# The Fourier transform of $\frac{\text{erf}(\omega x)}{x}$

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.

• 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 $pv.(\frac1x)$ – reuns Nov 18 '18 at 6:56
• 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. – David G. Stork Nov 18 '18 at 7:03
• 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? – Yang Nov 18 '18 at 7:28
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:
$$\frac{\Gamma \left(0,\frac{\omega ^2}{4}\right)}{\sqrt{2 \pi }}$$