0

We know that in 1d, $\frac{1}{|x|^{3.5}}$ is not locally integrable around the origin. I learnt from this post that we could apply the following definition to define a tempered distribution:

$$\langle\frac{1}{|x|^{3.5}},\phi\rangle:=\int _{-\infty}^{\infty}\frac{\phi(x)-\sum_{m=0}^{2}\frac{\phi^{(m)}(0)}{m!}x^m}{|x|^{3.5}}dx,\quad \phi\in\mathbb{S}.$$

My question is, here are we actually considering the Cauthy principal value of the above integral?

If so, how do we actually show $\langle\frac{1}{|x|^{3.5}},\phi\rangle$ defined above is a well-defined tempered distribution?

Thank you very much in advance for any of your helps:-)

Chang
  • 341

1 Answers1

1

Around origin, $\phi(x)-\sum_{m=0}^{2}\frac{\phi^{(m)}(0)}{m!}x^m$ is $O(x^3)$ so $\frac{\phi(x)-\sum_{m=0}^{2}\frac{\phi^{(m)}(0)}{m!}x^m}{|x|^{3.5}}$ is $O(|x|^{-0.5}),$ which is integrable (around origin), so a principal value integral is not needed.

md2perpe
  • 30,042