Suppose that $a = [a_m,a_M]$ and $b = [b_m,b_M]$ are two non-empty and disjoint hyperrectangle inside the unit hypercube $\mathbb{I}^d$, that is :
$$\forall i \in \{1,...,d\},\, 0\le a_{m,i} < a_{M,i}\le 1 \text{ and } 0 \le b_{m,i} < b_{M,i} \le 1$$
If we denote by $\lambda$ the Lebesgue measure of a d-dimentional set, i.e for a :
$$\lambda(a) = \prod\limits_{i=1}^{d}(a_{M,i} - a_{m_i})\text{,}$$
and furthermore denote for all $u \in \mathbb{I}^d$ :
$$\lambda_a(u) = \frac{\lambda([0,u]\cap a)}{\lambda(a)}\text{.}$$
I want to compute the d-dimensional integral :
$$A = \int\limits_{\mathbb{I}^d} \lambda_a(u)\lambda_b(u) \,du$$
I tried to expend the $\lambda$ notation into sums with indicatrix but it seems complicated. Does someone have a better idea ?
Edit : For d=1, i.e in $\mathbb{R}$, this is easily done since the non-overlapping criterion leads to nice domain :
$$A\lambda(a)\lambda(b) = \int\limits_{a \times [b_M,1]} u \, du + \int\limits_{b \times [a_M,1]} u \, du + \int\limits_{[b_M,1]\times[a_M,1]} \, du$$
and therefore :
$$A = \frac{\left(\frac{a_M^2 - a_m^2}{2}+1-a_m\right)\mathbb{1}_{a_m > b_M} + \left(\frac{b_M^2 - b_m^2}{2}+1-b_m\right)\mathbb{1}_{b_m > a_M}}{\lambda(a)\lambda(b)}$$
But i'm stuck for dimensions $\ge 1$ since the cases are less clear for me..