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Suppose we have a distribution defined on $\mathbb{R}\setminus\{0\}$ (with a singularity at the origin). I'd like to do two things: extend by zero at the origin and show that the fourier transform is well defined on the extended distribution.

Concretely, let: $$f= \begin{cases} \frac{1}{e^{-x}-1} & x<0\\ -\frac{1}{e^{x}-1} & x>0\\ \end{cases} $$

and for a test function $\phi$ our distribution is $\theta=\int\phi(x)f(x)dx$

Question 1: it seems to me that, by a cauchy principal value argument, there should be some type of extension by zero of $\theta$ to all of $\mathbb{R}$. What is the rigorous way to do this?

Question 2: How does one then show that the Fourier Transform is well defined on the extended distribution? (does one simply show that the distribution is tempered?)

I've come across pieces of this question, but never in full detail: How to interpret $|x|^{-1}$ as a distribution, or Extension of a distribution

APIs
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  • Your $f$ is not locally integrable near $0$ since it blows up like $-1/x$, so $\langle f, \phi \rangle := \int f\phi ,dx$ probably isn't well defined. – Mason Sep 06 '22 at 19:22
  • @Mason $\phi$ is assumed to be supported on $\Bbb{R}^*$, the OP wants to extend to all test functions. – reuns Sep 06 '22 at 19:23
  • Perhaps you can write $f(x) = -\frac{1}{x} + \frac{1}{2}\operatorname{sign}x + b(x)$ where $b$ is smooth? – md2perpe Sep 06 '22 at 21:27

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The distributions extending $f$ are of the form $$\langle F,\phi \rangle = \sum_{k=0}^K c_k \phi^{(k)}(0) + \int_{-\infty}^\infty f(x) (\phi(x)-\phi(0) 1_{|x| < 1})dx$$ Of course a natural choice is $K=-1$.

"Principal value and extend by $0$ at the origin" doesn't make much sense here.

reuns
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  • Thank you for the answer. i have a few followups: 1. what is the rationale for the subscript $|x|<1$? 2. is this a general construction for extending other odd distributions with similar singularity (eg $f=\frac{1}{x}$)? 3.How does one show that the Fourier Transform is well defined on the extended function? – APIs Sep 06 '22 at 19:41
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    You can replace $g=1_{|x|<1}$ by any (integrable) function smooth around $0$ with $g(0)=1$, it is the same as changing the value of $c_0$. It is a tempered distribution, its Fourier transform is well-defined. @MaxJames – reuns Sep 06 '22 at 20:23
  • thank you. this is very helpful – APIs Sep 06 '22 at 20:30
  • iv'e spend some time with this and am still curious about the purpose of the summation over k? why is it needed? thank you in advance – APIs Sep 10 '22 at 13:12
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    It is not needed, just that there are many different extensions from $C^\infty(\Bbb{R}^*)$ to $C^\infty(\Bbb{R})$ and they are all of this form. – reuns Sep 10 '22 at 15:08