I'm working on characteristic functions, and in this process I have came across the following step in a proof
$$ \frac{i}{2\pi}e^{-\frac{1}{2}\tau_k\mu^2/\sigma^2}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\sigma^2v^2\tau_k}\frac{\text{d} v}{v - i\mu / \sigma^2} + 1 = 1 - \frac{1}{2}\operatorname{erfc}\Bigg[{\frac{\mu}{\sigma}\sqrt{\frac{1}{2}\tau_k}}\Bigg] $$
I received the proof from which this step is taken from an 'integrals guru' my professor contacted. So I guess it is correct, but I do not see what substitution would be used to go from left to right in one step. For reference, the complementary error function $\operatorname{erfc}[.]$ is defined as:
$$ \operatorname{erfc}[x] = \frac{2}{\sqrt\pi} \int_x^\infty e^{-t^2}\,\mathrm dt \\ $$
Thanks in advance!!
The easiest way is to look up some special functions, in particular the integral representation of the Fadeeva function.
The Fadeeva function is defined (for complex arguments) as $w(z)=e^{-z^2} \operatorname{erfc}(-i z)$.
The integral representation of the Faddeva function is
$w(z)=\frac{i}{\pi} \int_{-\infty}^{\infty}\frac{e^{-t^2}}{z-t}dt$
which can be transformed to your problem with some simple substitutions.