Editing ThreeDimensionalConfigurations/RiemannStype (section)

====Evaluating A<sub>11</sub>====

As has been stated in [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Evaluating_Aℓℓ|a separate derivation]], quite generally,
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<math>\frac{A_{\ell\ell}}{a_\ell a_m a_s}</math>
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<math>=</math>
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<math>
\int_0^\infty \biggl[ (a_\ell^2 + u)^5(a_m^2 + u)(a_s^2 + u) \biggr]^{-1 / 2} du \, .
</math>
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Now, in the case of Maclaurin spheroids, <math>a_m = a_\ell</math>, so this integral expression becomes,
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<math>\frac{A_{\ell\ell}}{a_\ell^2 a_s}</math>
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<math>=</math>
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<math>
\int_0^\infty \biggl[ (a_\ell^2 + u)^6 (a_s^2 + u) \biggr]^{-1 / 2} du
=
\int_0^\infty \frac{du}{(a_\ell^2 + u)^3 (a_s^2 + u)^{1 / 2} } 
=
\int_0^\infty \frac{du}{v^m \sqrt{w}} \, ,
</math>
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where, <math>m = 3, v = (a_\ell^2 + u), w = (a_s^2 + u), b = d = 1, a = a_s^2, c = a_\ell^2, k = (a_s^2 - a_\ell^2).</math>  Before applying the limits, carrying out the integration gives,
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<math>\int \frac{du}{v^m \sqrt{w}}</math>
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<math>=</math>
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<math>
-~\frac{1}{(m-1)(a_s^2 - a_\ell^2)} \biggl[
\frac{\sqrt{w}}{v^{m-1}} + \biggl(m - \frac{3}{2}\biggr) \int \frac{du}{v^{m-1} \sqrt{w}}
\biggr]
</math>
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[<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>] chapter on INTEGRALS, p. 408, Eq. (154)<br />
[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. 91, Eq. (2.248.8)</td>
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&nbsp;
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<math>=</math>
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<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v^{2}} + \biggl(\frac{3}{2}\biggr) \int \frac{du}{v^{2} \sqrt{w}}
\biggr]
</math>
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&nbsp;
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<math>=</math>
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<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)} 
\biggl\{
\frac{1}{(a_\ell^2 - a_s^2)} \biggl[
\frac{\sqrt{w}}{v} + \biggl(\frac{1}{2}\biggr) \int \frac{du}{v \sqrt{w}}
\biggr]\biggr\}
</math>
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&nbsp;
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<math>=</math>
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<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)^2} \biggl[\frac{\sqrt{w}}{v}\biggr] 
+
\frac{3}{8(a_\ell^2 - a_s^2)^2} 
\biggl\{
\int \frac{du}{v \sqrt{w}}
\biggr\}
</math>
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[<b>[[Appendix/References#CRC|<font color="red">CRC</font>]]</b>] chapter on INTEGRALS, p. 407, Eq. (148)<br />
[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. 90, Eq. (2.246)
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&nbsp;
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<math>=</math>
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<math>
\frac{1}{2(a_\ell^2 - a_s^2)} \biggl[\frac{\sqrt{w}}{v^{2}}\biggr] 
+
\frac{3}{4(a_\ell^2 - a_s^2)^2} \biggl[\frac{\sqrt{w}}{v}\biggr] 
+
\frac{3}{8(a_\ell^2 - a_s^2)^2} 
\biggl\{
\frac{2}{(a_\ell^2 - a_s^2)^{1 / 2}} \tan^{-1}\biggl[ \frac{\sqrt{w}}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}\, .
</math>
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Applying the limits then gives,

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<math>\int_0^\infty \frac{du}{v^m \sqrt{w}}</math>
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<math>=</math>
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<math>
\frac{3\pi}{8(a_\ell^2 - a_s^2)^{5 / 2}} 
-~\frac{a_s}{2(a_\ell^2 - a_s^2)a_\ell^4}  
-
\frac{3a_s}{4(a_\ell^2 - a_s^2)^2 a_\ell^2}  
-
\frac{3}{4(a_\ell^2 - a_s^2)^{5 / 2}} 
\biggl\{
\tan^{-1}\biggl[ \frac{a_s}{(a_\ell^2 - a_s^2)^{1 / 2}} \biggr]
\biggr\}
</math>
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&nbsp;
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<math>=</math>
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<math>
\frac{3\pi}{8 a_\ell^5 e^5} 
-~\frac{(1-e^2)^{1 / 2}}{2a_\ell^5 e^2}  
-
\frac{3(1-e^2)^{1 / 2}}{4a_\ell^5 e^4 }  
-
\frac{3}{4a_\ell^5 e^5} 
\biggl\{
\sin^{-1}(1-e^2)^{1 / 2}
\biggr\}
</math>
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&nbsp;
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<math>=</math>
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<math>
1.49957 - 0.16703 - 0.27593 - 0.29421 = 0.76239
</math>
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<math>\Rightarrow ~~~ \frac{A_{\ell\ell} }{a_\ell^3 (1 - e^2)^{1 / 2} }</math>
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<math>=</math>
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<math>
\frac{1}{4a_\ell^5 e^5}\biggl\{
\frac{3\pi}{2 } 
-3 \biggl[\sin^{-1}(1-e^2)^{1 / 2}\biggr]
-~2e^3(1-e^2)^{1 / 2}  
- 3e(1-e^2)^{1 / 2}
\biggr\}  
</math>
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<math>\Rightarrow ~~~ a_\ell^2 A_{\ell\ell} </math>
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<math>=</math>
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<math>
\frac{3(1-e^2)^{1 / 2}}{4 e^5}
\underbrace{\biggl[ \frac{\pi}{2 } - \sin^{-1}(1-e^2)^{1 / 2}\biggr]}_{\sin^{-1}e}
-  
\frac{1}{4 e^4}( 2e^2   + 3)(1-e^2)  
</math>
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&nbsp;
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<math>=</math>
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<math>
0.36562 - 0.13436 = 0.23126
</math>
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