# Apparent contradiction between the theory of hyperfunctions and the table of Fourier transforms

Graf's book on hyperfunction theory says (page $$36$$) that

$$\frac1{(x-i0)^n}=\frac{(-1)^{n-1}\pi i}{(n-1)!}\delta^{(n-1)}(x)+\operatorname{fp}\frac1{x^n},$$

while the table of Fourier transforms says that the Fourier transform of $$x^n$$ is $$2\pi i^n\delta^{(n)}(x)$$.

The apparent contradiction is in the factorial factor $$n!$$ Based on the second formula the first one should be

$$\frac1{(x-i0)^n}=(-1)^{n-1}n\pi i\delta^{(n-1)}(x)+\operatorname{fp}\frac1{x^n}$$

Where is the mistake?

• @Noah Schweber the one by Urs Graf, page 36 – Anixx Oct 12 at 20:36
• @Noah Schweber yes – Anixx Oct 12 at 20:39
• I've edited to incorporated the citation. – Noah Schweber Oct 12 at 20:39
• Your formula for the Fourier transform of $x^n$ needs parentheses: $(2\pi i)^n\delta^{(n)}$. The cited formula from Graf seems correct to me. What did you do to get the alternative? That is, how did you use the Fourier transform? – paul garrett Oct 12 at 21:40
• @paul garrett Wikipedia gives that formula with no parentheses: en.wikipedia.org/wiki/Fourier_transform – Anixx Oct 12 at 21:50