How to compute $E[Xr / (Xr +1 - X)] $ where $X$ follows a Beta distibution?

Question

I would like to compute $E[Xr / (Xr +1 - X)] $ where $X$ follows a Beta distribution $\operatorname{Beta}(\alpha, \beta)$ with $\alpha, \beta > 1$, $\alpha < \beta$ and $r \in (0,1)$. This is the same as computing the integral: $$\int_0^1 \frac{xr}{xr +1 -x} \frac{x^{\alpha-1} (1-x) ^ {\beta - 1}}{\operatorname{Beta}(\alpha,\beta)} \, dx$$ where $Beta$ is the beta function.

I'm looking for a closed formula that would be a "simple" function of $r$ and known functions, such as the $Beta$ or $\Gamma$ functions.

For instance, to solve $E[X] =\int_0^1 x \frac{x^{\alpha-1} (1-x) ^ {\beta - 1}}{\operatorname{Beta}(\alpha,\beta)} \, dx$, noticing that:

$\int_0^1 x^\alpha (1-x)^{\beta - 1} \, dx = \operatorname{Beta}(\alpha+1,\beta) = \operatorname{Beta}(\alpha,\beta) \frac{\alpha}{\alpha + \beta} $ (via the $\Gamma$ function)

gives the answer. But I can't find such a trick or a nice change of variable for $E[Xr / (Xr +1 - X)] $.

Thanks in advance

Answer 1

Have you tried using a computer algebra system to help see if a 'closed-form' might exist? For example, given $X \sim \text{Beta}(a,b)$ with pdf $f(x)$:

... you seek:

This is a 'closed-form' solution, but perhaps not the simple solution you might have hoped for.

Notes:

1. The Expect function used above is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors.

2. For more detail on the Hypergeometric2F1Regularized function, see, for instance: http://reference.wolfram.com/language/ref/Hypergeometric2F1Regularized.html