Some properties of coherent spin states? I am reading an article of J.M. Radcliffe (Some properties of coherent spin states, 
J. Phys. A: Gen. Phys. 4, 313 (1970)), and I have found some difficulties in replicating some of the calculations.
In the article, a coherent spin state is defined as 
$$ \vert \mu \rangle = (1+ \vert \mu \vert ^2 )^{-s} \exp^{\mu \hat{S_{-}}} \vert 0\rangle \, .$$
To prove the completeness of this states Radcliffe proceeds:  

The states $\vert \mu \rangle$ do form a complete set, although it is
  necessary to include a weight function $ m(\vert \mu \vert ^2) \geq 0
> $ in the integral. We require: 
  $$ \int d^2 \mu \;\; \vert \mu \rangle\:  m(\vert \mu \vert ^2) \:\langle \mu \vert = \sum^{2S}_{p=0} \;\; \vert
 p\rangle \langle p\vert = 1 \, .$$ 
  By doing the angular integration and putting $ \vert \mu \vert = \rho $, one finds 
  $$ \int d^2 \mu \;\;
\vert \mu \rangle \: m(\vert \mu \vert ^2) \: \langle \mu \vert = 2 \pi
 \sum^{2S}_{p=0} \vert p\rangle \langle p\vert \frac{(2S)!}{p!(2S-p)!}
 \int^{\infty}_{0} d \rho \frac{\rho^{2p}}{(1+\rho^2)^{2S}} m(\rho^2)$$
  $$ = \sum^{2S}_{p=0} \vert p\rangle \langle p\vert
\frac{(2S)!}{p!(2S-p)!} I(p,S) \, .$$

Until now nothing so suspicious, but then he writes:

Where $$I(p,S) = \pi \int^{\infty}_{0} d \sigma
 \frac{\sigma^p}{(1+\sigma)^{2S}} m(\sigma).$$
Now one seeks a form for $m(\sigma)$ such that  $ I(p,S) =
 \frac{p!(2S-p)!}{(2S)!} $ . A little thought shows that a suitable
  choice is : $$ m(\sigma) = \frac{2S+1}{\pi} \frac{1}{1+\sigma^2}. $$
  So, finally, the completeness relation is $$ \frac{2S+1}{\pi} \int
 \frac{d^2\mu}{(1+ \vert \mu \vert ^2 )^2} \vert \mu\rangle
 \langle\mu\vert =  \sum^{2S}_{p=0} \;\; \vert p\rangle \langle p\vert
 = 1. $$

In the last box, he replaced $m(\sigma)$ by its value but how did he solve the integral?
 A: This appears to be a typo in the original paper. In its essence, here the author is just making the rather boring claim that
$$
 \int^{\infty}_{0} \frac{\sigma^p}{(1+\sigma)^{2S}} \frac{\mathrm d \sigma}{1+\sigma^2}
= 
\frac{p!(2S-p)!}{(2S+1)!},
\tag{claim}
$$
which should normally get omitted from a paper, because it's (i) a boring calculation, which is nevertheless (ii) uniquely specified, so (iii) it can be checked directly if the reader is that interested, but really (iv) it doesn't add much to the argument, so actually (v) putting in the gory details would in fact harm the readability of the paper.
Here the snag, of course, is that the claim as written down is incorrect, since e.g. putting in $p=0$, $S=1$ returns
$$
 \int^{\infty}_{0} \frac{\mathrm d \sigma}{(1+\sigma)^{2}(1+\sigma^2)}
= 
\frac12
$$
where the intended right-hand side evaluates to $\frac{p!(2S-p)!}{(2S+1)!}=\frac13$, so the claim simply cannot be true.
Fortunately, the fix is simple: just change the denominator in $\mu(\sigma)$ from $1+\sigma^2$ to $(1+\sigma)^2$ (which is the first thing to try given the other factor in the denominator in the integral) and you get the amended 
$$
 \int^{\infty}_{0} \frac{\sigma^p\:\mathrm d \sigma}{(1+\sigma)^{2S+2}} 
= 
\frac{p!(2S-p)!}{(2S+1)!},
\tag{amended claim}
$$
which does check out. You can then confirm that the expression in $(3.11)$ is a typo because it is inconsistent with $(3.12)$, but the fixed denominator makes that inconsistency go away.

As to how you actually calculate the integral
$$
\int^{\infty}_{0} \frac{\sigma^p\:\mathrm d \sigma}{(1+\sigma)^{2S+2}} 
$$
in question? Nowadays, you normally use integration by computer algebra system, or if you're really stubborn you look in e.g. §3.241 of Gradshteyn and Ryzhik, or whatever your favourite integrals table is. 
If you really want to go the masochistic route and calculate every integral for yourself, then you can do what Mathematica does behind the scenes (i.e. transform everything to hypergeometric functions, simplify the resulting product, integrate that, and simplify the result), or you can do a change of variables back to $\rho=\sqrt\sigma,$ symmetrize to $\rho\in(-\infty,\infty)$, loop back with a half-circle at infinity and then do a residue calculation at one of the two poles at $\rho=\pm i$, or you can do repeated integration by parts through the chain
\begin{align}
\int^{\infty}_{0} \frac{\sigma^p\:\mathrm d \sigma}{(1+\sigma)^{2S+2}} 
& =
\int^{\infty}_{0}  
\sigma^p \frac{d}{d\sigma}\mathopen{}\left(\frac{-1/(2S+1)}{(1+\sigma)^{2S+1}} \right)\mathclose{}
\mathrm d \sigma
\\ & =
\int^{\infty}_{0} \left[
\frac{d}{d\sigma}\left(\sigma^{p}\frac{-1/(2S+1)}{(1+\sigma)^{2S+1}} \right)
-
\frac{d}{d\sigma}\left(\sigma^{p}\right)\frac{-1/(2S+1)}{(1+\sigma)^{2S+1}} 
\right]\mathrm d \sigma
\\ & =
-\frac{1}{2S+1} \left[
\frac{\sigma^{p}}{(1+\sigma)^{2S+1}} \right]^{\infty}_{0}
+
\frac{p}{2S+1} \int^{\infty}_{0}
\frac{\sigma^{p-1}}{(1+\sigma)^{2S+1}} 
\mathrm d \sigma
\end{align}
down to more elementary integrals. But really, why would you want to?

A much more interesting question is whether the (amended) $ m(\sigma) = \frac{2S+1}{\pi} \frac{1}{(1+\sigma)^2}$ is a unique solution to the moment problem
$$
\int^{\infty}_{0} \frac{\sigma^p}{(1+\sigma)^{2S}} m(\sigma) \: \mathrm d \sigma
=
\frac{1}{\pi} \frac{p!(2S-p)!}{(2S)!},
$$
and here things are rather less clear: with respect to $p$ it's a standard form of the moment problem but you only have a finite set $p=0,1,\ldots,2S$ of moments, but you also have an infinite set of less-standard moments on $S$ (which themselves give you more $p$ moments), so it's reasonable to look for a uniqueness proof.
To do that, the easiest way is to flip things around by taking the unbounded $S$ index as the positive power, i.e. by changing the independent variable to $x=(1+\sigma)^{-2}$. To do this, start by cleaning up the notation slightly and changing to $f(\sigma) = \frac{\pi}{2S+1} {(1+\sigma)^2}m(\sigma)$, so you get the moment problem in the form
$$
\int^{\infty}_{0} \frac{\sigma^p}{(1+\sigma)^{2S+2}} f(\sigma) \: \mathrm d \sigma
=
\frac{p!(2S-p)!}{(2S+1)!},
$$
and you're testing the uniqueness of the solution $f(\sigma)=1$, and then change to $x=(1+\sigma)^{-2}$ (so that $\sigma=-1+1/\sqrt{x}$ and $\mathrm d\sigma = -\frac12 \mathrm dx/x^{3/2}$), which rephrases the integral as
$$
\int^{1}_{0} 
x^{S-\lfloor p/2\rfloor}
x^{-\{p/2\}}
 \frac{\left(1-\sqrt{x}\right)^p}{\sqrt{x}}f(x) \:  \mathrm dx
=
2\frac{p!(2S-p)!}{(2S+1)!}.
$$
That is then a standard Hausdorff moment problem, with the $p$ dependence factoring out (i.e. it's a set of independent moment problems of the form $\int^{1}_{0} x^{n}g(x) \:  \mathrm dx=2\frac{p!(2S-p)!}{(2S+1)!}$ in 
$$
g(x)=x^{-\{p/2\}-1/2}\left(1-\sqrt{x}\right)^pf(x),
$$ 
each for a different $p$, which can then be solved independently, and whose unicity follows from the properties of the standard Hausdorff problem.
