Consider $n\times n$ symmetric Cauchy-like matrix $M$ with elements $(M\_{ij})\_{i,j=1}^{n}$ given by


$$M\_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int\_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x.$$


Is there a way to compute the elements of the inverse $(M^{-1})\_{ij}$ analytically?



Inverse of a Cauchy-like matrix
I was able to figure this out by viewing $M$ as a scaled Cauchy matrix.



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> **Theorem.** $\left(M^{-1}\right)\_{ij} = \dfrac{(n-i)!(n-j)!}{2n-i-j+1}\dfrac{\displaystyle\prod\_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod\_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod\_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)}$.
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**Proof.** Define $n\times 1$ vector $\alpha$ with elements $(\alpha)\_{i} = 1/(n-i)!$. Then $M = \text{diag}(\alpha) N \text{diag}(\alpha)$, where $N\_{ij} := 1/(2n-i-j+1)$. Hence $\left(M^{-1}\right)\_{ij} = (n-i)!(n-j)!\left(N^{-1}\right)\_{ij}$.


Now, write $N$ as a Cauchy matrix: $N\_{ij} = 1/(a\_i + b\_j)$ where $a\_i := n-i$, $b\_j := n-j+1$. Then using the known result [1, Sec. 1.2.3, Exercise 41] for Cauchy matrix inverse:
$$\left(N^{-1}\right)\_{ij} = \dfrac{1}{2n-i-j+1}\dfrac{\displaystyle\prod\_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod\_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod\_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)},$$
the result follows.


[1] D. E. Knuth, The Art of Computer Programming. Volume 1: Fundamental Algorithms, 3rd ed. Addison- Wesley, 1997.


