - u^2 x^2 Sin[θ]^2]/u) < x Cos[θ] < Sqrt[1 - u^2 x^2 Sin[θ]^2]/u && 
-Sqrt[1 - u^2 x^2] < y < Sqrt[1 - u^2 x^2] && 
-Sqrt[1 - u^2 x^2 - y^2] < z < Sqrt[1 - u^2 x^2 - y^2] && 0 < θ < 2 π && 
1 > x > 0

That is,

Integrate[
  (π^2 x (1 + (-1 + u^2) x^2 - y^2 - z^2) (-1 + u^2 x^2 + y^2 + z^2)) / 
    (2 (-1 + y^2 + z^2))
    Boole[
      u > 1 && -(1/u) < x Sin[θ] < 1/u && 
      -(Sqrt[1 - u^2 x^2 Sin[θ]^2]/u) < x Cos[θ] < Sqrt[1 - u^2 x^2 Sin[θ]^2]/u && 
      -Sqrt[1 - u^2 x^2] < y < Sqrt[1 - u^2 x^2] && 
      -Sqrt[1 - u^2 x^2 - y^2] < z < Sqrt[1 - u^2 x^2 - y^2]], 
  {x, 0, 1}, {θ, 0, 2 Pi}, {y, -1, 1}, {z, -1, 1}]

I integrated out the $z$ variable, arriving at the two-dimensional problem (the angular variable $\theta$ being absent from the resulting integrand).

I am trying to obtain the $\tilde{\chi_2}(\varepsilon)$ counterpart to $\tilde{\chi_1}(\varepsilon)$, reported in eq. (9) of https://arxiv.org/pdf/1610.01410.pdf.

I am somewhat puzzled as to the complex-valued nature of the two-dimensional integration result. (For $u=1$, one obtains $1-\frac{16 i}{15 \pi }$.)

Answer 1

(Not an answer, but an extended comment.)

Is this the integral your are looking at?

iFunc = (π^2 x (1 + (-1 + u^2) x^2 - y^2 - z^2) (-1 + u^2 x^2 + y^2 + z^2)) /(2 (-1 + y^2 + z^2))

u = 1;
NIntegrate[
 iFunc*Boole[
   u >= 1 && -(1/u) < x Sin[θ] < 
     1/u && -(Sqrt[1 - u^2 x^2 Sin[θ]^2]/u) < x
      Cos[θ] < 
     Sqrt[1 - u^2 x^2 Sin[θ]^2]/u && -Sqrt[1 - u^2 x^2] < y < 
     Sqrt[1 - u^2 x^2]], {x, 0, 1}, {y, -1, 1}, {θ, 0, 
  2 π}, {z, -Sqrt[1 - u^2 x^2 - y^2], Sqrt[1 - u^2 x^2 - y^2]}, 
 PrecisionGoal -> 6, AccuracyGoal -> 4, MinRecursion -> 8, 
 MaxRecursion -> 20]

(* Out[83]= 8.11743 *)

Answer 2

The answer to the real part of the evaluation of the integral is \begin{equation} f(u)= \frac{4 u^2-1}{3 u^4}, \end{equation} it almost certainly seems. (See my very extended answer to the highly-related mathematics.stack.exchange question https://math.stackexchange.com/questions/2267678/solve-a-constrained-4d-integration-problem-using-either-cartesian-or-paired-po/2271420#2271420, the 2D integrand given at the beginning here being derived from the 4D integrands there.)

The complex part of the evaluation certainly bears investigation, apparently also being simple in nature, as indicated by the $-\frac{16 i}{15 \pi}$ result for $u=1$, but it was not germane to the original, underlying research question, and seems to have been introduced rather inadvertently through my various manipulations.

Answer 3

fn = (\[Pi]^2 x (1 + (-1 + u^2) x^2 - y^2 - z^2) (-1 + u^2 x^2 + y^2 +
       z^2))/(2 (-1 + y^2 + z^2))
Boole[u > 1 && -(1/u) < x Sin[\[Theta]] < 
    1/u && -(Sqrt[1 - u^2 x^2 Sin[\[Theta]]^2]/u) < x Cos[\[Theta]] < 
    Sqrt[1 - u^2 x^2 Sin[\[Theta]]^2]/u && -Sqrt[1 - u^2 x^2] < y < 
    Sqrt[1 - u^2 x^2] && -Sqrt[1 - u^2 x^2 - y^2] < z < 
    Sqrt[1 - u^2 x^2 - y^2]];

fzx = Integrate[fn, {x, 0, 1}, {z, -1, 1}]

(\[Pi]^2 (2 (u^2 - 1) u^2 ArcCot[Sqrt[y^2 - 1]] + (1 - 6 y^2) Sqrt[
y^2 - 1]))/(12 Sqrt[y^2 - 1])

fzxt = Integrate[fzx, {\[Theta], 0, 2 \[Pi]}]

(\[Pi]^3 (2 (u^2 - 1) u^2 ArcCot[Sqrt[y^2 - 1]] + (1 - 6 y^2) Sqrt[
y^2 - 1]))/(6 Sqrt[y^2 - 1])

Integrate[fzxt, {y, -1, 1}]

1/6 \[Pi]^3 (-2 + I (\[Pi]^2 + 8 I Catalan) u^2 (u^2 - 1))