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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?

Kagaratsch
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    Do you know what is a distribution, the principal value $pv.(\frac{1}{x})$, and the periodic distributions ? – reuns Dec 04 '17 at 20:01
  • @reuns I do know distributions, the Cauchy principal value and can imagine how to compose periodic distributions out of simple ones. Is there a way to apply this to get the Fourier transform in this case? If so, could you please elaborate? – Kagaratsch Dec 04 '17 at 20:06
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    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 $pv. \int_{-\pi}^\pi \frac{e^{-inx}}{\sin(x)}dx = pv. \int_{|z|=1} \frac{2 z^{1-n}}{z-z^{-1}}dz = \int_{|z|=1+\epsilon} \frac{ z^{1-n}}{z-z^{-1}}dz+ \int_{|z|=1-\epsilon} \frac{ z^{1-n}}{z-z^{-1}}dz$, – reuns Dec 04 '17 at 20:11
  • @reuns From my calculations, the Fourier transform of $f(x)=1/\sin(\pi x)$ comes out to be $\tilde{f}(w)=i\sum_{m=-\infty}^\infty e^{i m(\pi+w)}$, simply from taking anti-derivative of $e^{i x w}/\sin(\pi x)$ in $x$ (which is $-\frac{2e^{i(\pi+w)x}}{\pi+w}{}_2F_1(1,\frac{\pi+w}{2\pi},\frac{3\pi+w}{2\pi},e^{2\pi i x})$) and summing principal value differences at $x=m+0^+$ and $x=m+0^-$. Could you explain why you expect the result to be sum of delta distributions? – Kagaratsch Dec 04 '17 at 20:24
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    The Dirac comb is its own Fourier transform. – reuns Dec 04 '17 at 20:36
  • @reuns Wow! But... if the Dirac comb is its own Fourier transform, and it is equal to the Fourier transform of $1/\sin(\pi x)$, then $1/\sin(\pi x)$ must be the Dirac comb! That does not seem right since each $\delta$ is symmetric while $\sin$ is anti-symmetric. Am I missing something? – Kagaratsch Dec 04 '17 at 20:50
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    The correct result should be $-i|a| \sum_n \text{sign}(n) e^{i a n x}$. See my 2nd comment – reuns Dec 04 '17 at 21:07
  • @reuns So you are saying that $1/\sin(a x)=-i|a|\sum_{n=-\infty}^\infty \text{sign}(n)e^{ianx}$? This is a very interesting formula! I assume it should be known in the literature (unless you were the first one to derive it right now!), but trying to search online I can't seem to find it. Perhaps you know of a reference? – Kagaratsch Dec 04 '17 at 21:23
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    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. – reuns Dec 04 '17 at 21:35
  • @reuns Turns out I lost a $\text{sign}(w)$ in my anti-derivative calculation somewhere. But with the appropriate sum representation of $1/\sin(x)$ the Fourier transform is very easy to obtain! See my answer below. – Kagaratsch Dec 04 '17 at 23:26
  • @reuns OK, the sum representation trick was aesthetically a bit unsatisfactory. I added a derivation via residues to the answer, in case if you're interested. – Kagaratsch Dec 05 '17 at 16:01

1 Answers1

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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

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

Kagaratsch
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    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 $\Im(z)$. – reuns Dec 05 '17 at 16:16
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    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. ... – reuns Dec 05 '17 at 16:19
  • @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 $pv.$). – Kagaratsch Dec 05 '17 at 16:22
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    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. – reuns Dec 05 '17 at 16:35