I have a Fourier-Laplace Transform over space and time that I need to compute. But before this, I'd like to average over angle $\theta$. I think averaging before the FLT will be easier than after. The integrals are: $$\hat g(\vec \omega,s) = \int \int_0^\infty \int_{-\pi}^\pi \delta(x-vt\cos\theta)\delta(y-vt\sin\theta)\lambda e^{-\lambda t} e^{i\vec \omega\cdot\vec r-st}d\theta dt d^2r$$.
So my question is how to do the $\theta$ integral? $\theta$ is defined to be uniformly distributed between $-\pi$ and $\pi$. I've tried doing the spatial and time integrals first but this results in a $\theta$ integral: $$\int_{-\pi}^\pi\frac{d\theta}{s+\lambda-i\omega_xv\cos\theta-i\omega_yv\sin\theta}$$ which is not 'easy'. Is there a way to do the integral: $$\int_{-\pi}^\pi\delta(x-vt\cos\theta)\delta(y-vt\sin\theta) d\theta$$ Thanks
I see now that the way to do this was in vector form:
$$\int\int\int\delta(\vec{r}-\vec{v}t)e^{-i\vec{q}\cdot\vec{r}}e^{-t(s+\lambda)}d^2rdtd\theta = 2\int_0^\pi\frac{d\theta}{(s+\lambda)-iqv\cos\theta} $$
Where the dot product gave a $\cos\theta$. The final answer being:
$$ \frac{2\pi\lambda}{\sqrt{(s+\lambda)^2+q^2v^2}}$$