I am having trouble getting from one line to the next from this wiki page. http://en.wikipedia.org/wiki/Screened_Poisson_equation. I am referring to " Green's function in r is therefore given by the inverse Fourier transform" where
$$G(r) = \frac{1}{(2\pi)^3} \int\int\int d^3k \frac{e^{ik*r}}{k^2+\lambda^2}$$
goes to
$$G(r) = \frac{1}{2\pi^2r} \int^{+\infty}_0 dk_r \frac{k_r \sin(k_r r)}{k_r^2+\lambda^2}$$
where does the $\frac{1}{r}$ term come from and what is $k_r$. How did they simplify the tripple integral? Divergence theorem? Stokes? Detailed steps would be much appreciated.
They used spherical coordinates. $k_1 = k \sin{\theta} \cos{\phi}$, $k_2 = k \sin{\theta} \sin{\phi}$, $k_3 = k \cos{\theta}$, and
$$\mathrm{d^3}\vec{k} = k^2 \sin{\theta} \, dk \, d \theta \, d \phi $$
So
$$\begin{align}G(r) &= \frac{1}{(2 \pi)^3} \int_{\mathbb{R^3}} d^3 \vec{k} \frac{e^{i \vec{k} \cdot \vec{r}}}{k^2 + \lambda^2} \\ &= \frac{1}{(2 \pi)^3} \int_0^{\infty} dk \: k^2 \int_0^{\pi} d \theta \sin{\theta} \: \int_0^{2 \pi} d \phi \frac{e^{i k r_{\perp} \sin{\theta} \cos{(\phi - \phi')}} e^{i k z \cos{\theta}} }{k^2 + \lambda^2} \\ &= \frac{1}{(2 \pi)^2} \int_0^{\infty} dk \: \frac{k^2}{k^2 + \lambda^2} \int_0^{\pi} d \theta \sin{\theta} \: e^{i k z \cos{\theta}} J_0(k r_{\perp} \sin{\theta}) \end{align}$$
where $J_0$ is the Bessel function (zeroth order, 1st kind), and $r_{\perp} = \sqrt{x^2+y^2}$. Now, it seems to me that your integral above was derived using $z=0$. In this case, the integral over $\theta$ is
$$\int_0^{\pi} d \theta \sin{\theta} \: J_0(k r_{\perp} \sin{\theta}) = \frac{\sin{k r_{\perp}}}{k r_{\perp}}$$
(Go here and here for details.) Therefore, we get
$$G(r_{\perp}) = \frac{1}{(2 \pi)^2 r_{\perp}} \int_0^{\infty} dk \: \frac{k \sin{(k r_{\perp})}}{k^2 + \lambda^2} $$
which is what you had above. This in turn simplifies to
$$G(r_{\perp}) = \frac{e^{- |\lambda| r_{\perp}}}{8 \pi r_{\perp}}$$