Mind the sum

My approach

Given integral $I= 4\int_{0}^{1} \sqrt {1-x^2}\,dx$ represent the total area of circle of radius $r=1$ . so $I=\pi$ . Also the integral can be solved by beta function.

Evaluting the sum as $\begin{aligned} \sum_{n=1}^{\infty} \dfrac{((2n-1)!)^2}{(4n-1)!}&=\sum_{n=1}^{\infty}\int_{0}^{1} t^{2n-1}\,(1-t)^{2n-1}\,dt = \int_{0}^{1}\dfrac{t^2-t}{(t^2-t-1)(t^2-t+1)}\,dt\\& =\dfrac{1}{2} \int_{0}^{1}\left(\dfrac{1}{t^2-t-1}+\dfrac{1}{t^2-t+1}\right)\,dt\\&= \dfrac{1}{2}\int_{0}^{1}\left(\dfrac{1}{\left(t-\frac{1}{2}\right)^2-\frac{5}{4}} +\dfrac{1}{\left(t-\frac{1}{2}\right)^2+\frac{3}{4}}\right)\,dt\end{aligned}$ By the use of standard integration formula we have $=\dfrac{2}{\sqrt 5} \tan^{-1}\left(\dfrac{2t-1}{\sqrt 3}\right)+\dfrac{1}{2\sqrt 3}\ln\left[\dfrac{|2t-1-\sqrt 5|}{|2t-1+\sqrt 5|}\right]$ now setting then limits we deduce $=\dfrac{\pi}{3\sqrt 3}+\dfrac{1}{2\sqrt 5}\ln\left[\dfrac{3-\sqrt 5}{3+\sqrt 5}\right]$ Thus $\begin{aligned}\Omega & = 3\sqrt 3\left(\log\phi +\dfrac{\pi}{3\sqrt 3}+\dfrac{1}{2\sqrt 5}\ln\left[\dfrac{3-\sqrt 5}{3+\sqrt 5}\right]\right) -\pi\\ & =3\sqrt 3\log \phi +\dfrac{3\sqrt 3}{2\sqrt 5}\ln\left[\dfrac{3-\sqrt 5}{3+\sqrt 5}\right]=3\sqrt 3\left(\log \phi +\dfrac{1}{2\sqrt 5}\ln\left[\dfrac{7-3\sqrt 5}{2}\right]\right)\end{aligned}$ gives us $a+b+c+d+e=18$