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When I was trying to evaluate a definite integral (given below the question for those who're curious), I came across a paper of Ramanujan pertaining to the evaluation of the very same integral. Amidst the paper's proofs, Ramanujan mentions the following:

It's easy to see that: $$\prod^{n=\alpha}_{n=1}\left\{\frac{\left(1+\frac{\alpha+2\beta}{n}\right)\left(1+\frac{\beta+2\alpha}{n}\right)}{\left(1+\frac{\alpha}{n}\right)^3\left(1+\frac{\beta}{n}\right)^3}\right\}=\frac{[\Gamma(1+\alpha)\Gamma(1+\beta)]^3}{\Gamma(1+\alpha+2\beta)\Gamma(1+\beta+2\alpha)}$$

Right. So how do I go about proving this?

Note: The integral I was attempting to evaluate: $$\int^{\infty}_0\frac{\tan^{-1}x^3}{e^{4\pi x}-1}dx$$

Panglossian Oporopolist
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  • Is the upper limit of the product $\alpha$ or $\infty$? – robjohn Jun 15 '17 at 04:59
  • If the upper limit is $\alpha$, just plug in $\alpha=\beta=1$ and we get that $\frac14=\frac1{36}$. If the upper limit is $\infty$, use either of the answers given so far. – robjohn Jun 15 '17 at 13:43
  • In some older textbooks the $\infty$ symbol is written with the right loop open and that may look like a Greek $\alpha $. The identity involves upper limit as $\infty$. – Paramanand Singh Jun 18 '17 at 05:54

2 Answers2

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Use the infinite product formula for $\Gamma$: $$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n}$$ which yields $$\frac1{\Gamma(z+1)}=\frac1{z\Gamma(z)}=e^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-z/n}$$ etc.

Angina Seng
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2

I assume that the upper limit of the product is supposed to be $\infty$.

Hint:

By Gautschi's Inequality, $\lim\limits_{m\to\infty}\frac{\Gamma(m+\alpha+1)}{m^\alpha\,\Gamma(m+1)}=1$. Therefore, $$ \begin{align} \lim_{m\to\infty}\frac1{m^\alpha}\prod_{n=1}^m\left(1+\frac{\alpha}n\right) &=\lim_{m\to\infty}\frac1{m^\alpha}\prod_{n=1}^m\left(\frac{\color{#C00}{n+\alpha}}{\color{#090}{n}}\right)\\ &=\lim_{m\to\infty}\frac1{m^\alpha}\color{#C00}{\frac{\Gamma\left(m+\alpha+1\right)}{\Gamma(\alpha+1)}}\color{#090}{\frac{\Gamma(1)}{\Gamma\left(m+1\right)}}\\ &=\lim_{m\to\infty}\frac1{\Gamma(\alpha+1)}\frac{\Gamma\left(m+\alpha+1\right)}{m^\alpha\,\Gamma(m+1)}\\ &=\frac1{\Gamma(\alpha+1)} \end{align} $$

robjohn
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