Integration over the Marchenko-Pastur distribution Problem Statement:
I want to find closed form expression of the following definite integral
\begin{equation}
\int_{\alpha_{-}}^{\alpha_{+}}  \ln x \, \frac{1}{2\pi \alpha x} \sqrt{ (\alpha_{+}- x )(x- \alpha_{-} ) }\,  dx 
\end{equation}
where $0 < \alpha < 1$, $\alpha_{\pm} = (1 \pm \sqrt{ \alpha} )^2$. This is no singularity in the integrand and this integral is well defined. 
The quarter circle part is the Marchenko-Pastur distribution
\begin{equation}
\rho_{\rm MP}(x) =  \frac{1}{2\pi \alpha x} \sqrt{ (\alpha_{+}- x )(x- \alpha_{-} ) }
\end{equation}
The integral basically calculate $\mathbb{E}[\ln x]$
Moments Calculation:
One can calculate its moments by variable substitution $x = \frac{\alpha_{+} + \alpha_{-} }{2} +\frac{\alpha_{+} - \alpha_{-} }{2}t $ as follows
\begin{equation}
\begin{aligned}
  \mathbb{E}[x^n] &= \int_{\alpha_{-}}^{\alpha_{+}}   x^{n-1}\, \frac{1}{2\pi \alpha } \sqrt{ (\alpha_{+}- x )(x- \alpha_{-} ) }\,  dx  \\
  &= \frac{1}{2\pi \alpha} \left( \frac{\alpha_{+} - \alpha_{-}}{2}\right)^2 \int_{-1}^{1}  \left[ \frac{\alpha_{+} + \alpha_{-}}{2} + \frac{\alpha_{+} - \alpha_{-}}{2}t \right]^{n-1} \sqrt{ 1 - t^2 } \, dt \\
  &= \frac{2}{\pi} ( 1 + \alpha)^{n-1} \int_{-1}^{1} \, dt \left[ 1 + \frac{ 2 \sqrt{\alpha}}{ 1 + \alpha }t \right]^{n-1} \sqrt{ 1 - t^2 } \, dt \\
  &= ( 1 + \alpha)^{n-1}  {}_2F_1\left[ \frac{1 - n}{2}, 1- \frac{n}{2}; 2;  \left( \frac{ 2 \sqrt{\alpha}}{ 1 + \alpha }\right)^2 \right]
\end{aligned}
\end{equation}
The $n = 0$ case is the normalization. Since $ {}_2F_1\left[\frac{1}{2}, 1; 2; z \right] = - \frac{2(\sqrt{1-z}-1 ) }{z}$, we have
\begin{equation}
E[x^0] = \frac{1}{1 + \alpha} \frac{( 1+ \alpha)^2}{4 \alpha}  \frac{4 \alpha}{ 1+ \alpha} = 1
\end{equation}
The $n = 1$ case corresponds to the fixed trace in the original Wishart-Laguerre ensemble. Since $ {}_2F_1\left[0, \frac{1}{2}; 2; z \right] = 1$, we have
\begin{equation}
E[x^1] = 1 
\end{equation}
both are in accordance with the requirements. 
The expectation value of the $\ln x$ corresponds to the derivative
\begin{equation}
\mathbb{E}[\ln x ] = \partial_{n} \mathbb{E}[x^n]\Big|_{n = 0} 
\end{equation}
But calculating the term with the derivative hitting on the hypergeometric function seems to be equally hard as evaluating $\mathbb{E}[\ln x ]$ itself. 
Perhaps the knowledge of the Marchenko-Pastur distribution will lead to a quick answer, and I will appreciate to accept relevant references. 
Warning: Unfortunately, Mathematica 8 gives the wrong symbolic answer, because it contradicts with numerical value at particular $\alpha$. Both Mathematica 9 and 10 fail to produce the symbolic answer. 
Numerical Hint: For small value of $\alpha$, the integral seems to be close to $- \frac{\alpha}{2}$.
 A: We use identity (9.134.2) in Table of Integrals, Series, and Products
\begin{equation}
{}_2F_1(2a, 2a+1 - c; c; z ) = ( 1 +z)^{-2a} {}_2F_1(a, a+1; \gamma; \frac{4z}{(1+z)^2} )
\end{equation}
rewrite the moments as a simpler hypergeometric function 
\begin{equation}
\mathbb{E}[x^n] = {}_2F_1( 1- n, -n; 2; \alpha )
\end{equation}
The average $\mathbb{E}[\ln x]$ can be obtained by analytically continuing $\partial_n \mathbb{E}[x^n]$ to $n = 0$. 
Instead of resorting to the integral representation of the hypergeometric function and do an almost equally hard integral (with log inside), we choose to write down the equation satisfied by the moments
\begin{equation}
\alpha( 1 - \alpha) \frac{\partial ^2}{ \partial \alpha^2 }\mathbb{E}[x^n] + [ 2 - ( 2 - 2n ) \alpha ]\frac{\partial }{ \partial \alpha }\mathbb{E}[x^n] - n ( n - 1 ) \mathbb{E}[x^n] = 0
\end{equation}
assuming that the $n$ derivative and $\alpha$ derivative commute, we have 
\begin{equation}
\alpha( 1 - \alpha) \frac{\partial ^2}{ \partial \alpha^2 } \frac{\partial }{ \partial n }\mathbb{E}[x^n] + 2(1 -  \alpha )\frac{\partial }{ \partial \alpha } \frac{\partial }{ \partial n }\mathbb{E}[x^n] + (2\alpha  \frac{\partial }{ \partial \alpha } + 1)\mathbb{E}[x^n]  \Big|_{n=0} = 0
\end{equation}
Plug in the known quantities
\begin{equation}
\mathbb{E}[x^n]  \Big|_{n=0} = 1, \quad  \frac{\partial }{ \partial \alpha } \mathbb{E}[x^n]  \Big|_{n=0} = 0
\end{equation}
we get an ordinary differential equation satisfied by $\mathbb{E}[\ln x]$,
\begin{equation}
\alpha( 1 - \alpha) \frac{\partial ^2}{ \partial \alpha^2 } \mathbb{E}[\ln x] + 2(1 -  \alpha )\frac{\partial }{ \partial \alpha } \mathbb{E}[\ln x] + 1 = 0
\end{equation}
whose general solution is
\begin{equation}
c_1 + \frac{c_2}{\alpha} + ( 1 - \frac{1}{\alpha} )  \ln ( 1- \alpha ) - 1 
\end{equation}
By matching to $\mathbb{E}[\ln x]\big|_{\alpha = 0}  = 0$, we have $c _1 = c_2 = 0$. Therefore
\begin{equation}
\mathbb{E}[\ln x] = ( 1 - \frac{1}{\alpha} )  \ln ( 1- \alpha ) - 1 
\end{equation}
the small parameter expansion is indeed $-\frac{\alpha}{2}$. 
