- #1
GoutamTmv
- 13
- 0
Hello everyone,
I came across this identity while browsing Wikipedia, and I decided to try to prove it for myself. ( It was discovered by S Ramanujan)
[tex]\int_0^\infty \cfrac{1+{x}^2/({b+1})^2}{1+{x}^2/({a})^2} \times\cfrac{1+{x}^2/({b+2})^2}{1+{x}^2/({a+1})^2}\times\cdots\;\;dx = \frac{\sqrt \pi}{2} \times\frac{\Gamma(a+\frac{1}{2})\Gamma(b+1)\Gamma(b-a+\frac{1}{2})}{\Gamma(a)\Gamma(b+\frac{1}{2}) \Gamma(b-a+1)}[/tex]
I would like to ask two questions regarding this:
1) Is the product in the integral on the left hand side an infinite product or a finite one?
2) I personally think I can derive this by finding the right substitution. Would I be wrong? Are there more mathematics in play behind this, aside from calculus?
Thanks a lot
I came across this identity while browsing Wikipedia, and I decided to try to prove it for myself. ( It was discovered by S Ramanujan)
[tex]\int_0^\infty \cfrac{1+{x}^2/({b+1})^2}{1+{x}^2/({a})^2} \times\cfrac{1+{x}^2/({b+2})^2}{1+{x}^2/({a+1})^2}\times\cdots\;\;dx = \frac{\sqrt \pi}{2} \times\frac{\Gamma(a+\frac{1}{2})\Gamma(b+1)\Gamma(b-a+\frac{1}{2})}{\Gamma(a)\Gamma(b+\frac{1}{2}) \Gamma(b-a+1)}[/tex]
I would like to ask two questions regarding this:
1) Is the product in the integral on the left hand side an infinite product or a finite one?
2) I personally think I can derive this by finding the right substitution. Would I be wrong? Are there more mathematics in play behind this, aside from calculus?
Thanks a lot