I need to compute the following integral with three indicator functions, $I(x)$, $I_1(x_1)$ and $I_2(x_2)$,
$$ \int_0^1\int_0^1I(x)\left( \ I_1(x_1)\,x_{1} + I_2(x_2)\,x_{2}-1 \right) dx_1 dx_2 $$
where
$$ I(x)= \begin{cases} 1 & \text{if } I_1(x_1)x_1 + I_2(x_2)x_2 \geq s \\ 0 & \text{otherwise}% \end{cases} $$
$$ I_i(x_i)=% \begin{cases} 1 & \text{if } x_i \geq t_i \\ 0 & \text{otherwise} \end{cases} \qquad i=1,2 $$
I try to write it as
Integrate[
(x1 + x2 - 1)*(Boole[x1 + x2 >= s && x1 >= t1 && x2 >= t2]),{x1,0,1},{x2,0,1}]
It does generate some superficially long output, which makes me further suspect the syntax.
I found at this place a similar problem about multiple indicator functions was also discussed , however, it did not include the above "compound" indicator case.
PS: The original integral is with $n+1$ indicator functions. I simplified it to the above $2+1$ indicator functions with $\textit{U}(0,1)$ case.
UnitStep[x-t]
$\endgroup$Integrate[ (x1 + x2 - 1)*(Boole[x1 + x2 >= s)UnitStep[x1 - t1]UnitStep[x1 - t2],{x1,0,1},{x2,0,1}]
generate exact the same output. However, if changing toIntegrate[ (x1 + x2 - 1)UnitStep[x1 + x2 -s]UnitStep[x1 - t1]UnitStep[x1 - t2],{x1,0,1},{x2,0,1}]
, it seems not working, and I am not sure the reason for the later. $\endgroup$