Properties of $f(p,\nu ,\xi )$

Let us establish a few important properties of $f(p,\nu ,\xi )$, starting by its behavior under the inversion of the sign of $\xi$. We start with the analogous property of $J_{\nu}(\xi)$, for which we have, using the Maclaurin series for these functions [3], which converges over the whole complex plane,

$$J_{\nu}(-\xi) = (-1)^{\nu} J_{\nu}(\xi).$$ (5)

Using this in the expression for $f(p,\nu ,\xi )$ we get

$$f(p,\nu,-\xi) = -f(p,\nu,\xi),$$ (6)

that is, $f(p,\nu ,\xi )$ is an odd function of $\xi$, for all $p$ and all $\nu$.

Next we show that $f(p,\nu ,\xi )$ has a simple pole at $\xi=0$. Since $J_{\nu+p}(\xi)$, $\xi^{p+1}$ and $J_{\nu}(\xi)$ are analytical functions over the whole complex-$\xi$ plane, it follows that $f(p,\nu ,\xi )$ is analytical over the whole plane except for those points where the denominator vanishes, where it has poles. These are the origin $\xi=0$ and the zeros $\xi_{\nu k}$ of the Bessel function in the denominator. Note that while for non-integer $\nu$ and $p$ the functions involved have branching points at $\xi=0$, the function $f(p,\nu ,\xi )$ never does. In order to determine the residue of $f(p,\nu ,\xi )$ at $\xi=0$ we consider the limit

$$r_{0}(p,\nu) = \lim_{\xi\to 0} \xi f(p,\nu,\xi)$$

$$= \frac{\Gamma(\nu+1)}{2^{p}\Gamma(\nu+p+1)},$$ (7)

where $\Gamma(z)$ is the gamma function and we used once more the Maclaurin series for $J_{\nu}(\xi)$. Since the limit is finite and non-zero, it follows that $f(p,\nu ,\xi )$ has a simple pole at $\xi=0$, and that $r_{0}(p,\nu)$ is the corresponding residue. Turning to the poles at $\xi_{\nu k}$, since $\xi_{\nu k}$ is a simple zero of $J_{\nu}(\xi)$, at which its derivative $J'_{\nu}(\xi)$ is different from zero, it follows that $f(p,\nu ,\xi )$ has a Taylor expansion around this point, with the form

$$J_{\nu}(\xi) = (\xi-\xi_{\nu k})J'_{\nu}(\xi_{\nu k}) + \sum_{i=2}^{\infty} c_{i}(\xi-\xi_{\nu k})^{i},$$ (8)

for certain finite coefficients $c_{i}$. In order to determine the residue of $f(p,\nu ,\xi )$ at $\xi_{\nu k}$ we consider then the limit

$$r_{k}(p,\nu) = \lim_{\xi\to\xi_{\nu k}} (\xi-\xi_{\nu k})f(p,\nu,\xi)$$

$$= \frac{J_{\nu+p}(\xi_{\nu k})}{\xi_{\nu k}^{p+1}J'_{\nu}(\xi_{\nu k})},$$ (9)

where we used this Taylor expansion. Since the derivative is finite and non-zero at $\xi_{\nu k}$, this limit also is finite and non-zero, and hence it follows that $f(p,\nu ,\xi )$ has a simple pole at $\xi_{\nu k}$, and that $r_{k}(p,\nu)$ is the corresponding residue. We can simplify this expression using the well-known identity [4]

$$\xi\frac{\partial}{\partial\xi}J_{\nu}(\xi) = -\xi J_{\nu+1}(\xi)+\nu J_{\nu}(\xi),$$ (10)

which applied at $\xi=\xi_{\nu k}$, since $J_{\nu}(\xi_{\nu k})=0$, results in

$$J'_{\nu}(\xi_{\nu k}) = -J_{\nu+1}(\xi_{\nu k}).$$ (11)

It follows therefore that we have for the residues of the poles at $\xi_{\nu k}$,

$$r_{k}(p,\nu) = -\,\frac{J_{\nu+p}(\xi_{\nu k})}{\xi_{\nu k}^{p+1}J_{\nu+1}(\xi_{\nu k})}.$$ (12)

We will now establish the behavior of the absolute value of the ratio of Bessel functions which is contained in the expression of $f(p,\nu ,\xi )$ in Eq. (3), for large values of $R$. In order to do this we use the asymptotic expansion of the Bessel functions [5], valid in the whole complex plane so long as $\arg(\xi)\neq\pm\pi$, written in terms of $R=\vert\xi\vert$ and $\theta=\arg(\xi)$, to the lowest orders, and with the trigonometric functions expressed as complex exponentials,

$$J_{\nu}(\xi) = \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}} \left\{ \left[ 1 + \frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}} \right] \right. \times$$

$$\hspace{4.5em} \times \left. \left[ e^{-R\sin(\theta)}\,e^{\imath... ...,\nu)} + e^{R\sin(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)} \right] \right. +$$

$$\hspace{2.7em} + \imath\, \left. \left[ \frac{4\nu^{2}-1}{8R\,e^{... ...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}} \right] \right. \times$$

$$\hspace{4.5em} \times \left. \left[ e^{-R\sin(\theta)}\,e^{\imath... ...n(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)} \right] \rule{0em}{4ex} \right\},$$ (13)

where ${\cal R}_{c}(\nu,\xi)$ and ${\cal R}_{s}(\nu,\xi)$ are certain limited functions of $\xi$ and $\alpha(R,\theta,\nu)$ is a certain real number, given by

$$\alpha(R,\theta,\nu) = R\cos(\theta)-\pi\,\frac{2\nu+1}{4}.$$ (14)

The behavior of the expression in Eq. (13) for large values of $R$ depends on the sign of $\theta$, and the particular case $\theta=0$ has to be examined separately. In this particular case we have

$$J_{\nu}(\xi) = \sqrt{\frac{2}{\pi R}} \left\{ \left[ 1 + \frac{{\cal R}_{c}(\nu,\xi)}{R^{2}} \right] \cos[\alpha(R,0,\nu)] \right. +$$

$$\hspace{3.95em} - \left. \left[ \frac{4\nu^{2}-1}{8R} + \frac{{\cal R}_{s}(\nu,\xi)}{R^{3}} \right] \sin[\alpha(R,0,\nu)] \right\},$$ (15)

where all the functions involved are now limited, so that for large values of $R$ we have for the dominant part of $J_{\nu}(\xi)$,

$$J_{\nu}(\xi) \approx \sqrt{\frac{2}{\pi R}}\, \cos[\alpha(R,0,\nu)].$$ (16)

Note now that the points where $\cos[\alpha(R,0,\nu)]=0$ are the zeros of $J_{\nu}(\xi)$, expressed in the asymptotic limit. We will now choose a way to take the $R\to\infty$ limit such that these zeros are avoided. We may simply chose for the passage of the circuit across the real axis that point between two zeros where $\cos[\alpha(R,0,\nu)]=\pm 1$ and $\sin[\alpha(R,0,\nu)]=0$. Since for $\theta=0$ we have

$$\alpha(R,0,\nu) = R-\pi\,\frac{2\nu+1}{4},$$ (17)

and we must have $\alpha(R,0,\nu)=j\pi$ for some integer $j$, we conclude that Eq. (4) holds, which will cause the crossing of the circuit and the real axis to avoid the zeros. This defines the $R\to\infty$ limit in full detail. It follows that for our purposes here we may write the asymptotic expansion in the case $\theta=0$ as

$$J_{\nu}(\xi) = \pm \sqrt{\frac{2}{\pi R}} \left[ 1 + \frac{{\cal R}_{c}(\nu,\xi)}{R^{2}} \right].$$ (18)

In the case $\theta>0$ we put the dominant real exponential in evidence and obtain

$$J_{\nu}(\xi) = \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}}\,e^{R\sin(\theta)} \l... ...\frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}} \right] \right. \times$$

$$\hspace{8em} \times \left. \left[ e^{-2R\sin(\theta)}\,e^{\imath\,\alpha(R,\theta,\nu)} + e^{-\imath\,\alpha(R,\theta,\nu)} \right] \right. +$$

$$\hspace{6.05em} + \imath\, \left. \left[ \frac{4\nu^{2}-1}{8R\,e^... ...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}} \right] \right. \times$$

$$\hspace{8em} \times \left. \left[ e^{-2R\sin(\theta)}\,e^{\imath\... ...eta,\nu)} - e^{-\imath\,\alpha(R,\theta,\nu)} \right] \rule{0em}{4ex} \right\},$$ (19)

where all the functions within the brackets are now limited or go to zero in the $R\to\infty$ limit. Finally, we do the same thing for the case $\theta<0$, obtaining

$$J_{\nu}(\xi) = \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}}\,e^{-R\sin(\theta)} \... ...\frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}} \right] \right. \times$$

$$\hspace{8.5em} \times \left. \left[ e^{\imath\,\alpha(R,\theta,\nu)} + e^{2R\sin(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)} \right] \right. +$$

$$\hspace{6.6em} + \imath\, \left. \left[ \frac{4\nu^{2}-1}{8R\,e^{... ...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}} \right] \right. \times$$

$$\hspace{8.5em} \times \left. \left[ e^{\imath\,\alpha(R,\theta,\n... ...n(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)} \right] \rule{0em}{4ex} \right\},$$ (20)

where once more all the functions within the brackets are now limited or go to zero in the $R\to\infty$ limit. We are now in a position to analyze the behavior of the absolute value of the ratio of two Bessel functions which appears in the definition of $f(p,\nu ,\xi )$. The factors which do not depend on $\nu$ are common to the numerator and denominator, and cancel out. In the case $\theta=0$ we get

$$\left\vert\frac{J_{\nu+p}(\xi)}{J_{\nu}(\xi)}\right\vert =... ...e {\cal R}_{c}(\nu,\xi)}{\displaystyle R^{2}} } \right\vert,$$ (21)

so that in the $R\to\infty$ limit we get

$$\lim_{R\to\infty} \left\vert\frac{J_{\nu+p}(\xi)}{J_{\nu}(\xi)}\right\vert = 1.$$ (22)

It is not difficult to verify that for both the case $\theta>0$ and the case $\theta<0$ we get this same value for this limit. We see therefore that the $R\to\infty$ limit of the absolute value of this ratio is simply $1$, for all values of $\theta$ in $(-\pi,\pi)$.