The response of a simple visual neural cell to a counterphase grating stimulus is given by $L(t) = L_sL_t(t)$ where:
$$L_s = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{A}{2\pi\sigma^2}\exp({\frac{-x^2 - y^2}{2\sigma^2}})\cos(kx - \phi)\cos(Kx\cos\Theta + Ky\sin\Theta - \Phi) dxdy $$
These appear in the book Theoretical Neuroscience: Computational and Mathematical Modeling of Neural Systems by Peter Dayan.
The answer for $L_s$ is given as
$$ L_s = \frac{A}{2}\exp(-\frac{\sigma^2(k^2 + K^2)}{2})(\cos(\phi - \Phi)\exp(\sigma^2kK\cos(\Theta)) + \cos(\phi + \Phi)\exp(-\sigma^2kK\cos(\Theta))) $$
I have been trying to solve $L_s$ for a while now to no success. Any guidance or hints in the right direction would be highly appreciated.
My first idea was to do a polar coordinate transformation which led to the integral:
$$L_s = \frac{A}{4\pi\sigma^2}\int_{0}^{\infty} \int_{0}^{2\pi} r \exp(-\frac{r^2}{2\sigma^2})(\cos(r\cdot\alpha(\theta) - (\phi + \Phi)) + \cos(r\cdot\beta(\theta) - (\phi - \Phi))) drd\theta $$
with $\alpha(\theta) = k\cos\theta + K \cos(\theta - \Theta)$ and $\beta (\theta) = k \cos \theta - K \cos(\theta - \Theta)$
with my next intention to somehow integrate out the $r$ variable. However, I can not see how I can continue from here.