For my physics class I have to calculate the following integral:
$$ \frac{\text{d}\sigma}{\text{d}\omega}=|f(\vec{k}, \vec{k'})|^2 $$
where
$$ f(\vec{k}, \vec{k'})=-\frac{m}{2\pi \hbar^2} \int e^{i(\vec{k}, \vec{k'})\cdot \vec{r}}\cdot V(\vec{r})\text{d}^3r $$
and
$$ V(r)=\frac{qQ}{4\pi\epsilon_r \epsilon_0 r}e^{-r/\lambda_D} $$
The result should be
$$ \frac{\text{d}\sigma}{\text{d}\omega}=\left(\frac{2mqQ}{\hbar^24\pi\epsilon_r\epsilon_0}\right)^2 \frac{1}{(2k^2(1-\cos\theta)+\lambda_D^{-2})^2} $$
So far I have rearranged the integrals to
$$ \frac{\text{d}\sigma}{\text{d}\omega}=C \cdot \int e^{i \cdot \Delta \vec{k} \cdot \vec{r}} \cdot e^{-r/\lambda_D}\cdot \frac{1}{r}\text{d}^3r \cdot \int e^{-i \cdot \Delta \vec{k} \cdot \vec{r}} \cdot e^{-r/\lambda_D}\cdot \frac{1}{r}\text{d}^3r $$
with
$$ C=\frac{m^2q^2Q^2}{64\pi^4\hbar^4\epsilon_0^2\epsilon_r^2} $$
Can anyone tell me how I can get to the solution?
Thanks a lot in advance.