$$\sqrt{\frac{\pi }{2}} e^{-\frac{t^2}{2}} \text{erfi}\left(\frac{t}{\sqrt{2}}\right) \rightarrow -\frac{1}{2} e^{\frac{s^2}{2}} \text{Ei}\left(-\frac{s^2}{2}\right)$$ or $$ e^{-\frac{t^2}{2}}\int^t_0e^{\frac{q^2}{2}}dq \rightarrow e^{\frac{s^2}{2}}\int^{\infty}_{s}\frac{e^{\frac{-q^2}{2}}}{q}dq $$
how do I laplace transform these? Which is:
$$ \int^\infty_0 e^{-\frac{t^2}{2}}\bigg(\int^t_0e^{\frac{q^2}{2}}dq\bigg)e^{-st}dt = e^{\frac{s^2}{2}}\int^{\infty}_{s}\frac{e^{\frac{-q^2}{2}}}{q}dq $$
There is a wealth of other transformations here. (I could not recognize the particular one that I am asking here though.) Any insight on how those could be performed would also be appreciated.
Use change of order of the double integral (which can be checked to be valid by Fubini's theorem) in the LHS (i.e. the Laplace transform integral) to get $$\int_{0}^{\infty} e^{-st-t^2/2}\left(\int_{0}^t e^{q^2/2}dq\right)dt=\int_{0}^\infty e^{q^2/2}\left(\int_{q}^\infty e^{-st-t^2/2}dt\right)dq$$ I assume $s$ to be real. Then, $$\int_{q}^\infty e^{-st-t^2/2}dt=e^{s^2/2}\int_{q+s}^\infty e^{-t^2/2}dt=e^{s^2/2}e^{-q^2/2}\int_{s}^\infty e^{-qt-t^2/2}dt\\\implies LHS=e^{s^2/2}\int_{0}^\infty \int_{s}^\infty e^{-qt-t^2/2}dtdq\\=e^{s^2/2}\int_{s}^\infty e^{-t^2/2}\int_{0}^\infty e^{-qt}dq dt=RHS$$ The last two steps follow because the integration limits are decoupled and one can easily check using Fubini's theorem that the change of order of integration is valid.