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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?

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