Characteristic Function of Sum and Difference of Non-Identical Exponential Random Variables This question is related to an answer of this question.
In the linked question, three independent (non-identical) exponential random variables $X,Y,Z$ with means $\mu_X,\mu_Y,\mu_Z>0$ are considered. The characteristic function of $X+Y-Z$ is for $t\in\mathbb{R}$
$$\mathbb{E}\exp(it(X+Y-Z))=\frac{1}{(1-i\mu_Xt)(1-i\mu_Yt)(1+i\mu_Zt)},$$ where $i$ is the imaginary unit. In one of the answers to the linked question the characteristic function is rewritten as 
$$\frac{A_X}{1-i\mu_Xt}+\frac{A_Y}{1-i\mu_Yt}+\frac{A_Z}{1+i\mu_Zt}$$
for some real coefficients $A_X,A_Y,A_Z$.
My question is how to find these coefficients in a systematic way. In the above example one can find them by simple calculation, but what if one considers more than three random variables. Let me give a specific, more complicated example.
Let $X_j,Y_j$, $j\in\mathbb{N}$, be independent exponential random variables with mean $j$. Let $n,x,y\in\mathbb{N}$. The characteristic function of $D:=\sum_{j=x}^n X_j - \sum_{j=y}^n Y_j$ is
$$ \mathbb{E}\exp(itD)= \left(\prod_{k=x}^n \frac{1}{1-ikt}\right)\left(\prod_{l=y}^n \frac{1}{1+ikt}\right).$$ I want to rewrite this as a sum
$$\sum_{k=x}^n \frac{A_k}{1-ikt} + \sum_{k=y}^n \frac{B_k}{1+ikt}$$
for some coefficients $A_x,A_{x+1},\ldots,A_n,\,B_y,B_{y+1},\ldots,B_n\in\mathbb{R}$.
Are there any techniques that one can try out, to find these coefficients (if they indeed exist)? I tried turning the product into a sum using the logarithm, but that does not seem to help here.

Answering my own question using the hint by Robert Israel.
Let $$f(t):= \left(\prod_{k=x}^n \frac{1}{1-ikt}\right)\left(\prod_{l=y}^n \frac{1}{1+ikt}\right),\quad t\in\mathbb{R}.$$
We know from partial fraction decomposition theory that 
$$A_k=f(t)(1-ikt)|_{t=-i/k},\quad k=x,x+1,\ldots,n$$ 
and
$$B_k=f(t)(1+ikt)|_{t=i/k},\quad k=y,y+1,\ldots,n.$$
Hence, 
$$A_k=\left(\prod_{j=x:\, j\neq k}^n \frac{k}{k-j}\right)\left(\prod_{j=y}^n \frac{k}{k+j}\right)=\frac{k^{2n+1-x-y}(-1)^{n-k}(k+y-1)!}{(n-k)!(k-x)!(k+n)!}$$
and 
$$B_k=\left(\prod_{j=x}^n \frac{k}{k+j}\right)\left(\prod_{j=y:\, j\neq k}^n \frac{k}{k-j}\right)=\frac{k^{2n+1-x-y}(-1)^{n-k}(k+x-1)!}{(n+k)!(n-k)!(k-y)!}.$$
