In this computer dominated era, the Tables of Integrals have lost their attraction. Fortunately, they are still around, and they can get you out of many jams. I was looking for some compact descriptions of some integrals $\newcommand{\ii}{{\boldsymbol{i}}}$
$$ F^+_\nu(\xi, c) :=\int_c^\infty (x^2-c^2)^{ \nu-\frac{1}{2} } e^{-\ii\xi x} dx,$$
$$ F^-_\nu(\xi,c) :=\int_{-\infty}^{-c} (x^2-c^2)^{\nu-\frac{1}{2}} e^{-\ii\xi x} dx,$$
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.
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