Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.

Using residues, try the contour below with $R \rightarrow \infty$ and $$\lim_{R \rightarrow \infty } \int_0^R \frac{1}{1+r^n} dr \rightarrow \int_0^\infty \frac{1}{1+x^n} dx$$ I've attempted the residue summation, but my sum did not converge.

• 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. – Zaataro Nov 30 '12 at 7:29

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 $\frac{\pi}{\sin(\pi)}=\frac\pi0$.
• If we consider the case where n = 7, this method fails? I am doing it for this particular case and I am getting a negative in front of the division and thus I am off by a $-1$ factor. – Bayerischer May 19 '16 at 6:29
• It is unclear where you are getting the negative. $(6)$ says that $$\int_0^\infty\frac1{1+x^7}\,\mathrm{d}x=\frac{\pi/7}{\sin(\pi/7)}$$ – robjohn May 19 '16 at 6:37