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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.

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You wrote $z^2-z^2$ in the denominator. Perhaps you should fix that. – Patrick Da Silva Feb 20 '12 at 21:25
How do you evaluate the integral ? – Joel Cohen Feb 20 '12 at 21:34
@joelcohen See this question. – Pedro Tamaroff Feb 20 '12 at 21:35
Let $S(a)=\sum_{k=-\infty}^\infty \frac{(-1)^k}{a-k}$. Doesn't $S(a)=-S(-a)$ and $S(a+2)=S(a)$ immediately extend your result to all non-integer reals? – Zander Jun 27 '12 at 23:44
up vote 2 down vote accepted

Write $$S(a)=\sum_{k=-\infty}^\infty \frac{(-1)^k}{a-k}$$ for your sum. Note that $S(-a) =-S(a)$ which extends your validity to $-1<a<0$, and $S(a+2)=S(a)$ which extends it to all $\mathbb R$ except for the integers.

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