$\newcommand{\sinc}{\operatorname{sinc}}$
Let $\sinc:\mathbb{R}\to\mathbb{R}$, $\sinc(x) = \frac{\sin (x)}{x}$ with $\sinc(0)=1$. A classical result is that the sum and integral of $\sinc$ are the same, namely $$ \sum_{n=-\infty}^{\infty} \sinc(n) = \int_{-\infty}^{\infty} \sinc(x)\,dx = \pi $$The integral is clearly translation invariant: were we to replace $x$ with $x+c$, a quick substitution $y=x+c$ gets us right back where we started. More surprising to me is that the series is also translation invariant: for any $c\in\mathbb{R}$, $\sum_{n\in\mathbb{Z}} \sinc(n) = \sum_{n\in\mathbb{Z}} \sinc(n+c).$ If $c$ is an integer, a similar reindexing trick makes it obvious, but it's unclear to me why this should work for any fixed $c$. I asked Mathematica why this might be true and it returned the following: $$ \sum_{n\in\mathbb{Z}}\sinc(n+c) = \frac{1}{2} i \left(B_{e^{-i}}(1-c,0)+B_{e^{-i}}(c,0)-B_{e^i}(1-c,0)-B_{e^i}(c,0)\right); $$here $B_z(a,b)$ is the incomplete beta function, the analytic continuation of $\int _0^z t^{a-1}(1-t)^{b-1}\,dt$.
I'm looking for one of two things: either a clean and sensible explanation for why the sum is translation invariant or, failing that, a derivation of Mathematica's output and a walkthrough of the functional identities to show that it collapses to $\pi$ for $c\in\mathbb{R}$.
With the residue theorem or the Fourier inversion theorem, for $a> 0$ $$\int_{-\infty}^\infty \frac{\sin(\pi a t)}{\pi t} e^{-2i\pi kt}dt = 1_{k\in [-a/2,a/2]}$$ For $a\in (0,2)$ let $$f(t) = \sum_n \frac{\sin(\pi a (t+n))}{\pi (t+n)} $$ It converges, it is $1$-periodic and it is equal to its Fourier series $$f(t)=\sum_k c_k e^{2i\pi kt}, \qquad c_k = \int_0^1 f(t)e^{-2i\pi kt}dt=\int_{-\infty}^\infty \frac{\sin(\pi a t)}{\pi t} e^{-2i\pi kt}dt = 1_{k\in [-a/2,a/2]}$$ ie. $$f(t)=1$$