Using Boole to integrate over region

Question

I have a complex and long piecewise function to integrate, so I have written a Mathematica script that cuts the whole range of integration, evaluates the integrand in each region and then integrates it.

In order to integrate the evaluated function on the right range, I set the integration on the whole original range and I use Boole to define the appropriate region.

Now, it happens that both the integrand and the region depends on other parameters, say a and b, so I should get an integral which depends on these, as in the following WORKING example

Assuming[0 < 2 a < 1, 
 Integrate[
  Exp[-t] Boole[(b < 0.5 && a < t < 2 a) || (b > 0.5 && 
       0 < t < a)], {t, 0, 1}]]

which gives

$\begin{cases} e^{-2. a} \left(e^a-1.\right) & 0.<a<0.5\land b<0.5 \\ e^{-1. a} \left(e^a-1.\right) & 0.<a<0.5\land b>0.5 \\ 0. & \text{True} \end{cases}$

However, when the range of integration becomes too complex, Mathematica stops to evaluate the integral and give me back an implicit solution. For example, consider the following code

r = (0 < b < 1/
     2 && ((0 < a < b && 
        b + a < z < 
         1 && ((z - b < x <= z - a && 0 < y < a) || (z - a < x < z && 
            0 < y < z - x))) || (a == b && b + a < z < 1 && 
        z - a < x < z && 0 < y < z - x) || (b < a < 1 - b && 
        b + a < z < 1 && z - b < x < z && 0 < y < z - x))) || (1/2 <= 
     b < 1 && 0 < a < 1 - b && 
    b + a < z < 
     1 && ((z - b < x <= z - a && 0 < y < a) || (z - a < x < z && 
        0 < y < z - x)))
int = Exp[-0.5 y1^2 y - 
    2 (y2^2 (z - b - a) + y1^2 (-z + b + x + a))] y1^2 Boole[r]
Integrate[int, {z, a, 1}, {x, a, z}, {y, 0, a}]

The last line gives me $\text{y1}^2 \int _a^1\int _a^z\int _0^a\exp \left(-2 \text{y1}^2 (a+b+x-z)+2 \text{y2}^2 (a+b-z)-0.5 y \text{y1}^2\right) \text{Boole}[...]dydxdz$

The same happens if I isolate intervals with respect to a,b , as in

r = LogicalExpand[(0 < b < 
      1/2 && ((0 < a < b && 
         b + a < z < 
          1 && ((z - b < x <= z - a && 0 < y < a) || (z - a < x < z &&
              0 < y < z - x))) || (a == b && b + a < z < 1 && 
         z - a < x < z && 0 < y < z - x) || (b < a < 1 - b && 
         b + a < z < 1 && z - b < x < z && 0 < y < z - x))) || (1/2 <=
       b < 1 && 0 < a < 1 - b && 
     b + a < z < 
      1 && ((z - b < x <= z - a && 0 < y < a) || (z - a < x < z && 
         0 < y < z - x)))]
s = Reduce[r[[2]], {x, y}]
int = Exp[-0.5 y1^2 y - 
    2 (y2^2 (z - b - a) + y1^2 (-z + b + x + a))] y1^2 Boole[s]
Integrate[int, {z, a, 1}, {x, a, z}, {y, 0, a}]

or if I split the Integrate function

Integrate[
 Integrate[Integrate[int, {y, 0, a}], {x, a, z}], {z, a, 1}]

Why Mathematica cannot evaluate the integral?

Answer 1

It helps to tell Integrate a little about your parameters:

int = Exp[-1/2 y1^2 y - 2 (y2^2 (z - b - a) + y1^2 (-z + b + x + a))] y1^2 Boole[r];
Integrate[int, {z, a, 1}, {x, a, z}, {y, 0, a}, 
 Assumptions -> 0 < a < b < 1/2 && y1 > 0 && y2 > 0]
(*
   -(1/(6 y1^2 y2^2))
    E^(-((5 a y1^2)/2) - 2 b y1^2 - 2 (a + b) y1^2 - 
      2 y2^2) (-1 + E^((a y1^2)/2)) (E^((5 a y1^2)/2 + 2 b y1^2) + E^(
       3 a y1^2 + 2 b y1^2) + E^((7 a y1^2)/2 + 2 b y1^2) - 
       3 E^(4 a y1^2 + 2 b y1^2) + 3 E^(2 a y1^2 + 2 (a + b) y1^2) - 
       3 E^(2 b y1^2 + 2 (a + b) y1^2)) (E^(2 y2^2) - E^(2 (a + b) y2^2))
*)


int = Exp[-1/2 y1^2 y - 2 (y2^2 (z - b - a) + y1^2 (-z + b + x + a))] y1^2 Boole[s];
Integrate[int, {z, a, 1}, {x, a, z}, {y, 0, a}, 
 Assumptions -> 0 < a < b < 1/2 && y1 > 0 && y2 > 0]
(*
   (E^(-2 (a + b) y1^2 - 
     2 y2^2) (-1 + E^((a y1^2)/2))^2 (1 + 2 E^((a y1^2)/2) + 
      3 E^(a y1^2)) (E^(2 y2^2) - E^(2 (a + b) y2^2)))/(6 y1^2 y2^2)
*)

(I'm just guessing that y1 and y2 are real.)