How much makes $\sum\limits_{i=-\infty}^{\infty} \frac{1}{i2\pi+x}$? I believe $\sum\limits_{i=-\infty}^{\infty} \frac{1}{i2\pi+x} = \frac{1+\cos x}{2 \sin x}$ and that it is possible to prove it in a very indirect way (using filtering, Fourier Series and transforms. But is there a simpler way to get to this result ?

Edit : 
Here is the outline of the proof I had in mind. it consists in matching the perfect lowpass filter in fourier transforms with the equivalent in Fourier series : 
The cardinal sine function ($s(t) = {{\sin t} \over t}$) is the impulse response of a square filter with no phase shifting and cutoff pulsation $\omega_c=1$ (and cutoff frequency $f_c=1/2\pi$). Its Fourier Transform would be $F(\omega) = \left. \begin{cases} C^{(*)}, & \text{for } -\omega_c \le \omega \le \omega_c \\ 0, & \text{otherwise }\end{cases} \right\}$
(*): According to the version of the Fourier Transform 
So, the impulse response for a cutoff frequency of $f_c = 1$ would be $s(t)={{\sin 2\pi t} \over {2 \pi t}}$
Converting this to a Fourier series pattern would require making both the filtered signals and the filter impulse response periodic (let's say, of period 1). This means our impulse response will become $s(t)=\sum\limits_{i=-\infty}^\infty {{\sin 2\pi t} \over {2\pi (t+i)}}$
However, the equivalent filter in the Fourier Series domain is the one which accepts both the constant component and the fundamental frequency with a certain amplification factor (A) and no phase shifting, and reject all other frequencies. - i.e. $ s(t) = A (1+\cos 2 \pi t) $, which should be equal to $\sum\limits_{i=-\infty}^\infty {{\sin 2\pi t} \over {2\pi (t+i)}}$ (since we know that $\lim\limits_{t \to 1}{{\sin 2 \pi t} \over {2 \pi t}} = 1$, then A=1/2).
This leads to $\sum\limits_{i=-\infty}^\infty {{\sin 2\pi t} \over {2\pi (t+i)}} = {{1+\cos 2 \pi t} \over 2}$ which means that $\lim\limits_{N\to\infty}\sum\limits_{i=-N}^N {1 \over {2\pi (t+i)}} = {{1+\cos 2 \pi t} \over {2 \sin 2\pi t}}$ or for $x=2\pi t$, $\lim\limits_{N\to\infty}\sum\limits_{i=-N}^N {1 \over {2\pi i + x}} = {{1+\cos x} \over {2 x}}$ 
 A: METHODOLOGY $1$:
The product representation of the sine function is
$$\sin( x)= x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)\tag1$$
Taking the logarithmic derivative of $(1)$ reveals
$$\begin{align}
\cot(x)&=\frac1x +\sum_{n=1}^\infty \frac{2x}{\left(n^2\pi^2-x^2\right)}\\\\
&=\frac1x+\sum_{n=1}^\infty \left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\\\\
&=\frac1x+\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1{x+n\pi}+\sum_{n=1}^N\frac{1}{x-n\pi}\right)\\\\
&=\frac1x+\lim_{N\to \infty}\left(\sum_{n=1}^N \frac1{x+n\pi}+\sum_{n=-N}^{-1}\frac{1}{x+n\pi}\right)\\\\
&=\lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n\pi}
\end{align}$$
Now note that 
$$\frac{\cos(x)+1}{2\sin(x)}=\frac12\cot(x/2)$$
Can you wrap this up?

METHODOLOGY $2$:
In the same approach used in the Appendix of THIS ANSWER to derive the partial fraction expansion of the secant and cosecant functions, we begin by expanding the function $\cos(px)$ in the Fourier series
$$\cos(px)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag2$$
for $x\in [-\pi/\pi]$.  The Fourier coefficients are given by
$$\begin{align}
a_n&=\frac{2}{\pi}\int_0^\pi \cos(px)\cos(nx)\,dx\\\\
&=\frac1\pi (-1)^n \sin(\pi p)\left(\frac{1}{p +n}+\frac{1}{p -n}\right)\tag 3
\end{align}$$
Substituting $(3)$ into $(2)$, setting $x=\pi$, and dividing by $\sin(\pi p)$ reveals
$$\begin{align}
\pi \cot(\pi p)&=\frac1p +\sum_{n=1}^\infty \left(\frac{1}{p -n}+\frac{1}{p +n}\right)\tag4\\\\
&=\sum_{n=0}^\infty \left(\frac1{n+p}-\frac1{n-p+1}\right)
\end{align}$$
Now, letting $p=x/\pi$ in $(4)$, we find that 
$$\begin{align}
\cot(x)&=\frac1x+\sum_{n=1}^\infty\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)\\\\
&=\lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n\pi}
\end{align}$$
as was to be shown!
A: Complex Analytic Approach
From this answer we get
$$
\sum_{k\in\mathbb{Z}}\frac1{k+x}=\pi\cot(\pi x)
$$
Thus,
$$
\begin{align}
\sum_{k\in\mathbb{Z}}\frac1{2\pi k+x}
&=\frac1{2\pi}\sum_{k\in\mathbb{Z}}\frac1{k+\frac{x}{2\pi}}\\
&=\frac12\cot\left(\frac x2\right)
\end{align}
$$

Real Analytic Approach
This may be about as much, or more, work as the complex analytic approach, but it only uses real Fourier analysis to derive the sum above for $\pi\cot(\pi x)$.
Lemma $\bf{1}$: For $x\in(0,2\pi)$,
$$
\sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}2
$$
Proof: $\frac{\pi-x}2$ is odd on $(0,2\pi)$ and the Fourier coefficients are
$$
\begin{align}
\frac1\pi\int_0^{2\pi}\frac{\pi-x}2\,\sin(kx)\,\mathrm{d}x
&=-\frac1{k\pi}\int_0^{2\pi}\frac{\pi-x}2\,\mathrm{d}\cos(kx)\\
&=\frac1k-\frac1{2k\pi}\int_0^{2\pi}\cos(kx)\,\mathrm{d}x\\[3pt]
&=\frac1k
\end{align}
$$
$\square$
Theorem $\bf{1}$:
$$
\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x=\pi
$$
Proof: Turn the integral into a Riemann Sum and apply Lemma $1$:
$$
\begin{align}
\int_{-\infty}^\infty\frac{\sin(x)}x\,\mathrm{d}x
&=2\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\\
&=2\lim_{n\to\infty}\sum_{k=1}^\infty\frac{\sin(k/n)}{k/n}\frac1n\\
&=\lim_{n\to\infty}\left(\pi-\frac1n\right)\\[12pt]
&=\pi
\end{align}
$$
$\square$
Lemma $\bf{2}$: For $n\ge1$,
$$
\int_0^1\pi\cot(\pi x)\sin(2\pi nx)\,\mathrm{d}x=\pi
$$
Proof: Case $n=1$:
$$
\begin{align}
\int_0^1\pi\cot(\pi x)\sin(2\pi x)\,\mathrm{d}x
&=\pi\int_0^12\cos^2(\pi x)\,\mathrm{d}x\\
&=\pi\int_0^1(\cos(2\pi x)+1)\,\mathrm{d}x\\[6pt]
&=\pi
\end{align}
$$
Using the identity
$$
\begin{align}
\cot(x)\sin(2(n+1)x)
&=\cot(x)\sin(2x)\cos(2nx)+\cot(x)\cos(2x)\sin(2nx)\\
&=2\cos^2(x)\cos(2nx)+\cot(x)\left(1-2\sin^2(x)\right)\sin(2nx)\\
&=(\cos(2x)+1)\cos(2nx)+(\cot(x)-\sin(2x))\sin(2nx)\\
&=\cot(x)\sin(2nx)+\cos(2nx)+\cos(2(n+1)x)
\end{align}
$$
we get the inductive step: for $n\ge1$,
$$
\int_0^1\pi\cot(\pi x)\sin(2(n+1)\pi x)\,\mathrm{d}x
=\int_0^1\pi\cot(\pi x)\sin(2n\pi x)\mathrm{d}x
$$
$\square$
Theorem $\bf{2}$:
$$
\sum_{k\in\mathbb{Z}}\frac1{k+x}=\pi\cot(\pi x)
$$
Proof: It is not difficult to show that the sum is an odd function with period $1$. Furthermore, Theorem $1$ says the Fourier coefficients of the sum are
$$
\begin{align}
\sum_{k\in\mathbb{Z}}\int_0^1\frac{\sin(2n\pi x)}{k+x}\,\mathrm{d}x
&=\int_{-\infty}^\infty\frac{\sin(2n\pi x)}x\,\mathrm{d}x\\[6pt]
&=\pi
\end{align}
$$
and according to Lemma $2$, the Fourier coefficients match.
$\square$
