Calculating the residues of $\prod_{i=1}^N\frac{1}{a_is+1}$

Question

I have a generic expression $$\prod_{i=1}^N \frac{1}{a_is+1}$$ where $a_i$ are constants and some of the $a_i$ elements necessarily not the same. The thing I want to obtain is I want to express this product $$ \prod_{i=1}^N \frac{1}{a_is+1} = \sum_{k=1}^t\sum_{\ell=1}^{v_t} \frac{\rho_{k,\ell}}{(a_ks+1)^j}$$ where $\sum_{k=1}^tv_t=N$. I have found that the elements $\rho_{k,\ell}$ can be obtained as $$ \rho_{k,\ell}= \frac 1 {(v_k-\ell)!a_\ell^{v_k-\ell} } \frac {\partial^{v_k-\ell}}{\partial s^{v_k -\ell}} \left[ \prod_{m=1,m\neq k}^t \frac 1 {1 + a_ks} \right]$$ but I could not perform the derivations to achieve $\rho_{k,l}$ constants. If you help me, I'll be so glad. Thank you!

Answer 1

Before we answer the question we also look somewhat closer at OP's expressions which contain some (technical) inaccuracies and which should be discussed beforehand.

We consider the product $\prod_{i=1}^s\frac{1}{a_is+1}$ where the numbers $a_i$ are not necessarily distinct. In fact we have precisely $t$ pairwise different elements which we denote as \begin{align*} \{a_1,a_2,\ldots,a_N\}=\{b_1,b_2,\ldots,b_t\}\qquad\text{with}\qquad |\{b_1,b_2,\ldots,b_t\}|=t \end{align*} Each $b_k, 1\leq k\leq t$ occurs in $\{a_1,a_2,\ldots,a_N\}$ with multiplicity $\nu_k$, so that $\sum_{k=1}^t\nu_k=N$. We obtain the following representation \begin{align*} \prod_{i=1}^N\frac{1}{a_is+1} = \prod_{k=1}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k}\tag{1} \end{align*}

Selecting $k_0, 1\leq k_0\leq t$ we have a pole $-\frac{1}{b_{k_0}}$ of order $\nu_{k_0}$ and a Laurent series expansion \begin{align*} \prod_{k=1}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k}&=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(s-\frac{1}{b_{k_0}}\right)^l+H(s)\\ &=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(-\frac{1}{b_{k_0}}\right)^l\left(b_{k_0}s+1\right)^l+H(s) \end{align*} where the right-hand sum is the principal part of the Laurent series and $H(s)$ is a holomorphic function. Multiplication with $(b_{k_0}s+1)^{\nu_{k_0}}$ gives \begin{align*} \prod_{{k=1}\atop{k\ne k_0}}^t\left(\frac{1}{b_ks+1}\right)^{\nu_k} &=\sum_{l=-\nu_{k_0}}^{-1}\alpha_l\left(-\frac{1}{b_{k_0}}\right)^l\left(b_{k_0}s+1\right)^{l+\nu_{k_0}}+H(s)\left(b_{k_0}s+1\right)^{\nu_{k_0}}\\ &=\sum_{l=1}^{\nu_{k_0}}\underbrace{\alpha_{-l}\left(-b_{k_0}\right)^l}_{\rho_{k_{0},l}}\left(b_{k_0}s+1\right)^{\nu_{k_0}-l}+\tilde{H}(s)\tag{2}\\ \end{align*} with $l$ substituted by $-l$ in the last line.

From (2) we derive OP's representation for $\rho_{k_{0},l_{0}}, 1\leq l_0\leq \nu_{k_{0}}$ as \begin{align*} \color{blue}{\rho_{k_{0},l_{0}}} &\color{blue}{=\frac{1}{\left(\nu_{k_{0}}-l_0\right)!b_{k_{0}}^{\nu_{k_{0}}-l_0}}\, \frac{d^{\nu_{k_{0}}-l_0}}{ds^{\nu_{k_{0}}-l_0}}\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(b_ks+1\right)^{-\nu_{k}}}\tag{3}\\ \end{align*} Using $\mu=\nu_{k_{0}}-l_0$ as short-hand we obtain: \begin{align*} \color{blue}{\rho_{k_{0},l_{0}}} &=\frac{1}{\mu!b_{k_0}^{\mu}}\, \frac{d^{\mu}}{ds^{\mu}}\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(b_ks+1\right)^{-\nu_{k}}\\ &=\frac{1}{\mu!b_{k_{0}}^{\mu}}\, \sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}\prod_{{k=1}\atop{k\ne k_{0}}}^t\frac{d^{q_k}}{ds^{q_k}}\left(b_ks+1\right)^{-\nu_{k}}\tag{4}\\ &=\frac{1}{\mu!b_{k_{0}}^{\mu}}\, \sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}\\ &\qquad\qquad\qquad\qquad\cdot\prod_{{k=1}\atop{k\ne k_{0}}}^t\left(-\nu_k\right)^{\underline{q_k}}b_k^{q_k}\left(b_ks+1\right)^{-\nu_{k}-q_k}\tag{5}\\ &\,\,\color{blue}{=\frac{(-1)^{\mu}}{\mu!b_{k_{0}}^{\mu}}\, \sum_{q_1+\cdots+q_{k_0}^{\prime}+\cdots+q_t=\mu}\binom{\mu}{q_1,\ldots,q_{k_0}^{\prime},\ldots,q_t}}\\ &\qquad\qquad\qquad\qquad \color{blue}{\cdot\prod_{{k=1}\atop{k\ne k_{0}}}^t\frac{(\nu+q_k-1)!}{(\nu-1)!}b_k^{q_k}\left(b_ks+1\right)^{-\nu_{k}-q_k}}\tag{6}\\ \end{align*}

Comment:

• In (4) we use the general Leibniz rule to find the $q_k$-th derivative of the product. We use the prime symbol for $q_{k_0}^{\prime}$ to indicate $q_{k_0}$ is not part of the sum resp. multinomial coefficient.

• In (5) and (6) we use falling factorials \begin{align*} \left(-\nu_k\right)^{\underline{q_k}}&=(-\nu)(-\nu-1)\cdots(-\nu-q_k+1)\\ &=(-1)^{q_k}\nu(\nu+1)\cdots(\nu+q_k-1)\\ &=(-1)^{q_k}\frac{(\nu+q_k-1)!}{(\nu-1)!} \end{align*}