Integration on an ordered domain

The standard technique if the bounds depend on one another is to use a change of variables. This is the usual, easy way to use hypercube-integration packages with a differently shaped domain. For example, see: How to Compute the integration of the following formula by HCubature

If I understand you correctly, this is:

\int_0^\infty dt_n \int_0^{t_n} dt_{n-1} \int_0^{t_{n-1}} dt_{n-2} \cdots \int_0^{t_2} dt_1 \,f(t_1, \ldots, t_n)

which can be transformed to a hypercube via the change of variables t_k = s_k t_{k+1} = t_n \prod_{j=k}^{n-1} s_j for s_{k} \in [0,1]:

\int_0^\infty dt_n \int_0^{1} ds_{n-1} \int_0^{1} ds_{n-2} \cdots \int_0^{1} ds_1\, f\left(t_n \prod_{j=1}^{n-1} s_j, t_n \prod_{j=2}^{n-1} s_j, \ldots, t_n\right) t_n^{n-1} \prod_{j=1}^{n-1} s_j^{j-1}

(Or add a finite upper bound on t_n if desired. If you want a semi-infinite t_n domain like this you may need an additional change of variables.)

2 Likes