Partial fraction decomposition of $\frac{1}{x^a(x+c)^b}$

Question

We have the partial fraction decomposition $$\frac{1}{x^a(x+c)^b}=\frac{d_a}{x^a}+\frac{d_{a-1}}{x^{a-1}}+...+\frac{d_{1}}{x}+\frac{e_b}{(x+c)^b}+\frac{e_{b-1}}{(x+c)^{b-1}}+...+\frac{e_{1}}{x+c},$$ where $a,b,c\in\mathbb{N}$.

Is there anything particular we can say about the nature of coefficients $d_i$ and $e_i$? Any references would be greatly appreciated.

P.S. Sorry for being so open-ended.

Answer 1

If $$f(a,b)=\dfrac1{x^a(x+c)^b},$$

$$f(a,b)=\dfrac1c\cdot\dfrac{x+c-x}{\cdots}=\dfrac{f(a,b-1)}c-\dfrac{f(a-1,b)}c$$

We can use this reduction formula repeatedly until at least one of $a,b$ becomes zero

Answer 2

Multiply both sides by $x^a$ and $(x+ c)^b$ to get $$1= d_a(x+ c)^b+ d_{a-1}x(x+ c)^b+ \cdots+ d_1x^{a-1}(x+ c)^b+ e_bx^a+ e_{b+1}x^a(x+ c)+ \cdots+e_1x^a(x+ c)^{b-1}~.$$

Taking $~x= 0~$ gives immediately $~1= c^b~d_a~$ so $~d_a= 1/c^b~$. Taking $~x= -c~$ immediately gives $~1= e_b~c^a~$ so $~e_b= 1/c^a~$.

That leaves $~a-1+ b-1= a+ b- 2~$ values to be determined. Although there are no more "easy" equations but taking $~a+ b- 2~$ different values for $~x~$ will give $~a+ b- 2~$ equations for the $~a+ b- 2~$ unknown values.