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Using residues, try the contour below with $R \rightarrow \infty$ and $$\lim_{R \rightarrow \infty } \int_0^R \frac{1}{1+r^n} dr \rightarrow \int_0^\infty \frac{1}{1+x^n} dx$$

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I've attempted the residue summation, but my sum did not converge.

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Thanks EuYu, you beat me to it. :) –  Ben Nov 30 '12 at 6:39
The sum will only have one term (there is only one pole inside the contour). –  mrf Nov 30 '12 at 7:19
I thought it would have more than one, because the poles at n=1,2,3,... are each at a certain angle from the real axis, however with each n the contour angle changes as well, keeping the poles within the contour. –  Ben Nov 30 '12 at 7:29

1 Answer 1

up vote 4 down vote accepted

The integral of $$ \int_\gamma\frac1{1+z^n}\mathrm{d}z\tag{1} $$ on the outgoing ray on the real axis tends to $$ \int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{2} $$ On the incoming ray parallel to $e^{2\pi i/n}$, the integral tends to $$ -e^{2\pi i/n}\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{3} $$ For $n\ge2$, the integral on the circular arc vanishes. Therefore, $$ \int_\gamma\frac1{1+z^n}\mathrm{d}z =\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{4} $$ There is one singularity contained in $\gamma$ at $z_0=e^{\pi i/n}$. The residue of $\frac1{1+x^n}$ at $z_0$ is $\frac1{nz_0^{n-1}}=-\frac{z_0}{n}$. Thus, $$ 2\pi i\left(-\frac{e^{\pi i/n}}{n}\right) =\left(1-e^{2\pi i/n}\right)\int_0^\infty\frac1{1+x^n}\mathrm{d}x\tag{5} $$ which resolves by division to $$ \int_0^\infty\frac1{1+x^n}\mathrm{d}x=\frac{\pi/n}{\sin(\pi/n)}\tag{6} $$ For $n=1$, the integral diverges and $\frac{\pi}{\sin(\pi)}=\frac\pi0$.

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Very nice (+1). –  Chris's sis Feb 13 '13 at 20:34

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