An identity involving partial fractions decompositions

Question

In Vladimir A. Smirnov's book Analytic Tools for Feynman Integrals (page 38), the following identity is suggested to perform partial fractions decompositions $$ \begin{split} \frac{1}{(x+x_1)^{a_1}(x+x_2)^{a_2}}&=\sum_{i=0}^{a_1-1}\binom{a_2-1+i}{a_2-1} \frac{(-1)^i}{(x_2-x_1)^{a_2+i}(x+x_1)^{a_1-i}} \\ &\qquad+\sum_{i=0}^{a_2-1} \binom{a_1-1+i}{a_1-1}\frac{(-1)^{a_1}}{(x_2-x_1)^{a_1+i}(x+x_2)^{a_2-i}} \end{split} $$ where $a_1$, $a_2>0$ and $\binom{n}{j}$ is a binomial coefficient. How to prove this identity?

Answer 1

We can write \begin{align} f(z)=\frac{1}{(z+r_1)^{n_1}(z+r_2)^{n_2}}=\sum^{n_1}_{k=1}\frac{a_k}{(z+r_1)^{k}}+\sum^{n_2}_{k=1}\frac{b_k}{(z+r_2)^{k}} \end{align} where $a_k$ and $b_k$ are coefficients to be determined.

To evaluate $a_k$, we simply note that $a_k$ is the coefficient of $\dfrac{1}{z+r_1}$ in $(z+r_1)^{k-1}f(z)$. That is, \begin{align} a_k &=\operatorname*{Res}_{z=-r_1}\frac{1}{(z+r_1)^{n_1-k+1}(z+r_2)^{n_2}}\\ &=\frac{1}{(n_1-k)!}\frac{d^{n_1-k}}{dz^{n_1-k}}\frac{1}{(z+r_2)^{n_2}}\Bigg{|}_{z=-r_1}\\ &=\frac{1}{(n_1-k)!}\frac{(-1)^{n_1-k}n_2(n_2+1)\cdots(n_2+n_1-k-1)}{(r_2-r_1)^{n_1+n_2-k}}\\ &=\frac{(-1)^{n_1-k}}{(r_2-r_1)^{n_1+n_2-k}}\binom{n_2+n_1-k-1}{n_2-1} \end{align} Similarly, $b_k$ is the coefficient of $\dfrac{1}{z+r_2}$ in $(z+r_2)^{k-1}f(z)$. Thus $$b_k=\frac{(-1)^{n_2-k}}{(r_1-r_2)^{n_1+n_2-k}}\binom{n_1+n_2-k-1}{n_1-1}$$ We arrive at the identity after an index shift and reversing the order of summation. \begin{align} f(z) =&\sum^{n_1}_{k=1}\binom{n_2+n_1-k-1}{n_2-1}\frac{(-1)^{n_1-k}}{(r_2-r_1)^{n_1+n_2-k}(z+r_1)^k}\\ &+\sum^{n_2}_{k=1}\binom{n_1+n_2-k-1}{n_1-1}\frac{(-1)^{n_2-k}}{(r_1-r_2)^{n_1+n_2-k}(z+r_2)^k}\\ =&\sum^{n_1-1}_{k=0}\binom{n_2+n_1-k-2}{n_2-1}\frac{(-1)^{n_1-k-1}}{(r_2-r_1)^{n_1+n_2-k-1}(z+r_1)^{k+1}}\\ &+\sum^{n_2-1}_{k=0}\binom{n_1+n_2-k-2}{n_1-1}\frac{(-1)^{n_2-k-1}}{(r_1-r_2)^{n_1+n_2-k-1}(z+r_2)^{k+1}}\\ =&\color{red}{\sum^{n_1-1}_{k=0}\binom{n_2+k-1}{n_2-1}\frac{(-1)^{k}}{(r_2-r_1)^{n_2+k}(z+r_1)^{n_1-k}}}\\ &\color{red}{+\sum^{n_2-1}_{k=0}\binom{n_1+k-1}{n_1-1}\frac{(-1)^{k}}{(r_1-r_2)^{n_1+k}(z+r_1)^{n_2-k}}}\\ \end{align} Link