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I would like to calculate the Fourier's inverse transform of the following function

$$F(\omega)=\frac{\omega-ia}{\omega-ib}, \ \ \ \ \ \ (1)$$

where a and b are real and positive.

Hence, I should evaluate the following integral:

$$f(t)=\int_{-\infty}^{+\infty}\frac{\omega-ia}{\omega-ib}e^{i\omega t} d\omega$$.

If a=b, then $f(t)=\delta(t)$.

If $a\neq b$, I can use residue theorem. The function has only one imaginary pole, at $\omega=ib$.

Hence, for $t<0$ I get $f(t)=0$. On the other hand, if $t>0$ I get:

$$f(t)=2\pi \left(a-b\right)e^{-bt}$$

Putting everything together, I have that

$$f(t)=2\pi \left(a-b\right)e^{-bt}\theta(t),$$

where $\theta(t)$ is the Heaviside Function.

However, I read on the books that the Fourier transform of $e^{-\alpha t}\theta(t)$ (with $\alpha$ real and positive) is:

$$\frac{1}{\alpha+i\omega}.$$

But then, according to the latter relation, the Fourier transform of $f(t)=2\pi \left(a-b\right)e^{-bt}$ would read:

$$F(\omega)=\frac{a-b}{b-i\omega}=\frac{ib-ia}{\omega-ib}. \ \ \ \ \ \ (2)$$

The latter expression is different with respect to the relation (1).

Why do I get this difference?

Can you help me?

Thank you very much for your help.

Best Regards.

Marco Gandolfi
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  • Hi Marco. I hope that you're doing well. I've sent you a couple of messages, but am unsure if you've received them. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer (the check mark symbol under the down arrow) as you see fit. ;-) – Mark Viola Feb 28 '21 at 18:57

1 Answers1

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Note that $F(\omega)=\frac{\omega-ia}{\omega-ib}=1-\frac{i(a-b)}{\omega-ib}$. Then,

$$\mathscr{F^{-1}}\{F\}(t)=\delta(t)-\frac{i(a-b)}{2\pi}\int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-ib}\,d\omega$$

From the residue theorem

$$\begin{align} \int_{-\infty}^\infty \frac{e^{i\omega t}}{\omega-ib}\,d\omega=\begin{cases}2\pi i e^{-bt}&,t>0\\\\0&,t<0\end{cases} \end{align}$$

Hence, we find that

$$\mathscr{F^{-1}}\{F\}(t)=\delta(t)+(a-b)e^{-bt}H(t)$$

Alternatively, we know that if $G(\omega)=\omega F(\omega)$, then

$$\mathscr{F^{-1}}\{G\}(t)=-i\frac{dg(t)}{dt}$$

where the derivative is in the sense of distributions. Hence, since the derivative of the Heaviside function is the Dirac Delta distribution, we find that

$$\begin{align} \mathscr{F^{-1}}\{F\}(t)&=\left(-i\frac{d}{dt}-ia\right)\left( ie^{-bt}H(t)\right)\\\\ &=\delta(t)+(a-b)e^{-bt}H(t) \end{align}$$

as expected!

Mark Viola
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  • Dear Mark, many thanks for your help. I think your answer is clear and useful. – Marco Gandolfi Jul 29 '20 at 14:01
  • Pleased to hear. – Mark Viola Jul 29 '20 at 15:17
  • @marcogandolfi Hi Marco. I hope that you're doing well. I've sent you a couple of messages, but am unsure if you've received them. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer (the check mark symbol under the down arrow) as you see fit. ;-) – Mark Viola Feb 28 '21 at 18:57