Numerics appear to indicate that the function \begin{equation} f(\alpha)= \end{equation} \begin{equation} \frac{\sqrt{\pi } 3^{-3 \alpha -1} \Gamma \left(2 \alpha +\frac{3}{2}\right) \, _5F_4\left(\frac{3 \alpha }{2},\alpha +\frac{1}{2},\frac{3 \alpha }{2}+\frac{1}{2},\frac{3 \alpha }{2}+\frac{11}{8},2 \alpha +\frac{1}{2};\frac{3 \alpha }{2}+\frac{3}{8},\frac{3 \alpha }{2}+\frac{3}{4},\frac{3 \alpha }{2}+\frac{5}{4},3 \alpha +1;1\right)}{\Gamma \left(\alpha +\frac{5}{6}\right) \Gamma \left(\alpha +\frac{7}{6}\right)} \end{equation} is a step-function, assuming the value $\frac{1}{2}$ for $\alpha \geq 0$ and 1, for $\alpha<0$.
Demonstrate this please, if possible.
As Nemo commented, there are things which look strange in the behaviour of $f(\alpha)$.
I computed its value for $-3\leq \alpha \leq 3$ by step of $0.01$ and obtained the following results $$\left( \begin{array}{cc} -3.00 & \color{red}{1.0} \\ -2.99 & 0.5 \\ \cdots & \cdots \\ -2.76 & 0.5 \\ -2.75 & \color{red}{\text{ComplexInfinity}} \\ -2.74 & 0.5 \\ \cdots & \cdots \\ -2.51 & 0.5 \\ -2.50 & \color{red}{1.5} \\ -2.49 & 0.5 \\ \cdots & \cdots \\ -2.26 & 0.5 \\ -2.25 & \color{red}{\text{ComplexInfinity}} \\ -2.24 & 0.5 \\ \cdots & \cdots \\ -2.01 & 0.5 \\ -2.00 & \color{red}{1.0}\\ -1.99 & 0.5 \\ \cdots & \cdots \\ -1.76 & 0.5 \\ -1.75 & \color{red}{\text{ComplexInfinity}} \\ -1.74 & 0.5 \\ \cdots & \cdots \\ -1.51 & 0.5 \\ -1.50 & \color{red}{1.5} \\ -1.49 & 0.5 \\ \cdots & \cdots \\ -1.26 & 0.5 \\ -1.25 & \color{red}{\text{ComplexInfinity}} \\ -1.24 & 0.5 \\ \cdots & \cdots \\ -1.01 & 0.5 \\ -1.00 & \color{red}{1.0} \\ -0.99 & 0.5 \\ \cdots & \cdots \\ -0.76 & 0.5 \\ -0.75 & \color{red}{\text{Indeterminate}} \\ -0.74 & 0.5 \\ \cdots & \cdots \\ -0.51 & 0.5 \\ -0.50 & \color{red}{1.5} \\ -0.49 & 0.5 \\ \cdots & \cdots \\ -0.26 & 0.5 \\ -0.25 & \color{red}{\frac{\sqrt{\pi }}{\sqrt[4]{3}\, \Gamma \left(\frac{7}{12}\right) \Gamma \left(\frac{11}{12}\right)}} \\ -0.24 & 0.5 \\ \cdots & \cdots \\ +3.00 & 0.5 \end{array} \right)$$
Looking around the places where non $0.5$ values occur$$f(-3.001)=0.5 \qquad f(-2.999)=0.5$$ $$f(-2.751)=0.5 \qquad f(-2.749)=0.5$$ $$f(-2.501)=0.5 \qquad f(-2.499)=0.5$$ $$f(-2.251)=0.5 \qquad f(-2.249)=0.5$$ $$\cdots$$ $$f(-0.501)=0.5 \qquad f(-0.499)=0.5$$ $$f(-0.251)=0.5 \qquad f(-0.249)=0.5$$
The reasons for these troubles is that $\Gamma(x)$ is undefined for negative integer values of the argument $x$ and we face three of them $$\Gamma \left(\alpha +\frac{5}{6}\right)\qquad,\qquad \Gamma \left(\alpha +\frac{7}{6}\right) \qquad,\qquad\Gamma \left(2 \alpha +\frac{3}{2}\right)$$ On the other hand, the involved hypergeometric function also present a lot of discontinuities for negative values of variable $\alpha$. This means that building the above table for more detailed values (smaller step sizes) would reveal much more problems that reported here.