Fourier transform on $\mathbb{R}^3$ I've been stuck on this one homework problem for nearly a day now, so I'd be really thankful for any pointers. The problem is to find
$$\int\int\int_{\mathbb{R}^3} \frac{1}{\mathbf{x}^2+1}e^{-2\pi i (x_1\zeta_1+x_2\zeta_2+x_3\zeta_3)}dx_1dx_2dx_3$$
where $\mathbf{x}\in \mathbb{R}^3$. I'm not sure how to go about this; I know that the Fourier transform of $1/(1+x^2)$ in $\mathbb{R}$ is $\pi e^{-2\pi |\zeta|}$, but I don't think that helps at all. I've tried changing it to spherical coordinates as well.
Also, if anyone could briefly explain how this relates to solving the equation $u(x)-\Delta u(x)=f(x)$, $x\in \mathbb{R}^3$ and $\Delta$ being the Laplacian operator, I'd be eternally grateful. 
 A: Let's compute the Fourier Transform of the surface measure on a sphere of radius $r$. We will do this by integrating in slices.
The singular measure supported on the sphere is the limit of the volume measure on a thin sphere of thickness $\mathrm{d}r$ divided by $\mathrm{d}r$. When integrating a slice, the angle of intersection of the slice with the sphere must be taken into account. In the diagram below, it is shown that a surface of thickness $\mathrm{d}r$
intersecting a surface of thickness $\mathrm{d}x$ at an angle of $\theta$ has a cross sectional area of $\mathrm{d}x\,\mathrm{d}r\sec(\theta)$.
$\hspace{2cm}$
When integrating along the slice whose intersection with the sphere is a circle of radius $r\cos(\theta)$, the angle of intersection of the slice with the surface of the sphere is $\theta$. Thus, the $\sec(\theta)$ from the angle of the intersection is cancelled by the $\cos(\theta)$ from the radius of the circle. Therefore,
$$
\begin{align}
\int_{rS^2}e^{-2\pi i\,x\cdot\xi}\,\mathrm{d}x
&=\int_{-r}^r2\pi re^{-2\pi i\,t|\xi|}\mathrm{d}t\\
&=\frac{r}{-i|\xi|}\left(e^{-2\pi i\,r|\xi|}-e^{2\pi i\,r|\xi|}\right)\\
&=\frac{2r}{|\xi|}\sin(2\pi r|\xi|)\tag{1}
\end{align}
$$
Computing the Fourier Transform
To compute the Fourier transform of $\dfrac{1}{r^2+1}$, we integrate against $(1)$:
$$
\begin{align}
\int_0^\infty\frac{2r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r
&=\int_{-\infty}^\infty\frac{r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r\\
&=\frac{1}{|\xi|}\int_{-\infty}^\infty\frac{r\sin(r)\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\\
&=\frac{1}{|\xi|}\Im\left(\int_{-\infty}^\infty\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\
&=\frac{1}{|\xi|}\Im\left(\int_\gamma\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\
&=\frac{1}{|\xi|}\Im\left(2\pi i\operatorname{Res}\left(\frac{re^{ir}}{r^2+4\pi^2|\xi|^2},2\pi i|\xi|\right)\right)\\
&=\frac{1}{|\xi|}\Im\left(2\pi i\frac{2\pi i|\xi|e^{-2\pi|\xi|}}{4\pi i|\xi|}\right)\\
&=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{2}
\end{align}
$$
where $\gamma$ is the limit of the path on the real axis from $-M$ to $M$ followed by the semi-circle in the upper half-plane centered at $(0,0)$ from $M$ back to $-M$ as $M\to\infty$.
Therefore,
$$
\int_{\mathbb{R}^3}\frac{1}{|x|^2+1}e^{-2\pi i\,x\cdot\xi}\;\mathrm{d}x=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{3}
$$

Relation to the Laplacian Equation
Taking the Fourier Transform of
$$
u(x)-\Delta u(x)=f(x)\tag{4}
$$
yields
$$
(1+4\pi^2|\xi|^2)\hat{u}(\xi)=\hat{f}(\xi)\tag{5}
$$
which becomes
$$
\hat{u}(\xi)=\dfrac{1}{1+4\pi^2|\xi|^2}\hat{f}(\xi)\tag{6}
$$
and taking the Inverse Fourier Transform yields
$$
\begin{align}
u(x)
&=\left(\dfrac{1}{1+4\pi^2|\xi|^2}\right)^\wedge(x)\ast f(x)\\
&=\frac{1}{4\pi|x|}e^{-|x|}\ast f(x)\tag{7}
\end{align}
$$
The convolution in $(7)$ is a Singular Integral.
A: Let's begin with the second part of the question. Take the Fourier transform on both sides of the equation to obtain
$$
(1+|\xi|^2)\hat u(\xi)=\hat f(\xi)\implies \hat u(\xi)=\frac{1}{|\xi|^2+1}\,\hat f(\xi).
$$
If $K(x)$ is the inverse Fourier transform of $(1+|\xi|^2)^{-1}$, then the solution of the equation is $u=K\ast f$.
The integral you want to compute (modulo some factor of $\pi$) is $K(\zeta)$. We know that the Fourier transform commutes with rotations and hence, that the Fourier transform of a radial function is radial. Then
$$
K(\zeta)=K(\zeta_1,\zeta_2,\zeta_3)=K(0,0,|\zeta|).
$$
This simplifies the computation of the integral (that I have not carried out).
