Solve this complex integral

Question

Solve this complex integral

$$\lim_{\varepsilon \rightarrow 0} \int_{-\infty}^{\infty} \frac{d\omega}{2\pi i}\frac{e^{-i\omega x}}{\omega + i\varepsilon}$$

Where $\varepsilon > 0$ and $x$ is real.

I tried to integrate it in a closed complex contour, apply Jordan's Lemma and solve it using Residue Theorem, but I could not apply Jordan's Lemma, it diverged.

Answer 1

For $x>0$, closing the contour in the lower-half plane, invoking Jordan's Lemma, and using the residue theorem, we find that

$$\begin{align} \frac{1}{2\pi i}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{\omega +i\varepsilon}\,d\omega&=-2\pi i \frac{1}{2\pi i}\text{Res}\left(\frac{e^{-i\omega x}}{\omega +i\varepsilon}, \omega=-i\varepsilon \right)\\\\ &=-2\pi i \frac{1}{2\pi i}e^{-\epsilon x}\\\\ &=-e^{-\varepsilon x} \tag 1 \end{align}$$

For $x<0$, the integrand is analytic in and on the integration contour comprised of the real line and a semicircle in the upper-half plane centered at the origin with radius $R\to \infty$. Invoking Jordan's Lemma and applying Cauchy's Integral Theorem, we find that

$$\frac{1}{2\pi i}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{\omega +i\varepsilon}\,d\omega=0 \tag 2$$

Putting $(1)$ and $(2)$ together yields

$$\lim_{\varepsilon\to 0^+}\frac{1}{2\pi i}\int_{-\infty}^\infty \frac{e^{-i\omega x}}{\omega +i\varepsilon}\,d\omega=-u(x)$$

where $u(x)$ is the unit step function.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\lim_{\varepsilon \to 0^{\pm}}\int_{-\infty}^{\infty} {\dd\omega \over 2\pi \ic}{\expo{-\ic\omega x} \over \omega + \ic\varepsilon}} = \mathrm{P.V.}\int_{-\infty}^{\infty}{\dd\omega \over 2\pi \ic} {\expo{-\ic\omega x} \over \omega} + \int_{-\infty}^{\infty}{\dd\omega \over 2\pi \ic} \expo{-\ic\omega x}\bracks{\mp\,\pi\ic\,\delta\pars{\omega}} \\[3mm] = &\ \bracks{\int_{0}^{\infty}{\dd\omega \over 2\pi \ic} \pars{{\expo{-\ic\omega x} \over \omega} + {\expo{\ic\omega x} \over -\omega}}} \mp \half = -\,{1 \over \pi}\int_{0}^{\infty} {\sin\pars{\omega x} \over \omega}\dd\omega \mp \half \\[3mm] = &\ -\,{1 \over \pi}\,{\pi \over 2}\,\mathrm{sgn}\pars{x} \mp \half = \half\bracks{-\mathrm{sgn}\pars{x} \mp 1} = \color{#f00}{\left\lbrace\begin{array}{ccrcl} \ds{{1 \mp 1 \over 2}} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\[2mm] \ds{{-1 \mp 1 \over 2}} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}\right.} \end{align}

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That is $$ \begin{array}{rcccl} \ds{\color{#f00}{\lim_{\varepsilon \to 0^{-}}\int_{-\infty}^{\infty} {\dd\omega \over 2\pi \ic}{\expo{-\ic\omega x} \over \omega + \ic\varepsilon}}} & \ds{=} & \color{#f00}{\braces{\begin{array}{rcrcl} \ds{1} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\[2mm] \ds{0} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}}} & = & \phantom{-}\Theta\pars{-x} \\[5mm] \ds{\color{#f00}{\lim_{\varepsilon \to 0^{+}}\int_{-\infty}^{\infty} {\dd\omega \over 2\pi \ic}{\expo{-\ic\omega x} \over \omega + \ic\varepsilon}}} & \ds{=} & \color{#f00}{\braces{\begin{array}{lcrcl} \ds{0} & \mbox{if} & \ds{x} & \ds{<} & \ds{0} \\[2mm] \ds{-1} & \mbox{if} & \ds{x} & \ds{>} & \ds{0} \end{array}}} & = & -\Theta\pars{x} \end{array} $$

$\ds{\Theta}$ is the Heaviside Step Function.