Question: Decompose the fraction$$\frac 1{\left\{1+X\right\}\left\{1+Xa\right\}\left\{1+Xa^2\right\}\cdots\left\{1+Xa^n\right\}}$$where $X=x^2$.

My work: Obviously, the right-hand side has to take the form$$\prod\limits_{j=0}^n\frac 1{1+Xa^j}=\sum\limits_{i=0}^n\frac {C_i}{1+Xa^i}$$Multiplying both sides by $1+Xa^i$ and evaluating at $X=-a^{-i}$ isolates the $C_k$ in the right-hand side. Hence$$\begin{align*}C_k & =\prod\limits_{j=0}^n\frac {1+Xa^i}{1+Xa^j}=\prod\limits_{j=0,\,j\neq k}^n\frac 1{1+Xa^{j-k}}\end{align*}$$Splitting up the product into two separate terms, we have$$C_k=\frac 1{\left(1-\frac X{a^k}\right)\left(1-\frac X{a^{k-1}}\right)\cdots\left(1-\frac Xa\right)}\frac 1{\left(1+aX\right)\left(1+a^2X\right)\cdots\left(1+a^nX\right)}$$However, I'm not sure what to do afterwards. The main goal here is to rewrite each product in terms of the rising/falling factorial and gamma function.

This is because the original problem is to integrate the fraction and I feel more comfortable integrating from $0$ to $\infty$ in terms of gamma functions and factorials, than with the product of terms.

The original problem was$$\int\limits_0^{\infty}dx\,\frac 1{(1+x^2)(1+ax^2)(1+a^2x^2)\ldots}$$And I plan to take the limit as $n\to\infty$.


The partial fraction decomposition in question reads: \begin{equation} \frac{1}{\prod\limits_{j=0}^n (1+ a^j X)}= \frac{1}{a^{\binom{n+1}{2}}}\cdot \sum\limits_{i=0}^n \frac{1}{X+\frac{1}{a^i}} \cdot \prod\limits_{i\neq j} \frac{1}{\frac{1}{a^j} - \frac{1}{a^i}} \end{equation}

and the integral in question reads: \begin{eqnarray} \int\limits_0^\infty \frac{1}{\prod\limits_{j=0}^n (1+a^j x^2)} dx&=&\frac{\pi}{2} \sum\limits_{i=0}^n \frac{a^{-\frac{i}{2}}}{\prod\limits_{j=0,j\neq i}^n (1-a^{j-i})}\\ &=&\frac{\pi}{2} \sum\limits_{i=0}^n \frac{a^{-\frac{i}{2}}}{(a^{-i};a)_i (a;a)_{n-i}} \end{eqnarray}



$$\dfrac{1}{(1+X)(1+Xa)(1+Xa^2)} =\dfrac{c_0}{1+X} + \dfrac{c_1}{1+aX} + \dfrac{c_2}{1+a^2}$$

Multiplying both sides by $(1+X)(1+Xa)(1+Xa^2)$, we get

$$c_0(1+Xa)(1+Xa^2) + c_0(1+X)(1+Xa^2) + c_0(1+X)(1+Xa) = 1$$

Let $X=-1$, and we get

$$c_0(1-a)(1-a^2) = 1$$

$$c_0 = \dfrac{1}{(1-a)(1-a^2)}$$

Let $X=-\dfrac{1}{a}$, and we get

$$c_1(1-\frac 1a)(1-a) = 1$$

$$c_1 = \dfrac{1}{(1-\frac 1a)(1-a)}$$

Let $X=-\dfrac{1}{a^2}$, and we get

$$c_0(1-\frac{1}{a^2})(1-\frac 1a) = 1$$

$$c_2 = (1-\frac{1}{a^2})(1-\frac 1a)$$

Finding the general answer is just a matter of bookkeeping.


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