Summation Theorem Derivative

Question

Suppose that $f(z)=p(z)/q(z)$ is a rational function with degree~$q(z)\geq2+$ degree~$p(z)$. Suppose that $q(z)$ has no roots at the integers, except possibly at 0. Show that $${\sum_{k=-\infty}^\infty}' \frac{p(k)}{q(k)}=-\pi \sum_j {\rm Res}_{z=z_j}f(z)\cot(\pi z)$$ where the $z_j$'s are the roots of $q(z)$ (including 0). Notice that the sum on the right-hand side has a finite number of terms. Here the prime on the left-hand sum indicates that the $k=0$ term is skipped.

I have proved the summation theorem $${\sum_{k=-\infty}^\infty} \frac{p(k)}{q(k)}=-\pi \sum_j {\rm Res}_{z=z_j}f(z)\cot(\pi z)$$. An example of the proof is available here.

Any ideas on how to prove the first equation that deals with the prime. Like, how omitting the k = 0 term does not make any change?

Answer 1

Partial proof, not entirely self-contained.

Both sides depend linearly of $f$, because $\mathrm{Res}(\lambda g+h,a) = \lambda \mathrm{Res}(g,a) + \mathrm{Res}(h,a)$ for all meromorphic functions $g,h$ on $\mathbb{C}$ and all $\lambda \in \mathbb{R}$ (by convention, the residue is $0$ if $a$ is not a pole). Note the condition $\mathrm{deg}(p) \le \mathrm{deg}(q)-2$ forces the sum of all residues of $p/q$ to be $0$.

The equality to be proved is already known when $q(0) \ne 0$. Using decomposition into simple elements and linearity, it suffices to check it for fractions $1/z^k$ with $k \ge 2$, and for fractions of the form $$\frac{1}{z-a} -\frac{1}{z} = \frac{a}{z(z-a)}, \text{ where } a \in \mathbb{C} \setminus \mathbb{Z}.$$

For fractions $1/z^k$ with $k \ge 2$, the equality is true (both sides are null if $k$ is odd, and for $k$ even, this is the classical relation between $\zeta(k)$ and the Bernoulli number $B_k$).

For fractions $\frac{a}{z(z-a)}$ \where $a \in \mathbb{C} \setminus \mathbb{Z}$, one may observe that \begin{eqnarray*} \sum_{k \in \mathbb{Z} \setminus\{0\}} \Big(\frac{1}{k-a} - \frac{1}{k}\Big) &=& \sum_{k=1}^{+\infty} \Big(\frac{1}{k-a} - \frac{1}{k} + \frac{1}{-k-a} - \frac{1}{-k}\Big) \\ &=& -\sum_{k=1}^{+\infty} \Big(\frac{1}{a-k} + \frac{1}{a+k}\big) \\ &=& \frac{1}{a} - \pi\cot(\pi a). \end{eqnarray*}

The residue of $z \mapsto \frac{a\cot(\pi z)}{z(z-a)}$ at $a$ is $\cot(\pi a)$.

The residue of $z \mapsto \frac{a\cot(\pi z)}{z(z-a)}$ at $0$ is $-1/(\pi a)$ since, as $z \to 0$,
\begin{eqnarray*} \frac{a\cot(z)}{z(z-a)} &=& -z^{-1}\cot(\pi z)\frac{a}{a-z} \\ &=& -\big(\pi^{-1}z^{-2} + O(1)\big) \sum_{n=0}^{+\infty}\frac{z^n}{a^n} \\ &=& -\pi^{-1}z^{-2} -\pi^{-1}a^{-1}z^{-1}+O(1). \end{eqnarray*} [1]: About the proof of $\sum\limits_{n=-\infty}^\infty f(n)=-\pi\sum\limits_{k=1}^m\text{res} [f(z)\cot(\pi z)]_{z=a_k}$?