I am reading Harmonic analysis written by Stein. In page 426, there is a result about the asymptotic of the fourier transform of $\frac{e^{2\pi i |\xi|}}{|\xi|^\alpha}$ at the unit sphere.
Suppose $\gamma(\xi)$ is a smooth function that equal $|\xi|^{-\alpha}$ for large $\xi$. Then the distribution $$f_\alpha=\int_{\mathbf{R}^n}e^{2\pi i |\xi|}e^{2\pi i x\cdot\xi}\gamma(\xi)d\xi$$ equal a function $f_\alpha(x)$ when $|x|=1$; this function is smooth there and moreover, $$|f_\alpha(x)|\approx|1-|x||^{\alpha-\frac{n+1}{2}} \quad as\quad |x|\rightarrow1$$ as long as $\alpha<\frac{n+1}{2}$.
In the book, Stein use the formula for raidial function:
If
$f(x)=f_0(|x|)$ is a radial function on $\mathbf{R}^n$, then
$$\widehat{f}(\xi)=2\pi|\xi|^{\frac{2-n}{2}}\int_{0}^{\infty}J_{\frac{n-2}{2}}(2\pi|\xi|r)f_0(r)r^\frac{n}{2}dr;$$
where $J_{\frac{n-2}{2}}$ is a Bessel function.
By using this formula, $f_\alpha$ is essentially $$2\pi|x|^{\frac{2-n}{2}}\int_{1}^{\infty}\gamma(\xi)e^{2\pi i u}J_{\frac{n-2}{2}}(2\pi|x|u)u^\frac{n}{2}du,$$ Then, by the asymptotic formula of Bessel function, the main contribution to the above integral is $$c|x|^{\frac{1-n}{2}}\int_{1}^{\infty}e^{2\pi i u(1-|x|)}u^{\frac{n-1}{2}-\alpha} du$$ from which the assertion follows.
But I don't know how to obtain the result from the last integral. By changing variable, we have $$\int_{1}^{\infty}e^{2\pi i u(1-|x|)}u^{\frac{n-1}{2}-\alpha} du= |1-|x||^{\alpha-\frac{n+1}{2}}\int_{|1-|x||}^{\infty}e^{2\pi i ysign(1-|x|)}y^{\frac{n-1}{2}-\alpha} dy$$ It suffice to show $$\int_{|1-|x||}^{\infty}e^{2\pi i ysign(1-|x|)}y^{\frac{n-1}{2}-\alpha} dy$$ is bounded uniformly in $x$. But I don't have any idea. Also, I actually want to use this result for $\alpha>\frac{n+1}{2}$ since I am reading a thesis written by Charles Fefferman, who is Stein's student, and he use this result for $\alpha>\frac{n+1}{2}$. How can I obtain this result for $\alpha>\frac{n+1}{2}$?
PS.The asymptotic formula for Bessel function mentioned is $$J_m(r)\sim r^\frac{-1}{2}e^{ir}\sum\limits_{j=0}^{\infty}a_jr^{-j}+ r^\frac{-1}{2}e^{-ir}\sum\limits_{j=0}^{\infty}b_jr^{-j}\quad as \quad r\rightarrow \infty$$
Another issue is: If $(2\pi |x|u)$ is very small such that the asymptotic formula does not apply, how can I get the result ?