Expanding formula in partial fractions I want to expand the following quotient into a sum of partial fractions. We have the following formula:
$$\frac{1}{(x-a)^k(x-b)^l}$$
How can I expand this in partial fractions without getting a huge sum? Or is there a series which represents this quotient?
 A: Hint: $$(x-a)^{-k}(x-b)^{-l}=\sum_{j=1}^{k}A_j(x-a)^{-j}+\sum_{j=1}^{l}B_j(x-b)^{-j},$$ where the "$A_j$" part can be obtained from the power series of $(x-b)^{-l}$ around $x=a$: $$(x-b)^{-l}=(a-b)^{-l}\left(1+\frac{x-a}{a-b}\right)^{-l},\qquad(1+z)^{-l}=\sum_{s=0}^{\infty}(-1)^s\binom{l-1+s}{l-1}z^s.$$ The "$B_j$" part follows by symmetry.
A: So the expansion has form:
$$ \frac{1}{(x-a)^k (x-b)^l }  = \sum_{r=0}^{r=k} \frac{\alpha_r}{(x-a)^{r} } + \sum_{t=0}^{t=l}   \frac{\beta_t}{ (x-b)^{t} } $$
w.l.o.g let $ k>l$, multiply both side by $(x-b)^l$
$$ \frac{1}{(x-a)^k} = (x-b)^l \sum_{r=0}^{r=k} \frac{\alpha_r}{(x-a)^{r} } + \sum_{t=0}^{t=l}\beta_t (x-b)^{l - t} $$
Apply $ \frac{d}{dx^j}$ on both sides.
$$  \frac{  (-1)^j (k+j-1)!}{(k-1)!(x-a)^{k+j} }= \sum_{t=0}^{l} \beta_t \frac{(l-t)!(x-b)^{l- t -j}}{(l-t-j)!} +Junk$$
If  we were to evaluate the sum on the right side at x=b, only the the term with index $ t= l-j$ would survive.
$$ \frac{ (-1)^j (k+j-1)!}{ (k-1)! (b-a)^{k+j} } = j!\beta_{l-j} $$
Rearranging,
$$ \beta_{l-j}  = \frac{ (-1)^j (k+j-1)!}{ j! (k-1)! (b-a)^{k+j} }$$
and, for $ r<l$
$$ \alpha_{k-j}  = \frac{ (-1)^j (l+j-1)!}{ j! (l-1)! (b-a)^{l+j} }$$
Not sure how to do it for:
$$ l < r \leq k$$
But it's something..
