Inverse of a "Vandermonde-like" matrix composed of power series Is there an analytical formula for the inverse of a complex matrix whose elements are sets of "power series" except the last term is scaled? 
Let $0<x_1<x_2<...<x_n$ be monotonically increasing. The matrix $A$ is formed by
$$A=\begin{bmatrix}\frac{1}{x_1} & \frac{1}{x^2_1} &\dots & \frac{1}{x^n_1}+I_1\\\frac{1}{x_2} & \frac{1}{x^2_2}&\dots & \frac{1}{x^n_2}+I_2\\\vdots&\vdots&\ddots&\vdots\\\frac{1}{x_n} & \frac{1}{x^2_n}&\dots & \frac{1}{x^n_n}+I_n\end{bmatrix}$$
The last column terms $[A_{in}]$ in the original problem are the sum of the highest power of $x_i^n$ and series of Laplace frequency shifting like this
$$[A_{in}]=\frac{1}{x^n_i}+\frac{\int_0^\infty f(t)e^{-(i-1)\epsilon}e^{-xt}dt}{x^n_i}$$
where $\epsilon=x_j-x_i,j=i+1$, and $f(t)$ is a very slowly convergent function like $\mathrm{Sinc}(t)$.
 A: As K.K.McDonald and Omnomnomnom point out, $A$ is simply the tranpose
of a Vandermonde matrix:
$$
A=\begin{bmatrix}\frac{1}{x_{1}} & \frac{1}{x_{1}^{2}} & \dots & \frac{1}{x_{1}^{n}}\\
\frac{1}{x_{2}} & \frac{1}{x_{2}^{2}} & \dots & \frac{1}{x_{2}^{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\frac{1}{x_{n}} & \frac{1}{x_{n}^{2}} & \dots & \frac{1}{x_{n}^{n}}
\end{bmatrix}=\begin{bmatrix}\frac{1}{x_{1}} & \frac{1}{x_{2}} & \dots & \frac{1}{x_{n}}\\
\left(\frac{1}{x_{1}}\right)^{2} & \left(\frac{1}{x_{2}}\right)^{2} & \dots & \left(\frac{1}{x_{n}}\right)^{2}\\
\vdots & \vdots & \ddots & \vdots\\
\left(\frac{1}{x_{1}}\right)^{n} & \left(\frac{1}{x_{2}}\right)^{n} & \dots & \left(\frac{1}{x_{n}}\right)^{n}
\end{bmatrix}^{\intercal}
$$
Since $(A^{-1})^{\intercal}=(A^{\intercal})^{-1}$, we can use the
expression on the page linked to by K.K.McDonald to get $A^{-1}=(b_{ij})$
where
$$
b_{ij}=\begin{cases}
\dfrac{x_{j}\left(-1\right)^{i-1}{\displaystyle \sum_{\substack{1\leq m_{1}<\cdots<m_{n-i}\leq n\\
m_{1},\ldots,m_{n-i}\ne j
}
}\frac{1}{x_{m_{1}}}\cdots\frac{1}{x_{m_{n-i}}}}}{{\displaystyle \prod_{\substack{1\le m\le n\\
m\ne j
}
}\left(\frac{1}{x_{m}}-\frac{1}{x_{j}}\right)}} & \text{if }1\le i<n;\\
\dfrac{x_{j}}{{\displaystyle \prod_{\substack{1\le m\le n\\
m\ne j
}
}\left(\frac{1}{x_{j}}-\frac{1}{x_{m}}\right)}} & \text{if }i=n.
\end{cases}
$$
You should check to make sure that my $b_{ij}$s are correct.
