Editing ThreeDimensionalConfigurations/HomogeneousEllipsoids (section)

===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>