I'm trying to do this integral essentially:
$\int^\infty_0 \frac{e^{-ax^2}\sin(kx)}{x}dx$ which I realized to be $\frac{1}{2}\operatorname{Re}\left[F\left(\frac{e^{-ax^2}}{x}\right)\right]$ where $F$ denotes the ordinary Fourier Transform (where I am using the convention $F[f] = \int^\infty_{-\infty}e^{ikx}f(x) \, dx)$.
This answer is supposed to be something related to the error function $\operatorname{erf}(k)$ but I don't have any ideas about how to get it to look like one. Could I somehow apply the convolution theorem?