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I have already known how to prove

$$ \int_0^{\infty}{\frac{dx}{1+x^n}}=\frac{\pi}{n \sin(\pi/n)} $$

when $n$ is a postive even integer with complex analysis. But I don't know how to show that the result is still valid when $n$ is an positive odd integer.

Please help enlighten me!

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  • $\begingroup$ duplicate ${}{}{}$ $\endgroup$ – N3buchadnezzar Sep 2 '14 at 14:02
  • $\begingroup$ Alternately, let $t=\dfrac1{1+x^n}$ and then recognize the expression of the beta function in the new integral. Then, after employing Euler's reflection formula for the $\Gamma$ function, the proof is complete. $\endgroup$ – Lucian Sep 2 '14 at 15:58
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The result holds for any $n > 1$, integer or not. To avoid suggesting an integer value, let me replace the exponent with $\alpha$.

Then we look at the angular sector $S(\alpha) = \left\{z : 0 < \arg z < \frac{2\pi}{\alpha}\right\}$, and for $R > 1$ consider the boundary of $S_R(\alpha) = S(\alpha) \cap D_R(0)$. The sector contains only one pole of the integrand, at $e^{\pi i/\alpha}$, hence

$$\int_{\partial S_R(\alpha)} \frac{dz}{1+z^\alpha} = 2\pi i \operatorname{Res} \left(\frac{1}{1+z^\alpha}; e^{\pi i/\alpha}\right) = 2\pi i \frac{1}{\alpha e^{\pi i(\alpha-1)/\alpha}} = \frac{-2\pi i}{\alpha}e^{\pi i/\alpha}.$$

For the integral on the part $\arg z = \frac{2\pi i}{\alpha}$ of the boundary, we have

$$\int_{Re^{2\pi i/\alpha}}^0 \frac{1}{1+ (re^{2\pi i/\alpha})^\alpha}\, d(re^{2\pi i/\alpha}) = - e^{2\pi i/\alpha}\int_0^R \frac{dr}{1+r^\alpha},$$

and hence, letting $R\to \infty$, since the integral over the circular arc in the boundary tends to $0$ then, we obtain

$$-\frac{2\pi i}{\alpha} e^{\pi i/\alpha} = \left(1 - e^{2\pi i/\alpha}\right)\int_0^\infty \frac{dx}{1+x^\alpha},$$

and rearranging leads to

$$\int_0^\infty \frac{dx}{1+x^\alpha} = -\frac{2\pi ie^{\pi i/\alpha}}{\alpha(1-e^{2\pi i/\alpha})} = \frac{2\pi i}{\alpha (e^{\pi i/\alpha} - e^{-\pi i/\alpha})} = \frac{\frac{\pi}{\alpha}}{\sin \frac{\pi}{\alpha}}.$$

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  • $\begingroup$ Hah, I think the selection of a boundary to integral is pretty tricky. I've never thought an angular sector would work. Many thanks. $\endgroup$ – Highman Sep 3 '14 at 1:31

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