Turning Incomplete Beta Integral to Complete Beta Integral On page four, equation (11) of this paper it is stated that when $0<\alpha<1$ the integral $$\int_0^{1/2} z^{-\alpha} (1-z)^{-2-\alpha} \;dz$$ can be "reduced to integrals for the complete beta function by elementary substitutions."
I have seen such types of substitutions say of the kind $t=(az-b)/(cz-d)$, but I cannot get those to work because the limits of integration constraint those substitutions. I assume this is why we need multiple Beta integrals but I cannot see how. 
Thanks in advance. 
 A: First, assuming $\alpha<0$ integrate by parts to obtain
$$
\int_0^{1/2} z^{-\alpha} (1-z)^{-2-\alpha} \;dz=\frac{z^{-\alpha} (1-z)^{-1-\alpha}}{1+\alpha}\Big{|}_{0}^{1/2}+\frac{\alpha}{1+\alpha}\int_0^{1/2} z^{-1-\alpha} (1-z)^{-1-\alpha} \;dz\\
=\frac{2^{1+2\alpha}}{1+\alpha}+\frac{\alpha}{1+\alpha}\ I.
$$
Then
\begin{align}
I&=\int_0^{1/2} z^{-1-\alpha} (1-z)^{-1-\alpha} \;dz\\
&=\int_0^{1/2}  (z(1-z))^{-1-\alpha} \;dz.
\end{align}
First approach:
\begin{align}
I&=\int_0^{1/2}  \left(\frac{1}{4}-\left(\frac{1}{2}-z\right)^2\right)^{-1-\alpha}\;dz\\
&=\int_0^{1/2}  \left(\frac{1}{4}-t^2\right)^{-1-\alpha}\;dt\\
&=2^{1+2\alpha}\int_0^{1}  \left(1-t^2\right)^{-1-\alpha}\;dt\\
&=2^{2\alpha}\int_0^{1}  \left(1-t\right)^{-1-\alpha}t^{-1/2}\;dt\\&=2^{2\alpha}B(-\alpha,1/2).
\end{align}
This is the same as making the substitution $1-2z=\sqrt{t}$.
Second approach:
\begin{align}
I&=\int_0^{1/2}  (z(1-z))^{-1-\alpha} \;dz\\&=\frac{1}{2}\left(\int_0^{1/2}  (z(1-z))^{-1-\alpha} \;dz+\int_{1/2}^1  (z(1-z))^{-1-\alpha} \;dz\right)\\&=\frac{1}{2}\int_0^{1}  (z(1-z))^{-1-\alpha} \;dz=\frac{1}{2}B(-\alpha,-\alpha).
\end{align}
Note that the initial integral converges for $\alpha<1$. So, once it has been calculated for $\alpha<0$ it can be analytically continued to $\alpha<1$. As a result we get

\begin{align}
\int_0^{1/2} z^{-\alpha} (1-z)^{-2-\alpha} \;dz&=\frac{2^{1+2\alpha}}{1+\alpha}+\frac{\alpha}{1+\alpha}2^{2\alpha}B(-\alpha,1/2)\\
&=\frac{2^{1+2\alpha}}{1+\alpha}+\frac{\alpha}{2(1+\alpha)}B(-\alpha,-\alpha),\quad \alpha<1.
\end{align}

