Expansion of elementary symmetric function in Jack's? Consider the expansion
$$
e_\mu(x) = \sum_\lambda c_{\mu\lambda}(\alpha) J_\lambda^{(\alpha)}(x)
$$
where $J_\lambda^{(\alpha)}(x)$ are the integral-form Jack polynomials (the ones with $n!$ as coefficient of $m_{1^n}(x)$).
Is there some result which proves that each $c_{\mu\lambda}(\alpha)$ is of the form $P(\alpha)/Q(\alpha)$, where $P,Q \in \mathbb{N}[\alpha]$?
It seems that $Q$ can be chosen as some $Q_\mu(\alpha)$ that factors in a nice way (some product of deformed hook values perhaps?) 
 A: Regarding my comment above, note that by induction we only need to show that $\langle J_\lambda, J_{1^n}J_\nu\rangle\in\mathbb{N}[\alpha]$. This follows from the dual Pieri formula for Jack polynomials (obtained by combining Theorems 3.3 and 6.1 of http://math.mit.edu/~rstan/pubs/pubfiles/73.pdf), so we do get $c_{\mu\lambda}(\alpha)\in\mathbb{N}[\alpha]$, as well as a factorization of $Q_\mu(\alpha)$ into linear factors.
A: I was told the following argument, due to V. Feray. 
First, note that $J^{(\alpha)}_\lambda$ expands in the monomial basis with rational coefficients, where both numerator and denominator have rational coefficients (this is due to the Knop-Sahi formula).
Now, since the homogeneous symmetric functions are dual to the monomial symmetric functions, and since the Jacks are self-dual, we should have that the complete homogeneous symmetric functions expands positively in the Jack basis.
Finally, one needs to apply the $\omega$ involution on both sides, and basically have the positivity (one needs to do some work to verify that $\omega$ on Jack polynomials behaves well).
