For $H>L$ , $p,q,\alpha,\beta>0$, and B(.,.) the beta functon, trying to solve this integral:
$$\mathbb{E}(X)_0^H=\frac{\alpha H }{\beta B(p,q)}\int_0^H \frac{x \left(\frac{-H \log \left(\frac{H-x}{H}\right)}{\beta }\right)^{\alpha p-1} \left(\left(\frac{-H \log \left(\frac{H-x}{H}\right)}{\beta }\right)^{\alpha }+1\right)^{-p-q}}{H-x} \, \mathrm{d}x$$
${\bf Motivation: }$ This is the partial expectation of the random variable $X \in[0,H]$, a transformation of a r.v. following the generalized Beta distribution of second kind (also known as the Beta prime distribution).
${\bf Note: }$ I simplified the question and changed the support from $X \in [L,H]$ to $X \in [0,H]$.
Best Answer
Well, I can get into a somewhat more tractable form by basic transformations. Taking $x=H-y$, $y=He^u$, $u = -\frac{\beta}{H} v$, we get \begin{equation} \frac{\alpha H}{B(p,q)} \int_0^{\infty} (1 - e^{-\frac{\beta}{H} v}) v^{\alpha p - 1} (v^\alpha + 1)^{-p-q} dv. \end{equation} Taking a further transformation $v = z^{1/\alpha}$ gives \begin{equation} \frac{H}{B(p,q)} \int_0^\infty (1-\exp(-\frac{\beta}{H}z^{1/\alpha})) z^{p-1} (z+1)^{-p-q} dz. \end{equation} I think the first form is better, and you could then try expanding the binomial and finding something in Gradshteyn and Ryzhik that will resum in a reasonable way. Alternatively, find a way to express some set of terms on the inside as a simple integral and see if you can exchange the order of integration?