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.
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.