1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @Noah Schweber the one by Urs Graf, page 36 $\endgroup$ – Anixx Oct 12 at 20:36
  • $\begingroup$ @Noah Schweber yes $\endgroup$ – Anixx Oct 12 at 20:39
  • $\begingroup$ I've edited to incorporated the citation. $\endgroup$ – Noah Schweber Oct 12 at 20:39
  • $\begingroup$ 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? $\endgroup$ – paul garrett Oct 12 at 21:40
  • $\begingroup$ @paul garrett Wikipedia gives that formula with no parentheses: en.wikipedia.org/wiki/Fourier_transform $\endgroup$ – Anixx Oct 12 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.