Integrating and showing equal to some quantity

Question

Define $G_N(z)=(\ \dfrac{\sin(\pi z)}{\pi } )\ (\sum_{n=-N}^{n=N}(-1 )^n sgn(n)(\ (z-n)^{-1}+n^{-1} )\ -\sum_{m=-N}^{m=N}(-1)^msgn(m)m^{-1} $

I want to show that $G_N(z)=\sum_{n=-N}^{n=N}sgn(n) \dfrac{\sin \pi(z-n)}{\pi (z-n)}$ Can you please give any suggestion?? Thanks in advance.

Answer 1

Note that \begin{align} G_N(z) & = \frac{\sin(\pi z)}{\pi}\left(\sum_{n=-N}^{N}\mathrm{sgn}(n)(-1)^n\left(\frac{1}{z-n}+\frac{1}{n}-\frac{1}{n}\right)\right)\\ & = \frac{\sin(\pi z)}{\pi}\sum_{n=-N}^{N}\mathrm{sgn}(n)(-1)^n\frac{1}{z-n}\\ & = \sum_{n=-N}^{N}\mathrm{sgn}(n)\frac{\sin(\pi z)}{\pi}(-1)^n\frac{1}{z-n}. \end{align} By noting that $$\sin(\pi(z-n))=\sin(\pi z) \cos(\pi n)-\cos(\pi z)\sin(\pi n),$$ and using $\cos(\pi n) = (-1)^n$, and $\sin(\pi n) = 0$, we arrive at $$G_N(z) = \sum_{n=-N}^{N}\mathrm{sgn}(n)\frac{\sin(\pi (z-n))}{\pi(z-n)}.$$