Show that $\operatorname{sinc}(x-y) = \sum_{n \in \mathbb{Z}}\operatorname{sinc}(x+n)\operatorname{sinc}(y+n)$

Question

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.

Answer 1

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.)