into a difference of two similar integrals $$I_2(\beta,\zeta)=:I^{-}_2(\beta,\zeta)-I^{+}_2(\beta,\zeta).$$ Observing that $$\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}(1+z^2)+\frac{{{\mathrm i}}\zeta z}{v}\right)=
\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}-\frac{\zeta^2}{4{{\mathrm i}}\beta{\bar{L}}}-
\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}\left(z+\frac{\zeta v}{2\beta{\bar{L}}}\right)^2\right),$$ we can write $$\begin{aligned}
  I_2^{-}(\beta,\zeta)&:=&\frac{1}{2\pi{{\mathrm i}}}\int_{-\infty}^{\infty}\exp\left(
-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}(1+z^2)+\frac{{{\mathrm i}}\zeta z}{v}\right)\frac{{{\mathrm d}}z}
{z-i}\\
&=&\frac{1}{2\pi{{\mathrm i}}}\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}+\frac{{{\mathrm i}}\zeta^2}
{4\beta{\bar{L}}}\right)
\int_{-\infty}^{\infty}\frac{\exp(-{{\mathrm i}}\beta{\bar{L}}y^2/v^2)}
{y-{{\mathrm i}}-\zeta v/2\beta{\bar{L}}}{{\mathrm d}}y\end{aligned}$$ via $y=z+\zeta v/2\beta{\bar{L}}$. We make the change of variable $$r=\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}y.$$ This is permitted, because it rotates the contour of integration into the second and fourth quadrants of the complex plane, where the analytic function ${{\mathrm e}}^{-{{\mathrm i}}z^2}$ decays rapidly.

In the case $\zeta v/2\beta{\bar{L}}> -1$ the new contour of integration avoids the pole at $y=\zeta v/2\beta{\bar{L}}+{{\mathrm i}}$ (figure \[fig:1\]) and Cauchy’s Theorem yields $$I_2^{-}(\beta,\zeta)=
\frac{1}{2\pi{{\mathrm i}}}\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}+\frac{{{\mathrm i}}\zeta^2}
{4\beta{\bar{L}}}\right)
\int_{-\infty}^{\infty}\frac{{{\mathrm e}}^{-r^2}}{r-\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}(
{{\mathrm i}}+\frac{\zeta v}{2\beta{\bar{L}}})}{{\mathrm d}}r.$$ This integral is standard, and may be found in, for example, [@abr:ste] (Equation [**7.1.4**]{}): $$\frac{{{\mathrm i}}}{\pi}\int_{-\infty}^{\infty} \frac{{{\mathrm e}}^{-t^2}}{z-t} {{\mathrm d}}t
={{\mathrm e}}^{-z^2}{\mathop{\rm erfc}}(-{{\mathrm i}}z),\qquad\mbox{for ${{\mathfrak{Im}}}\:z>0$.}$$ The result we get is $$\begin{aligned}
  I^{-}_2(\beta,\zeta)&=&\frac{1}{2}\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}+
\frac{{{\mathrm i}}\zeta^2}{4\beta{\bar{L}}}\right)\exp\left(-\frac{{{\mathrm i}}\beta{\bar{L}}}{v^2}
\left({{\mathrm i}}+\frac{\zeta v}{2\beta{\bar{L}}}\right)^2\right)\\
& &\times{\mathop{\rm erfc}}\left(
\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\left(1-\frac{{{\mathrm i}}\zeta v}{2\beta{\bar{L}}}
\right)\right)\\
&=&\frac{1}{2}\exp\left(\frac{\zeta}{v}\right){\mathop{\rm erfc}}\left(\frac{\zeta}{2\sqrt{
{{\mathrm i}}\beta{\bar{L}}}}+\frac{\sqrt{{{\mathrm i}}\beta{\bar{L}}}}{v}\right).\end{aligned}$$

(1,1) (0,0)[![Deforming the contour of integration avoiding pole[]{data-label="fig:1"}](contour1.eps "fig:"){width="7.0cm" height="6cm"}]{} (0.74,0.55)[$y=\frac{\zeta v}{2\beta{\bar{L}}}+{{\mathrm i}}$]{}

(1,1) (0,0)[![Deforming the contour of integration enclosing pole[]{data-label="fig:2"}](contour2.eps "fig:"){width="7.0cm" height="6cm"}]{}

If $\zeta v/2\beta{\bar{L}}<-1$ then the 