Nested integrals with products of Heaviside step functions

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I am interested in integrals of the form $$ I_n(r) = \int_0^rdr_1\cdots\int_0^rdr_n\,\Theta_A|\mathbf{H}_1\times\mathbf{H}_2|(\mathbf{H}_3\cdot\mathbf{H}_4)\cdots(\mathbf{H}_{2n-1}\cdot\mathbf{H}_{2n}), $$ where $\mathbf{H} = (a,b)$ is a two-dimensional vector, $\cdot$ denotes the Euclidean inner product, and $|\mathbf{H}_i\times\mathbf{H}_j| = a_ib_j - b_ia_j$. Here subscripts denote variable dependence, so that $\mathbf{H}_1 = \mathbf{H}(r_1)$ and so forth. Finally, $\Theta_A$ is a product of Heaviside step functions of the form \begin{align} \Theta_A &= \prod_{i \in A}\theta_{i,i+1}, & \theta_{i,j} = \begin{cases} 0 \qquad\text{if $r_i - r_j < 0$,}\\ 1 \qquad\text{if $r_i - r_j \geq 0$,} \end{cases} \end{align} where $A \subset \mathbb{Z}_{2n-1}$. In the end I am interested in all possible subsets $A$.

It is obvious that $I_n$ vanishes unless $1 \in A$ or $2 \in A$ (or both). Additionally if, for $m > 1$ we have $(2m-1) \in A$, but $(2m-2) \not\in A$ and $2m \not\in A$ then we can replace the factor $\theta_{2m-1,2m}$ by $\frac{1}{2}$ under integration. Similarly if $2m \in A$ but neither $(2m-2)$, $(2m-1)$, $(2m+1)$, nor $(2m+2)$ is, then we can replace the factor $\theta_{2m,2m+1}$ by $\frac{1}{2}$. In fact, if $$\left\{(2m-1),2m,(2m+1)\right\} \subset A,$$ but $(2m-2) \not\in A$ and $(2m+2) \not\in A$, then we can replace the factor $\theta_{2m-1,2m}\theta_{2m,2m+1}\theta_{2m+1,2m+2}$ by $\frac{1}{2^3}$. All of these can be found by considering the symmetry of the integral, or by doing integration by parts. However, other factors in $\Theta_A$ seem to yield little when attacked in this way.

Is there any way to further simplify $\Theta_A$, or otherwise work with this integral without explicit knowledge of $a$ and $b$? I am more than happy to read relevant material, if any exist. If it is relevant, I do know that $\mathbf{H} : [0,\infty) \to \mathbb{R}^2$ satisfies $\mathbf{H}(0) = (0,0)$, and we can safely assume that $\mathbf{H}$ is analytic.

Note that $$\int_0^rdr_{i+1}\,\theta_{i,i+1}\mathbf{H}_{i+1} = (\mathbf{H} * \bar{\theta})(r_i),$$ where the vector quantity is to be treated in the obvious manner, where $*$ denotes the convolution, and where $\bar{\theta}$ is the Heaviside step function (defined as a regular function, matching $\theta_{i,j} = \bar{\theta}(r_i-r_j)$). For this reason I have considered using the Laplace transform, but because there will, in general, be a lot of regular products in $I_n$, this does not seem to help.