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\begin{align}&\color{#66f}{\large\int_{-\infty}^{\infty}
{\dd x \over 1 + x + x^{2} + \cdots + x^{2m}}}
=\int_{-\infty}^{\infty}{1 - x \over 1 - x^{2m + 1}}\,\dd x
\\[5mm]&=\int_{0}^{\infty}
\pars{{1 - x \over 1 - x^{2m + 1}} + {1 + x \over 1 + x^{2m + 1}}}\,\dd x
=2\int_{0}^{\infty}{1 - x^{2m + 2} \over 1 - x^{4m + 2}}\,\dd x
\\[5mm]&=2\bracks{\int_{0}^{1}{1 - x^{2m + 2} \over 1 - x^{4m + 2}}\,\dd x
+ \int_{1}^{0}{1 - x^{-2m - 2} \over 1 - x^{-4m - 2}}
\,\pars{-\,{\dd x \over x^{2}}}}
\\[5mm]&=2\bracks{\int_{0}^{1}{1 - x^{2m + 2} \over 1 - x^{4m + 2}}\,\dd x
+ \int_{0}^{1}{x^{2m - 2} - x^{4m} \over 1 - x^{4m + 2}}\,\dd x}
\\[5mm]&=2\bracks{\int_{0}^{1}{1 - x^{2m + 2} \over 1 - x^{4m + 2}}\,\dd x
+ \int_{0}^{1}{1 - x^{4m} \over 1 - x^{4m + 2}}\,\dd x
- \int_{0}^{1}{1 - x^{2m - 2} \over 1 - x^{4m + 2}}\,\dd x}\tag{1}
\end{align}
Sin embargo,
\begin{align}&\dsc{\int_{0}^{1}{1 - x^{\mu} \over 1 - x^{\nu}}\,\dd x}
=\int_{0}^{1}{1 - x^{\mu/\nu} \over 1 - x}\,{1 \over \nu}\,x^{1/\nu - 1}\dd x
={1 \over \nu}\int_{0}^{1}{x^{1/\nu - 1} - x^{\pars{\mu - \nu + 1}/\nu}
\over 1 - x}\,\dd x
\\[5mm]&={1 \over \nu}\bracks{%
\int_{0}^{1}{1 - x^{\pars{\mu - \nu + 1}/\nu} \over 1 - x}\,\dd x
-\int_{0}^{1}{1 - x^{1/\nu - 1} \over 1 - x}\,\dd x}
\\[5mm]&={1 \over \nu}\braces{\Psi\pars{1 + {\mu - \nu + 1 \over \nu}}
-\Psi\pars{1 + \bracks{{1 \over \nu} - 1}}}
=\dsc{{1 \over \nu}\bracks{\Psi\pars{\mu + 1 \over \nu}
-\Psi\pars{1 \over \nu}}}
\end{align}
donde $\ds{\Psi}$ es la
Digamma Función.
La expresión de $\pars{1}$ se reduce a:
\begin{align}&\color{#66f}{\large\int_{-\infty}^{\infty}
{\dd x \over 1 + x + x^{2} + \cdots + x^{2m}}}
\\[5mm]&={1 \over 2m + 1}\braces{%
\bracks{\Psi\pars{2m + 3 \over 4m + 2} -\Psi\pars{2m - 1 \over 4m + 2}}+ \bracks{\Psi\pars{4m + 1 \over 4m + 2} - \Psi\pars{1 \over 4m + 2}}}
\end{align}
Con
Euler Reflexión Fórmula
$\ds{\Psi\pars{1 - z} - \Psi\pars{z} = \pi\cot\pars{\pi z}}$:
\begin{align}&\color{#66f}{\large\int_{-\infty}^{\infty}
{\dd x \over 1 + x + x^{2} + \cdots + x^{2m}}}
={\pi \over 2m + 1}\bracks{%
\cot\pars{{2m - 1 \over 4m + 2}\,\pi} + \cot\pars{{1 \over 4m + 2}\,\pi}}
\\[5mm]&=\color{#66f}{\large{\pi \over 2m + 1}\bracks{\cot\pars{\pi \over 4m + 2} + \tan\pars{\pi \over 2m + 1}}}\,,\qquad m = 1,2,3,\ldots
\end{align}
Para $\ds{m = 2}$ el resultado es:
$$
{\pi \sobre 5}\bracks{\cuna\pars{\pi \más de 10} + \tan\pars{\pi \sobre 5}}
={\raíz{10 + 2\raíz{5}} \over 5}\,\pi \aprox {\tt 2.3903}
$$
