Show that $\operatorname{sinc}(x-y) = \sum_{n \in \mathbb{Z}}\operatorname{sinc}(x+n)\operatorname{sinc}(y+n)$ My problem is:
Show that for every $x,y \in \mathbb{R}$
$$\operatorname{sinc}(x-y) = \sum_{n \in \mathbb{Z}}\operatorname{sinc}(x+n)\operatorname{sinc}(y+n)$$
Here is what I did so far:
Considering the function $g(t) = f(t + t_{0})$ I can show that the Fourier Transform is given by $\hat{g}(w) = e^{-iwt_{0}} \hat{f}(w)$.
Therefore considering $f(t) = sinc(t)$ I can show that
$$\hat{f}(w) = \begin{cases} 
      1, &  |w| \leq \pi \\
      0, &  |w| \gt \pi 
   \end{cases}
$$
So I can use Shannon's Theorem, where $\Delta t = 1$ and write
$$ f(t) = g(t-t_{0}) = \sum_{n \in \mathbb{Z}} \operatorname{sinc}(t-t_{0}-n)sinc(t_{0} + n) $$
Now thinking $t$ as $x$ and having an $t_{0}$ as $y$ for every $x$ I can write
$$ sinc(x-y) = \sum_{n \in \mathbb{Z}} \operatorname{sinc}(x-y-n)\operatorname{sinc}(y + n) $$
Which is close, but not the same to what I should prove.
Can anybody point where I have made any mistake or help me with the final steps?
Thanks in advance.
 A: For $n\in \mathbb{Z}$, we have
\begin{align}
\sin \pi (x+n) \cdot \sin \pi (y+n)
&= \frac{1}{2}[\cos \pi(x-y) - \cos \pi(x+y+2n)]\\
&= \frac{1}{2}[\cos \pi(x-y) - \cos \pi(x+y)]\\
&= \sin \pi x \cdot \sin \pi y.
\end{align}
Also, we have
\begin{align}
\sum_{n\in \mathbb{Z}} \frac{1}{(x+n)(y+n)}
&= \frac{1}{xy} + \sum_{n=1}^\infty \left(\frac{1}{(x+n)(y+n)} + \frac{1}{(x-n)(y-n)}\right)\\
&= \frac{1}{xy} + \frac{1}{x-y}\sum_{n=1}^\infty \left(\frac{2y}{y^2-n^2} - \frac{2x}{x^2-n^2}\right)\\
&= \frac{1}{xy} + \frac{1}{x-y}\sum_{n=1}^\infty \frac{2y}{y^2-n^2} - \frac{1}{x-y}\sum_{n=1}^\infty \frac{2x}{x^2-n^2}\\
&= \frac{1}{xy} + \frac{1}{x-y}\left(\pi \cot \pi y - \frac{1}{y} \right)
- \frac{1}{x-y}\left(\pi \cot \pi x - \frac{1}{x} \right)\\
&= \frac{\pi \cot \pi y - \pi \cot \pi x}{x-y}
\end{align}
where we have used the following identity
$$\pi \cot \pi z = \frac{1}{z} + 2z \sum_{n=1}^\infty \frac{1}{z^2-n^2}.$$
See: https://mathworld.wolfram.com/Cotangent.html, or https://people.reed.edu/~jerry/311/cotan.pdf, or Find the sum of $\sum 1/(k^2 - a^2)$ when $0<a<1$
Thus, we have
\begin{align}
\sum_{n\in \mathbb{Z}} \frac{\sin \pi(x+n) \cdot \sin \pi(y+n)}{\pi^2(x+n)(y+n)}
&= \frac{1}{\pi^2} \sin \pi x \cdot \sin \pi y \cdot \sum_{n\in \mathbb{Z}} \frac{1}{(x+n)(y+n)}\\
&= \frac{1}{\pi^2} \sin \pi x \cdot \sin \pi y \cdot \frac{\pi \cot \pi y - \pi \cot \pi x}{x-y}\\
&= \frac{\sin \pi x \cdot \cos \pi y - \cos \pi x \cdot \sin \pi y}{\pi (x-y)}\\
&= \frac{\sin \pi (x-y)}{\pi (x-y)}.
\end{align}
(Q. E. D.)
