Abstract. Let G be a real solvable exponential Lie group with Lie algebra g and let f ∈ g * . We take two polarizations p j , j = 1, 2, at f which meet Pukanszky's condition. Let P j := exp(p j ), j = 1, 2, be the associated subgroups in G. The linear functional f defines unitary characters χ j (exp(X)) := e −i f,X , X ∈ p j , of P j . Let τ j := ind G Pj χ j , j = 1, 2, be the corresponding induced representations, which are unitary and irreducible. It is well known that τ 1 and τ 2 are unitary equivalent. The description of the intertwining operator of such an equivalence is given via an abstract integral I p2,p1 over the homogeneous space P 2 /P 1 ∩ P 2 and one of the main problems is the convergence of such an integral. In this paper, we show that the product P 1 P 2 is closed in G. This then implies that our integral converges at least on a dense subspace of elements of the space of τ 1 and we can prove that this formal integral gives us a concrete intertwining operator. We can in this way avoid the use of a third Pukanszky polarization, which was necessary in the approach made in [1]. Finally, given three Pukanszky polarizations p i , i = 1, 2, 3 at f , we accurately determine the composition formula of I p3,p2 • I p2,p1 • I p1,p3 using Maslov's index.