I would like compute the following
$$I\_{t,x,y} = \int\_{\mathbb S^2} e^{-i\left<t,\omega\right>} \, e^{-i\left< A(\omega)x,y\right>} \, d\sigma(\omega); $$
where $\mathbb S^2$ is the two-dimensional sphere, $t\in \mathbb R^3, \, x,y\in \mathbb R^4$ and $A(.)$ is the matrix given by $$ A(s\_1,s\_2,s\_3):= \left(
 \begin{array}{cccc}
 0 & s\_{1} & s\_{2} & s\_{3} \\
 -s\_{1} & 0 & s\_{3} & -s\_{2} \\
 -s\_{2} & -s\_{3} & 0 & s\_{1} \\
 -s\_{3} & s\_{2} & -s\_{1}& 0
 \end{array}
\right)$$
for all $s\_1, s\_2, s\_3 \in \mathbb R$.



How to compute $\int_{\mathbb S^2} e^{-i\left<t,\omega\right>} \, e^{-i\left< A(\omega)x,y\right>} \, d\sigma(\omega)$
One can get the answer with elementary means by direct calculation.


Define $A^\*(x,y) \in \mathbb{R}^3$ such that $\langle A(\omega) x, y\rangle = \langle A^\*(x,y), \omega \rangle$. Also, let $z = t + A^\*(x,y) \in \mathbb{R}^3$. Now, parametrize the $2$-sphere in the usual spherical coordinates, such that the polar angle $\theta = 0$ is aligned with $z$ and hence $\langle z, \omega \rangle = |z| \cos\theta$:
$$
 I\_{t,x,y} = \int\_0^{2\pi}d\phi \int\_{-1}^{1} d(\cos\theta) \,
 e^{-i|z|\cos\theta}
 = 4\pi \frac{\sin|z|}{|z|} .
$$


