How do I evaluate this integral in the complex plane? I am studying a paper in which they define a function
$$D(\phi) = \frac{1}{\iota \pi} \frac{\exp(-\frac{\phi^2}{2})}{\phi - \iota \epsilon} = \delta(\phi) + \frac{\mathcal{P}}{\iota \pi} \frac{\exp(-\frac{\phi^2}{2})}{\phi}$$
where $\mathcal{P}$ is the principal value, $\epsilon$ is an infinitesimal number.
They then go on to claim that an elementary calculation reveals
$$ \int_{-\infty}^{\infty} d\phi D(\phi)\exp(\iota t \phi) = 2\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} \exp\left(-\frac{x^2}{2}\right)dx$$
I am not much conversant with complex contour integrals so I shall be grateful is someone can explain how the second equation holds and what could be meant by the principal value in the first equation.
Thanks
 A: $$\begin{align}
\int_{-\infty}^\infty D(\phi)e^{it\phi}\,d\phi&=\lim_{\epsilon\to0}\frac{1}{i\pi}\int_{-\infty}^\infty \frac{e^{-\phi^2/2}}{\phi-i\epsilon}e^{it\phi}\,d\phi\tag1\\\\
&=\lim_{\epsilon\to0}\frac{1}{i\pi}\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{-\frac12(x+i\epsilon)^2+it(x+i\epsilon)}}{x}\,dx\tag2\\\\
&=1+\frac{1}{i\pi}\lim_{\epsilon\to 0^+}\left(\int_{-\infty}^{-\epsilon}\frac{e^{-\frac12x^2+itx}}{x}\,dx+\int_{\epsilon}^{\infty}\frac{e^{-\frac12x^2+itx}}{x}\,dx\right)\tag 3\\\\
&=1+\text{PV}\left(\frac{1}{i\pi}\int_{-\infty}^\infty \frac{e^{-\frac12x^2+itx}}{x}\right)\tag 4
\end{align}$$

NOTES:
In going from $(1)$ to $(2)$, we enforced the substitution $\phi = x+i\epsilon$.
In going from $(2)$ to $(3)$, we used Cauchy's Integral Theorem to deform the contour to the real line.  In doing so, we need to avoid $x=0$ and hence the deformed contour has a semi-circular "indentation" around $x=0$.  This lead to the principal value integral plus a contribution of one-half the residue at $x=0$ of the integrand.  In the limit as $\epsilon \to 0^+$ of the extra contribution is $1$.
In arriving at $(4)$ we use the definition of the Cauchy Principal Value and the corresponding notation.

Next, we need to evaluate the principal value of the integral on the right-hand side of $(4)$.  To do so, we will use Feynman's Trick of differentiating under the integral.  To that end we now proceed.
Let $f(t)=\int_{-\infty}^{-\epsilon}\frac{e^{-\frac12x^2+itx}}{x}\,dx+\int_{\epsilon}^{\infty}\frac{e^{-\frac12x^2+itx}}{x}\,dx$.  Then, differentiating under the integral reveals
$$\begin{align}
f'(t)&=i\int_{-\infty}^\infty e^{-\frac12x^2+itx}\,dx\\\\
&=ie^{-t^2/2}\int_{-\infty}^\infty e^{-\frac12(x-it)^2}\,dx\\\\
&=i\sqrt{2\pi}e^{-t^2/2}\tag 4
\end{align}$$
Integrating $(5)$ and using $f(0)=0$, we find that 
$$\begin{align}
f(t)&=i\sqrt{2\pi}\int_0^t e^{-x^2/2}\,dx\\\\
&=i\sqrt{2\pi}\left(\int_{-\infty}^t e^{-x^2/2}\,dx-\sqrt{\pi/2}\right)\\\\
&=i\sqrt{2\pi}\int_{-\infty}^t e^{-x^2/2}\,dx-i\pi\tag6
\end{align}$$
Substituting $(6)$ into $(4)$ yields
$$\begin{align}
\int_{-\infty}^\infty D(\phi)e^{it\phi}\,d\phi&=\sqrt{\frac{2}{\pi}}\int_{-\infty}^t e^{-x^2/2}\,dx
\end{align}$$
as was to be shown!
