Editing ThreeDimensionalConfigurations/HomogeneousEllipsoids (section)

==When a<sub>m</sub> = a<sub>s</sub>==
When the length of the intermediate axis is the same as the length of the shortest axis &#8212; that is, when we are dealing with a prolate spheroid &#8212; the coefficient associated with the longest axis is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_s^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \frac{du}{ (a_s^2 + u)(a_\ell^2 + u)^{3 / 2}  } \, .
</math>
  </td>
</tr>
</table>
Changing the integration variable to <math>~x \equiv (a_s^2 + u)</math>, we obtain an integral expression that appears as equation (2.229.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>], namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_s^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{a_s^2}^\infty \frac{dx}{ x^2(a_\ell^2 - a_s^2 + x)^{1 / 2} } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2}{(a_\ell^2 - a_s^2) (a_\ell^2 - a_s^2 + x)^{1 / 2}} \biggr]_{a_s^2}^\infty
+
\frac{1}{(a_\ell^2 - a_s^2)} \int_{a_s^2}^\infty \frac{dx}{ x (a_\ell^2 - a_s^2 + x)^{1 / 2} } \, .
</math>
  </td>
</tr>
</table>
The remaining integral in this expression appears as equation (2.224.5) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ GR7<sup>th</sup>].  Its resolution depends on the sign of the constant term in the denominator, <math>~(a_\ell^2 - a_s^2)</math>.  Given that this term is positive, the integration gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_s^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2}{(a_\ell^2 - a_s^2) a_\ell} 
+
\frac{1}{(a_\ell^2 - a_s^2)} 
\biggl\{
\frac{1}{\sqrt{(a_\ell^2 - a_s^2)}} \ln \biggl[ \frac{ (a_\ell^2 - a_s^2 + x)^{1 / 2} - \sqrt{(a_\ell^2 - a_s^2)} }{(a_\ell^2 - a_s^2 + x)^{1 / 2} + \sqrt{(a_\ell^2 - a_s^2)}  } \biggr]
\biggr\}_{a_s^2}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ A_\ell</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2a_\ell a_s^2}{(a_\ell^2 - a_s^2) a_\ell} 
-
\frac{a_\ell a_s^2}{(a_\ell^2 - a_s^2)^{3 / 2}} 
\biggl\{
\ln \biggl[ \frac{ 1-e }{1+e}   \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (1-e^2)}{e^3} 
\biggl\{
\ln \biggl[ \frac{1+e}{ 1-e }   \biggr]
\biggr\}
- \frac{2(1-e^2)}{e^2 } 
\, ,
</math>
  </td>
</tr>
</table>
where, as above, <math>~e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>.  Now, given that <math>~A_m = A_s</math>, in this case we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A_\ell + A_m + A_s = A_\ell + 2A_s</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ A_s</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - \frac{A_\ell}{2} </math>
  </td>
</tr>

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

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