Let $J_\lambda^{(\alpha)}(x)$ be the Jack polynomials in $N$ variables, with a normalization such that the coefficient of the monomial polynomial $m_\lambda$ is equal to 1.

They satisfy the identity $$ \det(x)^s J_\lambda^{(\alpha)}(x^{-1})=J_{\widetilde{\lambda}}^{(\alpha)}(x),$$ where $s$ is a large enough integer and $\widetilde{\lambda}=(s-\lambda_N,s-\lambda_{N-1},...,s-\lambda_1)$.

For a single variable this is obvious, but not for many variables. I saw this in a paper. The paper referenced a book, but in the book this was left as an exercise.

What are ways to prove this identity?

(As specified by Jules Lamers in the comments, $x$ is an invertible $N\times N$ matrix whose eigenvalues are the arguments of the Jack polynomials)

[EDIT]: Jack polynomials in $N$ variables are eigenfunctions of the differential operator $$ D(\alpha)=\frac{\alpha}{2}\sum_{i=1}^Nx_i^2\frac{\partial ^2}{\partial x_i^2}+\sum_{j\neq i}\frac{x_i^2}{x_i-x_j}\frac{\partial}{\partial x_i},$$ the eigenvalue of $J_\lambda^{(\alpha)}(x)$ being $$e_\lambda=\alpha b(\lambda')-b(\lambda)+(N-1)|\lambda|,$$ with $b(\lambda)=\sum_i(i-1)\lambda_i$.

So I guess showing that $\det(x)^s J_\lambda^{(\alpha)}(x^{-1})$ is an eigenfunction of $D(\alpha)$ with eigenvalue $e_{\widetilde{\lambda}}$ would do it.

Another possibility is to work with the expression in terms of tableaux. We know that $J_\lambda^{(\alpha)}(x)=\sum_{T\in S(\lambda)}w_\alpha(T)x^T$, where the sum is over tableaux of shape $\lambda$ and $w_\alpha(T)$ is an appropriate weight. Now $\det(x)^sJ_\lambda^{(\alpha)}(x^{-1})=\sum_{T\in S(\lambda)}w_\alpha(T)x^{s^N-T},$ and then take it from here.

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