Editing Appendix/Mathematics/ToroidalSynopsis01 (section)

===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">

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  <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,
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  <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,

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<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.