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... enter image description here 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 ?



1 Answer 1


\begin{align} \lim_{\epsilon \to 0^{+}}\int_{\mathbb R} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x + {\rm i}\epsilon}\,{\rm d}x &= \int_{-\infty}^{\infty} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x + {\rm i}0^{+}}\,{\rm d}x = \int_{-\infty}^{\infty} {\rm e}^{-ax^{2}\ +\ {\rm i}bx} \left[{\cal P}{1 \over x} - {\rm i}\pi\delta\left(x\right)\right]\,{\rm d}x \\[3mm]&= {\cal P}\int_{-\infty}^{\infty} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x}\,{\rm d}x - {\rm i}\pi \\ -----------&--------------------------- \end{align}

\begin{align} &{\cal P}\int_{-\infty}^{\infty} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x}\,{\rm d}x = {\rm i}\int_{-\infty}^{\infty}{\rm e}^{-ax^{2}}\,{\sin\left(bx\right) \over x}\,{\rm d}x = {\rm i\,sgn}\left(b\right)\int_{-\infty}^{\infty}{\rm e}^{-ax^{2}/b^{2}}\,{\sin\left(bx\right) \over x}\,{\rm d}x \\[3mm]&= {\rm i\,sgn}\left(b\right)\int_{-\infty}^{\infty}{\rm e}^{-ax^{2}/b^{2}}\, {1 \over 2}\int_{-1}^{1}{\rm e}^{{\rm i}kx}\,{\rm d}k\,\,{\rm d}x = {1 \over 2}{\rm i\,sgn}\left(b\right)\int_{-1}^{1}{\rm d}k\int_{-\infty}^{\infty}{\rm e}^{-ax^{2}/b^{2}\ +\ {\rm i}kx}\,{\rm d}x \\[3mm]&= {1 \over 2}{\rm i\,sgn}\left(b\right)\int_{-1}^{1}{\rm d}k\int_{-\infty}^{\infty} \exp\left(-\,{b^{2} \over 4a}\,k^{2} - {a \over b^{2}}\,\left[x - {\rm i}\,{b^{2}k \over 2a}\right]^{2}\right) \,{\rm d}x \\[3mm]&= {1 \over 2}{\rm i\,sgn}\left(b\right)\int_{-1}^{1}{\rm d}k\, {\rm e}^{-b^{2}k^{2}/4a}\ \overbrace{\int_{-\infty}^{\infty}{\rm e}^{-ax^{2}/b^{2}}\,{\rm d}x} ^{\left\vert b\right\vert\,\sqrt{\vphantom{\Large A}\pi/a}} = {1 \over 2}\,{\rm i}b\,\sqrt{{\pi \over a}\,}2\int_{0}^{1}{\rm e}^{-b^{2}k^{2}/4a} \,{\rm d}k \\[3mm]&= {\rm i}b\,\sqrt{{\pi \over a}\,}\,{2\sqrt{a\,} \over \left\vert b\right\vert} \,{\sqrt{\pi\,} \over 2} \left(% {2 \over \sqrt{\pi\,}}\int_{0}^{b/2\sqrt{a\,}}{\rm e}^{-k^{2}}\,{\rm d}k \right) = {\rm i}\pi\,{\rm sgn}\left(b\right){\rm erf}\left(b \over 2\sqrt{a\,}\right) \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \lim_{\epsilon \to 0^{+}}\int_{\mathbb R} {{\rm e}^{-ax^{2}\ +\ {\rm i}bx} \over x + {\rm i}\epsilon}\,{\rm d}x \color{#000000}{\ =\ } {\rm i}\pi \left[{\rm sgn}\left(b\right){\rm erf}\left(b \over 2\sqrt{a\,}\right) - 1\right] \quad} \\ \\ \hline \end{array} $$

  • $\begingroup$ Very helpful post. Just one question (sorry for my ignorance) why $\mathcal{P} \int\limits_{-\infty}^{+\infty} \frac{e^{-ax^2+ibx}}{x} dx = i \int\limits_{-\infty}^{+\infty} e^{-ax^2} \frac{\sin(bx)}{x}$ ? $\endgroup$
    – MariNala
    Oct 9, 2021 at 13:27

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