How do I solve $\frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i(t-t')u}\,du}{-u^2 + \omega^2 -i\epsilon}$ to find Green's function?

Question

I want to find the Green's function defined by the following equation:

$$\left(\frac{d^2}{dt^2} + \omega^2 - i\epsilon \right)G(t,t') = \delta(t-t').$$

For this I performed the Fourier transform of both sides, using identities for the Fourier transform of a second derivative to get the algebraic equation:

$$\left(-u^2 + \omega^2 -i\epsilon \right) \tilde{G}(u,t') = \frac{1}{\sqrt{2\pi}} e^{-it'u}.$$

We can solve this algebraically and perform the inverse Fourier transform:

\begin{align} G(t,t') &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \frac{e^{-it'u}}{-u^2 + \omega^2 -i\epsilon} \: e^{itu}\,du \\ &= \frac{1}{2\pi}\int_{-\infty}^\infty \frac{e^{i(t-t')u}\,du}{-u^2 + \omega^2 -i\epsilon} \end{align}

I can solve this by identifying the poles and using the residue theorem, but it's not clear to me how to do this. Can someone explain? How do I identify the poles and what contour should I choose to get the result?

Answer 1

Let $G(t,t')$ be represented by the integral

$$\begin{align} G(t,t') &= \frac1{2\pi}\int_{-\infty}^\infty \frac{e^{i(t-t')u}}{-u^2 + \omega^2 -i\epsilon}\,du\tag1 \end{align}$$

We proceed to use the residue theorem to evaluate the integral in $(1)$.

---

The function $\displaystyle f(z)=\frac{e^{i(t-t')z}}{-z^2 + \omega^2 -i\epsilon}$ has poles at $\displaystyle z=\pm \sqrt{\omega^2-i\epsilon}$. We need to identify the location of these poles.

We assume that $\epsilon$ is chosen such that $2\text{Re}(\omega)\text{Im}(\omega)-\epsilon<0$, and that $-\pi<\arg(\omega^2-i\epsilon)<0$.

Hence, the pole at $\sqrt{\omega^2-i\epsilon}$ is in the lower-half plane while the pole at $-\sqrt{\omega^2-i\epsilon}$ is in the upper-half plane.

---

Let $C^{+}$ ($C^-$) be the closed contour in the complex plane comprised of $(i)$ the real line segment from $-R$ to $R$ and $(ii)$ the semicircular arc in the upper-half plane (lower-half plane), centered at the origin, from $R$ to $-R$.

Using the residue theorem, we have for $R>\left|\sqrt{\omega^2-i\epsilon}\,\right|$

$$\begin{align} \oint_{C^+} \frac{e^{i(t-t')z}}{-z^2 + \omega^2 -i\epsilon}\,dz&=\int_{-R}^R \,\,\frac{e^{i(t-t')u}}{-u^2 + \omega^2 -i\epsilon}\,du+\int_0^\pi \frac{e^{i(t-t')Re^{i\phi}}}{-R^2e^{i2\phi}+\omega^2-i\epsilon}\,iRe^{i\phi}\,d\phi\tag2\\\\ &=2\pi i\text{Res}\left(\frac{e^{i(t-t')z}}{-z^2 + \omega^2 -i\epsilon},z=-\sqrt{\omega^2-i\epsilon}\right)\\\\ &=2\pi\left( \frac i2 \,\frac{e^{-i\sqrt{\omega^2-i\epsilon}\,(t-t')}}{\sqrt{\omega^2-i\epsilon}}\right) \end{align}$$

As $R\to \infty$, the second integral on the right-hand side of $(2)$ vanishes for $t>t'$ and we find

$$G(t,t')=\frac i2 \,\frac{e^{-i\sqrt{\omega^2-i\epsilon}\,(t-t')}}{\sqrt{\omega^2-i\epsilon}} \tag3$$

---

Using the residue theorem, we have for $R>\left|\sqrt{\omega^2-i\epsilon}\,\right|$

$$\begin{align} \oint_{C^-} \frac{e^{i(t-t')z}}{-z^2 + \omega^2 -i\epsilon}\,dz&=\int_{-R}^R \,\,\frac{e^{i(t-t')u}}{-u^2 + \omega^2 -i\epsilon}\,du+\int_{2\pi}^\pi \frac{e^{i(t-t')Re^{i\phi}}}{-R^2e^{i2\phi}+\omega^2-i\epsilon}\,iRe^{i\phi}\,d\phi\tag2\\\\ &=-2\pi i\text{Res}\left(\frac{e^{i(t-t')z}}{-z^2 + \omega^2 -i\epsilon},z=+\sqrt{\omega^2-i\epsilon}\right)\\\\ &=2\pi\left( \frac i2 \,\frac{e^{i\sqrt{\omega^2-i\epsilon}\,(t-t')}}{\sqrt{\omega^2-i\epsilon}}\right) \end{align}$$

As $R\to \infty$, the second integral on the right-hand side of $(2)$ vanishes for $t<t'$ and we find

$$G(t,t')=\frac i2 \,\frac{e^{i\sqrt{\omega^2-i\epsilon}\,(t-t')}}{\sqrt{\omega^2-i\epsilon}} \tag4$$

---

Putting $(3)$ and $(4)$ together yields

$$G(t,t')=\frac i2 \,\frac{e^{-i\sqrt{\omega^2-i\epsilon}\,|t-t'|}}{\sqrt{\omega^2-i\epsilon}} $$

Taking the limit as $\epsilon\to 0$, we obtain

$$g(t,t')=\frac i2 \frac{e^{-i\omega |t-t'|}}{\omega}$$

where $g(t,t')$ is the solution to the ODE

$$\frac{d^2g(t,t')}{dt^2}+\omega^2 g(t,t')=\delta(t-t')$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{G}\pars{t,t'} & \equiv {1 \over 2\pi}\int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over -u^{2} + \omega^{2} - \ic\epsilon}\,\dd u = -\,{1 \over 2\pi}\int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u^{2} - \omega^{2} + \ic\epsilon}\,\dd u \\[5mm] & = -\,{1 \over 2\pi}\,\mrm{P.V.} \int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u^{2} - \omega^{2}} \,\dd u - {1 \over 2\pi} \int_{-\infty}^{\infty}\expo{\ic\pars{t - t'}u} \bracks{-\ic\pi\,\delta\pars{u^{2} - \omega^{2}}}\,\dd u \\[1cm] & = -\,{1 \over 4\pi\omega}\,\mrm{P.V.}\bracks{ \int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u - \omega}\,\dd u - \int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u + \omega}\,\dd u} \\[2mm] & + {\ic \over 2}\int_{-\infty}^{\infty}\expo{\ic\pars{t - t'}u} \bracks{{\delta\pars{u - \omega} \over 2\verts{\omega}} + {\delta\pars{u + \omega} \over 2\verts{\omega}}}\,\dd u \\[1cm] & = -\,{1 \over 4\pi\omega}\,\mrm{P.V.}\bracks{ \expo{\ic\omega\pars{t - t'}}\int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u}\,\dd u - \expo{-\ic\omega\pars{t - t'}}\int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u}\,\dd u} \\[2mm] & + {\ic \over 4\verts{\omega}}\bracks{\expo{\ic\omega\pars{t - t'}} + \expo{-\ic\omega\pars{t - t'}}} \\[1cm] & = -\,{\ic \over 2\pi\omega}\,\sin\pars{\omega\bracks{t - t'}}\, \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic\pars{t - t'}u} \over u}\,\dd u + {\ic \over 2\verts{\omega}}\,\cos\pars{\omega\bracks{t - t'}} \\[1cm] & = -\,{\ic \over 2\pi\omega}\,\sin\pars{\omega\bracks{t - t'}} \int_{0}^{\infty}{\expo{\ic\pars{t - t'}u} - \expo{-\ic\pars{t - t'}u} \over u}\,\dd u \\[2mm] & + {\ic \over 2\verts{\omega}}\,\cos\pars{\omega\bracks{t - t'}} \\ & = {\sin\pars{\omega\bracks{t - t'}} \over \pi\omega}\ \overbrace{\int_{0}^{\infty}{\sin\pars{\bracks{t - t'}u} \over u}\,\dd u} ^{\ds{\mrm{sgn}\pars{t - t'}\,{\pi \over 2}}}\ +\ {\ic \over 2\verts{\omega}}\,\cos\pars{\omega\bracks{t - t'}} \\[5mm] & = {\sin\pars{\omega\verts{t - t'}} \over 2\omega} + {\ic \over 2\verts{\omega}}\,\cos\pars{\omega\bracks{t - t'}} \\[5mm] & = {\sin\pars{\verts{\omega}\verts{t - t'}} \over 2\verts{\omega}} + {\ic \over 2\verts{\omega}}\,\cos\pars{\verts{\omega}\verts{t - t'}} \\[5mm] & = \bbx{\ic\,{\expo{-\ic\verts{\omega}\verts{t - t'}} \over 2\verts{\omega}}}\qquad \mbox{as}\ \epsilon\ \to\ 0^{+} \end{align}