Thursday, July 19, 2012

Some nagging Fourier transforms

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)

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


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

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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