Uniqueness of a moment problem over $\int^{\infty}_{0} \frac{\sigma^p}{(1+\sigma)^{2S}} m(\sigma) \: \mathrm d \sigma$ with a finite range in $p$

Question

A question over on the physics site asked about the moment problem $$ \int^{\infty}_{0} \frac{\sigma^p}{(1+\sigma)^{2S}} m(\sigma) \: \mathrm d \sigma = \frac{p!(2S-p)!}{(2S+1)!}, $$ where $S=0,1,2,\ldots$ and $p=0,1,2,\ldots,2S$, which turns out to have as a solution $$ m(\sigma) = \frac{1}{(1+\sigma)^2}, $$ where the solution can be verified by a direct calculation. Given its context in the original paper, it is reasonable to ask whether this solution is unique or not. Given how close this is to the standard moment problem over $p$, I'm inclined to think that it probably is, but the finite range in $p$ makes that more complicated. So: is this moment problem determinate? If so, how does one prove it? If not, what other solutions are there?

Answer 1

It's unique. Even if you only look at the $p=0$ data, and just ask for the $S$-th moments of $X = 1/(1+\sigma)^2$ you have enough data for the Hausdorff moment problem to give uniqueness. (The distribution of $X$ is supported on $[0,1]$, the moments of $X$ specify its distribution uniquely; from the distribution of $X$ the formula for $m(\cdot)$ drops out.)