Editing 2DStructure/ToroidalCoordinates (section)

===Integrate===

<!--  OLD WAY
Because the density is uniform throughout our test model torus, it can be pulled out of both integrals.  In combination with the limits of integration just derived, we can therefore write,
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<table border="0" cellpadding="5" align="center">

<tr>
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<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{1/2} a^{5/2}\rho_0
\int\limits_{\tanh^{-1}[a/(\varpi_t-\beta_-)]}^{\tanh^{-1}[a/(\varpi_t-\beta_+)]} \frac{K(\mu) \sinh x ~dx}{( \sinh x+\cosh x )^{1/2}}  
\int\limits_{\sin^{-1}(\xi_2|_-)}^{\sin^{-1}(\xi_2|_+)} 
\biggl[ \frac{d\theta}{(\xi_1 - \sin\theta)^{5/2}} \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>

Now, using WolframAlpha's online integrator, we find &hellip;
<div align="center">
[[File:TorusIntegration.png|450px|WolframAlpha Integration]]
</div>

Hence, the inner integral over our toroidal system's angular coordinate gives,

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

<tr>
  <td align="right">
<math>~\int\limits_{\sin^{-1}(\xi_2|_-)}^{\sin^{-1}(\xi_2|_+)} 
\biggl[ \frac{d\theta}{(\xi_1 - \sin\theta)^{5/2}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{2}{3(\xi_1^2-1)^2 (\xi_1-\sin\theta)^{3/2}}
\biggl[
\cos\theta ( -5\xi_1^2 + 4\xi_1\sin\theta + 1) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
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<math>~
+ (\xi_1+1)(\xi_1-1)^2 \biggl( \frac{\xi_1-\sin\theta}{\xi_1-1} \biggr)^{3/2} F\biggl(\frac{\pi - 2\theta}{4} \biggr| \frac{-2}{\xi_1-1}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 4\xi_1(\xi_1-1)(\sin\theta-\xi_1) \biggl( \frac{\xi_1-\sin\theta}{\xi_1-1} \biggr)^{1/2} E\biggl(\frac{\pi - 2\theta}{4} \biggr| \frac{-2}{\xi_1-1}  \biggr)
\biggr] \biggr\}_{\sin^{-1}(\xi_2|_-)}^{\sin^{-1}(\xi_2|_+)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ 
\biggl[ \frac{2[1-(\xi_2|_+)^2] ( -5\xi_1^2 + 4\xi_1\xi_2|_+ + 1)}{3(\xi_1^2-1)^2 (\xi_1-\xi_2|_+)^{3/2}} \biggr]
</math>
  </td>
</tr>

<tr>
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&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \tfrac{2}{3} (\xi_1-1)^{-3/2}\biggl[ (\xi_1+1) F\biggl(\frac{\cos^{-1}(\xi_2|_+)}{2} \biggr| \frac{2}{1-\xi_1}  \biggr)
- 4 \xi_1 E\biggl(\frac{\cos^{-1}(\xi_2|_+)}{2}\biggr| \frac{2}{1-\xi_1}  \biggr) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
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&nbsp;
  </td>
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&nbsp;
  </td>
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<math>~- \biggl\{ 
\biggl[ \frac{2[1-(\xi_2|_-)^2] ( -5\xi_1^2 + 4\xi_1\xi_2|_- + 1)}{3(\xi_1^2-1)^2 (\xi_1-\xi_2|_-)^{3/2}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \tfrac{2}{3} (\xi_1-1)^{-3/2}\biggl[ (\xi_1+1) F\biggl(\frac{\cos^{-1}(\xi_2|_-)}{2} \biggr| \frac{2}{1-\xi_1}  \biggr)
- 4 \xi_1 E\biggl(\frac{\cos^{-1}(\xi_2|_-)}{2}\biggr| \frac{2}{1-\xi_1}  \biggr) \biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
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<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{2}{3} (\xi_1-1)^{-3/2}\biggl\{ 
\biggl[ \frac{[1-(\xi_2|_+)^2] ( -5\xi_1^2 + 4\xi_1\xi_2|_+ + 1)}{(\xi_1-1)^{1/2} (\xi_1+1)^2 (\xi_1-\xi_2|_+)^{3/2}} \biggr]
- \biggl[ \frac{[1-(\xi_2|_-)^2] ( -5\xi_1^2 + 4\xi_1\xi_2|_- + 1)}{(\xi_1-1)^{1/2} (\xi_1+1)^2(\xi_1-\xi_2|_-)^{3/2}} \biggr]
</math>
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</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
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&nbsp;
  </td>
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<math>~
+ \biggl[ (\xi_1+1) F\biggl(\sin^{-1}\biggl[\frac{1-\xi_2|_+}{2} \biggr] \biggr| \frac{2}{1-\xi_1}  \biggr)
- 4 \xi_1 E\biggl(\sin^{-1}\biggl[\frac{1-\xi_2|_+}{2} \biggr]\biggr| \frac{2}{1-\xi_1}  \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
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&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[ (\xi_1+1) F\biggl(\sin^{-1}\biggl[\frac{1-\xi_2|_-}{2} \biggr]\biggr| \frac{2}{1-\xi_1}  \biggr)
- 4 \xi_1 E\biggl(\sin^{-1}\biggl[\frac{1-\xi_2|_-}{2} \biggr]\biggr| \frac{2}{1-\xi_1}  \biggr) \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

-->

Because the density is uniform throughout our test model torus, it can be pulled out of both integrals.  In combination with the limits of integration just derived, we can therefore write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2^{1/2} a^{5/2}\rho_0
\int\limits_{\tanh^{-1}[a/(\varpi_t-\beta_-)]}^{\tanh^{-1}[a/(\varpi_t-\beta_+)]} \frac{K(\mu) \sinh x ~dx}{( \sinh x+\cosh x )^{1/2}}  
\int\limits_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
\biggl[ \frac{d\theta}{(\xi_1 - \cos\theta)^{5/2}} \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>

Now, using WolframAlpha's online integrator, we find &hellip;
<div align="center">
[[File:TorusIntegration2.png|500px|WolframAlpha Integration]]
</div>

Hence, the inner integral over our toroidal system's angular coordinate gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int\limits_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
\biggl[ \frac{d\theta}{(\xi_1 - \cos\theta)^{5/2}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl\{ \frac{2}{3(\xi_1^2 - 1)^2 (\xi_1 - \cos \theta)^{3/2}} \biggr[
\sin \theta(-5\xi_1^2 + 4\xi_1 \cos \theta + 1) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
+ (\xi_1+1)(\xi_1-1)^2 \biggl( \frac{\xi_1 - \cos \theta}{\xi_1-1} \biggr)^{3/2}
F\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
+ 4\xi_1 (\xi_1-1)\biggl( \frac{\xi_1-\cos \theta}{\xi_1-1}\biggr)^{1/2} (\cos \theta - \xi_1) 
E\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) \biggr] \biggr\}_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3(\xi_1^2 - 1)^2 } \biggr[ \frac{
\sin \theta(5\xi_1^2 - 4\xi_1 \cos \theta - 1)}{(\xi_1 - \cos \theta)^{3/2}} \biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ + \frac{2(\xi_1-1)^{1/2}}{3(\xi_1^2 - 1)^2 } \biggr[ 4\xi_1 E\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) 
- (\xi_1+1) F\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) 
\biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
</math>
  </td>
</tr>
</table>
</div>

This last expression exactly matches the expression found in the integral tables published by [http://www.mathtable.com/gr/ Gradshteyn &amp; Ryzhik (1965)] &#8212; specifically, relation '''2.575'''(5) &#8212; after it is realized that the second argument of both elliptic integral functions, as published in GR65, is the square-root of the parameter, <math>~m = k^2</math>, provided by WolframAlpha.

We now have to deal with the realization that, for our particular problem, <math>~\xi_1 \geq 1</math>, which means that the second argument, <math>~m = 2/(1-\xi_1)</math>, of both elliptical integral functions is negative (if not zero).  Following [http://www.mymathlib.com/functions/elliptic_integrals.html the discussion provided at mymathlib.com], we define,

<div align="center">
<math>~k_1 \equiv \sqrt{\frac{2}{\xi_1-1}} </math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~k_1^' \equiv \sqrt{1-(ik_1)^2} = \sqrt{1 + \frac{2}{\xi_1-1}} = \sqrt{\frac{\xi_1 + 1}{\xi_1 - 1}} \, ,</math>
</div>

in which case,

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<math>~\frac{k_1}{k_1^'} = \sqrt{\frac{2}{\xi_1 + 1}} \, ,</math>
</div>

and we can write,
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<tr>
  <td align="right">
<math>~F\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) ~~\rightarrow ~~ F\biggl( \frac{\theta}{2} , ik_1\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{k_1^'} \biggl[K\biggl( \frac{k_1}{k_1^'} \biggr) - F\biggl( \frac{\pi-\theta}{2} \, , \frac{k_1}{k_1^'} \biggr) \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~E\biggl( \frac{\theta}{2} \biggr| \frac{-2}{\xi_1-1}\biggr) ~~\rightarrow ~~ E\biggl( \frac{\theta}{2} , ik_1\biggr) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~k_1^' \biggl[E\biggl( \frac{k_1}{k_1^'} \biggr) - E\biggl( \frac{\pi-\theta}{2} \, , \frac{k_1}{k_1^'} \biggr) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>

<!-- COMMENT MADE ABOUT NEGATIVE PARAMETERs

While I am pleased that I am able to obtain the same result after consulting two different sources (WolframAlpha, and GR65), I am troubled that the parameter found in both elliptic integrals, <math>~m = k^2</math>, is negative.  This leads me to believe that the integration still has not been carried out properly.  Another hint that the obtained result is not physically valid comes from the notes that accompany relation '''2.575'''(5) in GR65.  It is supposed to be valid for the following range of coefficient/parameter values:
<div align="center">
<math>0 \leq \theta \leq \pi</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\xi_1 > -1 > 0 \, .</math>
</div>
Clearly, this last condition is not met, as negative one is not greater than zero.
-->

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

<tr>
  <td align="right">
<math>~\int\limits_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
\biggl[ \frac{d\theta}{(\xi_1 - \cos\theta)^{5/2}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3(\xi_1^2 - 1)^2 } \biggl\{\biggr[ \frac{
\sin \theta(5\xi_1^2 - 4\xi_1 \cos \theta - 1)}{(\xi_1 - \cos \theta)^{3/2}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
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&nbsp;
  </td>
  <td align="left">
<math>~ + 4\xi_1(\xi_1-1)^{1/2} \sqrt{\frac{\xi_1 + 1}{\xi_1 - 1}}  \biggl[E\biggl( \sqrt{\frac{2}{\xi_1 + 1}} \biggr) - E\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
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&nbsp;
  </td>
  <td align="left">
<math>~ 
- (\xi_1-1)^{1/2} (\xi_1+1) \sqrt{\frac{\xi_1 - 1}{\xi_1 + 1}}  \biggl[K\biggl( \sqrt{\frac{2}{\xi_1 + 1}}  \biggr) - 
F\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) \biggr]\biggr\}_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(\xi_1+1)^{1/2} }{3(\xi_1^2 - 1)^2 } \biggr[ \frac{
\sin \theta(5\xi_1^2 - 4\xi_1 \cos \theta - 1)}{(\xi_1+1)^{1/2} (\xi_1 - \cos \theta)^{3/2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ - 4\xi_1 E\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) 
+ (\xi_1-1) F\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) \biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} \, .
</math>
  </td>
</tr>
</table>
</div>

<span id="Answer">We have therefore succeeded in deriving an expression for the gravitational potential of a uniform-density torus that requires numerical integration over only one dimension, that is, the "radial" dimension, <math>~\xi_1</math>.</span>  The final, combined expression is,

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<table border="1" cellpadding="8" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>~\Phi(a,Z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{5/2} G \rho_0 a^{2}}{3}
\int\limits_{\xi_1|_\mathrm{min}}^{\xi_1|_\mathrm{max}}  \frac{(\xi_1+1)^{1/2}K(\mu) d\xi_1}{(\xi_1^2 - 1)^2 [ (\xi_1^2 - 1)^{1/2}+\xi_1 ]^{1/2} } 
\biggr[ \frac{\sin \theta(5\xi_1^2 - 4\xi_1 \cos \theta - 1)}{(\xi_1+1)^{1/2} (\xi_1 - \cos \theta)^{3/2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ - 4\xi_1 E\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) 
+ (\xi_1-1) F\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) \biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} \, .
</math>
  </td>
</tr>
</table>

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


<!-- COMMENT OUT EARLIER EXPRESSION FOR PHI0
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi(a, Z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{5/2} G \rho_0 a^{2}}{3}
\int\limits_{\tanh^{-1}[a/(\varpi_t-\beta_-)]}^{\tanh^{-1}[a/(\varpi_t-\beta_+)]} \frac{K(\mu) ~dx}{\sinh^3 x( \sinh x+\cosh x )^{1/2}}  \cdot
\biggr[ \frac{\sin \theta(5\xi_1^2 - 4\xi_1 \cos \theta - 1)}{ (\xi_1 - \cos \theta)^{3/2}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ - 4\xi_1 (\xi_1+1)^{1/2} E\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) 
+ (\xi_1-1) (\xi_1+1)^{1/2} F\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\xi_1 + 1}} \biggr) \biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2^{5/2} G \rho_0 a^{2}}{3}
\int\limits_{\tanh^{-1}[a/(\varpi_t-\beta_-)]}^{\tanh^{-1}[a/(\varpi_t-\beta_+)]} \frac{K(\mu) ~dx}{\sinh^3 x( \sinh x+\cosh x )^{1/2}}  \cdot
\biggr[ \frac{4 \cosh x \sin \theta }{ (\cosh x - \cos \theta)^{1/2}} + \frac{\sinh^2 x \sin \theta}{ (\cosh x - \cos \theta)^{3/2}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ - 4\cosh x (\cosh x+1)^{1/2} E\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\cosh x + 1}} \biggr) 
+ \sinh x (\cosh x -1)^{1/2} F\biggl( \frac{\pi-\theta}{2} \, , \sqrt{\frac{2}{\cosh x + 1}} \biggr) \biggr]_{\cos^{-1}(\xi_2|_-)}^{\cos^{-1}(\xi_2|_+)} \, .
</math>
  </td>
</tr>
</table>
</div>

END DELETED COMMENT -->