Question statement
Evaluate the infinite product
$$\displaystyle{\prod_{n=1}^{\infty} \left ( 1 + \frac{x^2}{n^2+n-1} \right ) }$$
My try
Because of the square of $\displaystyle{x}$ , we can consider $\displaystyle{x \ge 0}$ . Initially (for convenience) we consider $\displaystyle{x<1}$ .
$$\displaystyle{f\left( x \right) = \Pi = \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{n^2} + n – 1}}} \right)} \Rightarrow g\left( x \right) = \log \left( {f\left( x \right)} \right) = } $$
$$\displaystyle{\sum\limits_{n = 1}^\infty {\log \left( {1 + \frac{{{x^2}}}{{{n^2} + n – 1}}} \right)} \Rightarrow g'\left( x \right) = \sum\limits_{n = 1}^\infty {\frac{{2x}}{{{n^2} + n + {x^2} – 1}}} }$$
If $\displaystyle{{\rho _1} < {\rho _2}}$ the roots (with respect to $\displaystyle{n}$ ) of the equation $\displaystyle{{n^2} + n + {x^2} – 1 = 0}$ , then
$$\rho_1 = \frac{{-1 – \sqrt{5 – 4x^2}}}{2},\quad \rho_2 = \frac{{-1 + \sqrt{5 – 4x^2}}}{2}$$
$$g'(x) = \sum\limits_{n = 1}^\infty \frac{{2x}}{{(n – \rho_1)(n – \rho_2)}} = \frac{{2x}}{{\rho_2 – \rho_1}} \cdot \sum\limits_{n = 1}^\infty \left( \frac{1}{{n – \rho_2}} – \frac{1}{{n – \rho_1}} \right)$$
$$\frac{{2x}}{{\rho_2 – \rho_1}} \cdot \sum\limits_{n = 1}^\infty \left( \left( \frac{1}{{n – \rho_2}} – \frac{1}{n} \right) – \left( \frac{1}{{n – \rho_1}} – \frac{1}{n} \right) \right)$$
$$\displaystyle{ = – \frac{{2x}}{{{\rho _1} \cdot {\rho _2}}} + \frac{{2x}}{{{\rho _2} – {\rho _1}}} \cdot \left( {{\psi _o}\left( { – {\rho _1}} \right) – {\psi _o}\left( { – {\rho _2}} \right)} \right)}$$ . However $$\displaystyle{\left( { – {\rho _1}} \right) + \left( { – {\rho _2}} \right) = 1 \Rightarrow \left( { – {\rho _2}} \right) = 1 – \left( { – {\rho _1}} \right) \Rightarrow {\psi _o}\left( { – {\rho _1}} \right) – {\psi _o}\left( { – {\rho _2}} \right) = }$$
$$\displaystyle{{\psi _o}\left( { – {\rho _1}} \right) – {\psi _o}\left( {1 – \left( { – {\rho _1}} \right)} \right) = \frac{\pi }{{\tan \left( {\pi \left( { – {\rho _1}} \right)} \right)}}}$$ . Therefore $$\displaystyle{g'\left( x \right) = – \frac{{2x}}{{{\rho _1} \cdot {\rho _2}}} + \frac{{2x}}{{{\rho _2} – {\rho _1}}} \cdot \left( {{\psi _o}\left( { – {\rho _1}} \right) – {\psi _o}\left( { – {\rho _2}} \right)} \right) }$$
Best Answer
We can solve this product by considering the Weirestrass factorization of $\cos(\pi z) = \prod_{n=0}^{\infty} \left(1-\frac{4z^2}{(2n+1)^2}\right)$ as hinted in the comments.
Denote the required product by $P$.
Specifically, your partial product product can be written as $$t_n = 1+\frac{4x^2}{4n^2+4n-4} = \frac{4n^2+4n-4+4x^2}{4n^2+4n-4}$$
In order to find a term that matches the partial product of the $\cos(\pi z)$ factor, we can try to divide and multiply by $(2n+1)^2$, giving $$t_n = \frac{(2n+1)^2}{(2n+1)^2-5}\left(1-\frac{5-4x^2}{(2n-1)^2}\right)= \left(\frac{1}{1-\frac{4\left(\frac{\sqrt{5}}{2}\right)^2}{(2n+1)^2}}\right) \left(1 - \frac{4\left(\sqrt{\frac{5}{4}-x^2}\right)^2}{(2n+1)^2}\right)$$
So the product becomes $$P = \prod_{n=0}^{\infty} t_n \cdot \frac{1}{1-x^2} = \boxed{\frac{\sec\left(\frac{\sqrt{5}}{2}\pi\right)\cos\left(\pi\sqrt{\frac{5}{4} - x^2}\right)}{1-x^2}}$$