How to compute a more general version of Vandermonde / Cauchy double alternant determinant Consider some variables $\{X_i\}_{1\le i \le n}$, $\{Y_i\}_{1\le i \le n}$, and $\{W_i\}_{1\le i \le n}$. Does anyone know how to compute the following determinant?
$$
\det ~ \left(\frac{W_j^{i-1}}{X_i+Y_j}\right)_{1\le i,j\le n}.
$$
Update: If you could also provide an answer for the case where $W_j=1$ for $j\ge2$, that would be sufficient for the problem I have encountered in my research. 
 A: Maybe you can get something out of the technique of displacement equations. It works as follows.
Notice, first, that a matrix $A$ is a Cauchy-like matrix if and only if it satisfies the so-called displacement equation $LA-AR = vu^T$, where $L$ and $R$ are diagonal matrices (containing the nodes) and $vu^T$ is a generic rank-1 term.
Suppose you are given a matrix $A$.

*

*Find (if you can!) two matrices $L$ and $R$ such that $LA-AR = vu^T$ has rank 1. If $A$ is a Cauchy matrix $A_{ij} = \frac{1}{X_i + Y_j}$, then $L = diag(X_i)$ and $R = diag(Y_j)$ work, while for a Vandermonde matrix $A = W_j^{i-1}$ then $L$ is a shift matrix and $R$ is $diag(W_j)$.


*Diagonalize (if you can do it explicitly) $L = VD_LV^{-1}$ and $R=UD_RU^{-1}$, and then with some algebra you get $D_L V^{-1}AU - V^{-1}AUD_R = V^{-1}vu^TU$.


*Then, $V^{-1}AU$ is a Cauchy matrix with nodes the entries of $D_L$ and $D_R$, because of that displacement equation. You can compute its determinant, and use it to get the determinant of $A$.
A: My conjecture is that the answer is close to this:
$$
\frac{\prod_{1\le i < j \le n} (X_j - X_i) (W_jY_j - W_iY_i)}
{\prod_{1\le i ,j \le n}(X_i + Y_j)}.
$$
