Recently I obtained an infinite series form involving Binomial coefficient and Gamma functions when calculating the following parametric integrals:
(Tip: $\displaystyle v\ \in\ \mathbb{R}$)
$\\ \\ \displaystyle \begin{aligned}\int_{0}^{1}{\dfrac{(x+\sqrt{1-x^2})^{2v}-1}{x}}\ \mathrm{d}x&=\int_{0}^{\large\frac{π}{2}}{\dfrac{((\sin x+\cos x)^{2v}-1)\ \cos x}{\sin x}\ \mathrm{d}x}\\\\&=\int_{0}^{\large\frac{π}{2}}{\dfrac{((1+\sin2x)^{v}-1)\ (1+\cos 2x)}{\sin 2x}\ \mathrm{d}x}\\\\&=\dfrac{1}{2}\ \int_{0}^{π}\dfrac{((1+\sin x)^v-1)(1+\cos x)}{\sin x}\ \mathrm{d}x\\\\&=\int_{0}^{\infty}{\dfrac{(1+x)^{2v}-(1+x^2)^v}{x\ (1+x^2)^{v+1}}\ \mathrm{d}x}\\\\&=\int_{0}^{\infty}\dfrac{\left(1+{\large\frac{2x}{1+x^2}}\right)^v-1}{x(1+x^2)}\ \mathrm{d}x\\\\&=\sum_{n=1}^{\infty}{\dfrac{(v)_{n}\ 2^{n}}{n!}}\int_{0}^{\infty}{\dfrac{x^{n-1}}{(1+x^2)^{n+1}}\ \mathrm{d}x}\\\\&=\dfrac{\sqrt{π}}{2}\ \sum_{n=1}^{\infty}{\dfrac{(v)_{n}\ \Gamma\left({\large\frac{n}{2}}\right)}{n!\ \Gamma\left({\large\frac{n+1}{2}}\right)}}\\\\&=\dfrac{\sqrt{π}}{2}\ \sum_{n=1}^{\infty}{\binom{v}{n}\dfrac{\Gamma\left({\large\frac{n}{2}}\right)}{\Gamma\left({\large\frac{n+1}{2}}\right)}}\end{aligned}$
Based on my experience I suspect a nice closed-form exists but have found none. Any kind of help will be appreciated.