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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

theads
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  • The curves $v t \cos \theta = x$ and $v t \sin \theta = y$ intersect at $(t_0, \theta_0) = (\sqrt {x^2 + y^2}/|v|, \arctan(x/v, y/v))$. Then, according to this formula, $$\int_0^\infty \int_{-\pi}^\pi \delta(x - v t \cos \theta) , \delta(y - v t \sin \theta) , \phi(t, \theta) , d\theta dt = \frac {\phi(t_0, \theta_0)} {|v| \sqrt {x^2 + y^2}}.$$ – Maxim May 28 '20 at 21:02
  • I see the formula you referenced but I'm not getting the same result as you. Specifically, I don't get the denominator. Can you elaborate? – theads Jun 08 '20 at 18:54
  • Actually in this case it's simpler to write the denominator as $$\left| \nabla f(t, \theta) \right| \left| \nabla g(t, \theta) \right| \sin \phi = \left| \det \frac {\partial(f(t, \theta), g(t, \theta))} {\partial(t, \theta)} \right| = v^2 |t|.$$ Then substitute $t = t_0$. – Maxim Jun 08 '20 at 20:20
  • Yes I see now it's simply a cross product. Thank you – theads Jun 09 '20 at 19:41

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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:

$$ \frac{2\pi\lambda}{\sqrt{(s+\lambda)^2+q^2v^2}}$$

theads
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