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


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.

  • $\begingroup$ Contrary to popular belief, there are no tricks in mathematics. There are rules, e.g. those for addition of fractions, people learn them in primary school, or never. One of them would be $$\frac1{(\omega^2+a^2)(\omega^2+b^2)}=\frac1{a^2-b^2}\left(\frac1{\omega^2+b^2}-\frac1{\omega^2+a^2}\right).$$ $\endgroup$
    – user436658
    Dec 29 '20 at 21:12

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.

  • $\begingroup$ Wonderful, thanks! $\endgroup$
    – Parseval
    Dec 30 '20 at 11:20
  • $\begingroup$ @Parseval You're welcome. $\endgroup$
    – S.H.W
    Dec 30 '20 at 11:30

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