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Permite a $\ds{r}$ $\ds{\bar{r}}$ de las raíces de
$\ds{x\pars{x + 1}/R^{2} + 1/L_{d}^{2} = 0}$ para los valores dados de
$\ds{R = 6371000}$ $\ds{L_{d} = 1000000}$.
Tenga en cuenta que
$\ds{r = -\,{1 \over 2} + {\raíz{40339641} \más de 1000}\,\ic \aprox
-\,{1 \over 2} + 6.3514\,\ic}$.
A continuación,
\begin{align}
G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over
\ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} =
-\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over
\pars{\ell - r}\pars{\ell - \bar{r}}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}
\\[5mm] & =
-\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}\pars{%
{2\ell + 1 \over \ell - r} + {2\ell + 1 \over \ell -\bar{r}}}
{1 \over r - \bar{r}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}
\\[5mm] & =
-\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\sum_{\ell = 0}^{\infty}
{2\ell + 1 \over \ell - r}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} =
-\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}%
\sum_{\ell = 0}^{\infty}{\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \over \ell - r}}
\\[5mm] & =
-\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}%
\sum_{\ell = 0}^{\infty}\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}
\int_{0}^{1}x^{\ell - r - 1}\,\dd x}
\\[5mm] & =
-\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}%
\int_{0}^{1}x^{-r - 1}
\sum_{\ell = 0}^{\infty}\mrm{P}_{\ell}\pars{\cos\pars{\gamma}}x^{\ell}\,\dd x}
\\[5mm] & =
-\,{1 \over 4\pi\,\Im\pars{r}}\,\Im\bracks{\pars{2r + 1}\int_{0}^{1}{x^{-r - 1} \over \root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,\dd x}
\end{align}
Resulta que $\ds{\Re\pars{r} = -1/2}$ tal que
\begin{align}
G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over
\ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} =
-\,{1 \over 2\pi}\,\Re\bracks{\int_{0}^{1}{x^{-1/2 - \Im\pars{r}\,\ic} \over \root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,\dd x}
\\[5mm] & =
-\,{1 \over 2\pi}\,\int_{0}^{1}{\cos\pars{\Im\pars{r}\ln\pars{x}} \over
\root{1 - 2x\cos\pars{\gamma} + x^{2}}}\,{\dd x \over \root{x}} =
\bbx{-\,{1 \over \pi}\,\int_{0}^{1}{\cos\pars{2\,\Im\pars{r}\ln\pars{x}} \over
\root{1 - 2x^{2}\cos\pars{\gamma} + x^{4}}}\,\dd x}
\end{align}
Ahora, usted puede probar algunos de cuadratura !!!.
Por otra parte,
\begin{align}
G & \equiv -\,{1 \over 4\pi}\sum_{\ell = 0}^{\infty}{2\ell + 1 \over
\ell\pars{\ell + 1}/R^{2} + 1/L_{d}^{2}}\,\mrm{P}_{\ell}\pars{\cos\pars{\gamma}} \\[5mm] & =
{1 \over \pi}\int_{0}^{1}\mrm{f}\pars{x,\gamma}\,\dd x -
{1 \over \pi}\,\ \underbrace{%
\int_{0}^{1}\cos\pars{2\,\Im\pars{r}\ln\pars{x}}\,\dd x}
_{\ds{1 \over 1 + 4\bracks{\Im\pars{r}}^{2}}}
\\[5mm] & \mbox{where}\quad
\mrm{f}\pars{x,\gamma} =
\left\{\begin{array}{ll}
\ds{\cos\pars{2\,\Im\pars{r}\ln\pars{x}}\bracks{%
1 - {1 \over \root{1 - 2x^{2}\cos\pars{\gamma} + x^{4}}}}\,,} &
\ds{0 < x \leq 1}
\\[2mm]
\ds{0\,,} & \mbox{otherwise}
\end{array}\right.
\end{align}
La siguiente imagen ilustra $\ds{G}$ como una función de
$\ds{\gamma \in \pars{-4\pi,4\pi}}$. La integración se realizó con un
$\ds{20}$De los puntos de la Regla Trapezoidal (TR). No hay un 'visible' mejora cuando se aumenta el número de puntos. Incluso con un $\ds{10}$-puntos de TR la imagen es muy similar. Espero no hacer ningún error !!!.
![enter image description here]()
