(Not sure if this question belongs here or on m.SE)
For a Lie group, $G$ (of dimension $n$), one can average over the group: $$ \Gamma = \int_{G} d\mu(g) ~g $$ (where $d\mu(g)$ is the left-Haar measure) to construct an object that is left-invariant: $G\Gamma=\Gamma$. If $\{T_i\}_{i=1}^n$ is a basis of the Lie algebra of $G$, and the exponential map is surjective, then we can write $\Gamma$ as an integral over some domain $D\subset\mathbb R^n$: \begin{equation} \Gamma = \int_{D} d\mu(\omega) \exp(\omega^iT_i) \tag{1} \end{equation} ($\{\omega^i\}_{i=1}^n$ here are a set of real numbers).
My question relates to a Lie algebroid and averaging over the groupoid associated with it. The Wikipedia article on Lie algebroids states that a Lie algebroid, $A$, can always be integrated to give a Weinstein groupoid, $W$, which is roughly the group of '$A$-paths' modulo $A$-homotopy, with group multiplication given by path concatenation.
Let us assume that the the vector bundle $A$ is trivial: $A=V\times M$, for a manifold $M$ and a vector space $V$. Let the vector space have basis $\{e_i\}_{i=1}^n$ where $n$ is the dimension of $M$. Then an $A$-path is given entirely by a 'history' $v^i(t)e_i$, $t\in[0,1]$, where $\{v^i(t)\}_{i=1}^n$ are a set of real numbers. The associated group element is given by the (time) ordered exponential $\mathcal T\{e^{\int_0^1 dt~ v^i(t)e_i}\}$. It would seem to me that the analogue of (1) is $$ \Gamma = \int_D d\mu(v^i(t))~\mathcal T \{e^{\int_0^1 dt ~v^i(t)e_i}\} $$ where the integral is over 'paths' $v^i(t)$ modulo the equivalence that descends from $A$-homotopy. My question is if there is an appropriate measure, $d\mu$, and integration domain, $D$, that assures $\Gamma$ is left-invariant in the sense that $\mathcal T\{e^{\int_0^1 dt~ v^i(t)e_i}\}\Gamma=\Gamma$ for all paths $v^i(t)$.