What is the sum over a shifted sinc function? What is the sum of a shifted sinc function:
$$g(y) \equiv \sum_{n=-\infty}^\infty \frac{\sin(\pi(n - y))}{\pi(n-y)} \, ?$$
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
 \begin{align}
\mrm{g}\pars{y} & \equiv
\sum_{n = -\infty}^{\infty}{\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} =
{\sin\pars{\pi y} \over \pi y} +
\sum_{n = 1}^{\infty}\bracks{%
{\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} +
{\sin\pars{\pi\bracks{-n - y}} \over \pi\pars{-n - y}}}
\\[5mm] & = 
-\,{\sin\pars{\pi y} \over \pi y} +
\sum_{n = 0}^{\infty}\bracks{%
{\sin\pars{\pi\bracks{n - y}} \over \pi\pars{n - y}} +
{\sin\pars{\pi\bracks{n + y}} \over \pi\pars{n + y}}}
\end{align}

Note that

  $\ds{\bracks{{\sin\pars{\pi\bracks{z - y}} \over \pi\pars{z - y}} +
{\sin\pars{\pi\bracks{z + y}} \over \pi\pars{z + y}}}
\expo{-2\pi\,\verts{\Im\pars{z}}}
\stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\large\sim}
\pm{\expo{-\pi\verts{\Im\pars{z}}} \over \pi\,\Im\pars{z}}
\stackrel{\mrm{as}\ \Im\pars{z}\ \to\ \pm\infty}{\large\to}{\large 0}}$

such that the sum can be evaluated by means of the
Abel-Plana Formula:
\begin{align}
\mrm{g}\pars{y} & =
-\,{\sin\pars{\pi y} \over \pi y} +
\int_{0}^{\infty}\bracks{%
{\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} +
{\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}\dd x
\\[2mm] &
+
{1 \over 2}
\bracks{{\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} +
{\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}_{\ x\ =\ 0}
\\[5mm] & =
-\,{\sin\pars{\pi y} \over \pi y} +
{1 \over 2}\int_{-\infty}^{\infty}\bracks{%
{\sin\pars{\pi\bracks{x - y}} \over \pi\pars{x - y}} +
{\sin\pars{\pi\bracks{x + y}} \over \pi\pars{x + y}}}\dd x
\\[2mm] &
+
{\sin\pars{\pi y} \over \pi y} =
{1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{\pi x} \over \pi x}\,\dd x +
{1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{\pi x} \over \pi x}\,\dd x =
\bbx{\Large 1}
\end{align}
