Recently I've found the following.
Let $n < m$. Then the integral
$$\int\limits_0^\infty {\frac{{{x^{n-1}}}}{{1 + {x^m}}}} dx$$
converges and its value is (using the $B$ and $\Gamma$ function to solve it)
$$\int\limits_0^\infty {\frac{{{x^{n - 1}}}}{{1 + {x^m}}}} dx = \frac{\pi }{m}\csc \frac{{n\pi }}{m}$$
Let $0 < a = \dfrac{n}{m}$. Then
$$\frac{1}{m}\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{1 + {e^t}}}} dt = \int\limits_0^\infty {\frac{{{x^{n - 1}}}}{{1 + {x^m}}}} dx$$
And finally, by means of infinite series we have that
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{1 + {e^t}}}} dt = \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} $$
Thus this should prove that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \frac{1}{a} + 2a\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{{a^2} - {k^2}}}} = \frac{\pi }{{\sin \pi a}}$$
How far can this result be accepted? I've been reading about the series I found the result is derived from partial fractions methods in complex analisys, or by Fourier analysis, but in this case, assuming $0 < a < 1$ suffices to show convergence so there shouldn't be any problem whatsoever and the methods used are rather elementary compared to the titan of complex analysis. My main concern is that the finding seems to be "bounded" by the assumption that $0 < a < 1$ but in complex analysis it seems
$$\frac{1}{z} + 2z\sum\limits_{k = 1}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{{z^2} - {k^2}}}} = \frac{\pi }{{\sin \pi z}}$$
is a much "stronger" result since it holds for any $z$ for which the series makes sense.
