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How can I calculate the Inverse Fourier transform of

$$f(\omega)=\frac{1}{(i\omega+a)(\omega^2-b^2)},\;a,b\in\mathbb{R}, a>0.$$

I guess that I cannot use the residual theorem, since the function has $2$ real poles.

Thank you very much for your help.

Best Regards!

Marco Gandolfi
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  • You should definitely use the residue theorem, however I think the answer depends on $a$, because if it is positive or not the residue change. For the integration path you choose one which skips the poles by tiny semicircles. – Baol Oct 01 '18 at 13:28
  • Dear Baol, thank you for your reply. In my case a is positive, so the pole in omega=ia has a positive imaginary part. However, I have one doubt about your idea. I have 2 possibilities to skip each real poles: a tiny semicircle around the pole in the plane im(omega)>0 (above the pole) or in the plane im(omega)<0 (below the pole). This choice will influence the final result. Is there a rule to decide which semicircle to choose? Thank you very much for your attention. – Marco Gandolfi Oct 02 '18 at 09:08
  • In truth you do: remember you are integrating $e^{i \xi x}$, depending on the sign of $\xi$ you have only one path availabe in order to skip the poles. So you have the cases $\xi<0, \xi \geq 0$. – Baol Oct 02 '18 at 09:15
  • Dear Baol, thank you for your reply. So, if I understand well, I have to choose the integration path not to include the poles in the integration loop. In the case of $\xi<0$ then I have to choose the circle with im(omega<0) and then f($\xi$)=0. On the other hand if $xi>0$ I should choose the path so that only omega=ia is included in the integration loop, so the result with residue theorem should be $f(\xi)=i*exp(-a\xi)/(b^2-a^2)$. But then performing back the direct Fourier tranform I am not able to get the initial $f(\omega)$, because I obtain $i/[(b^2-a^2)(a+i\omega)]$. Where am I wrong? – Marco Gandolfi Oct 02 '18 at 10:01
  • i think you should shape the "real" path so as to include the poles in the real line as well. – Baol Oct 02 '18 at 10:10
  • Dear Baol, thank you for your reply. This is what I am trying to understand. In the case of $\xi>0$ you should include the real poles, in the case $\xi<0$ you should not to include them. But in my opinion this is an arbitrary choice. Do you know if there is a rule? Otherwise you should pass through the distribution theory, but I am not an expert. – Marco Gandolfi Oct 02 '18 at 10:21
  • Sorry, I'll try to be clearer. In either case you include all the poles in the real line you can, however when you evaluate the residue, ONLY for the real line poles, you take half of their value. So for example if $\xi <0 $ the integral is $2\pi i $ $(1/2$ $Res(b,-b)) +Res(ia))$. The reason of this is that you would like to define a distribution which is half way in between the two path you can draw: the one containing real poles an the one not containing them. – Baol Oct 02 '18 at 10:27
  • for the latter explanation see this. link I hope to have been clear. – Baol Oct 02 '18 at 10:33
  • Dear Baol, thank you very much for your suggestions. I understood and I was able to obtain the result of my integral. Soon I will post the result. Best wishes for everything! – Marco Gandolfi Oct 02 '18 at 15:23

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