Computing the Fourier transform of a certain function Let $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ be the function defined by $f(x)=(1+|x|^2)^{-1}$.

Problem: How can I find $\hat{f}$ by direct computation?

Remark: This is exercise 8.5 in Rudin's Functional Analysis. Rudin suggests also an alternative solution avoiding direct computation, but I'm specifically interested in the direct solution (so far it seems that my calculations lead nowhere).
 A: You're after something called the Yukawa potential in 3 dimensions.  It'll be something like $g(\xi) = C |\xi|^{-1} e^{-|\xi|}$ for an appropriate $C>0$.  See, for example, Theorem 6.23 of Analysis by Lieb and Loss for some hints on how to compute it.  It's presumably also in one of the Stein books.
A: Write the FT as
$$\hat{f}(k) = \int_0^{\infty} dr \, r^2 \frac{1}{1+r^2} \int_0^{\pi} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi \: e^{i \mathbf{k} \cdot \mathbf{r}}$$
Note that the FT of a radially symmetric function is radially symmetric, so that we may write $ \mathbf{k} \cdot \mathbf{r} = k r \cos{\theta}$ where $k \ge 0$; we then have
$$\begin{align}\hat{f}(k) &= \int_0^{\infty} dr \, r^2 \frac{1}{1+r^2} \int_0^{\pi} d\theta \, e^{i k r \cos{\theta}} \, \int_0^{2 \pi} d\phi\\ &= \frac{4 \pi}{k} \int_0^{\infty} dr \frac{r^2}{1+r^2} \frac{\sin{k r}}{r}\\&=\frac{4 \pi}{k} \left [\int_0^{\infty} dr \frac{\sin{k r}}{r} - \int_0^{\infty} \frac{dr}{1+r^2} \frac{\sin{k r}}{r} \right ]\\ &= \frac{4 \pi}{k} \left [\int_0^{\infty} dr \frac{\sin{k r}}{r} - \frac12 \int_{-\infty}^{\infty} \frac{dr}{1+r^2} \frac{\sin{k r}}{r} \right ]\end{align}$$
The first integral is simply $\pi/2$.  The second integral may be evaluated via Parseval's theorem:
$$\int_{-\infty}^{\infty} dx \, g(x) h^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dv \, \hat{g}(v) \hat{h}^*(v)$$
where $\hat{g}$ and $\hat{h}$ are one-dimensional Fourier transforms. The second integral may then be evaluated as follows:
$$\begin{align}\int_{-\infty}^{\infty} \frac{dr}{1+r^2} \frac{\sin{k r}}{r} &= \frac{1}{2 \pi}\int_{-\infty}^{\infty} dv\, \pi\, e^{-|v|} \pi \mathbf{1}_{[-k,k]} \\ &=\frac{\pi}{2} \int_{-k}^k dv \, e^{-|v|} \\ &= \pi \int_0^k dv \, e^{-v} \\ &= \pi (1-e^{-k}) \end{align}$$
Therefore, putting this back into the previous expression,
$$\hat{f}(k) = 2 \pi^2 \frac{e^{-k}}{k}$$
