While studying the book *Vorticity and Incompressible Flow* by A. Majda and A. Bertozzi I came across the following Biot-Savart theorem in 3D:
The solution to:
$$div \ v=0$$
$$curl(v)=\omega$$
is
$$v=\int K(x-y) \omega(y,t) dy$$ where $$K(x)\omega=\frac{1}{4 \pi} \frac{x \times \omega}{|x|^3}$$


I tried to prove it using Fourier transform but I came across the following function for which I couldn't find the Inverse Fourier Transform:
$$\frac{h\_1}{h\_1^2+h\_2^2+h\_3^2}$$
To my knowledge, there is no way to find the transform as this function is not in $L^2$. After doing some research I found some hints that solving this set of equations by combining distributions and fourier transform is possible. 
The problem is that the math behind it is still pretty complicated and the calculations for this must've been done somewhere. So the question - does anyone know where can I find how this is calculated or alternatively - how to calculate the inverse transform?



Proof of Biot-Savart theorem in 3D using distributions and fourier transform
no distributions are needed to invert this Fourier transform:


$$(2\pi)^{-3}\int\_{-\infty}^\infty dk\_x\int\_{-\infty}^\infty dk\_y\int\_{-\infty}^\infty dk\_z\, e^{-i\vec{k}\cdot\vec{r}} \,\frac{\vec{k}}{|\vec{k}|^2}\\
=
(2\pi)^{-3}i\frac{\partial}{\partial\vec{r}}\int\_{-\infty}^\infty dk\_x\int\_{-\infty}^\infty dk\_y\int\_{-\infty}^\infty dk\_z\, e^{-i\vec{k}\cdot\vec{r}} \,|\vec{k}|^{-2}\\
=(2\pi)^{-3} i\frac{\partial}{\partial\vec{r}} \int\_0^\infty k^2 dk\int\_0^{2\pi}d\phi \int\_0^\pi \sin\theta d\theta\, e^{-ikr\cos\theta}k^{-2}\\
=2(2\pi)^{-2} i\frac{\partial}{\partial\vec{r}}\,|\vec{r}|^{-1}\int\_0^\infty dq\, \frac{\sin q}{q}\quad\quad({\rm with}\;q=kr)\\
=\frac{i}{4\pi} \frac{\partial}{\partial\vec{r}}\,|\vec{r}|^{-1}
=-\frac{i}{4\pi}\frac{\vec{r}}{|\vec{r}|^3}
$$
so the inverse Fourier transform of $4\pi i\, \vec{k}\,|\vec{k}|^{-2}$ equals $\vec{r}\,|\vec{r}|^{-3}$.




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UPDATE AUGUST 2015
------------------


The OP asks how the similar calculation would proceed in two spatial dimensions. Here it goes (with a formal interchange of integration and differentiation):


$$(2\pi)^{-2}\int\_{-\infty}^\infty dk\_x\int\_{-\infty}^\infty dk\_y\, e^{-i\vec{k}\cdot\vec{r}} \,\frac{\vec{k}}{|\vec{k}|^2}\\
=
(2\pi)^{-2}i\frac{\partial}{\partial\vec{r}}\int\_{-\infty}^\infty dk\_x\int\_{-\infty}^\infty dk\_y\, e^{-i\vec{k}\cdot\vec{r}} \,|\vec{k}|^{-2}\\
=(2\pi)^{-2} i\frac{\partial}{\partial\vec{r}} \int\_0^\infty k dk\int\_0^{2\pi}d\phi \, e^{-ikr\cos\phi}k^{-2}\\
=(2\pi)^{-1} i\frac{\vec{r}}{|\vec{r}|}\int\_0^\infty k^{-1}dk\, \frac{d}{dr}J\_0(kr)\\
=-(2\pi)^{-1} i\frac{\vec{r}}{|\vec{r}|}\int\_0^\infty dk\, J\_1(kr)\\
=-\frac{i}{2\pi}\frac{\vec{r}}{|\vec{r}|^2}
$$


