How to prove the identity involving Sinc-series? Here the Sinc function is defined as: ${\rm sinc} (x):= \sin(x)/x$.
I found the following identity by numerical experiments, but how to prove it?

Let:$$
f(x;x_0):={\rm sinc}\ x_0+\sum_{n=1}^\infty {\rm sinc}\ (nx+x_0)+\sum_{n=1}^\infty {\rm sinc}\ (nx-x_0), 0<x<2\pi, x_0\in\mathbf{R}
$$
  We have:
  $$
f(x;x_0)\equiv f(x;0)
$$

which means the function $f(x;x_0)$ has nothing to do with $x_0$ at all!
Anyone can help me?

I believe instead that $f$ depends on the value of $x_0$. Here is what I thought: let $x = \pi \in (0,2\pi)$ and let $x_0 \in (0, \pi)$.
For $N \geq 1$, we have:
\begin{align}
\sum_{k = 1}^N \frac{\sin (k \pi + x_0)}{k \pi + x_0} + \frac{\sin (k \pi - x_0)}{k \pi - x_0}
&= \cos x_0 \cdot \left( \sum_{k = 1}^{N} ({-1})^{k} \frac{1}{k \pi + x_0} - \sum_{k = 1}^N ({-1})^k \frac{1}{k \pi - x_0} \right) \\
&= - 2 x_0 \cdot \cos x_0 \cdot \sum_{k = 1}^N ({-1})^k \frac{1}{k^2 \pi^2 - x_0^2} \\
&= - 2 x_0 \cdot \cos x_0 \cdot S_N (x_0)
\end{align}
For all $k \geq 1$, the convergent series for even and odd indices satisfy
\begin{align}
0 < \sum_{k \geq 0} \frac{1}{(2k + 1)^2 \pi^2 -x_0^2} &< \sum_{k \geq 0} \frac{1}{(2k + 1)^2 \pi^2} = \frac{1}{8} \\
0 < \sum_{k \geq 1} \frac{1}{4k^2 \pi^2 - x_0^2} &< \sum_{k \geq 1} \frac{1}{4k^2 \pi^2} = \frac{1}{24}
\end{align}
so that
\begin{equation}
\frac{1}{8} > S_N (x_0) > - \frac{1}{8}
\end{equation}
Taking $y \in (0, \pi)$ close enough to $\pi$ so that $1 / 8 < \mathrm{sinc}\ y < 1 - 1 / 8$, we find that:
\begin{align}
0 &< f (\pi ; y) < 1 \\
f (\pi ; 0) &= \mathrm{sinc}\ 0 = 1
\end{align}
so that $f (\pi ; y) \neq f (\pi ; 0)$.
