Evaluate integral with complex exponential. Inverse Fourier transform.

Question

I'm trying to follow and understand a bigger problem in the book but at one step they simply write out the answer of this integral and take it as a given but it seems to be tricky to compute this integral. I have $g(\omega)=2a/((\omega^2+a^2)(\omega^2+b^2))$ where $a,b>0$ and I want to evaluate

$$\int\limits_{-\infty}^{\infty}g(\omega)e^{j\omega \tau} \ d\omega.$$

This is supposed to evaluate to

$$\frac{1}{(a^2-b^2)b}\left(ae^{-b|\tau|}-be^{-a|\tau|}\right).$$

Anyone who has a trick up their sleeve? I'm pretty sure it's possible to rewrite $g(\omega)$ in a way that allows us to use some neat trick.

Answer 1

By partial fraction decomposition, we have $$g(\omega)=\frac{2a}{(\omega^2+a^2)(\omega^2+b^2)} = \frac{2a}{b^2 - a^2}(\frac{1}{\omega^2 + a^2} - \frac{1}{\omega^2 + b^2}) = \frac{2a}{b^2 - a^2}(\frac{1}{2ia}(\frac{1}{\omega-ia} - \frac{1}{\omega+ia}) - \frac{1}{2ib}(\frac{1}{\omega-ib} - \frac{1}{\omega+ib}))$$If we define $$\mathcal{F}\{x(t)\} = X(i\omega) = \int_{-\infty}^{+\infty} x(t)e^{-i\omega t}dt$$ The inverse transform is given by $$\mathcal{F}^{-1}\{X(i\omega)\} = x(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} X(i\omega)e^{i\omega t}d\omega$$ It's easy to show that if $x(t) = e^{-at}u(t), \ \ a\gt0$ then $X(i\omega) = \frac{1}{a+ i\omega}$. Using this result, $$\mathcal{F}^{-1}\{\frac{1}{\omega - ia}\} = ie^{-at}u(t) \\ \mathcal{F}^{-1}\{\frac{1}{\omega + ia}\} = -ie^{at}u(-t)$$We should put these facts together.

We are interested in $h(t) = 2\pi\mathcal{F}^{-1}\{g(\omega)\}$. Using linearity property, we have $$h(t) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(\mathcal{F}^{-1}\{\frac{1}{\omega-ia}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ia}\}) - \frac{1}{2ib}(\mathcal{F}^{-1}\{\frac{1}{\omega-ib}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ib}\})) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(ie^{-at}u(t) +ie^{at}u(-t)) - \frac{1}{2ib}(ie^{-bt}u(t) +ie^{bt}u(-t))) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{e^{-a|t|}}{2a} - \frac{e^{-b|t|}}{2b})$$

Also we can check the result using Mathematica. Entering

Assuming[a > 0 && b > 0, 
InverseFourierTransform[(2 a/((w^2 + a^2) (w^2 + b^2))) , w , t , 
FourierParameters -> {1, -1}]]

gives $$\frac{1}{b(b^2 - a^2)}((b e^{a t} - a e^{b t})u(-t) + (b e^{-a t} - a e^{-b t}) u(t))$$this can be simplified to $$\frac{1}{b(b^2 - a^2)}(be^{-a|t|} - ae^{-b|t|}) = \frac{1}{(b^2 - a^2)}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$ which agrees with our previous result since $$h(t) = \frac{2\pi}{b^2 - a^2}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$ Note that there are various ways for calculating $\mathcal{F}^{-1}\{\frac{1}{\omega^2 + a^2}\}$. See for example this and this.