I am trying to find the Fourier transform $ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(t) \cdot e^{i\cdot\omega\cdot t} dt $ of the following function:
$$ f(t) = e^{-a\cdot t}\cdot \mathcal{N}\left( \frac{b\cdot t -c}{d\cdot\sqrt t} \right).$$
Where $\mathcal{N}$ is the cumulative standard normal distribution and $a,c,d>0$ and $b,w \in \Re$. In terms of the error function $erf$,
$$ f(t) = e^{-a\cdot t}\cdot \left[ \frac{1}{2} + erf\left( \frac{b\cdot t -c}{d\cdot\sqrt 2\sqrt{t}} \right) \right].$$
I am not quite sure how to even start evaluating this integral, but Mathematica readily says that the indefinite integral
$$\begin{eqnarray} \frac{1}{\sqrt{2\pi}} \int f(t) \cdot e^{i\cdot\omega\cdot t} \, dt = \frac{1}{ 2 d (a-i w) \sqrt{\frac{c^2}{d^2}} \sqrt{2 a+\frac{b^2}{d^2}-2 i w}}\cdot \left[ e^{\frac{b c}{d^2}-\sqrt{\frac{c^2}{d^2}} \sqrt{2 a+\frac{b^2}{d^2}-2 i w}} \left(\left(b \sqrt{\frac{c^2}{d^2}}-c \sqrt{2 a+\frac{b^2}{d^2}-2 i w}\right) \left(e^{2 \sqrt{\frac{c^2}{d^2}} \sqrt{2 a+\frac{b^2}{d^2}-2 i w}} \text{erf}\left(\frac{\sqrt{\frac{c^2}{d^2}}}{\sqrt{2} \sqrt{t}}+\sqrt{t} \sqrt{a+\frac{b^2}{2 d^2}-i w}\right)-e^{2 \sqrt{\frac{c^2}{d^2}} \sqrt{2 a+\frac{b^2}{d^2}-2 i w}}+1\right)- \\ \left(b \sqrt{\frac{c^2}{d^2}}+c \sqrt{2 a+\frac{b^2}{d^2}-2 i w}\right) \text{erf}\left(\frac{\sqrt{\frac{c^2}{d^2}}}{\sqrt{2} \sqrt{t}}-\sqrt{t} \sqrt{a+\frac{b^2}{2 d^2}-i w}\right)\right) \right]+\frac{1}{2\sqrt{2\pi}}\cdot\frac{e^{-t (a-i w)} \text{erf}\left(\frac{c-b t}{\sqrt{2} d \sqrt{t}}\right)}{a-i w}-\frac{1}{2\sqrt{2\pi}}\cdot\frac{e^{-t (a-i w)}}{a-i w} \end{eqnarray}$$