Is there a closed-form expression for the $z$-transform of $sinc[k]$?

Question

Eventually, it is just this question:

Define $\text{sinc}(x):=\frac{\sin(ax)}{x}$ if $x\neq0$

$\text{sinc}(x):=a$ if $x=0$

How to simplify this series: $$\sum_{k=-\infty}^\infty\text{sinc}(k)z^{-k}$$ Is there any closed form for it?

Answer 1

Assume $a$ is real, and is not an integer multiple of $\pi$. Then $$ \sum_{k=1}^\infty \frac{\sin(ak)}{k} z^{-k} $$ diverges if $|z| < 1$, converges absolutely if $|z|>1$, and converges conditionally (and not absolutely) for $z$ on the unit circle $|z|=1$. [Divergence is because the term does not go to zero.]

On the other hand, $$ \sum_{k=-\infty}^{-1} \frac{\sin(ak)}{k} z^{-k} $$ converges absolutely if $|z|<1$, diverges if $|z|>1$, and converges conditionally (and not absolutely) for for $z$ on the unit circle $|z|=1$.

So, the series $$ \sum_{k \ne 0} \frac{\sin(ak)}{k} z^{-k} $$ diverges for all $z$ except the unit circle $|z|=1$. On the unit circle, the series does not converge absolutely, so rearrangement of the series matters. For such $z$, the principal value in the sense $$ a+\lim_{N \to \infty} \left(\sum_{k=-N}^{-1}+\sum_{k=1}^{N}\right) \frac{\sin(ak)}{k} z^{-k} $$ exists.

For example, if $z=1$ Maple gives me the following stairstep function:

added
Maple says the closed form is $$ a-\arctan \left( {\frac {\sin \left( a \right) }{\cos \left( a \right) -z}} \right) +\arctan \left( {\frac {z\sin \left( a \right) }{1-z\cos \left( a \right) }} \right) $$