0
$\begingroup$

In the question, Fourier transformation of error function is derived as follows: $$ \int_{-\infty}^\infty \operatorname{erf}(x)e^{-ixk}dx=\frac{2}{ik} \exp \left(\frac{-k^{2}}{4}\right). $$ However, in the paper, following equation holds for any $a,b\in\mathbb{C}$: $$ \int \operatorname{erf}(a z) e^{b z} d z=\frac{1}{b} e^{b z} \operatorname{erf}(a z)-\frac{1}{b} \exp \left(\frac{b^{2}}{4 a^{2}}\right) \operatorname{erf}\left(a z-\frac{b}{2 a}\right). $$ According this formula, Fourier transformation of error function is not well-defined because $$ \int_{-\infty}^\infty \operatorname{erf}(x)e^{-ixk}dx=-\frac{1}{ik} e^{-ik \infty}-\frac{1}{ik} e^{ik \infty}+\frac{2}{ik} \exp \left(\frac{-k^{2}}{4}\right). $$ Which part of the above discussion is wrong?

$\endgroup$
1
  • $\begingroup$ The Fourier transform of $\operatorname{erf}x$ is not defined in the usual sense since the defining integral diverges. But it is well-defined in the distributional sense. $\endgroup$
    – metamorphy
    Jan 24 '21 at 14:31
1
$\begingroup$

The first equation gives a correct formula for the Fourier transform of the error function. As metamorphy writes, the integral on the left hand side is not defined, but the Fourier transform in a distributional sense is defined and equals $$ \frac{2}{i} \exp\left(-\frac{k^2}{4}\right) \operatorname{pv}\frac{1}{k} $$ where $\operatorname{pv}\frac{1}{k}$ is the principal value distribution.

The second equation gives a primitive function of the error function, but since the integral over all the reals is not defined, this formula cannot be used to determine the Fourier transform.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.