# Identity for the sum of products of Sinc functions

The Sinc function is defined as follows: $$\mathrm{sinc}(t) = \begin{cases} \frac{\sin(\pi t)}{ \pi t} & \mathrm{if} \quad t \neq 0, \\ 1 & \mathrm{otherwise.} \end{cases}$$ I want to show the following identity, $$\sum_{n=-\infty}^\infty \mathrm{sinc }(2Bt-n) \mathrm{sinc }(2Bs-n) = \mathrm{sinc }(2B(t-s)).$$ where $B > 0.$ I have tried the Poisson summation formula and Fourier Series approach but they didn't work.

It is just a convolution identity: $$\sum_{n\in\mathbb{Z}}\text{sinc}(A-n)\text{sinc}(n-B) = (\widehat{u}*\widehat{u})(A-B)=\left(\widehat{u\cdot u}\right)(A-B),$$ but since the inverse Fourier transform $u$ of the $\text{sinc}$ function is just the indicator function of an interval, $u^2=u$, hence the RHS of the previous line equals $\widehat{u}(A-B)=\text{sinc}(A-B)$. If $C(x)$ is the Dirac comb, the LHS equals:
$$\begin{eqnarray*}\int_{-\infty}^{+\infty}\text{sinc}(A-x)\text{sinc}(x-B)C(x)\,dx &=& \int_{-\infty}^{+\infty}\mathcal{F}\left(\text{sinc}(A-x)\text{sinc}(x-B)\right)\mathcal{F}^{-1}(C(x))\,dx\\&=&\int_{\mathbb{R}}\left[\mathcal{F}(\text{sinc}(A-x))*\mathcal{F}(\text{sinc}(x-B))\right]C\left(-\frac{x}{2\pi}\right)\,dx\\&=&\frac{1}{|B-A|}\int_{-2\pi}^{2\pi}C\left(-\frac{x}{2\pi}\right)\text{sign}(x)e^{\frac{B+A}{2}ix}\sin\left(\frac{|B-A|}{2}\,x\right)\,dx\end{eqnarray*}$$ that is just $\text{sinc}(B-A)$ by $e^{iz}+e^{-iz}=2\cos z$ and the sine duplication formula.
• Jack, since $n$ is an integer, how does one justify $$\sum_{n\in\mathbb{Z}}\text{sinc}(A-n)\text{sinc}(n-B)=\sum_{n\in\mathbb{Z}} \text{sinc}(n)\text{sinc}(n-(A-B))$$This results seems predicated on the index translation $n\to A-n$. Inasmuch as $A$ is not restricted to the integers, I don't quite see that this is valid. -Mark Apr 26, 2016 at 18:39
• @Dr.MV: you're right, but I have an escape route, that is just the Poisson summation formula. The Fourier transform of $\text{sinc}$ is easy to manage (multiply or convolve) with respect to a similar function. May 4, 2016 at 0:23
• @Dr.MV: exactly, and the RHS is now the Fourier transform of a simple function, since $\mathcal{F}(\text{sinc})$ is the characteristic function of an interval (times some constant depending on the definition of the Fourier transform) May 4, 2016 at 2:35