What is the sum over a shifted sinc function?

We use the Poisson summation formula. Define $f(x) \equiv \sin(\pi x) / (\pi x)$. Then the sum we are trying to solve is $$g(y) = \sum_{n=-\infty}^\infty f(n-y) \, .$$ The Poisson summation formula converts the sum over values of $f$ to a sum over values of the Fourier transform of $f$.

Poisson summation

Note that $g(y)$ is periodic with period $1$. The Fourier series coefficients of $g$ are by definition \begin{align} g_\nu &= \int_0^1 dy \, g(y)e^{-i 2 \pi \nu y} \\ &= \int_0^1 dy \, \sum_{n=-\infty}^\infty f(n-y) e^{-i 2 \pi \nu y} \\ (\text{Let }x\equiv n-y) \qquad &= \sum_{n=-\infty}^\infty \int_{n-1}^n dx \, f(x) e^{-i 2 \pi \nu (n-x)} \\ &= \int_{-\infty}^\infty dx \, f(x) e ^{i 2 \pi \nu x} \\ &= \tilde{f}(-\nu) \, . \end{align} where $\tilde{f}$ is the Fourier transform of $f$.

By definition of the Fourier series, \begin{align} g(y) &= \sum_{\nu = -\infty}^\infty e^{i 2 \pi \nu y} g_\nu \\ \text{so} \qquad \sum_{n=-\infty}^\infty f(n-y) &= \sum_{\nu=-\infty}^\infty e^{-i 2 \pi \nu y} \tilde{f}(\nu) \end{align} which is the Poisson summation formula

Solution to the problem

Using the Poisson summation formula, we can write $$g(y) = \sum_{n=-\infty}^\infty f(n-y) = \sum_{\nu=-\infty}^\infty \tilde{f}(\nu) e^{-i 2 \pi \nu y} \, .$$ What is $\tilde{f}$? We can easily compute that the Fourier transform of the tophat function $$ T(x) = \left\{ \begin{array}{l} 1, \qquad -1/2 < x <1/2 \\ 0, \qquad \text{otherwise} \end{array} \right. $$ is $\tilde{T}(\nu)=\sin(\pi \nu) / (\pi \nu)$. By duality of the Fourier transform, that means that $\tilde{f}$ is the tophat function $T$. Therefore we have $$g(y) = \sum_{\nu=-\infty}^\infty T(\nu) e^{-i 2 \pi \nu y} = 1 \, .$$ This is a remarkable result: no matter how much you shift your sample points on a sinc function, the sum of those samples is constant.