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


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


Can you wrap this up?


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!

  • $\begingroup$ 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

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.


  • $\begingroup$ 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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