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I'm trying to calculate the inverse of the following 3D Fourier transform. $$ \widetilde{f}= \frac{1}{(k^6-\alpha*k^2-\alpha*k_3^2)} $$ where $k = (k_1^2+k_2^2+k_3^2)^{1/2}$

the fourier transform is defined as follows: $$ \widetilde{f} = F(f) = \int_{R^3} f(\vec{x})e^{-i \vec{k}\vec{x}} \mathrm{d}^3\vec{x} $$ and the inverse is: $$ f =F^{-1}(\widetilde{f})=\frac{1}{8 \pi^3}\int_{R^3} \widetilde{f}(\vec{k})e^{-i \vec{k}\vec{x}} \mathrm{d}^3\vec{k} $$ where $\vec{k} =k_1 \vec{e_1}+k_2 \vec{e_2}+k_3 \vec{e_3}$ and $\vec{x} =x_1 \vec{e_1}+x_2 \vec{e_2}+x_3 \vec{e_3}$.

If any one could help me, i would be very greatful.

Thank you !

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Use spherical coordinates: $k_1 = k \sin{\theta} \cos{\phi}$, $k_2 = k \sin{\theta} \sin{\phi}$, $k_3 = k \cos{\theta}$, and

$$\mathrm{d^3}\vec{k} = k^2 \sin{\theta} \, dk \, d \theta \, d \phi $$

Also use the fact that

$$\int_0^{2 \pi} e^{i y \cos{(\phi - \phi')}} = 2 \pi J_0(y)$$

where $J_0$ is a Bessel function (zeroth order, 1st kind), to obtain

$$f(\vec{x}) = \frac{1}{(4 \pi)^2} \int_0^{\infty} dk \: \int_0^{\pi} d \theta \: \sin{\theta} \frac{J_0(k r_{\perp} \sin{\theta}) e^{-i k z \cos{\theta}}}{k^4 - \alpha \sin^2{\theta}}$$

where $r_{\perp} = \sqrt{x^2+y^2}$. That's about all I can do for now.

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Thank you for your help. Actually i have tried the following steps: if we take the vector in the $e_3$ direction ($\vec{x}= r \vec{e_3}$) and by using the following variables change $y=cos(\theta)$, we can find $$f =\frac{1}{4\pi^2}\int_0^{\infty} dk \int_{-1}^1 \frac{cos(r k y)}{k^4-\alpha (1-y^2)} dy$$. I think this final result is nearly the same you gave me. So we are at the same point. By the way thank you very much for your time. – user60947 Feb 8 '13 at 22:41
@user60947: you're welcome. If you found my solution useful, please think about accepting it and/or upvoting it so that the problem will at least be marked as answered. And, yes, the on-axis case may be doable. – Ron Gordon Feb 8 '13 at 22:51
i found your solution very intersting but unfortunately the problem still unsolved. – user60947 Feb 9 '13 at 10:49
@user60947: Look, try it some other way. But even in the simplest case, the on-axis one, the integral is a bear. Doing the angular one first, it is a convolution of a sinc and a Lorentzian, which is awful and still needs to be integrated over $k$. You can do the integral over $k$ first and get something like $(1-y^2)^{-3/4} \exp{[-a y \sqrt{1-y^2}]}$ where $a$ is complex, but the resulting integral over $y$ may also prove to be a bit much - I had no luck after working at it a while, and I'm pretty good at breaking these down. Doesn't mean it's impossible, but I doubt it. – Ron Gordon Feb 9 '13 at 14:28

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