Solving an Infinite Complex Integral with a Singularity and Oscillatory Behavior

I have derived the expression for the motion of a bead in an infinite elastic medium in frequency space, and now I am attempting to take the inverse Fourier Transform of my expression in order to recover the time domain behavior of the bead. Using $$ h(t) = \frac{1}{2 \pi}\int_{-\infty}^{\infty}h(\omega)e^{i \omega t}\mathrm{d}\omega $$ as the definition of the inverse fourier transform, I obtain some terms of the form $$ \int_{-\infty}^{\infty}\frac{i e^{i \omega t}}{(\omega-\omega_0)g(\omega)}\mathrm{d}\omega $$ where $g(\omega)$ is a complex polynomial of the form $\alpha \omega^2 + i \beta \omega + \gamma$. The constants $\alpha$, $\beta$, $\gamma$, and $\omega_0$ are all real.

Can you help me figure out how to solve this integral?

Attempted Solution

Thanks to the advice of Andy Walls, I have learned that the approach to use is to convert this into a contour integral, and to solve it using the Cauchy Residue Theorem. The contour is an infinite semicircle in the upper half of the complex plane, with the flat side on the real axis, and a small semicicular cut-out around (and infinitely close to) $\omega_0$.

According to the Cauchy Residue Theorem, if C is a simple, closed contour then when a function $f$ is analytic inside and on C except for a finite number of singular points inside C,

$$ \oint_C f(z)\,\mathrm{d}z = 2 \pi i \sum_{k = 1}^n \mathrm{Res}[f(z),z_k]. $$

In my case, I have just 1 singular point (at $\omega_0$... I think?), and my function is a function of $\omega$ rather than $z$: $$ f(\omega) = \frac{i e^{i \omega t}}{(\omega-\omega_0)g(\omega)} $$ Writing this function as a Laurent series, i.e. $$ f(\omega) = \sum_{n=0}^{\infty}a_n(\omega-\omega_0)^n + \frac{b_1}{\omega-\omega_0}+\frac{b_2}{(\omega-\omega_0)^2}+ \cdot \cdot \cdot $$ turns out to be trivial. The residue at $\omega_0$, is the value of coefficient $b_1$, so $$ \mathrm{Res}[f(\omega),\omega_0] = \frac{i e^{i \omega_0 t}}{g(\omega_0)}. $$ Putting this into the definition, I get that $$ \oint_C f(\omega)\,\mathrm{d}\omega = \frac{-2 \pi e^{i \omega_0 t}}{g(\omega_0)} $$

So now I have two follow-up questions:

1. My complex analysis skills are exceptionally rusty -- did I do this contour integral correctly?
2. Given the solution to this contour integral, how do I go about using it to get the solution to my original definite integral?

Solution 1:

First let's find the roots and factor $g(\omega)$ so the whole denominator of the integrand is factored completely. We'll need it factored to find residues anyway.

Completing the square to find the roots:

$$\begin{align*}g(\omega) = \alpha\omega^2 + i\beta\omega + \gamma &= 0\\ \\ \omega^2 + i\dfrac{\beta}{\alpha}\omega + \dfrac{\gamma}{\alpha}&= 0\\ \\ \omega^2 + i\dfrac{\beta}{\alpha}\omega - \left(\dfrac{\beta}{2\alpha}\right)^2 &= - \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}\\ \\ \left(\omega + i\dfrac{\beta}{2\alpha}\right)^2 &= - \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}\\ \\ \omega_{1,2} &= - i\dfrac{\beta}{2\alpha} \pm \sqrt{- \left(\dfrac{\beta}{2\alpha}\right)^2-\dfrac{\gamma}{\alpha}}\\ \\ \omega_{1,2} &= i\left[-\dfrac{\beta}{2\alpha} \pm \dfrac{\sqrt{ \beta^2+4\alpha\gamma}}{{2\alpha}}\right]\\ \\ \end{align*}$$

Thus

$$g(\omega)=\alpha\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)$$

and I'll stop using the $g(\omega)$ notation.

You are interested in integrals of the form

$$P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega$$

With $\alpha$, $t$, $\omega$, and $\omega_0$ real; and $\omega_1$ and $\omega_2$ complex.

I am specifically excluding cases where $\gamma$ is $0$, as that puts another singularity on the real axis (either $\omega_1$ or $\omega_2$ becomes 0) that needs to be handled specially. You can extend what I do later in this answer to handle that case, if you desire.

Consider the complex contour integral

$$\oint_{C}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz$$

This contour integral, the Residue Theorem, Jordan's Lemma, and a properly selected contour can be used to find the Principal Value of your improper integral on the real line.

You need to consider cases for $t >0$, $t<0$, and $t=0$, as they require different contours or different handling altogether.

For $t>0$, select closed contour $C_P$, that is a semi-circular contour in the upper half plane, that includes the real axis as its straight edge with a small semi-circular arc excursion into the upper half plane around $\omega_0$.

From the Residue Theorem

$$\begin{align*}&2\pi i \sum_{\omega_k \in \mathrm{UHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] } = \oint_{C_P}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz\\ \\ &=\lim_{{r \to 0},{R \to \infty}}{\left[\int_{-R}^{\omega_0-r}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\ \\ \int_{\omega_0+r}^{R}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\ \\ \int_{\pi}^{0}{\dfrac{ie^{it\left(\omega_0+re^{i\omega}\right)}ire^{i\omega}}{\alpha\left(\omega_0+re^{i\omega}-\omega_0\right)\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega +\\ \\ \int_{0}^{\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega\right]} \\ \\ &= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\\ \\ &\lim_{r \to 0}{\left[ \dfrac{e^{i\omega_0 t}}{\alpha}\int_{0}^{\pi}{\dfrac{e^{itre^{i\omega}}}{\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega \right]}+\\ \\ &\lim_{R \to \infty}{\left[\int_{0}^{\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega \right]}\\ \\ &= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\dfrac{\pi e^{i\omega_0 t}}{\alpha\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\space + 0\\ \end{align*}$$

So for $t>0$

$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= -\dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} - \dfrac{2\pi}{\alpha}\sum_{\omega_k \in \mathrm{UHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{e^{iz t}}{\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] }\\ \end{align*}$$

And, for example, for $t>0$, if both $\omega_1$ and $\omega_2$ are in the upper half plane, then working out the residues, one gets

$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= -\dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} - \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} - \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\\ \end{align*}$$

\

For $t<0$, select closed contour $C_N$, that is a semi-circular contour in the lower half plane, that includes the real axis as its straight edge with a small semi-circular arc excursion into the lower half plane around $\omega_0$.

From the Residue Theorem

$$\begin{align*}&-2\pi i \sum_{\omega_k \in \mathrm{LHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] } = \oint_{C_N}{\dfrac{ie^{iz t}}{\alpha\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}}dz\\ \\ &=\lim_{{r \to 0},{R \to \infty}}{\left[\int_{-R}^{\omega_0-r}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\ \\ \int_{\omega_0+r}^{R}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega +\\ \\ \int_{-\pi}^{0}{\dfrac{ie^{it\left(\omega_0+re^{i\omega}\right)}ire^{i\omega}}{\alpha\left(\omega_0+re^{i\omega}-\omega_0\right)\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega +\\ \\ \int_{0}^{-\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega\right]} \\ \\ &= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space+\\ \\ &\lim_{r \to 0}{\left[ -\dfrac{e^{i\omega_0 t}}{\alpha}\int_{-\pi}^{0}{\dfrac{e^{itre^{i\omega}}}{\left(\omega_0+re^{i\omega}-\omega_1\right)\left(\omega_0+re^{i\omega}-\omega_2\right)}}d\omega \right]}+\\ \\ &\lim_{R \to \infty}{\left[\int_{0}^{-\pi}{\dfrac{ie^{itRe^{i\omega}}iRe^{i\omega}}{\alpha\left(Re^{i\omega}-\omega_0\right)\left(Re^{i\omega}-\omega_1\right)\left(Re^{i\omega}-\omega_2\right)}}d\omega \right]}\\ \\ &= P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \space-\dfrac{\pi e^{i\omega_0 t}}{\alpha\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\space + 0\\ \end{align*}$$

So for $t<0$

$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= \dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} + \dfrac{2\pi}{\alpha}\sum_{\omega_k \in \mathrm{LHP}} {\underset{z=\omega_k}{\mathrm{Res}}\left[\dfrac{e^{iz t}}{\left(z-\omega_0\right)\left(z-\omega_1\right)\left(z-\omega_2\right)}\right] }\\ \end{align*}$$

And, for example, for $t<0$, if both $\omega_1$ and $\omega_2$ are in the lower half plane, then working out the residues, one gets

$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= \dfrac{\pi}{\alpha}\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)} + \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} + \dfrac{2\pi}{\alpha}\dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\\ \end{align*}$$

Update 2

For $t=0$, we'll need to use the following result, which I won't prove here. But it can be proven using $u$ substitution, the definition of Cauchy Principal Value, and two different contour integrations that have closed contours that are infinitely long rectangles, with one edge along the ral axis with a infintely small semi-circular excursion around the pole on the real axis:

$$ P.V. \int_{-\infty}^\infty{\dfrac{1}{x-z_k}}dx = \pi i \space\mathrm{sgn}\left(\Im\left[z_k\right]\right)$$

We'll attack the $t=0$ case by performing Partial Fraction Expansion of the integrand and then applying the above result to resultant three simpler integrals

$$\begin{align*} & P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega 0}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= P.V. \int_{-\infty}^{\infty}{\dfrac{i}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= P.V. \dfrac{i}{\alpha} \int_{-\infty}^\infty {\left[\dfrac{1}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)\left(\omega-\omega_0\right)} \\ \\ + \dfrac{1}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)\left(\omega-\omega_1\right)} \\ \\ + \dfrac{1}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)\left(\omega-\omega_2\right)}\right]}d\omega\\ \\ &= \pi i\dfrac{i}{\alpha}\left[\dfrac{1}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\mathrm{sgn}\left(\Im\left[\omega_0\right]\right)\\ \\ + \dfrac{1}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)}\mathrm{sgn}\left(\Im\left[\omega_1\right]\right)\\ \\ + \dfrac{1}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\mathrm{sgn}\left(\Im\left[\omega_2\right]\right)\right]\\ \end{align*}$$

\

Collecting all of the above results, we have

$$\begin{align*}&P.V. \int_{-\infty}^{\infty}{\dfrac{ie^{i\omega t}}{\alpha\left(\omega-\omega_0\right)\left(\omega-\omega_1\right)\left(\omega-\omega_2\right)}}d\omega \\ \\ &= 2\pi i\dfrac{i}{\alpha}\left[\dfrac{ e^{i\omega_0 t}}{\left(\omega_0-\omega_1\right)\left(\omega_0-\omega_2\right)}\dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_0]\right)\right] + \\ \\ \dfrac{ e^{i\omega_1 t}}{\left(\omega_1-\omega_0\right)\left(\omega_1-\omega_2\right)} \dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_1]\right)\right]+ \\ \\ \dfrac{ e^{i\omega_2 t}}{\left(\omega_2-\omega_0\right)\left(\omega_2-\omega_1\right)}\dfrac{1}{2}\left[\mathrm{sgn}(t)+\mathrm{sgn}\left(\Im[\omega_2]\right)\right]\right]\\ \end{align*}$$