# 2 dimensional Fourier transform integral

I'm trying to calculate the 2D fourier transform of this function:

$$\frac{1}{(x^2+y^2+z^2)^{3/2}}$$

I only want to do the fourier transform for x and y (and leave z as it is).

So far, I've tried using the method of residues (assuming $k_x$ > 0 and closing the contour on the top), but I'm left with $$\frac{e^{ik_x x}}{[(x - ic)(x+ic)]^{3/2}}$$

(where c depends only on y and z). And because of the fractional power on the bottom I'm not sure how to get the residue. I've also tried using polar coordinates but that didn't get me anywhere. Does anyone know how I should proceed?

Also, I know the answer should look like $e^{- \sqrt{k_x^2 + k_y^2}z}$.

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 Instead, try looking at the 1D Fourier transform of $1/(x^2+\alpha^2)$. Now imagine using that for $\alpha^2=y^2+z^2$ inside of the integrand of the 2D transform... – anon Aug 3 '11 at 20:28 I think I understand that part. What I'm confused about is the 3/2 power. How do I go from the transform of 1/($x^2+y^2$) to the transform of 1/$(x^2+y^2)^{3/2}$? – user14192 Aug 4 '11 at 2:05

First, change variables $x = r \cos \varphi$ and $y= r \sin \varphi$. Then

$$\frac{1}{(2 \pi)^2} \int \int r dr d \varphi e^{i k_x r \cos \varphi + i k_y r \sin \varphi} (r^2+z^2)^{-3/2}$$

Integrate with respect to $\varphi$ first. Reducing $k_x \cos \varphi + k_y \sin \varphi = \sqrt{k_x^2+k_y^2} \cos ( \varphi + \varphi_0)$. Then integration yields $\frac{1}{2\pi} \frac{r}{(r^2+z^2)^{3/2}} J_0(r \sqrt{k_x^2+k_y^2})$.

Integration with respect to $r$ can now be carried out using convolution technique with respect to Mellin transform, giving

$$\frac{1}{2 \pi \vert z \vert} e^{- \vert z \vert \sqrt{k_x^2+k_y^2}}$$

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If you are acquainted with the notion of fundamental solution for PDE, it is possible to obtain the answer without integration. Namely, let $z=(z_1,z_2,z_3)\$ and the function $$\Gamma(x,y,z_1,z_2,z_3)=\frac1{3\sigma_5(x^2+y^2+|z|^2)^{3/2}}\ \$$ be a fundamental solution for 5-dimensional Laplace equation: $$\Delta \Gamma(x,y,z_1,z_2,z_3)= \delta(x,y,z_1,z_2,z_3).\ \$$ Here $\sigma_5={8 \pi ^2}/{3}$ is the volume of the unit sphere in $\mathbb R^5$. Note that the original function $$f(x,y,|z|)=(x^2+y^2+|z|^2)^{-3/2}=3\sigma_5 \Gamma(x,y,z_1,z_2,z_3).$$ Now Fourier transform in $(x,y)$ of the equation for $\Gamma$ is (if we norm F.t. to get $F(\delta(x,y))=1$) $$\Delta \tilde \Gamma(z_1,z_2,z_3)-k^2\tilde\Gamma(z_1,z_2,z_3)= \delta(z_1,z_2,z_3),$$ where $k^2=k_x^2+k_y^2\$. So the function $\tilde\Gamma$ is a fundamental solution to the 3-dimensional equation $\Delta u-k^2u=0\$. The f.s. for this equation are known, they are $\frac{e^{k|z|}}{4\pi |z|}$ and $\frac{e^{-k|z|}}{4\pi |z|}$. To obtain them it is enough to change sign of $k^2$ in the fundamental solutions of the Helmholtz equation $\Delta u+k^2u=0\$. The Fourier transform of $f$ decreases at infinity, so $\tilde \Gamma(|z|)=\frac{e^{-k|z|}}{4\pi |z|}$. It leads to the result written above by Sasha.

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 Thanks for your help. – user14192 Aug 5 '11 at 21:20