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The function $1/\sin(a x)$ with parameter $a\in\mathbb{R}$ is periodic in the argument $x\in\mathbb{R}$. Does a Fourier transform for it exist, and is it known?

The infinite values might pose a problem. If so, how about the function $1/(\sin(a x)+ib)$ with a complex shift by an extra parameter $b\in\mathbb{R}$ which makes the function finite at all $x$.

What is the Fourier transform of any of these functions?

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    $\begingroup$ Do you know what is a distribution, the principal value $pv.(\frac{1}{x})$, and the periodic distributions ? $\endgroup$ – reuns Dec 4 '17 at 20:01
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    $\begingroup$ The Fourier transform is $\sum_n c_n \delta(\omega-an)$. You can find the $c_n$ directly by solving $(\delta(\omega-a)-\delta(\omega+a)) \ast \sum_n c_n \delta(\omega-an) = \delta(\omega)$, or you can use some complex analysis to evaluate the Fourier series coefficients $pv. \int_{-\pi}^\pi \frac{e^{-inx}}{\sin(x)}dx = pv. \int_{|z|=1} \frac{2 z^{1-n}}{z-z^{-1}}dz = \int_{|z|=1+\epsilon} \frac{ z^{1-n}}{z-z^{-1}}dz+ \int_{|z|=1-\epsilon} \frac{ z^{1-n}}{z-z^{-1}}dz$, $\endgroup$ – reuns Dec 4 '17 at 20:11
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    $\begingroup$ The Dirac comb is its own Fourier transform. $\endgroup$ – reuns Dec 4 '17 at 20:36
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    $\begingroup$ The correct result should be $-i|a| \sum_n \text{sign}(n) e^{i a n x}$. See my 2nd comment $\endgroup$ – reuns Dec 4 '17 at 21:07
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    $\begingroup$ There was a typo. And no. I'm saying that $pv.(\frac{1}{\sin(x)})=-i\sum_{n=-\infty}^\infty \text{sign}(2n+1)e^{i(2n+1)x}$ in the sense of distributions, ie. $pv. \int_{-\pi}^\pi \frac{\varphi(x)}{\sin(x)}dx = -i\sum_{n=-\infty}^\infty \text{sign}(2n+1)b_{2n+1}$ where $b_n = \frac{1}{2\pi}\int_{-\pi}^\pi \varphi(x)e^{-inx}dx$ whenever $\varphi$ is $C^\infty$ (or $C^1$) and $2\pi$-periodic. $\endgroup$ – reuns Dec 4 '17 at 21:35
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We can use the following identity

$$\begin{align} \csc (x)=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{x-n\pi} \end{align}$$

which was proven in another discussion here on stackexchange.

With this it is trivial to take the Fourier transform term by term:

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dx \csc(x)e^{iwx}=i\sqrt{\frac{\pi}{2}}\text{sign}(w)\sum_{n=-\infty}^{\infty}e^{i\pi n(1+w)}$$

Note that due to the $\text{sign}(w)$ in front this is actually not the Dirac comb.

EDIT

There is also a very simple way to derive the result from residues:

The integral $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dx \csc(x)e^{iwx}$ naturally has holes at all positions $x=\pi n$ with $n\in\mathbb{Z}$, to even be defined (since integers are a set of zero measure with respect to the real line, it is still the full Fourier transform).

If $w>0$ (or $w<0$) we can close the contour at its far ends via the upper (or lower) half plane. This adds zero contribution to the integral due to Jordan's lemma.

Then deforming the half circle contour back to the real line, we cancel all contributions from the initial integral and are left with just tiny half-circles around each $x=\pi n$ point in counter-clockwise (or clockwise) direction.

The half-circles contribute half a residue each. Counter-clockwise gives positive residue while clockwise gives negative residue, therefore we immediately find a $\text{sign}(w)$ term to be multiplied with the result.

Since $1/\sin(x)$ only has simple poles, we trivially have:

$$\frac{2\pi i}{2}\text{res}_{x=\pi n}\left(\frac{1}{\sqrt{2\pi}} \csc(x)e^{iwx}\right)=i\sqrt{\frac{\pi}{2}}e^{i\pi n(1+w)}$$

So that again, the full result properly is

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty dx \csc(x)e^{iwx}=i\sqrt{\frac{\pi}{2}}\text{sign}(w)\sum_{n=-\infty}^{\infty}e^{i\pi n(1+w)}.$$

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    $\begingroup$ You need to mention the principal values. And you can obtain the result directly from $pv.(\frac{1}{\sin(x)}) = \frac{1}{2}\lim_{a \to 0^+} \frac{1}{\sin(x+ia)}+\frac{1}{\sin(x-ia)}$ and expand $\frac{2i}{\sin(z)}=\frac{1}{e^{z} - e^{-z}}$ as a geometric series in $e^{z}$ or $e^{-z}$ depending on $\Im(z)$. $\endgroup$ – reuns Dec 5 '17 at 16:16
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    $\begingroup$ An equivalent method is to start from the Fourier series $pv.(\frac{1}{\cos(x)}) = \sum_n c_n e^{i n x}$ (which converges in the sense of distributions). Then $1=\frac{e^{ix}+e^{-ix}}{2} pv.(\frac{1}{\cos(x)}) =\sum_n \frac{c_{n-1}+c_{n+1}}{2} e^{i n x}$ thus For $n \ne 0$ : $c_{n+1} = - c_{n-1}$ and for $n= 0$ : $c_1+c_{-1} = 2$, ie. ... $\endgroup$ – reuns Dec 5 '17 at 16:19
  • $\begingroup$ @reuns Interesting! How come the principal values are taken from a complex direction when the contour is purely real though? (I believe saying that the initial contour has holes at the pole locations accounts for $pv.$). $\endgroup$ – Kagaratsch Dec 5 '17 at 16:22
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    $\begingroup$ If $F$ has a pole at $t= 0$ then $pv. \int_{-1}^1 F(t)g(t)dt \overset{def}= \lim_{a \to 0^+} \int_{-1}^{-a}+\int_a^1 F(t)g(t)dt$ or when $h$ has many poles. It works the same way for $pv.\int_\gamma f(z)dz = pv.\int_0^1 f(\gamma(t)) \gamma'(t)dt$. And if $f$ is meromorphic then $pv.\int_\gamma f(z)dz = \frac{1}{2} \int_{\gamma_a}+\int_{\gamma_{-a}} f(z)dz$ where $\gamma_a,\gamma_{-a}$ are contours with a small enough identation around the pole. $\endgroup$ – reuns Dec 5 '17 at 16:35

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