How to get the $(d\sigma)^{\lor}(x)\lesssim_{d} |x|^{-(d-1)/2}$? In Tao's blog, if we define
$$(d\sigma)^{\lor}(x):=\int_{S^{d-1}}e^{2\pi i x\cdot\xi}d\sigma(\xi)$$
which is the Fourier transform of the measure on a sphere $S^{d-1}:=\{x\in R^d: |\xi|=1\}$.
How to get the following estimation: 

$$(d\sigma)^{\lor}(x)\lesssim_{d} |x|^{-(d-1)/2}$$
  by the method of stationary phase or Bessel function asymptotic. 

 A: First notice that you can split the integral over $S^{d-1}$ as follows: fix $e\in S^{d-1}$ and let $S_\theta = \{ x \in S^{d-1} : e\cdot x = \cos\theta \}$, for $\theta \in [0,\pi]$. The set $S_\theta$ is a $(d-2)$-dimensional sphere of radius $\sin\theta$. Let $d\sigma^{d-2}_{\sin(\theta)}$ be the uniform measure on $S_\theta$, with total mass $k\cdot (\sin(\theta))^{d-2}$ (for some constant $k$ that only depends on $d$. Then we have 
$$\int_{S^{d-1}}e^{i 2\pi x \xi}  d\sigma^{d-1} = \int_0^\pi \int_{S_\theta} e^{i 2\pi |\xi| cos(\theta)} \, d\sigma^{n-2}_{\sin(\theta)}(x) d\theta = \int_0^\pi e^{i 2\pi |\xi| cos(\theta)} d\theta $$
Now apply the change of variable $\cos(\theta)\mapsto t$ and use the following expression for the Bessel function $J_m$
$$ J_m(u) = c(d)\, u^m \int_{-1}^1 e^{iut} (1-t^2)^{m-1/2} dt$$
where $c(n)$ is a constant that only depends on the dimension $d$ of the space. You then get that the transform of the uniform measure on $S^{d-1}$ goes like $|\xi|^{(2-d)/2}J_{(d-2)/2)}(2\pi |\xi|)$ and the decay follows from the one of the Bessel function. 
This derivation was taken from Mattila, Fourier Analysis and Hausdorff dimension, Chap. 3.3, but you can find more details on Stein and Weiss, Introduction to Fourier Analysis on Euclidean spaces, Chap IV.3.
