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Does anyone know the Fourier transform of

$\Large\frac{\text{erf}(\omega x)}{x}$?

I think it should be something like $\frac{4\pi}{k^2}\exp{(-k^2/4\omega^2)}$.

Is this right? How can one go about deriving this? Any hints are much appreciated.

Thank you in advance!

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  • $\begingroup$ You mean the Fourier transform of the distribution $pv.(\frac{erf(\omega x)}{x})= \lim_{\epsilon \to 0} \frac{erf(\omega x)}{x} 1_{|x| > \epsilon}$. The method is the same as for $pv.(\frac1x)$ $\endgroup$ – reuns Nov 18 '18 at 6:56
  • $\begingroup$ If $x$ is your primal variable, what is the transform variable? It surely cannot be $\omega$ because you have used that in the original function. $\endgroup$ – David G. Stork Nov 18 '18 at 7:03
  • $\begingroup$ The transform variable is just $k$, isn't it? The transform would be $\int_0^\infty dx\frac{\text{erf}(\omega x)}{x}\text{e}^{-ikx}$, right? $\endgroup$ – Yang Nov 18 '18 at 7:28
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I think you have an error in your question: $\omega$ is typically used for the transform variable, and hence certainly shouldn't be in the untransformed function.

Anyway, the Fourier transform of ${\rm{Erf}(x) \over x}$ (of your title) is:

$$\frac{\Gamma \left(0,\frac{\omega ^2}{4}\right)}{\sqrt{2 \pi }}$$

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