I am going through a paper:
Nijimbere, Victor, Evaluation of some non-elementary integrals involving sine, cosine, exponential and logarithmic integrals. I, Ural Math. J. 4, No. 1, 24-42 (2018). ZBL1455.33016.
and then on page three, I have the following equation:
I am uncertain on how the equation surrounded by the curved red lines, is derived from the previous equation?
For example, I have when $$(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$$ and so I have tried: $$\frac{\left(\frac{\Gamma(n+\frac{1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{3}{2})\Gamma(1)n!}\right)}{\left(\frac{\Gamma(2n+2)\Gamma(n+\frac{3}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{3}{2})\Gamma(1)n!}\right)}=\frac{\left(\frac{(\frac{1}{2})_n}{\Gamma(\frac{3}{2})\Gamma(1)n!}\right)}{\left(\frac{\Gamma(2n+2)(\frac{3}{2})_n}{\Gamma(\frac{1}{2})\Gamma(1)n!}\right)}$$ However, I cannot foresee from my approach the solution. I recognise the $(\frac{1}{4})^n$ will come from factoring out the $2, \Gamma(2n+2)$, but I cannot see how the $n!$ is cancelled out from thereon (even when attempting with the duplication theorem).
By using the duplication formula $$\Gamma(n)\Gamma(n+\frac{1}{2})=2^{1-2n}\Gamma(\frac{1}{2})\Gamma(2z) \\ \Gamma(n+1)\Gamma(n+\frac{3}{2})=2^{-2n-1}\Gamma(\frac{1}{2})\Gamma(2z+2)$$ By applying the gamma identity $$\Gamma(x+1)=x\Gamma(x)$$
So I have
$$\frac{\left(\frac{\Gamma(n+\frac{1}{2})}{n!}\right)}{\left(\frac{\Gamma(2n+2)\Gamma(n+\frac{3}{2})}{n!}\right)}=\frac{\left(\frac{(\frac{1}{2})_n\Gamma(\frac{1}{2})}{n!}\right)}{\left(\frac{\Gamma(n+1)\Gamma(n+\frac{3}{2})(\frac{3}{2})_n \Gamma(\frac{3}{2})}{2^{-2n-1}\Gamma(\frac{1}{2})n!}\right)}=\frac{\left(\frac{(\frac{1}{2})_n\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2})n!}\right)}{\left(\frac{\Gamma(n+\frac{3}{2})(\frac{3}{2})_n }{2^{-2n}\Gamma(\frac{3}{2})}\right)}=\frac{\left(\frac{(\frac{1}{2})_n}{\frac{1}{2}n!}\right)}{\left(\frac{(\frac{3}{2})_n (\frac{3}{2})_n }{2^{-2n}}\right)}=\frac{2^{-2n-1}\left(\frac{1}{2}\right)_n}{(\frac{3}{2})_n (\frac{3}{2})_n n!}$$
However, not so sure the indices are correct for the powers of $2$ because I still have $\frac{x}{2}$ to take into account, so it becomes $\frac{1}{4}\left(\frac{1}{4}\right)^{n}$ however, the factor of $\frac{1}{4}$ seems to be missing from the equation in the article, unless I have made a mistake