Fourier transform that is almost a brick wall - but why? Let $$g(x) := \sqrt{1+x^2},$$ and $$h(x) := g^{-3/2}(x) \exp(-i2\pi g(x)).$$
I can observe that the Fourier transform $|H(f)|$ is almost flat if $|f|<1$, and $H(f)\approx 0, \; |f|>1$.
This observation is important for my work, but I cannot understand why it happens. I tried for quite some time to figure out why, but failed. It basically means that $h(x)$ is a $\sin(x)/x$ function.
 A: The Fourier transform $H_p(f)$ of $h_p(x)=g^{-p}(x)\exp[-2\pi ig(x)]$, with $g(x)=\sqrt{1+x^2}$ has a closed form expression for $p=1$:
$$H_{1}(f)=\int_0^\infty h_{1}(x)\cos(2\pi f x)dx=K_0\left[2\pi\sqrt{f^2-1}\right],$$
see page 17 of Erdelyi's "Tables of Integral Transforms" (Volume I).
The Fourier transform of $1/\sqrt g$ is also a Bessel function,
$$G(f)=\int_0^\infty g^{-1/2}(x)\cos(2\pi f x)dx=\frac{(\pi/f)^{1/4}}{\Gamma(\tfrac{1}{4})}K_{1/4}(2\pi f).$$
The key thing to note at this point is that $G(f)$ is basically a broadened delta function. The function $H_{1}$ if real for $f>1$ and decays rapidly to zero. This is unaffected by the convolution with $G$. For small $f$ there is a plateau at $|K_0(2\pi i)|=0.4992$, not exactly $1/2$ but close.

Plot of $|H_{1}(f)|$ (blue) and $|H_{3/2}(f)|$ (gold).

Plot of $|H_{3/2}(f)|$ for $f<1$, to show that it is almost but not quite flat, and almost but not quite $1/2$ for $f\rightarrow 0$. The sharp peak at $f=1$ that was present in $|H_{1}(f)|$ has been greatly suppressed by the convolution with $G$.
