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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 $x=2\pi t$, $\lim\limits_{N\to\infty}\sum\limits_{i=-N}^N {1 \over {2\pi i + x}} = {{1+\cos x} \over {2 x}}$

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    $\begingroup$ You need to be careful with the order of summation otherwise it diverges. In complex analysis : show the difference is an entire periodic bounded function (easier to look at the derivative). In Fourier analysis : show $e^{-y}1_{y > 0}$'s Fourier transform is $\frac{1}{x+2i \pi}$ thus $e^{-y} 1_{y>0}$ is the FT of $\frac{1}{x-2i \pi}$ and $e^{-n}1_{n > 0}$ are the Fourier series coefficients of the $1$-periodization $\lim_{N \to \infty}\sum_{n=-N}^N\frac{1}{x+n-2i \pi}$ $\endgroup$ – reuns Mar 9 '19 at 3:28
  • $\begingroup$ Otherwise look at the Fourier series of $\sum_n h(x+n)$ where $h(x)=e^{-ax}1_{x> 0}$ $\endgroup$ – reuns Mar 9 '19 at 4:04
  • $\begingroup$ @reuns, I had to search to understand your notation «$1_{x>0}$». Would you confirm it is a variant of the heavyside function ? (which, according to wikipedia, is 1/2 for x=0) $\endgroup$ – Camion Mar 9 '19 at 12:59
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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!

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  • $\begingroup$ Shame on me for not having been able to see that $\frac{\cos(x)+1}{2\sin(x)}=\frac12\cot(x/2)$ $\endgroup$ – Camion Mar 9 '19 at 15:01
  • $\begingroup$ (for future reference : $\frac{\cos(x)+1}{2\sin(x)}=\frac{\cos²(x/2)-\sin²(x/2)+1}{4\sin(x/2)\cos(x/2)}=\frac{2\cos²(x/2)}{4\sin(x/2)\cos(x/2)}=\frac12\cot(x/2)$) $\endgroup$ – Camion Mar 9 '19 at 15:02
  • $\begingroup$ @Camion: Here is a proof without words. $\endgroup$ – robjohn Mar 10 '19 at 14:25
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    $\begingroup$ @Mark Viola : Because since one of its summits is the center of the circle, the triangle is then isosceles. consequently the sum of its angles which makes 180° is A+A+(180°-2A) $\endgroup$ – Camion Mar 11 '19 at 3:32
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    $\begingroup$ @Camion There was a typographical error ($n\to N$, which should be $N\to\infty$) that you mentioned, which I have edited accordingly. The reason it appeared twice is likely a consequence of "cutting-and-pasting." $\endgroup$ – Mark Viola Mar 11 '19 at 16:58
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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.

$\square$

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  • $\begingroup$ If the complex analytic proof given in the cited answer is out of range, I can provide a real-only proof (using Fourier series). $\endgroup$ – robjohn Mar 9 '19 at 9:42
  • $\begingroup$ I've caught it, but I'm still interested in your other proof :-) $\endgroup$ – Camion Mar 11 '19 at 3:19
  • $\begingroup$ @Camion: I've added a real analytic approach deriving the sum for $\pi\cot(\pi x)$, then we proceed as before. $\endgroup$ – robjohn Mar 11 '19 at 3:56
  • $\begingroup$ your last proof is quite long and I haven't wrapped it up, yet. I just added the outline of the one i did think about. I'm not sure yet how similar/different both of them are. $\endgroup$ – Camion Mar 14 '19 at 4:28
  • $\begingroup$ Hi Rob. I posted an alternative way forward using Fourier Series which complements yours. $\endgroup$ – Mark Viola Mar 18 '19 at 16:31

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