I have fractions of the form
$$F(p) = \prod_{i=1}^{n} \frac{1}{1+\alpha_i p} $$
and it appears that the partial fraction expansion is
$$F(p) = \sum_{i=1}^{n} \frac{1}{(1+\alpha_i p)} \prod_{j=1,j\neq i }^{n} \frac{\alpha_i}{\alpha_i-\alpha_j}$$
at least for $n\in{1,2,3,4}.$ Is that a general relationship? Or did I just get lucky?
When you are looking at the partial fraction expansion, you want to solve this problem :
$$F(p)=\sum_{i=0}^{n}\frac{A_i}{1+\alpha_ip} $$ And solve for the $A_i$. First, you can note that $-\frac{1}{\alpha_i}$ is a root of $F(p)$, but it is not a root of $(1+\alpha_ip)F(p) $ Hence, let : $G(p)=(1+\alpha_ip)F(p) $
$$\forall i \leq n, G\left(-\frac{1}{\alpha_i}\right)=\underbrace{A_i}_{\text{from the sum}}=\underbrace{\prod_{j=0,i\neq j}^n\frac{1}{1-\alpha_j/\alpha_i}}_{\text{from the product}}\\ =\prod_{j=0,i\neq j}^n\frac{\alpha_i}{\alpha_i-\alpha_j} $$
Hence, you get the result :
$$ F(p)=\sum_{i=0}^n\frac{1}{1+\alpha_ip} \prod_{j=0,i\neq j}^n\frac{\alpha_i}{\alpha_i-\alpha_j}$$
