Let $n\geq m$ be non negative integers, and consider a list of $(n+m+1)$ distinct numbers (complex or real). I am interested in getting a closed form formula for the following determinant: $\det\left[\sum\_{k=i}^{n+i}s\_k^{m+1-j}e^{-s\_k}\left[\prod\_{l=i,l\neq k}^{n+i}(s\_k-s\_l)\right]^{-1}\right]\_{i,j=1\ldots m+1}$, i.e.


$$
\det\left(\begin{array}[ccc]
 \,\sum\_{k=1}^{n+1}s\_k^me^{-s\_k}\left[\prod\_{l=1,l\neq k}^{n+1}(s\_k-s\_l)\right]^{-1}& \ldots &\sum\_{k=1}^{n+1}e^{-s\_k}\left[\prod\_{l=1,l\neq k}^{n+1}(s\_k-s\_l)\right]^{-1} \\ 
\vdots & \ddots & \vdots \\ 
\sum\_{k=m+1}^{n+m+1}s\_k^me^{-s\_k}\left[\prod\_{l=m+1,l\neq k}^{n+m+1}(s\_k-s\_l)\right]^{-1}& \ldots &\sum\_{k=m+1}^{n+m+1}e^{-s\_k}\left[\prod\_{l=m+1,l\neq k}^{n+m+1}(s\_k-s\_l)\right]^{-1}
\end{array}\right)
$$.


This determinant can be rewritten as the determinant of an alternant matrix $\left(F\_{j}(S\_i)\right)\_{i,j=1\ldots m+1}$, for suitable multivariate functions $F\_j:\mathbb{C}^{n+1}\rightarrow\mathbb{C}$.


If the exponential are not there, then this determinant should some king of Schur polynomial. I am wondering if this determinant can be interpreted in terms of representation theory.


A related post : [A class of matrix determinants between Wronskians and Vandermondes](https://mathoverflow.net/questions/180974/a-class-of-matrix-determinants-between-wronskians-and-vandermondes)



A Vandermonde like determinant with exponentials
After slight renumbering $(m,n)\mapsto (m,n)-1$ and horizontal reflection $j\mapsto m+1-j$, let's call the $m\times m$ matrix in the OP's determinant
$$
A=\left[\sum\_{k=i}^{i+n-1}s\_k^{j-1}e^{-s\_k}\prod\_{l=i,l\neq k}^{i+n-1}(s\_k-s\_l)^{-1}\right]\_{\,i,j=1}^{\,m},\tag{1}
$$
and define $r=m+n-1$. Note that this answer also holds for $n<m$.


$A$ can be factorized similar to [this posting](https://mathoverflow.net/questions/236323) using the $r \times m$ Vandermonde matrix
$$
V\_{r,m}=\left[ s\_\rho^{\mu-1}\right]\_{\rho,\mu=1}^{r,m},\tag{2}
$$
the diagonal matrix
$$
E\_r = \mathrm{diag}\,[e^{-s\_\rho}]\_{\rho=1}^r\tag{3}
$$
and the matrix of inverse derivatives
$$
D\_{m,r}^{(o)}=\left[ \frac{1}{P\_{\mu,o}'(s\_\rho)}
 \begin{cases} 
 1 & \mu \leq \rho < m+\mu \\
 0 & \text{else}
 \end{cases}
\right]\_{\,\mu,\rho=1}^{\,m,r},\tag{4}
$$
of the polynomials
$$
P\_{l,o}(s)=\prod\_{\rho=l}^{l+o-1}(s-s\_\rho)\tag{5}
$$
according to
$$
A = D\_{m,r}^{(m)} E\_{r} \, V\_{r,m}.\tag{6}
$$
Note that $D$ is not diagonal in this case, e.g., for $m=3$, $n=3$, where $r=5$:
$$
D\_{3,5}^{(3)}=
\left(
\begin{array}{ccccc}
 \frac{1}{\left(s\_1-s\_2\right) \left(s\_1-s\_3\right)} &
 \frac{1}{\left(s\_2-s\_1\right) \left(s\_2-s\_3\right)} &
 \frac{1}{\left(s\_3-s\_1\right) \left(s\_3-s\_2\right)} & 0 & 0 \\
 0 & \frac{1}{\left(s\_2-s\_3\right) \left(s\_2-s\_4\right)} &
 \frac{1}{\left(s\_3-s\_2\right) \left(s\_3-s\_4\right)} &
 \frac{1}{\left(s\_4-s\_2\right) \left(s\_4-s\_3\right)} & 0 \\
 0 & 0 & \frac{1}{\left(s\_3-s\_4\right) \left(s\_3-s\_5\right)} &
 \frac{1}{\left(s\_4-s\_3\right) \left(s\_4-s\_5\right)} &
 \frac{1}{\left(s\_5-s\_3\right) \left(s\_5-s\_4\right)} \\
\end{array}
\right).\tag{7}
$$


As the matrices $D$ and $V$ are not square, the determinant of $A$ does not factorize in a simple way. However, one could try to enlarge both matrices to $r \times r$ by inserting appropriate matrix elements $O(\epsilon)$, factorize and let $\epsilon\to 0$. Alternatively, one can naturally enlarge $D$ and $V$ and define the $r \times r$ matrix
$$
B=D\_{r,r}^{(m)} E\_{r} \, V\_{r,r},\tag{8}
$$
where $\det B$ factorizes. The matrix $A$ then is the upper left $m \times m$ submatrix of $B$.


