Editing Appendix/Mathematics/ToroidalSynopsis01 (section)

==New Insight==

===Identical Green's Function Expressions===

[https://authors.library.caltech.edu/43491/ Caltech's electronic version] of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions''; in particular, &sect;3.11, p. 169 of Volume I gives,
{{ Math/EQ_Toroidal01 }}

If we make the association, <math>~t \leftrightarrow \coth\eta</math>, then we also have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\sinh\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{t^2 - 1} \, ,</math>
  </td>
</tr>
</table>
</div>

in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \Chi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
t t^' - (t^2-1)^{1 / 2}(t^{'2}-1)^{1 / 2}\cos(\theta^' - \theta) \, .
</math>
  </td>
</tr>
</table>
</div>

Put together, then, these expressions mean,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
Q_{m-1 / 2}(\coth\eta) P_{m - 1 / 2}(\coth\eta^') + 2\sum_{n=1}^\infty (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)] \, .
</math>
  </td>
</tr>
</table>
</div>

Also, from [[Appendix/Mathematics/ToroidalConfusion#QPrelation|our derived <math>~Q-P</math> relation]],
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{m - \frac{1}{2}} (\coth\eta)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
P^{-n}_{m - \frac{1}{2}} (\coth\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta) 
\, .
</math>
  </td>
</tr>
</table>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta) 
\biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\} 
\cos[n(\theta^' - \theta)] 
</math>
  </td>
</tr>
</table>
</div>

Next, we pull from the [[Appendix/Mathematics/ToroidalConfusion#Gil|accompanying discussion of the Gil et al. (2000) expression]],

{{ Math/EQ_Toroidal02 }}

Identifying  <math>~x</math> with <math>~\cosh\eta</math>, in which case we have <math>~\lambda = \coth\eta</math>, and, switching index notation, <math>~n \leftrightarrow m</math>, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{m-1 / 2}^n (\coth\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(m-n+\frac{1}{2})} (\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta) \, ,
</math>
  </td>
</tr>
</table>
</div>

where, this last step also incorporates the [[Appendix/Mathematics/ToroidalConfusion#Proper_Interpretation_of_DLMF_Expression|"Euler reflection formula for gamma functions"]], namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\Gamma(m-n+\tfrac{1}{2})} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\Gamma(n-m+\frac{1}{2}) }{\pi (-1)^{m+n}} \, .</math>
  </td>
</tr>
</table>
</div>

So we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n 
\biggl\{(-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta)\biggr\}
\biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\} 
\cos[n(\theta^' - \theta)] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\sinh\eta^'} \sqrt{\sinh\eta} 
\sum_{n=0}^\infty \epsilon_n  (-1)^m
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} 
P_{n-1 / 2}^m(\cosh\eta) 
Q^m_{n-\frac{1}{2}}(\cosh\eta^') 
\cos[n(\theta^' - \theta)] \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, the CT99 Green's function may be rewritten as,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{1}{|\vec{x} - \vec{x}^{~'}|}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{a\pi} 
[ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2}
\sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] 
\sum_{n=0}^\infty \epsilon_n  (-1)^m
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} 
P_{n-1 / 2}^m(\cosh\eta) 
Q^m_{n-\frac{1}{2}}(\cosh\eta^') 
\cos[n(\theta^' - \theta)] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{1}{a\pi} 
[ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2}
\sum_{m=0}^{\infty} \sum_{n=0}^\infty \epsilon_m\epsilon_n  (-1)^m   
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} 
\cos[m(\psi - \psi^')] \cos[n(\theta^' - \theta)]
P_{n-1 / 2}^m(\cosh\eta) 
Q^m_{n-\frac{1}{2}}(\cosh\eta^') \, .
</math>
  </td>
</tr>
</table>

Let's compare this with Wong's (1973) Green's function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{|~\vec{r} - {\vec{r}}^{~'} ~|} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi a} \biggl[ (\cosh\eta - \cos\theta)(\cosh \eta^' - \cos\theta^') \biggr]^{1 /2 }
\sum\limits_{m,n} (-1)^m \epsilon_m \epsilon_n ~\frac{\Gamma(n-m+\tfrac{1}{2})}{\Gamma(n + m + \tfrac{1}{2})}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times \cos[m(\psi - \psi^')][\cos[n(\theta - \theta^')] ~\begin{cases}P^m_{n-1 / 2}(\cosh\eta) ~Q^m_{n-1 / 2}(\cosh\eta^') ~~~\eta^' > \eta \\P^m_{n-1 / 2}(\cosh\eta^') ~Q^m_{n-1 / 2}(\cosh\eta)~~~\eta^' < \eta
\end{cases}\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.53)
  </td>
</tr>
</table>
</div>
[June 10, 2018]  <font color="red"><b>Amazing! </b></font> The two expressions match precisely!

===Integral Over Polar Angle===

On p. 293 of his article, [http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)] references [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' and states, "It can be shown that &hellip;"
<div align="center">
<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int^\pi_{-\pi} d\theta (\cosh\eta - \cos\theta)^{-\tfrac{5}{2}} \cos[n(\theta - \theta^')]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(8\sqrt{2}/3) Q^2_{n - \frac{1}{2}} (\cosh\eta) \cos (n\theta^')/\sinh^2\eta \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.56)
  </td>
</tr>
</table>
</td></tr></table>
</div>

Let's see if we can replicate this integration result.  (We tried using WolframAlpha's integration tool, but were unsuccessful.)  We presume that Wong initially took the following steps to simplify the left-hand-side of this integral expression:

<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_{-\pi}^{\pi} \frac{\cos[n(\theta - \theta^')] d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} </math>
  </td>
  <td align="right">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos(n\theta^') \int_{-\pi}^{\pi} \frac{ \cos(n\theta) ~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} 
+ \sin(n\theta^') \cancelto{0}{ \int_{-\pi}^{\pi} \frac{ \sin(n\theta)  d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \cos(n\theta^') \int_{0}^{\pi} \frac{ \cos(n\theta)~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} \, .
</math>
  </td>
</tr>
</table>
</div>
That is to say, given that the limits of the integration are <math>~-\pi</math> to <math>~+\pi</math>: &nbsp; The second integral on the right-hand-side will go to zero because the numerator of its integrand &#8212; ''i.e.,'' <math>~\sin(n\theta)</math> &#8212; is an odd function; and, with regard to the first integral on the right-hand-side, the lower integration limit can be set to zero and the result doubled because the numerator of its integrand &#8212; ''i.e.,'' <math>~\cos(n\theta)</math> &#8212; is an even function.

Now, examining Wong's reference to [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'', we find:
<ul>
  <li>
Equation (5) in &sect;3.7, p. 155 of Volume I gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_\nu^\mu(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{i \mu \pi} ~2^{-\nu - 1} \frac{\Gamma(\nu + \mu + 1) }{\Gamma(\nu + 1) } (z^2 - 1)^{-\mu/2} \int_0^\pi (z+\cos t)^{\mu - \nu - 1} (\sin t)^{2\nu + 1} dt \, .
</math>
  </td>
</tr>
</table>

This is valid for,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathrm{Re} ~\nu > -1</math>&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; and &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\mathrm{Re} (\nu + \mu + 1) > 0 \, .</math>
  </td>
</tr>
</table>

  </li>
  <li>
Equation (10) in &sect;3.7, p. 156 of Volume I gives,

{{ Math/EQ_Toroidal03 }}

  </li>
</ul>

Focusing in on this second integral definition of the Legendre function, <math>~Q^\mu_\nu</math>, let's set  <math>~z = \cosh\eta</math>, <math>~t = \theta</math>, <math>~\mu = 2</math>, and, <math>~\nu = n - \tfrac{1}{2}</math>, where <math>~n</math> is zero or a positive integer.  in this case we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^2 (\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi)^{-\frac{1}{2}} (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~\biggl\{
\int_0^\pi (\cosh\eta - \cos \theta)^{-\frac{5}{2}} \cos(n\theta) ~d\theta
- \cancelto{0}{\cos[(n-\tfrac{1}{2})\pi] }~~\int_0^\infty (\cosh\eta + \cosh \theta)^{- \frac{5}{2}} e^{-n\theta} ~d\theta
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
where the prefactor of the second term &#8212; that is, <math>~\cos[(n-\tfrac{1}{2})\pi] </math> &#8212; goes to zero for all allowable values of the integer, <math>~n</math>.  Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\cos(n\theta^')
\int_0^\pi \frac{ \cos(n\theta)~d\theta }{ (\cosh\eta - \cos \theta)^{\frac{5}{2}} }
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 2(2\pi)^{\frac{1}{2}} Q_{n - \frac{1}{2}}^2 (\cosh\eta) \cos(n\theta^')}{ (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~ } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{ 2^3 \sqrt{2} }{ 3 } \biggr] \frac{  Q_{n - \frac{1}{2}}^2(\cosh\eta) \cos(n\theta^')}{ \sinh^2\eta  }
\, .
</math>
  </td>
</tr>
</table>
where we have set, 
<div align="center">
<math>~
\Gamma(\tfrac{5}{2}) = \Gamma(\tfrac{1}{2} + 2) = \frac{ \sqrt{\pi} \cdot 4! }{4^2 \cdot 2!}
= \frac{\sqrt{\pi} \cdot 2^3\cdot 3}{  2^5 } = \frac{3 \sqrt{\pi}}{2^2} \, .
</math>
</div>

The right-hand-side of this last expression exactly matches the result published by Wong (1973) and [[#Integral_Over_Polar_Angle|rewritten inside the box, above]].  

Q.E.D.

===Integral Over Radial Coordinate===

On p. 294 of his article, [http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)] references [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' and states that,
<div align="center">
<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int^x dt \Biggl[ \frac{Q^2_{n - \frac{1}{2}}(t) X_{n-\frac{1}{2}}(t)}{t^2 - 1} \Biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl[
(n-\tfrac{3}{2}) X_{n-\frac{1}{2}}(x)~ Q^2_{n+\frac{1}{2}}(x)
- (n+\tfrac{1}{2}) X_{n+\frac{1}{2}}(x) ~Q^2_{n-\frac{1}{2}}(x)
\biggr] \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.58)
  </td>
</tr>
</table>
</td></tr></table>
</div>

Let's see if we can replicate this integration result.  Let's start with the "Key Equation",

{{ Math/EQ_Toroidal05 }} 

<!--numbered equation under &sect;3.12 (p. 169) of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi (1953)], which is prefaced by the statement, "If <math>~w_\nu^\mu(z)</math> and <math>w_\sigma^\rho(z)</math> denote any solutions of Legendre's differential equation &hellip; with the parameters <math>~\nu, \mu</math> and <math>~\sigma, \rho</math>, respectively, then it follows &hellip; that,"
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int_a^b\biggl[(\nu - \sigma)(\nu + \sigma + 1) + (\rho^2 - \mu^2)(1 - z^2)^{-1} \biggr] w_\nu^\mu ~w_\sigma^\rho ~dz
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
z(\nu-\sigma) w_\nu^\mu ~w_\sigma^\rho + (\sigma+\rho) w_\nu^\mu ~ w_{\sigma-1}^\rho - (\nu + \mu) w_{\nu - 1}^\mu ~w_\sigma^\rho
\biggr]_a^b \, .
</math>
  </td>
</tr>
</table>
-->

In order to match the left-hand side of Wong's expression, we should adopt the associations: &nbsp; <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math>, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, <math>~\rho \rightarrow 0</math>, and <math>~\sigma \rightarrow ( n - \tfrac{1}{2})</math>.  In which case, [https://authors.library.caltech.edu/43491/1/Volume%201.pdf Erd&eacute;lyi's (1953)] expression becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int_a^b\biggl[ -4(1 - t^2)^{-1} \biggr] Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) ~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
(n - \tfrac{1}{2} ) Q_{n - \frac{1}{2}}^2(t) ~ X_{n - \frac{3}{2}}(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t)
\biggr]_a^b \, .
</math>
  </td>
</tr>
</table>
This is very similar to, but does not appear to match, Wong's expression.  

<font color="maroon">'''Reconciliation Attempt #1:'''</font> &nbsp; Keeping in mind that,

{{ Math/EQ_Toroidal04 }}

which means, after making the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 0</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2n t X_{n-\frac{1}{2}}(t) - (n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~(n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>
  </td>
</tr>
</table>

the integral can be rewritten as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int_a^b\biggl[  \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
\biggl[ 2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) \biggr]  Q_{n - \frac{1}{2}}^2(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t)
\biggr\}_a^b 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
\biggl[ 2n t  Q_{n - \frac{1}{2}}^2(t) 
- (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) \biggr] X_{n - \frac{1}{2}}(t)
-  (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t) 
\biggr\}_a^b 
</math>
  </td>
</tr>
</table>
Returning to the same ''recurrence'' "Key Equation," but this time adopting the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n t Q^2_{n - \frac{1}{2}}(t) - (n + \tfrac{3}{2}) Q^2_{n - \frac{3}{2}} (t) \, ,
</math>
  </td>
</tr>
</table>
in which case the integral becomes,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int_a^b\biggl[  \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
(n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t) ~ X_{n - \frac{1}{2}}(t)
-  (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t) 
\biggr\}_a^b \, .
</math>
  </td>
</tr>
</table>
Hooray!  This ''does'' indeed match Wong's relation (2.58)!

===Evaluating Q<sup>2</sup><sub>&nu;</sub>===

How do we evaluate an "order 2" associated Legendre function, such as, <math>~Q_\nu^2</math> ?

Start by recognizing that, from our identified set of "Key Equations,"

{{ Math/EQ_Toroidal07 }}

Hence, after adopting the association, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, we have, when <math>~\mu = 0</math>,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^{1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-\tfrac{1}{2}) (z^2-1)^{-\frac{1}{2}} [z Q_{n - \frac{1}{2}}(z) - Q_{n - \frac{3}{2}}(z)]
</math>
  </td>
  <td allign="center">&nbsp; &nbsp; &hellip; &nbsp; &nbsp;</td>
  <td align="left">
for <math>~n \ge 1 \, ,</math>
  </td>
</tr>
</table>

and, when <math>~\mu = 1</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2-1)^{-\frac{1}{2}} \{ (n-\tfrac{3}{2}) z Q^1_{n - \frac{1}{2}}(z) - (n+\tfrac{1}{2})Q^1_{n - \frac{3}{2}}(z)\}
</math>
  </td>
  <td allign="center">&nbsp; &nbsp; &hellip; &nbsp; &nbsp;</td>
  <td align="left">
for <math>~n \ge 1 \, .</math>
  </td>
</tr>
</table>

All we are missing, then, is expressions for the index, <math>~n=0</math>, that is, we need independent expressions for <math>~Q^1_{-\frac{1}{2}}</math> and for <math>~Q^2_{-\frac{1}{2}}</math>.

From [https://dlmf.nist.gov/14.6#E2 DLMF] &sect;14.6.2, or from  &sect;3.6.1, eq. (7) (p. 149) of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathsf{Q}^{m}_{\nu}\left(x\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^{m}\left(1-x^{2}\right)^{m/2}\frac{{\mathrm{d}}^{m}\mathsf{Q}_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>
  </td>
</tr>
</table>

or, from [https://dlmf.nist.gov/14.6#E4 DLMF] &sect;14.6.4, and from &sect;3.6.1, eq. (5) (p. 148) of  [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^{m}_{\nu}\left(x\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\left(x^{2}-1\right)^{m/2}\frac{{\mathrm{d}}^{m}Q_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>
  </td>
</tr>
</table>

Leaning on the latter of the two possible expressions, we therefore have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1)^{1 / 2} \frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, .
</math>
  </td>
</tr>
</table>

Therefore, starting from the "Key Equation",
{{ Math/EQ_QminusHalf01 }}
we associate <math>~k^2 \leftrightarrow 2/(z+1)</math>, which implies,
<div align="center">
<math>~\frac{dk}{dz} = - [2(z+1)^3]^{-1 / 2} \, .</math>
</div>

Hence, drawing on the [https://dlmf.nist.gov/19.4#i DLMF's &sect;19.4 expressions for the derivatives of complete elliptic integrals] and appreciating that, <math>~(k')^2 \equiv (1-k^2)</math>, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dk}{dz} \cdot \frac{d}{dk} \biggl[ k K(k) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dk}{dz} \cdot \biggl[ K(k) + \frac{E(k) - (k')^2 K(k) }{(k')^2} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} \biggl[ \frac{E(k)  }{(k')^2} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} E(k)  \biggl[ \frac{z+1}{z-1} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)(z-1)^2} \biggr]^{1 / 2} E(k) \, .</math>
  </td>
</tr>
</table>
As a result,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (z^2 - 1)^{1 / 2} \biggl[\frac{1}{2(z^2-1)(z-1)} \biggr]^{1 / 2} E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E\biggl( \sqrt{ \frac{2}{z+1} } \biggr)  \, .
</math>
  </td>
</tr>
</table>

With the exception of the leading negative sign, this appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [<b>[[Appendix/References#MF53|<font color="red">MF53</font>]]</b>].

Now, let's evaluate the second derivative &hellip;

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 2^{-\frac{1}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}E(k) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 2^{-\frac{1}{2}} \biggl\{ 
-\frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k)
- (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-1} \frac{dk}{dz} \cdot \frac{dE(k)}{dk}
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} \biggl\{ 
\frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-1} [2(z+1)^3]^{-1 / 2} \biggl[ \frac{E(k) - K(k)}{k} \biggr]
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ 
\frac{1}{2} (z-1) E(k)
+ (z+1)  E(k)
+ (z+1)(z-1) [2(z+1)^3]^{-1 / 2} \biggl[ E(k) - K(k)\biggr]\biggl[ \frac{z+1}{2} \biggr]^{\frac{1}{2}}
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ 
\frac{1}{2} (z-1) E(k)
+ (z+1)  E(k)
+ \frac{1}{2} (z-1)  \biggl[ E(k) - K(k)\biggr] 
\biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ 
4 z E(k) - (z-1) K(k)
\biggr\} \, .</math>
  </td>
</tr>
</table>

Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}\biggl\{ 
4 z E(k) - (z-1) K(k)
\biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>
This ''also'' appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [<b>[[Appendix/References#MF53|<font color="red">MF53</font>]]</b>].

<div align="center" id="Q1Q2Summary">
<table border="1" align="center" cellpadding="8">
<tr>
  <th align="center">Summary</th>
</tr>
<tr>
  <td align="left">

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} 
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
where: &nbsp; <math>~k = \sqrt{ \frac{2}{z+1}} \, .</math>
  </td>
</tr>
</table>

<table align="center" cellpadding="5" border="1">
<tr>
  <td align="center"><math>~| ~Q^1_{-\frac{1}{2}}(z) ~|</math></td>
  <td align="center"><math>~Q^2_{-\frac{1}{2}}(z)</math></td>
</tr>
<tr>
  <td align="center">
[[File:ABSQ1minusHalf.png|250px|ABSQ1minusHalf]]
  </td>
  <td align="center">
[[File:Q2minusHalf.png|250px|Q2minusHalf]]
  </td>
</tr>
<tr>
  <td align="left" colspan="2">
See [[Appendix/SpecialFunctions##Caption_for_Plots|relevant caption]].
  </td>
</tr>
</table>

  </td>
</tr>
</table>
</div>


Let's push forward a bit more; specifically, let's find the expressions that are relevant when <math>~n = +1</math>.  When <math>~\mu = 0</math>,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z Q_{+\frac{1}{2}}(z) - Q_{ - \frac{1}{2}}(z) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z^2 k~K ( k ) ~-~ [2z^2(z+1)]^{1 / 2} E( k ) ~-~ k K(k)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}  \biggl\{ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl[ \frac{2z^2(z+1)}{z^2-1} \biggr]^{1 / 2} E( k ) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}  \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr] \, .
</math>
  </td>
</tr>
</table>

And, when <math>~\mu = 1</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z Q^1_{+\frac{1}{2}}(z) ~+~3Q^1_{- \frac{1}{2}}(z) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ \frac{z}{2} \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr] 
~-~3 \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z (z^2 - 1)^{1 / 2} k~K ( k )  
~-~(z^2+3) \biggl( \frac{2}{z-1} \biggr)^{1 / 2} E(k)\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}\biggl\{ z k~K ( k )  
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, .
</math>
  </td>
</tr>
</table>