A Cauchy principal value integral, using contour integration and Plemelj.

I came across the following integral $$\lim_{\epsilon\to0^+}\int_\mathbb{R}\frac{e^{-ax^2+ibx}}{x+i\epsilon}dx$$ with $$a,b>0$$. Using Plemelj's formula led me to evaluating $$P.V\int_\mathbb{R}e^{ibx}\frac{e^{-ax^2}}{x}dx$$ I thought then of using a closed contour in the complex upper half plane to evaluate this integral. I would be thrilled if $$e^{-az^2}/z$$ would vanish on the infinite semicircle $$\theta\in[0,\pi]$$, since then I could apply Jordan's lemma on this function and retrieve my improper integral with just the residue around 0... After plotting to get a feel, it seems that it does indeed vanish for an "infinite" half circle... However it is just a plot, hardly prove anything... I would like pointer on how to give a definite proof of that, since I don't really want to waste another 3 pages...

An additional thing: since the integral does "look like" a F.T, is there an easier way to reach a final result than the contour integral way I chose ?...if it is indeed as simple as looking for the FT of $$e^{-ax^2}/x$$ which as it seems equals $$i\pi \text{Erf}\frac{b}{2\sqrt a}$$, then how could I retrieve it using the complex tools ?

Thanks

$$\begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \lim_{\epsilon \to 0^{+}}\int_{\mathbb R} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x + {\rm i}\epsilon}\,{\rm d}x \color{#000000}{\ =\ } {\rm i}\pi \left[{\rm sgn}\left(b\right){\rm erf}\left(b \over 2\sqrt{a\,}\right) - 1\right] \quad} \\ \\ \hline \end{array}$$
• Very helpful post. Just one question (sorry for my ignorance) why $\mathcal{P} \int\limits_{-\infty}^{+\infty} \frac{e^{-ax^2+ibx}}{x} dx = i \int\limits_{-\infty}^{+\infty} e^{-ax^2} \frac{\sin(bx)}{x}$ ? Oct 9 '21 at 13:27