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,


\begin{displaymath}
J_{\nu}(-\xi)
=
(-1)^{\nu}
J_{\nu}(\xi).
\end{displaymath} (5)

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


\begin{displaymath}
f(p,\nu,-\xi)
=
-f(p,\nu,\xi),
\end{displaymath} (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


$\displaystyle r_{0}(p,\nu)$ $\textstyle =$ $\displaystyle \lim_{\xi\to 0}
\xi f(p,\nu,\xi)$  
  $\textstyle =$ $\displaystyle \frac{\Gamma(\nu+1)}{2^{p}\Gamma(\nu+p+1)},$ (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


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


$\displaystyle r_{k}(p,\nu)$ $\textstyle =$ $\displaystyle \lim_{\xi\to\xi_{\nu k}}
(\xi-\xi_{\nu k})f(p,\nu,\xi)$  
  $\textstyle =$ $\displaystyle \frac{J_{\nu+p}(\xi_{\nu k})}{\xi_{\nu k}^{p+1}J'_{\nu}(\xi_{\nu k})},$ (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]


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

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


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

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


\begin{displaymath}
r_{k}(p,\nu)
=
-\,\frac{J_{\nu+p}(\xi_{\nu k})}{\xi_{\nu k}^{p+1}J_{\nu+1}(\xi_{\nu k})}.
\end{displaymath} (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,


$\displaystyle J_{\nu}(\xi)$ $\textstyle =$ $\displaystyle \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}}
\left\{
\left[
1
+
\frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{4.5em}
\times
\left.
\left[
e^{-R\sin(\theta)}\,e^{\imath...
...,\nu)}
+
e^{R\sin(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\right.
+$  
    $\displaystyle \hspace{2.7em}
+
\imath\,
\left.
\left[
\frac{4\nu^{2}-1}{8R\,e^{...
...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{4.5em}
\times
\left.
\left[
e^{-R\sin(\theta)}\,e^{\imath...
...n(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\rule{0em}{4ex}
\right\},$ (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


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


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

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


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


\begin{displaymath}
\alpha(R,0,\nu)
=
R-\pi\,\frac{2\nu+1}{4},
\end{displaymath} (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


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

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


$\displaystyle J_{\nu}(\xi)$ $\textstyle =$ $\displaystyle \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}}\,e^{R\sin(\theta)}
\l...
...\frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{8em}
\times
\left.
\left[
e^{-2R\sin(\theta)}\,e^{\imath\,\alpha(R,\theta,\nu)}
+
e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\right.
+$  
    $\displaystyle \hspace{6.05em}
+
\imath\,
\left.
\left[
\frac{4\nu^{2}-1}{8R\,e^...
...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{8em}
\times
\left.
\left[
e^{-2R\sin(\theta)}\,e^{\imath\...
...eta,\nu)}
-
e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\rule{0em}{4ex}
\right\},$ (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


$\displaystyle J_{\nu}(\xi)$ $\textstyle =$ $\displaystyle \sqrt{\frac{\,e^{-\imath\,\theta}}{2\pi R}}\,e^{-R\sin(\theta)}
\...
...\frac{{\cal R}_{c}(\nu,\xi)}{R^{2}\,e^{\imath\,2\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{8.5em}
\times
\left.
\left[
e^{\imath\,\alpha(R,\theta,\nu)}
+
e^{2R\sin(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\right.
+$  
    $\displaystyle \hspace{6.6em}
+
\imath\,
\left.
\left[
\frac{4\nu^{2}-1}{8R\,e^{...
...\frac{{\cal R}_{s}(\nu,\xi)}{R^{3}\,e^{\imath\,3\theta}}
\right]
\right.
\times$  
    $\displaystyle \hspace{8.5em}
\times
\left.
\left[
e^{\imath\,\alpha(R,\theta,\n...
...n(\theta)}\,e^{-\imath\,\alpha(R,\theta,\nu)}
\right]
\rule{0em}{4ex}
\right\},$ (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


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

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


\begin{displaymath}
\lim_{R\to\infty}
\left\vert\frac{J_{\nu+p}(\xi)}{J_{\nu}(\xi)}\right\vert
=
1.
\end{displaymath} (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 $(-\pi,\pi)$.