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Difficult question from some test somewhere (I forget).


$x$ is, of course, an integer.

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There are some relevant closed forms here but it still isn't helping me understand the solution. :( – carokann9 Dec 11 '12 at 22:53
When $x=1$, $x^4-1 = 0$. So, the whole product is zero. Are you sure there isn't a typo?? – dexter04 Dec 11 '12 at 22:54
The OP likely means for the bottom index to be $2$, especially since the closed form in the link provided above also defines the bottom index to be $2$. – bzprules Dec 11 '12 at 22:56
Isn't this just (11) at that link? And doesn't it give a reference to a publication of Borwein? – Gerry Myerson Dec 11 '12 at 23:05
@GerryMyerson I guess I'm a little unwilling to pay money to solve one problem. Also, on the directly referenced pages, Borwein just lists Maple's results for these computations. I imagine he explains them a few pages later, but that's not free to preview. – carokann9 Dec 11 '12 at 23:32

Write $$\frac{x^4 - 1}{x^4+1} = \frac{(x-a_1)\ldots(x-a_4)}{(x-b_1)\ldots(x-b_4)}$$ where $a_j$ are the roots of $x^4-1$ and $b_j$ are the roots of $x^4+1$. Then the partial product $$ \prod_{x=2}^n \frac{x^4 - 1}{x^4+1} = \frac{\Gamma(n+1-a_1)\ldots \Gamma(n+1-a_4) \Gamma(2-b_1) \ldots \Gamma(2-b_n)}{\Gamma(2-a_1) \ldots \Gamma(2-a_4) \Gamma(n+1-b_1) \ldots \Gamma(n+1-b_4)}$$ Now (carefully) take the limit as $n \to \infty$.

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