Editing ThreeDimensionalConfigurations/HomogeneousEllipsoids (section)

=Derivation of Expressions for A<sub>i</sub>=

Let's carry out the integrals that appear in the definition of the <math>~A_i</math> coefficients,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~A_i
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} ,
</math>
  </td>
</tr>
</table>
where,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~\Delta
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~\biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u)  \biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>
Here, we are adopting the <math>~(\ell, m, s)</math> subscript notation to identify which semi-axis length is the (largest, medium-length, smallest).  

==Evaluating A<sub>&#8467;</sub>==
First, let's focus on the coefficient associated with the longest axis <math>~(i = \ell)</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \biggl[ (a_\ell^2 + u)^3(a_m^2 + u)(a_s^2 + u) \biggr]^{-1 / 2} du 
</math>
  </td>
</tr>
</table>
Changing the integration variable to <math>~x \equiv -u</math>, we obtain a definite integral expression that appears as equation (3.133.1) in [https://www.academia.edu/36550954/I_S_Gradshteyn_and_I_M_Ryzhik_Table_of_integrals_series_and_products_Academic_Press_2007_ I. W. Gradshteyn &amp; I. M. Ryzhik (2007; 7<sup>th</sup> Edition)], ''Table of Integrals, Series, and Products'' &#8212; hereafter, GR7<sup>th</sup> &#8212; namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)^3(a_m^2 - x)(a_s^2 - x) \biggr]^{-1 / 2} dx 
</math>
  </td>
  <td align="left">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} 
\biggl[ F(\alpha, p) - E(\alpha, p) \biggr]
</math>
  </td>
  <td align="left">&nbsp; &nbsp; &nbsp; &hellip; valid for <math>[a_\ell > a_m > a_s \ge 0]</math></td>
</tr>
<tr>
<td align="center" colspan="4">[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>], p. 255, Eq. (3.133.1)</td>
</tr>
</table>
where (see p. 254 of [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>]),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin^2\alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{a_\ell^2 - a_s^2}{a_\ell^2 - 0} = 1 - \frac{a_s^2}{a_\ell^2} \, ,</math>
  </td>
<td align="center" width="20%">&nbsp;</td>
  <td align="right">
<math>~p</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr]^{1 / 2} \, ,</math>
  </td>
</tr>
</table>
and where, <math>E(\alpha, p)</math> and <math>F(\alpha, p)</math> are [https://dlmf.nist.gov/19.2#ii Legendre incomplete elliptic integrals of the first and second kind], respectively.  (Note that in the notation convention adopted by [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>], the order of the argument list, <math>~(\alpha, p)</math>, is flipped relative to the convention that we have adopted [[#Evaluation_of_Coefficients|above]] and elsewhere throughout our online, MediaWiki-based chapters.) Recognizing that,

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

<tr>
  <td align="right">
<math>~p^2 \sin^3\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{a_\ell^2 - a_s^2}{a_\ell^2 }\biggr]^{3 / 2}\biggl[ \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} \biggr]
=
\frac{(a_\ell^2 - a_s^2)^{1 / 2}}{a_\ell^3 } \biggl[ a_\ell^2 - a_m^2 \biggr] \, ,
</math>
  </td>
</tr>
</table>
we see that the expression for <math>~A_\ell</math> can be rewritten as,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{2}{a_\ell^3 ~p^2 \sin^3\alpha} 
\biggl[ F(\alpha, p) - E(\alpha, p) \biggr] \, .
</math>
  </td>
</tr>
</table>
This matches the expression that we have provided for <math>~A_1</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]].

==Evaluating A<sub>m</sub>==
Next, let's evaluate the coefficient associated with the axis of intermediate length <math>~(i = m)</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_m}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \biggl[ (a_\ell^2 + u)(a_m^2 + u)^3(a_s^2 + u) \biggr]^{-1 / 2} du \, .
</math>
  </td>
</tr>
</table>
This time, by changing the integration variable to <math>~x \equiv -u</math>, we obtain a definite integral expression that appears as equation (3.133.7) 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_m}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)(a_m^2 - x)^3(a_s^2 - x) \biggr]^{-1 / 2} dx 
</math>
  </td>
  <td align="left">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{2\sqrt{a_\ell^2 - a_s^2}}{(a_\ell^2 - a_m^2) (a_m^2 - a_s^2)} E(\alpha, p) 
-
\frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} F(\alpha, p)
-
\frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_s^2}{a_\ell^2 a_m^2} \biggr]^{1 / 2}
</math>
  </td>
  <td align="left">&nbsp; &nbsp; &nbsp; &hellip; valid for <math>[a_\ell > a_m > a_s \ge 0]</math></td>
</tr>
<tr>
<td align="center" colspan="4">[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>], p. 256, Eq. (3.133.7)</td>
</tr>
</table>

(Here, the parameters, <math>~\alpha</math> and <math>~p</math>, have the same definitions as in our [[#Evaluating_A.E2.84.93|above evaluation of]] <math>~A_\ell</math>.)  This time it is useful to recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1 -p^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{a_\ell^2 - a_m^2}{a_\ell^2 - a_s^2} 
=
\frac{a_m^2 - a_s^2 }{a_\ell^2 - a_s^2}
</math>
  </td>
</tr>
</table>
in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~p^2(1-p^2) \sin^3\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ ( a_\ell^2 - a_m^2 )( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>
So the coefficient, <math>~A_m</math>, may be rewritten as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~p^2 (1-p^2) \sin^3\alpha\biggl[ \frac{A_m}{a_\ell a_m a_s} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ ( a_\ell^2 - a_m^2 )( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}}
\biggl\{
\frac{2\sqrt{a_\ell^2 - a_s^2}}{(a_\ell^2 - a_m^2) (a_m^2 - a_s^2)} E(\alpha, p) 
~-~
\frac{2}{(a_\ell^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} F(\alpha, p)
~-~\frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_s^2}{a_\ell^2 a_m^2} \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl\{
E(\alpha, p) 
\biggr\}
-~\frac{ 2( a_m^2 - a_s^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)}
\biggl\{
F(\alpha, p)
\biggr\}
-~\frac{ 2( a_\ell^2 - a_m^2 )}{a_\ell^3 (a_\ell^2 - a_s^2)^{1 / 2}}
\biggl[ \frac{a_s}{a_\ell a_m} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl\{
E(\alpha, p) 
-~(1-p^2)
F(\alpha, p)
-~p^2\sin\alpha
\biggl[ \frac{a_s}{a_m} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{A_m}{a_\ell a_m a_s} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl[ \frac{
E(\alpha, p) 
-~(1-p^2)
F(\alpha, p)
-~(a_s/a_m)p^2\sin\alpha}{p^2 (1-p^2)\sin^3\alpha}
\biggr] \, .
</math>
  </td>
</tr>
</table>
This matches the expression that we have provided for <math>~A_2</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]].


==Evaluating A<sub>s</sub>==
Finally, let's evaluate the coefficient associated with the shortest axis, <math>~(i = s)</math>:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_s}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u)^3 \biggr]^{-1 / 2} du \, .
</math>
  </td>
</tr>
</table>
By changing the integration variable to <math>~x \equiv -u</math>, this time we obtain a definite integral expression that appears as equation (3.133.13) 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_s}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{-\infty}^0 \biggl[ (a_\ell^2 - x)(a_m^2 - x)(a_s^2 - x)^3 \biggr]^{-1 / 2} dx 
</math>
  </td>
  <td align="left">&nbsp;</td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\frac{2}{(a_s^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} E(\alpha, p) 
+
\frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_m^2}{a_\ell^2 a_s^2} \biggr]^{1 / 2}
</math>
  </td>
  <td align="left">&nbsp; &nbsp; &nbsp; &hellip; valid for <math>[a_\ell > a_m > a_s > 0]</math></td>
</tr>
<tr>
<td align="center" colspan="4">[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>], p. 256, Eq. (3.133.13)</td>
</tr>
</table>

(And, again, the parameters, <math>~\alpha</math> and <math>~p</math>, have the same definitions as in our [[#Evaluating_A.E2.84.93|above evaluation of]] <math>~A_\ell</math>.)  Recognizing that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-p^2) \sin^3\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (a_\ell^2 - a_s^2)^{1 / 2}( a_m^2 - a_s^2 )}{a_\ell^3 } \, ,
</math>
  </td>
</tr>
</table>
the coefficient, <math>~A_s</math>, may be rewritten as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-p^2) \sin^3\alpha\biggl[ \frac{A_s}{a_\ell a_m a_s} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (a_\ell^2 - a_s^2)^{1 / 2}( a_m^2 - a_s^2 )}{a_\ell^3 }
\biggl\{
~\frac{2}{(a_s^2 - a_m^2) \sqrt{a_\ell^2 - a_s^2}} E(\alpha, p) 
+
\frac{2}{a_m^2 - a_s^2} \biggl[ \frac{a_m^2}{a_\ell^2 a_s^2} \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl\{
- E(\alpha, p) 
\biggr\}
~+~
\frac{ 2(a_\ell^2 - a_s^2)^{1 / 2}}{a_\ell^3 }
\biggl[ \frac{a_m}{a_\ell a_s} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl\{
\biggl(\frac{a_m}{a_s}\biggr) \sin\alpha
- E(\alpha, p) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{A_s}{a_\ell a_m a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 2}{a_\ell^3 }
\biggl[\frac{
(a_m/a_s) \sin\alpha
- E(\alpha, p)}{ (1-p^2) \sin^3\alpha } 
\biggr] \, .
</math>
  </td>
</tr>
</table>
This matches the expression that we have provided for <math>~A_3</math>, [[#Triaxial_Configurations|above in the context of triaxial configurations]].

==When a<sub>m</sub> = a<sub>&#8467;</sub>==

When the length of the intermediate axis is the same as the length of the longest axis &#8212; that is, when we are dealing with an oblate spheroid &#8212; we can write, 
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
\Delta
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
\biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u)  \biggr]^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ (a_\ell^2 + u)^2(a_s^2 + u)  \biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>

===Index Symbols of the 1<sup>st</sup> Order===

Keeping in mind that, generically, the 1<sup>st</sup>-order index symbol is given by the expression,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
~A_i
</math>
  </td>
  <td align="center">
<math>
~\equiv
</math>
  </td>
  <td align="left">
<math>
~a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, p. 54, Eq. (103)</font> 
  </td>
</tr>
</table>

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^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \frac{du}{ (a_\ell^2 + u)^2(a_s^2 + u)^{1 / 2} } \, .
</math>
  </td>
</tr>
</table>
Changing the integration variable to <math>~x \equiv (a_\ell^2 + u)</math>, we obtain an integral expression that appears as equation (2.228.1) 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^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{a_\ell^2}^\infty \frac{dx}{ x^2(a_s^2 - a_\ell^2 + x)^{1 / 2} } 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>-
\frac{a_s}{(a_\ell^2 - a_s^2 ) a_\ell^2} + \frac{1}{2(a_\ell^2 - a_s^2)}
~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^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_s^2 - a_\ell^2)</math>.  Given that this term is negative, the integration gives,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{A_\ell}{a_\ell^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-
\frac{a_s}{(a_\ell^2 - a_s^2) a_\ell^2} + \frac{1}{2(a_\ell^2 - a_s^2)}
~\biggl\{
\frac{2}{ (a_\ell^2 - a_s^2)^{1 / 2} } \tan^{-1}\bigg[ \frac{(a_s^2 - a_\ell^2 + x)^{1 / 2} }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}_{a_\ell^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{a_s^2}{(a_\ell^2 - a_s^2) } + 
~\frac{a_\ell^2 a_s}{ (a_\ell^2 - a_s^2)^{3 / 2} } 
\biggl\{\frac{\pi}{2} -
\tan^{-1}\bigg[ \frac{a_s }{ (a_\ell^2 - a_s^2)^{1 / 2}} \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^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } 
\biggl\{\frac{\pi}{2} -
\tan^{-1}\bigg[ \frac{ (1-e^2)^{1 / 2}}{ 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^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl\{\frac{\pi}{2} -
\cos^{-1}e
\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^2} + 
~\frac{(1-e^2)^{1 / 2}}{ e^3 } \biggl\{
\sin^{-1}e
\biggr\} \, ,
</math>
  </td>
</tr>
</table>

where, <math>~e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>.  Similarly, the coefficient associated with the shortest axis is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{A_s}{a_\ell^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_0^\infty \frac{du}{ (a_\ell^2 + u)(a_s^2 + u)^{3 / 2} } \, .
</math>
  </td>
</tr>
</table>
This time, after changing the integration variable to <math>~x \equiv (a_\ell^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_s}{a_\ell^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{3 / 2} } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{2}{(a_s^2 - a_\ell^2) \sqrt{a_s^2 - a_\ell^2 + x}}\biggr]_{a_\ell^2}^\infty 
+ \frac{1}{(a_s^2 - a_\ell^2)}
~\int_{a_\ell^2}^\infty \frac{dx}{ x(a_s^2 - a_\ell^2 + x)^{1 / 2} } \, .
</math>
  </td>
</tr>
</table>

As before, 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>]; and, as before, the sign of the constant term in the denominator, <math>~(a_s^2 - a_\ell^2)</math>,  is negative.  Hence, the integration gives,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{A_s}{a_\ell^2 a_s}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{(a_\ell^2 - a_s^2) a_s} 
- \frac{1}{(a_\ell^2 - a_s^2)}
~\biggl\{
\frac{2}{ (a_\ell^2 - a_s^2)^{1 / 2} } \tan^{-1}\bigg[ \frac{(a_s^2 - a_\ell^2 + x)^{1 / 2} }{ (a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}_{a_\ell^2}^\infty 
</math>
  </td>
</tr>

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

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

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

</table>

Because we are evaluating the case where <math>~A_m = A_\ell</math>, we alternatively should have been able to obtain the expression for <math>~A_s</math> immediately from our derived expression for <math>~A_\ell</math> via the known relation,
<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 = 2A_\ell + A_s \, .</math>
  </td>
</tr>
</table>
This approach gives,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{e^2} - 
~\frac{2(1-e^2)^{1 / 2}}{ e^3 } \biggl[
\sin^{-1}e
\biggr] \, ,
</math>
  </td>
</tr>
</table>
which, indeed, matches our separately derived expression for <math>~A_s</math>.

===Index Symbols of the 2<sup>nd</sup> Order===

Keeping in mind that, generically, the 2<sup>nd</sup>-order index symbol is given by the expression,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
A_{ij}
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )(a_j^2 + u )} ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], <font color="#00CC00">Chapter 3, p. 54, Eq. (103)</font> 
  </td>
</tr>
</table>
and, in addition, for oblate-spheroidal configurations,
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
\Delta
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ (a_\ell^2 + u)^2(a_s^2 + u)  \biggr]^{1/2} \, ,
</math>
  </td>
</tr>
</table>
we have the following three independent expressions:

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
A_{\ell \ell} = A_{m m} = A_{m \ell}
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a_\ell^2 a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)^3(a_s^2 + u)^{1 / 2} } \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
A_{s\ell} = A_{sm}
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a^2_\ell a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)^2(a_s^2 + u)^{3 / 2} } \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
A_{ss}
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a_\ell^2 a_s \int_0^\infty \frac{du}{(a_\ell^2 + u)(a_s^2 + u)^{5  / 2} } \, .
</math>
  </td>
</tr>
</table>

Setting <math>x \equiv (a_\ell^2 + u)</math> in each expression gives:

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
\frac{A_{\ell \ell}}{a_\ell^2 a_s } = \frac{A_{m m}}{a_\ell^2 a_s } = \frac{A_{m \ell}}{a_\ell^2 a_s }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x^3(a_s^2 - a_\ell^2 + x)^{1 / 2} }
</math>
  </td>
  <td align="right">
<math>\cdots</math> &nbsp; See equation (2.228.2) 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>]
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{A_{s \ell}}{a_\ell^2 a_s } = \frac{A_{s m}}{a_\ell^2 a_s }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x^2(a_s^2 - a_\ell^2 + x)^{3 / 2} }
</math>
  </td>
  <td align="right">
<math>\cdots</math> &nbsp; See equation (2.229.2) 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>]
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{A_{s s}}{a_\ell^2 a_s } 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x(a_s^2 - a_\ell^2 + x)^{5 / 2} }
</math>
  </td>
  <td align="right">
<math>\cdots</math> &nbsp; See equation (2.227) 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>]
  </td>
</tr>
</table>

Completing these integrals one at a time &#8212; while realizing that <math>z \equiv (a + bx)</math>, with <math>b=1</math> and <math>a \equiv (a_s^2 - a_\ell^2) < 0</math> &#8212; we have:

<font color="red">FIRST INTEGRAL &hellip;</font>
<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
\frac{A_{\ell \ell}}{a_\ell^2 a_s } = \frac{A_{m m}}{a_\ell^2 a_s } = \frac{A_{m \ell}}{a_\ell^2 a_s }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x^3 z^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[
\biggl( - \frac{1}{2ax^2} + \frac{3b}{4a^2 x}\biggr) z^{1 / 2}
\biggr]_{a_\ell^2}^\infty
+
\frac{3b^2}{8a^2}\int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[
\biggl( - \frac{1}{2ax^2} + \frac{3b}{4a^2 x}\biggr) [(a_s^2 - a_\ell^2) + x]^{1 / 2}
\biggr]_{a_\ell^2}^\infty
+
\frac{3b^2}{8a^2} \biggl\{ 
\frac{2}{(-a)^{1 / 2}} \tan^{-1}  \biggl[ \frac{ z^{1 / 2} }{ (-a)^{1 / 2} } \biggr] 
\biggr\}_{a_\ell^2}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[
\biggl( - \frac{3}{4a^2 a_\ell^2} + \frac{1}{2a a_\ell^4} \biggr) [(a_s^2 - a_\ell^2) + a_\ell^2]^{1 / 2}
\biggr]
+
\frac{3}{4(-a)^{5 / 2}} 
\biggl\{\tan^{-1}\biggl[\infty\biggr] - \tan^{-1}  \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2a - 3a_\ell^2 }{4a^2 a_\ell^4} \biggr) a_s
+
\frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{\frac{\pi}{2} - \tan^{-1}  \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[\frac{2a_s^2 - 5a_\ell^2 }{4(a_\ell^2 - a_s^2)^2 a_\ell^4} \biggr] a_s
+
\frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{\frac{\pi}{2} - \tan^{-1}  \biggl[ \frac{ a_s }{ (a_\ell^2 - a_s^2)^{1 / 2} } \biggr] 
\biggr\}
\, .
</math>
  </td>
</tr>
</table>
Given that the eccentricity of an oblate-spheroidal configuration  is <math>e \equiv (1 - a_s^2/a_\ell^2)^{1 / 2}</math>, this latest expression can be rewritten as,

<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
A_{\ell \ell} = A_{m m} = A_{m \ell}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{2(1-e^2) - 5}{4e^4 } \biggr] \frac{a_s^2}{a_\ell^4}
+
\frac{3 }{4  e^{5}} 
\biggl\{\frac{\pi}{2} - \tan^{-1}  \biggl[ \frac{ a_s/a_\ell }{ e } \biggr] 
\biggr\} \frac{a_s}{a_\ell^3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ a_\ell^2 A_{\ell \ell} = a_\ell^2 A_{m m} = a_\ell^2 A_{m \ell}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{-2e^2-3}{4e^4 } \biggr] (1-e^2)
+
\frac{3 (1 - e^2)^{1 / 2}}{4  e^{5}} 
\biggl\{
\sin^{-1}e
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{4e^4}\biggl\{
- (3 + 2e^2) (1-e^2)
+
3 (1 - e^2)^{1 / 2} 
\biggl[
\frac{\sin^{-1}e}{e}
\biggr] \biggr\}
\, .
</math>
  </td>
</tr>
</table>

<font color="red">SECOND INTEGRAL &hellip;</font> remembering that <math>z \equiv (a + bx)</math>, with <math>b=1</math> and <math>a \equiv (a_s^2 - a_\ell^2) < 0</math> &hellip;

<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
\frac{A_{s \ell}}{a_\ell^2 a_s } = \frac{A_{s m}}{a_\ell^2 a_s }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x^2 z^{3 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[
\biggl(- \frac{1}{ax} - \frac{3b}{a^2}\biggr)\frac{1}{z^{1 / 2}}
\biggr]_{a_\ell^2}^\infty 
-
\frac{3b}{2a^2} \int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl[
\biggl(- \frac{1}{ax} - \frac{3b}{a^2}\biggr)\frac{1}{z^{1 / 2}}
\biggr]_{a_\ell^2}^\infty 
-
\frac{3b}{2a^2} 
\biggl\{
\frac{2}{(-a)^{1 / 2}} \tan^{-1}\biggl[ \frac{z^{1 / 2}}{(-a)^{1 / 2}} \biggr]
\biggr\}_{a_\ell^2}^\infty 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl\{
\biggl[ \frac{1}{(a_\ell^2 - a_s^2)x} - \frac{3}{(a_\ell^2 - a_s^2)^2}\biggr]\frac{1}{[a_s^2 - a_\ell^2 + x]^{1 / 2}}
\biggr\}_{a_\ell^2}^\infty 
-
\frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\tan^{-1}\biggl[ \frac{[a_s^2 - a_\ell^2 + x]^{1 / 2}}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}_{a_\ell^2}^\infty 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\biggl\{
\biggl[ \frac{3}{(a_\ell^2 - a_s^2)^2} - \frac{1}{(a_\ell^2 - a_s^2)a_\ell^2} \biggr]\frac{1}{a_s}
\biggr\} 
-
\frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\frac{\pi}{2}
-
\tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{a_s a_\ell^2}\biggl[ \frac{2a_\ell^2 + a_s^2}{(a_\ell^2 - a_s^2)^2 } \biggr]
-
\frac{3}{(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

That is to say,

<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
A_{s \ell} = A_{s m}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2a_\ell^2 + a_s^2}{(a_\ell^2 - a_s^2)^2 } 
-
\frac{3a_\ell^2 a_s}{(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2 + (1-e^2)}{a_\ell^2~ e^4 } 
-
\frac{3a_s}{a_\ell^3 e^5} 
\biggl\{
\frac{\pi}{2} - \tan^{-1}\biggl[ \frac{a_s/a_\ell}{e} \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{a_\ell^2 e^4} \biggl\{
(3-e^2) 
-
\frac{3 (1-e^2)^{1 / 2}}{e} \biggl[\sin^{-1}e\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
<div align="center"><font color="red">CROSSCHECK</font></div>

Now, according to the relations stated in [[ParabolicDensity/GravPot#Parabolic_Density_Distribution_2|an accompanying discussion]], we should find that, for <math>i = j</math>,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>2A_{ii} + \sum_{j=1}^3 A_{ij}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{a_i^2} \, .</math>
  </td>
</tr>
</table>
Let's check.  Specifically, for an oblate spheroid when <math> i = \ell</math>, the summation takes the form,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>2A_{\ell \ell} + (A_{\ell s} + A_{\ell m} + A_{\ell \ell})</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4A_{\ell \ell} + A_{\ell s} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{a_\ell^2e^4}\biggl\{
\biggl[ -3 -2e^2 \biggr] (1-e^2)
+
3 (1 - e^2)^{1 / 2} 
\biggl[
\frac{\sin^{-1}e}{e}
\biggr] \biggr\}
+
\frac{1}{a_\ell^2 e^4} \biggl\{
(3-e^2) 
-
3 (1-e^2)^{1 / 2} \biggl[\frac{\sin^{-1}e}{e}\biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{a_\ell^2e^4}\biggl\{
( - 3 - 2e^2 ) 
+
(  3 + 2e^2 )e^2
+
(3-e^2)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{a_\ell^2e^4}
\biggl\{
2e^4
\biggr\}
=
\frac{2}{a_\ell^2} \, .
</math>
  </td>
</tr>
</table>
Q. E. D. &nbsp; &nbsp; &nbsp; <font color="red">Yeah!</font>
</td></tr></table>


<font color="red">THIRD INTEGRAL &hellip;</font>

<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
\frac{A_{s s}}{a_\ell^2 a_s } 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\int_{a_\ell^2}^\infty \frac{dx}{x z^{5 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\sum_{k=0}^{+1}\biggl[\frac{2}{ (2k+1)a^{2-k} z^{k+1 / 2}  } \biggr]_{a_\ell^2}^\infty
+
\frac{1}{a^2}\int_{a_\ell^2}^\infty \frac{dx}{x z^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">
&hellip; on my whiteboard I completed the evaluation<br />of this integral and obtained &hellip;
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{3}{2} a_\ell^2 A_{ss} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{( 4e^2 - 3 )}{e^4(1-e^2)} 
+
\frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] 
\, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
<div align="center"><font color="red">CROSSCHECK</font></div>

Again, according to the relations stated in [[ParabolicDensity/GravPot#Parabolic_Density_Distribution_2|an accompanying discussion]], we should find that, for <math>i = j</math>,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>2A_{ii} + \sum_{j=1}^3 A_{ij}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{a_i^2} \, .</math>
  </td>
</tr>
</table>
This time, specifically for an oblate spheroid, when <math> i = s</math>, this gives,

<table border="0" align="center" cellpadding="8">

<tr>
  <td align="right">
<math>3A_{ss} + 2A_{s \ell} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2}{a_s^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \frac{3}{2} a_\ell^2 A_{ss} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{(1-e^2)} - a_\ell^2 A_{s \ell} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(1-e^2)} - 
\frac{1}{e^4} \biggl\{
(3-e^2) 
-
3 (1-e^2)^{1 / 2} \biggl[\frac{\sin^{-1}e}{e}\biggr] 
\biggr\} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{( 4e^2 - 3 )}{e^4(1-e^2)} 
+
\frac{3 (1-e^2)^{1 / 2}}{e^4} \biggl[\frac{\sin^{-1}e}{e}\biggr] 
</math>
  </td>
</tr>
</table>
Q. E. D. &nbsp; &nbsp; &nbsp; <font color="red">Yeah!</font>
</td></tr></table>

==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>


==For Spheres (a<sub>&#8467;</sub> = a<sub>m</sub> = a<sub>s</sub>)==

In the case of a sphere, where <math>a_\ell = a_m = a_s</math>, 
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>\Delta</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
\biggl[ (a_\ell^2 + u)(a_m^2 + u)(a_s^2 + u)  \biggr]^{1 / 2} 
=
(a_i^2 + u)^{3 / 2} 
\, ,
</math>
  </td>
</tr>
</table>

and the definition of all three <math>A_i</math> coefficients is obtained from the integral,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
A_i
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a_\ell a_m a_s \int_0^\infty \frac{du}{\Delta (a_i^2 + u )} 
=
a_i^3 \int_0^\infty\frac{du}{(a_i^2 + u)^{5 / 2}} \, .
</math>
  </td>
</tr>
</table>
Analogously,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
A_{ii}
</math>
  </td>
  <td align="center">
<math>
\equiv
</math>
  </td>
  <td align="left">
<math>
a_i^3 \int_0^\infty\frac{du}{(a_i^2 + u)^{7 / 2}} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
According to <font color="#00CC00">p. 406, Eq. (139)</font> of [<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>],
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
\int (a+bu)^{-n / 2} du
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2(a+bu)^{(2-n)/2}}{b(2-n)}
\, .
</math>
  </td>
</tr>
</table>

</td></tr></table>
Hence,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>\frac{A_i}{a_i^3} \equiv 
\int_0^\infty (a_i^2 + u)^{-5 / 2} du
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2}{3(a_i^2 + u)^{3 / 2}} \biggr]_0^\infty
=
\frac{2}{3a_i^3} \, ;
</math>
  </td>
</tr>
</table>
and,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>\frac{A_{ii}}{a_i^3} \equiv 
\int_0^\infty (a_i^2 + u)^{-7 / 2} du
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2}{5(a_i^2 + u)^{5 / 2}} \biggr]_0^\infty
=
\frac{2}{5a_i^5} \, .
</math>
  </td>
</tr>
</table>