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An identity of Ramanujan's |
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| Aug28-12, 09:26 AM | #1 |
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An identity of Ramanujan's
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 |
| Aug29-12, 09:12 AM | #2 |
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I'm pretty sure it is infinite. Good luck, ram was a beast
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| Aug29-12, 03:30 PM | #3 |
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Thanks. Is there, then, an easy way to find the partial fractional decomposition of the infinite product?
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