Midway through computing a result I come across the following integral:
$\displaystyle I = \int_{|\vec{x}| \in [a,b]} \frac{e^{-2\pi i (\vec{x} \cdot \vec{v}_1)}}{(\vec{x} \cdot \vec{v}_2)(\vec{x} \cdot \vec{v}_3)} d\vec{x}$
where $\vec{x} \in \mathbb{R}^3$ and $\vec{v}_i \in \mathbb{R}^3$. I attempt to solve this by moving to spherical coordinates, $\vec{x} = (r, \theta, \phi)$ aligned along $\vec{v}_1$, in which case the dot products take the form:
$\vec{x}_1\cdot \vec{x}_2 = r_1 r_2 (\sin \theta_1 \sin \theta_2 \cos(\phi_1 - \phi_2) + \cos\theta_1 \cos\theta_2)$ and $\vec{x}\cdot \vec{v}_1 = r|\vec{v}_1| \sin \phi$ thus,
$I = \int_{\theta \in [0, 2\pi]} \int_{\phi \in [0, \pi]} \int_{r \in [a,b]} \frac{e^{-2\pi i r |\vec{v}_1|\sin \phi}}{r^2 |\vec{v}_2| |\vec{v}_3| (\sin \theta \sin \theta_2 \cos(\phi - \phi_2) + \cos\theta \cos\theta_2) (\sin \theta \sin \theta_3 \cos(\phi - \phi_3) + \cos\theta \cos\theta_3)} r^2 \sin \phi dr d\phi d\theta $
The radial integral is straightforward, $ = \int_{\theta \in [0, 2\pi]} \int_{\phi \in [0, \pi]} \frac{{ e^{-2\pi i a |\vec{v}_1| \sin \phi} - e^{-2\pi i b |\vec{v}_1| \sin \phi }}}{{2\pi i |\vec{v}_1| }|\vec{v}_2| |\vec{v}_3| | (\sin \theta \sin \theta_2 \cos(\phi - \phi_2) + \cos\theta \cos\theta_2) (\sin \theta \sin \theta_3 \cos(\phi - \phi_3) + \cos\theta \cos\theta_3)} d\phi d\theta $
However, now I am left with a nasty trigonometric integral. Is there another smarter way to do this?