Simpler expresion for the integral $\int_{\mathbb S^{2}}\operatorname e^{-i\left<x,\omega \right>} d\sigma(\omega)$

Question

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 $\mathbb R^3$.

Answer 1

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 $$ I(x) = 2\pi\int_{-1}^{1}e^{i|x|a}da = \left.2\pi\frac{e^{i|x|a}}{i|x|}\right|_{a=-1}^{1} = 4\pi\frac{\sin |x|}{|x|}. $$