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.
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