$$ F^+_\nu(\xi, c) :=\int_c^\infty (x^2-c^2)^{ \nu-\frac{1}{2} } e^{-\ii\xi x} dx,$$

$$ F^-_\nu(\xi,c) :=\int_{-\infty}^{-c} (x^2-c^2)^{\nu-\frac{1}{2}} e^{-\ii\xi x} dx,$$

where $c>0$, $\xi>0$ and $\nu \in (-\frac{1}{2},\frac{1}{2})$. Fortunately, the venerable

*Gradshteyn and Ryzik (G& R)*had the answer. (If you're younger than forty it is very likely you haven't heard of this relique of the cold war.) Here is the beautiful answer (G & R, 6th Edition, 3.77.17, 3.771.9, page 436)

$$ F^+_\nu(\xi, c)= -\frac{\sqrt{\pi}}{2}\left(\frac{2c}{\xi}\right)^\nu\Gamma\Bigl(\nu+\frac{1}{2}\Bigr)\bigl(\;Y_{-\nu}(c \xi ) + \ii J_{-\nu}(c \xi )\;\bigr), $$

and

$$ F_\nu^-(\xi, c) =-\frac{\sqrt{\pi}}{2}\left(\frac{2c}{\xi}\right)^\nu\Gamma(\Bigl(\nu+\frac{1}{2}\Bigr)\bigl(\; Y_{-\nu}(c\xi)-\ii J_{-\nu}(c\xi)\;\bigr), $$

where $J_\nu$ is the Bessel function of the first kind and order $\nu$,

$$ J_\nu(z) := \frac{z^\nu}{2^\nu}\sum_{k=0}^\infty (-1)^k\frac{z^{2k}}{2^{2k}k!\Gamma(\nu+k+1)},\;\;|\arg z|<\pi, $$

and $Y_\nu(z)$ is the Bessel function of the second kind and order $\nu$,

$$ Y_\nu(z)=\frac{1}{\sin\nu\pi}\bigl( \;\cos\nu\pi J_\nu(z) -J_{-\nu}(z) \;\bigr).$$

The above expression makes sense only for non-integral $\nu$. To obtain the expression for integral $\nu$ we let $\nu$ approach an integer in the above equality. See G & R (6th Edition) Section 8.4-8.5.

We deduce that

$$-\ii F_\nu^+(\xi,c)+\ii\xi F^-_\nu(\xi,c)= -\sqrt{\pi}\Gamma\Bigl(\nu+\frac{1}{2}\Bigr)\xi\left( \frac{2c}{\xi} \right)^\nu J_{-\nu}(c\xi). $$

We can rewrite this as

$$-\ii F_\nu^-(\xi,c)+\ii\xi F_\nu^-(\xi, c) = Const(\nu) c^\nu \xi^{-\nu+1}J_{-\nu}(c\xi). $$

Define $\newcommand{\pa}{\partial}$

$$\chi_\pm^a(x_+^2-c^2):=\frac{1}{\Gamma(a+1)}(x_\pm^2-c^2)^a_+, $$

$$A_\nu(x)=\pa_x\Bigl(\chi_+^{\nu-1/2}(x^2_+-c^2)-\chi_+^{\nu-1/2}(x_-^2-c^2)\;\Bigr). $$

We conclude that

$$\widehat{A}_\nu(\xi)=-\sqrt{\pi}(2c)^\nu |\xi|^{1-\nu}J_{-\nu}(c\xi). $$

**July 21, 2012.**The term $\xi^{-\nu+1}J_{-\nu}(c\xi)$ is fortunately the kernel of the modified

*Hankel transform*discussed in these notes of Michael Taylor. This is good news for my project.