Contradiction in two ways to derive Fourier transformation of error function

Question

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?

Answer 1

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.