In search of $π$ [ $P:3$ ]

$\textrm{The above series can be transformed into a closed form}$ $\color{#20A900}\sum_{n=0}^∞ \frac{(\frac{1}{2})_n (\frac{1}{2})_n}{(1)_n .n!}(\frac{1}{2})^n = _2F_1(\frac{1}{2},\frac{1}{2};1;\frac{1}{2})$

$_2F_1(a,b;c;z)= \frac{\Gamma{(c)}}{\Gamma{(c-b)}\Gamma{(b)}} \int_0^1 t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a}dt$

$\textrm{Here }$ $z= \frac{1}{2}$ $\textrm{and}$ $a=\frac{1}{2}$ , $b= \frac{1}{2}$ , $c=1$

$\textrm{Then it is}$ $\color{#3D99F6}_2F_1(\frac{1}{2} ,\frac{1}{2} ;1;\frac{1}{2})= \frac{\sqrt{2} \Gamma{(1)} }{\Gamma^2 {(\frac{1}{2})}} \int_0^1 t^{-1/2}(1-t)^{-1/2} (2-t)^{-1/2} dt$

$\textrm{Using}$ $\int_{\phi}^{\zeta} f(x) dx = \int_{\phi}^{\zeta}f(\phi+\zeta-x)dx$

$\textrm{The integral becomes }$ $\implies \frac{\sqrt{2}}{π} \int_0^1 t^{-1/2} (1-t)^{-1/2} (1+t)^{-1/2} dt = \frac{\sqrt{2}}{π}\int_0^1 (1-t^2)^{-1/2} t^{-1/2} dt$

$\textrm{Take}$ $t^2 = u$ $\textrm{the integral becomes}$ $\frac{1}{\sqrt{2}π}\int_0^1 u^{-\frac{3}{4}} (1-u)^{-1/2} du = \frac{1}{\sqrt{2}π} \Beta(\frac{1}{4} , \frac{1}{2})$ $\color{#E81990}\implies \frac{1}{\sqrt{2}π} \frac{\Gamma{(\frac{3}{4})}\Gamma{(\frac{1}{4})} \Gamma{(\frac{1}{2})}}{\Gamma^2{(\frac{3}{4})}}= \frac{\sqrt{π}}{\Gamma^2{(\frac{3}{4})}}$