Simpler expresion for the integral $\int_{\mathbb S^{2}}\operatorname e^{-i\left} d\sigma(\omega)$ Given $x\in \mathbb R^3$, there exists a simpler expresion for the following integral ?
$$I_x = \int_{\mathbb S^{2}}\operatorname e^{-i\left<x,\omega \right>} d\sigma(\omega),$$
where $\mathbb S^{2}$ is the two-dimentional sphere of $\mathbb R^3$.
 A: The links in the comments are great. The whole story is there. I thought I would post the solution using those ideas, starting from basics.
The spherical coordinate system works well for this problem.
$$
           x = r\sin\theta\cos\varphi,\;\; y=r\sin\theta\sin\varphi,\;\; z=r\cos\theta.
$$
The volume element is $r^2\sin\theta dr d\theta d\varphi$. The area element on the unit sphere is $dS=\sin\theta d\theta d\varphi$. The expression you have is invariant under rotations. That is $I(x)=I(Ux)$ where $U$ is any rotation about the origin. You can reduce $I(x)$ to $I(x)=I(|x|\hat{z})$, which simplifies your integral, after writing $w=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)$:
\begin{align}
    I(x) & = \int_{0}^{2\pi}\int_{0}^{\pi}e^{-i |x|\cos\theta}\sin\theta d\theta d\varphi \\
     & = 2\pi\int_{0}^{\pi}e^{-i|x|\cos\theta}\sin\theta d\theta.
\end{align}
Let $a=-\cos\theta$. Then $da=\sin\theta d\theta$ and
$$
        I(x) = 2\pi\int_{-1}^{1}e^{i|x|a}da = \left.2\pi\frac{e^{i|x|a}}{i|x|}\right|_{a=-1}^{1} = 4\pi\frac{\sin |x|}{|x|}.
$$
