Fourier transform of $\,e^{-ax}\!\operatorname{erfc}(x)$

Question

Can someone help with calculating the Fourier transform of $e^{-ax}\!\operatorname{erfc}(x)$, where $a < 0$. There are some posts talking about how to calculate the Fourier transform of $\operatorname{erfc}(x)$, which is derived from distribution theory. The problem I am asking will not a distribution. I'm not sure how to solve this. Thank you!

Answer 1

I believe the Fourier transform of

$$f(x)=e^{-a x}\, \text{erfc}(x)\tag{1}$$

only converges for $\Re(a)<0$:

$$F(\omega)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} f(x)\, e^{i x \omega}\,dx=-\frac{\sqrt{\frac{2}{\pi}}\ e^{\frac{1}{4} (a-i \omega)^2}}{a-i \omega}\text{ if }\Re(a)<0\tag{2}$$

The result illustrated in formula (2) above was given by Mathematica.