Inverse fourier transform 3 dimensions Hoi, I want to show that for $n=3$ that $$\mathcal{F}^{-1}\left(\frac{1}{1+|s|^2}\right) = \frac{1}{4\pi |x|}e^{-|x|} $$
As a hint I've been given: Its the unique solution to the equation $(1-\Delta)u = \delta$ 
(which is easy to show) where $\Delta$ is the Laplace operator. 
From the theory for $n=3$ I have some sort of standard formula $|s|^{-2} = \mathcal{F} (\frac{1}{4\pi |x|})$ (not sure if its any useful at all).
Further, I have simply not the slightest clue how to approach this. Any help is greatly appreciated. Higher dimensional Fourier transforms, and convolutions don't make much sense to me. 
 A: The inverse FT in 3D of a spherically symmetric function is given by
$$f(x) = \frac{1}{(2 \pi)^3} \int_0^{\infty} dk \: k^2 \hat{f}(k) \int_0^{\pi} d\theta \, \sin{\theta} \: \int_0^{2 \pi} d\phi$$
For the function $\hat{f}(k) = 1/(1+k^2)$:
$$\begin{align}f(x) &= \frac{2 \pi}{(2 \pi)^3} \int_0^{\infty} dk \: \frac{k^2}{1+k^2} \: \int_0^{\pi} d\theta \, \sin{\theta} \, e^{-i k x \cos{\theta}}\\ &= \frac{1}{4 \pi^2} \int_0^{\infty} dk \frac{k^2}{1+k^2} \frac{i}{k x} (e^{-i k x} - e^{i k x}) \\ &= \frac{1}{4 \pi^2 x}\int_0^{\infty} dk \frac{k^2}{1+k^2} \frac{\sin{k x}}{k} \\ &= \frac{1}{4 \pi^2 x}\left ( \pi - \int_0^{\infty} \frac{dk}{1+k^2} \frac{\sin{k x}}{k} \right )\\ &= \frac{1}{4 \pi^2 x}\left ( \pi - \frac{1}{2 \pi} \int_{-x}^x dx' \: \pi \cdot \pi \, e^{-|x'|} \right)\\ &= \frac{1}{4 \pi x} [1-(1-e^{-x})] \end{align}$$
Therefore
$$f(x) = \frac{e^{-x}}{4 \pi x}$$
as was to be shown.  Reader should note that I used the fact that
$$\frac{k^2}{1+k^2} = 1-\frac{1}{1+k^2}$$
in the 4th line,
$$\int_{-\infty}^{\infty} dk \frac{\sin{k x}}{k} = \pi$$
when $x>0$, in the 4th line, as well as Parseval's Theorem in the 5th line.
