TITLE: Ramanujan's eccentric Integral formula
QUESTION [19 upvotes]: The wikipedia page on Srinivasa Ramanujan gives a very strange formula: 

Ramanujan: If $0 < a < b + \frac{1}{2}$ then, $$\int\limits_{0}^{\infty} \frac{ 1 + x^{2}/(b+1)^{2}}{  1 + x^{2}/a^{2}} \times \frac{ 1 + x^{2}/(b+2)^{2}}{ 1 + x^{2}/(a+1)^{2}} \times \cdots \ \textrm{dx} = \frac{\sqrt{\pi}}{2} \small{\frac{ \Gamma(a+\frac{1}{2}) \cdot \Gamma(b+1)\: \Gamma(b-a+\frac{1}{2})}{\Gamma(a) \cdot \Gamma(b+\frac{1}{2}) \cdot \Gamma(b-a+1)}}$$


Question I would like to pose to this community is: What could be the Intuition behind discovering this formula.
Next, I see that Ramanujan has discovered a lot of formulas for expressing $\pi$ as series. May I know what is the advantage of having a same number expressed as a series in a different way. Is it useful at all?
From what I know Ramanujan basically worked on Infinite series, Continued fractions, $\cdots$ etc. I have never seen applications of continued fractions, in the real world. I would also like to know if continued fractions has any applications.

Hope I haven't asked too many questions. As I was posting this question the last question on application of continued fractions popped up and I thought it would be a good idea to pose it here, instead of posing it as a new  question.

REPLY [8 votes]: Another proof of the Ramanujan's identity is given in https://arxiv.org/abs/1309.3455 (On gamma quotients and infinite products, by Marc Chamberland, Armin Straub). It is based on the formula
$$\prod_{k=0}^\infty \frac{(k+\alpha_1)\cdots (k+\alpha_n)}{(k+\beta_1)\cdots (k+\beta_n)}=\frac{\Gamma(\beta_1)\cdots\Gamma(\beta_n)}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_n)}, \tag{1} $$
where $n\ge 1$ is an integer, $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ are nonzero complex numbers, none of which are negative integers, and $\alpha_1+\ldots +\alpha_n=\beta_1+\ldots +\beta_n$.
Using the factorization
$$\frac{1+x^2/(b+k)^2}{1+x^2/(a+k)^2}=\frac{(k+a)^2(k+b+ix)(k+b-ix)}{(k+b)^2(k+a+ix)(k+a-ix)},$$
and (1), Ramanujan's product transforms to
$$ \prod_{k=0}^\infty\frac{1+x^2/(b+k)^2}{1+x^2/(a+k)^2}=\frac{\Gamma(b)^2\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(a)^2\Gamma(b+ix)\Gamma(b-ix)},$$
and then the Ramanujan's identity is equivalent to the Mellin–Barnes integral (in this respect, see also David Hansen's answer below)
$$\frac{1}{2\pi i}\int_\limits{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(a-s)}{\Gamma(b+s)\Gamma(b-s)}ds=\frac{1}{2\sqrt{\pi}}\frac{\Gamma(a)}{\Gamma(b)}\frac{\Gamma\left(a+1/2\right)\Gamma\left(b-a-1/2\right)}{\Gamma\left(b-1/2\right)\Gamma(b-a)}.$$