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I want to calculate the following inverse Fourier transform:

$f (t) = \mathcal{F}^{-1} \lbrace \frac{ e^{ - \vert g( \omega ) \vert } }{ \omega g( \omega ) } \rbrace (t) \, ,$

where

$g ( \omega ) = \sqrt{ a + ( a + b ) \left( \frac{ c }{ \omega } \right)^{2} } \, ,$

with $a \in \mathbb{R} > 0$, $b \in \mathbb{R} < 0$, $a + b > 0$ and $c \in \mathbb{R}$.

I am using the convention whereby

$f(t) = \frac{ 1 }{ 2 \pi } \int_{- \infty}^{+ \infty} \widetilde{f} ( \omega ) \, e^{ i \omega t } \, \textrm{d}\omega \, ,$

where $\widetilde{f} ( \omega )$ is the Fourier transform of $f ( t )$, given by

$\widetilde{f} ( \omega ) = \int_{- \infty}^{+ \infty} f ( t ) \, e^{ - i \omega t } \, \textrm{d} t \, .$

Due to the fact that $\lim_{\omega \to \pm \infty} g ( \omega ) = \sqrt{a}$ I can see that the integral should converge since this implies that $\lim_{\omega \to \pm \infty} \frac{ e^{ - \vert g( \omega ) \vert } }{ \omega g( \omega ) } = 0$.

However, as someone with limited experience working with Fourier transforms, I am unsure as to how to proceed - I can't see a way to do this integral directly, and am not sure of any tricks I could use to get this to work out nicely. I ask the question here in the hope that someone will have seen something similar before and will know whether or not this inverse Fourier transform can be calculated analytically.

I am also aware of the fact that $\mathcal{F} \lbrace \widetilde{f} ( \omega ) \rbrace (t) = f ( -t )$ (via the Fourier inversion theorem), so that the problem could be tackled by taking a Fourier transform ($\mathcal{F}$) as opposed to an inverse Fourier transform ($\mathcal{F}^{-1}$). Again, I can't see if or how this would help.

Any pointers or suggestions would be very welcome, and much appreciated. Thanks for taking the time to read my question.

Disclaimer: This isn't a homework problem or anything, so please go easy on me for my lack of progress on finding a solution! I have tried manipulating the integrand, as well as looking up some tables of integrals, but have not had any success via either route.

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  • $\begingroup$ I don't see how to calculate the result directly (though this does not mean that it is impossible), so my question is: In what context do you need this Fourier transform? Furthermore, that the integrand vanishes for $|x|\to\infty $ is neither necessary nor sufficient for existence of the integral. Worse, your argument shows behaves like $1/\omega $ as $\omega \to \infty$, which is not integrable. $\endgroup$ – PhoemueX Oct 28 '15 at 17:45
  • $\begingroup$ That's a good point about the behaviour of the integrand. As I continue to work on this, I get the impression that it is not integrable. As for the context, this is related to a problem in the theory of elasticity. I have cast it in the more familiar $t \leftrightarrow \omega$ language to make the question easier for people to understand. $\endgroup$ – white_ink Oct 28 '15 at 17:57
  • $\begingroup$ But at least, the integrand is $L^2$, so the Fourier transform will exist in the $L^2$ sense. $\endgroup$ – PhoemueX Oct 28 '15 at 18:08
  • $\begingroup$ That's what gave me hope, but unless someone amazes me here I don't see there being a neat closed form solution. $\endgroup$ – white_ink Oct 28 '15 at 18:19
  • $\begingroup$ you might notice that your integrand has two branch points. You may try to use contour integration, with a contour avoiding the origin and bending aroundone of the cuts in the upper or lower hp. Which half plane you choose depens on the sign of $t$. I don't think this will give you a closed form, but it will make various approximations more obvious $\endgroup$ – tired Oct 28 '15 at 19:43

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