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From Wikipedia, I learnt that on $\mathbb{R},$ the Fourier transform of $|x|^\alpha$ is given by $-\frac{2\sin(\frac{\pi \alpha}{2})\Gamma(\alpha+1)}{2\pi|\xi|^{\alpha+1}}.$ But this formula is only valid for $-1<\alpha<0,$ because in this case $|x|^\alpha$ is locally integrable in and is a tempered function, I think this makes sense to me.

However, Wikipedia also said that the function $\alpha \mapsto |x|^\alpha$ is a holomorphic function from the right half plane to the space of tempered functions, what does that mean? It seems that the Fourier transform of $|x|^\alpha$ also has some connections with homogeneous distributions, what is their relation?

In particular, how should we define the Fourier transform of $|x|^{-1.5}$ for $x\in \mathbb{R},$ since $|x|^{-1.5}$ is not in $L^1,$ and is not a tempered distribution? Thank you very much in advance for any of your ideas and helps!

Chang
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    $|x|^{-1/2}$ is also not in $L^1$. I believe the $-1<\alpha$ part of the condition $-1<\alpha<0$ is because $\Gamma(\alpha+1)$ has a pole at $\alpha=-1$. Mathematica gives the result $$\mathcal{F}_x\left|x|^{-3/2}\right=-4 \pi \sqrt{|\xi|}$$ which is consistent with the Wikipedia formula, so perhaps the formula is more generally valid for $\alpha<0\land \alpha\notin\mathbb{Z}$ which avoids the poles of $\Gamma(\alpha+1)$ at negative integer values of $\alpha$. – Steven Clark Apr 16 '24 at 00:26
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    You need to define that distribution for $\psi(x)=|x|^{-3/2}$ such that for any $\phi\in \mathbb{S}$, we have $$\langle \psi,\phi\rangle = \int_{-\infty}^\infty \frac{\phi(x)-\phi(0)}{|x|^{3/2}},dx$$. – Mark Viola Apr 16 '24 at 18:24
  • I think that makes sense! Now I see why $\alpha$ should avoid to take negative integer values. Thanks a lot for your comment! @StevenClark – Chang Apr 23 '24 at 11:41
  • Thanks for your comment! Now I understood how we could define a tempered distribution for $\psi(x)$ and the Fourier transformation could be defined. @MarkViola – Chang Apr 23 '24 at 11:43
  • @Chang You're welcome. My pleasure. – Mark Viola Apr 23 '24 at 14:41
  • @MarkViola I thought $f(x)\in L^1$ requires $$\int\limits_{-\infty}^{\infty} |f(x)| , dx<\infty$$ (see this answer) whereas for $f(x)=|x|^{-1/2}$ one has $$\int\limits_{-\infty}^{\infty} \left||x|^{-1/2}\right| , dx=\infty.$$ Also, how do you define the Fourier transform of $$f(x)=|x|^{-1}?$$ – Steven Clark Apr 23 '24 at 16:29
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    @StevenClark Apology. I meant that $\frac1{|x|^{1/2}}$ is locally integrable and thus has a Fourier Transform. As for the Fourier Transform of $\psi(x)=\frac1{|x|}$, we define it as a Tempered Distribution where for any $\phi \in \mathbb{S}$, the distribution $$\langle \psi,\phi\rangle = \int_{-\infty}^\infty \frac{\phi(x)-\phi(0) \xi_{[-1,1]}(x)}{|x|},dx$$Please see THIS ANSWER in which I developed the Fourier Transform for the distribution $|x|^\alpha$ for all $\alpha \in \mathbb{R}$. – Mark Viola Apr 23 '24 at 19:30
  • @StevenClark ... Continued from previous comment ... So, if we use that aforementioned definition of the tempered distribution for $\frac1{|x|}$, its Fourier Transform is $-2(\gamma +\log(|k|))$, where $\gamma $ is the Euler-Mascheroni constant. – Mark Viola Apr 23 '24 at 19:39

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