Evaluate the infinite product $ \prod_{n=1}^{\infty} \left ( 1 + \frac{x^2}{n^2+n-1} \right )$

Question

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 $$\displaystyle{g'\left( x \right) = - \frac{{2x}}{{{\rho _1} \cdot {\rho _2}}} + \frac{{2x}}{{{\rho _2} - {\rho _1}}} \cdot \left( {{\psi _o}\left( { - {\rho _1}} \right) - {\psi _o}\left( { - {\rho _2}} \right)} \right) }$$

Answer 1

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 $$P = \prod_{n=0}^{\infty} t_n \cdot \frac{1}{1-x^2} = \boxed{\frac{\sec\left(\frac{\sqrt{5}}{2}\pi\right)\cos\left(\pi\sqrt{\frac{5}{4} - x^2}\right)}{1-x^2}}$$

Answer 2

One can also use identities related to the Gamma function as follows. Write $$ \prod_{n=1}^{\infty} \left(1+\frac{x^2}{n^2+n-1}\right) = \prod_{n=1}^{\infty} \left(\frac{n^2+n+x^2-1}{n^2+n-1}\right) = \prod_{n=1}^{\infty} \left(\frac{(n+l)(n-l+1)}{(n+\phi)(n-\phi+1)}\right), $$ where $\phi = \frac{1+\sqrt{5}}{2}$ and $l = \frac{1+\sqrt{5-4x^2}}{2}$ (which could be possibly complex, but we may choose either complex root). For convenience, we multiply and divide out the corresponding $n=0$ term which is $1-x^2$, so that $$ \prod_{n=1}^{\infty} \left(\frac{(n+l)(n-l+1)}{(n+\phi)(n-\phi+1)}\right) = \frac 1{(1-x^2)} \prod_{n=0}^{\infty} \frac{(n+l)}{(n+\phi)}\prod_{n=0}^{\infty} \frac{n-l+1}{n-\phi+1}. $$

Each of these products is evaluable in terms of the Gamma function, using the identity $\Gamma(z+1) = z \Gamma(z)$. For example, for finite $m$, $$ \prod_{n=0}^{m} (n+l) = \frac{\Gamma(m+l+1)}{\Gamma(l)}\quad ;\quad \prod_{n=0}^m (n+\phi) = \frac{\Gamma(m+\phi+1)}{\Gamma(\phi)} $$ Taking their ratio we have $$ \prod_{n=0}^m \frac{(n+l)}{(n+\phi)} = \frac{\Gamma(\phi)}{\Gamma(l)} \frac{\Gamma(m+l+1)}{\Gamma(m+\phi+1)}. $$ Similarly we have $$ \prod_{n=0}^{m} \frac{(n-l+1)}{(n-\phi+1)} = \frac{\Gamma(1-\phi)}{\Gamma(1-l)} \frac{\Gamma(m+2-l)}{\Gamma(m+2-\phi)}, $$ whence $$ \prod_{n=0}^m \frac{(n+l)}{(n+\phi)}\frac{(n-l+1)}{(n-\phi+1)}=\left(\frac{\Gamma(\phi)}{\Gamma(l)} \frac{\Gamma(1-\phi)}{\Gamma(1-l)}\right)\left(\frac{\Gamma(m+l+1)}{\Gamma(m+\phi+1)}\frac{\Gamma(m+2-l)}{\Gamma (m+2-\phi)}\right). $$

As $m \to \infty$, turns out that second term (depending on $m$) goes to $1$. Readers are urged to prove this using equation $(5.11.12)$ in the link here. Thus we have $$ \prod_{n=1}^{\infty} \left(\frac{(n+l)(n-l+1)}{(n+\phi)(n-\phi+1)}\right) = \frac {1}{1-x^2}\frac{\Gamma(\phi)}{\Gamma(l)} \frac{\Gamma(1-\phi)}{\Gamma(1-l)} $$

Here the Euler reflection formula comes to our assistance, and we have $$ \frac{\Gamma(\phi)}{\Gamma(l)} \frac{\Gamma(1-\phi)}{\Gamma(1-l)} = \frac{\frac{\pi}{\sin(\pi \phi)}}{\frac{\pi}{\sin(\pi l)}} = \frac{\sin(\pi l) }{\sin(\pi \phi)},$$ whence the answer is $$ \frac{1}{1-x^2}\left(\frac{\sin\left(\pi\left(\frac{1+\sqrt{5-4x^2}}{2}\right)\right)}{\sin(\pi \phi)}\right) = \frac{1}{1-x^2}\left(\frac{\cos\left(\frac{\pi \sqrt{5-4x^2}}{2}\right)}{\cos\left(\pi \frac{\sqrt{5}}{2}\right)}\right) $$ where we used $\sin(\pi/2+y) = \cos(y)$ for all $y$ to obtain the last identity. This furnishes the required closed form.