Closed forms for definite integrals involving error functions I have been working for a while with these kinds of integrals
$$\int_0^\infty dx\,\text{erfc}\left(c +i x\right)\exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$
$$\int_\Lambda^\infty dx\,\frac{1}{x}\text{erfc}\left(c +i x\right) \exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$
where $c$ and $d$ are just real constants and $\Lambda>0$.
I have also been working with other similar integrals that have a closed-form expression, but I can't figure out the form of these ones. Does anyone know if these integrals have a closed-form solution?

If $u =\operatorname{erfc}(c+ix)$ then
$$du = \dfrac {-2} {\sqrt\pi} e^{-(c+ix)^2} i \, dx = \dfrac {-2} {\sqrt\pi} e^{-(c^2-2icx - x^2)^2} i \, dx. \tag 1$$
We need to work with
$$
\exp\left( \frac {-1}2 d^2x^2 + icx\right).
$$
The exponent is
$$
\frac{-d^2} 2 \left( x^2 - \frac{2icx}{d^2} - \frac{c^2}{d^4} \right) - \frac{c^2}{2d^2} = \frac{-d^2}2\left( x - \frac{ic}{d^2} \right)^2 - \frac{c^2}{2d^2} = \frac{-d^2}2 w^2 - \frac{c^2}{2d^2}. 
$$
Where we find $x$ in $(1)$, replace it with
$$
x = w + \frac{ic}{d^2}.
$$
Then do routine algebra and go on from there.
