How can I calculate Fourier transform of this 3D function? Calculate Fourier transform of function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ defined by this formula
$$f(\mathbf{x})=\frac{1}{1+|\mathbf{x}|^2}.$$
 A: From Bracewell, the n-dimensional Fourier Transform
$$F(s_1, \dots, s_n) = \int_{-\infty}^{\infty} \dots \int_{-\infty}^{\infty} f(x_1,\dots, x_n) e^{-2\pi i (x_1 s_1 + \dots + x_n s_n)} \space dx_1 \dots dx_n $$
when there is n-dimensional symmetry, can be manipulated into
$$F(q) = \dfrac{2\pi}{q^{\frac{1}{2}n-1}} \int_{0}^{\infty} f(r) J_{\frac{1}{2}n-1}(2\pi qr)\space r^{\frac{1}{2}n} \space dr$$
Where $r$ and $q$ are the radii from the origin in the original domain and transform domain respectively.
For the case of $n=3$ note that
$$J_{\frac{1}{2}}(x) = \sqrt{\dfrac{2}{\pi x}}\sin(x)$$
So you need to compute
$$\begin{align*}F(q) &= \dfrac{2\pi}{\sqrt{q}} \int_{0}^{\infty} \dfrac{1}{1+r^2} J_{\frac{1}{2}}(2\pi qr)\space r^{\frac{3}{2}} \space dr\\
\\
&=\dfrac{2\pi}{\sqrt{q}} \int_{0}^{\infty} \dfrac{1}{1+r^2} \sqrt{\dfrac{2}{\pi (2\pi qr)}}\sin(2\pi qr)\space r^{\frac{3}{2}} \space dr\\
\\
&=\dfrac{2}{q} \int_{0}^{\infty} \dfrac{r}{1+r^2} \sin(2\pi qr) \space dr\\
\end{align*}$$ 
For which, Wolfram Alpha returns
$$ F(q) = \dfrac{\pi}{q}e^{-2\pi q}$$
