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

enter image description here

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

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  • The sum will only have one term (there is only one pole inside the contour). – mrf Nov 30 '12 at 07:19
  • 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 07:29

2 Answers2


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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  • 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 06:29
  • 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 06:37
  • @robjohn How did you find the residue of $\frac{1}{1+z^n}$ at $z_0$? – Twink Aug 29 '20 at 20:33
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    @Twink: $z_0$ is a simple root of $1+z^n$. To find the coefficient of $\frac1{z-z_0}$ in the expansion of $\frac1{1+z^n}$, we multiply by $z-z_0$ and take the limit as $z\to z_0$. That is, using L'Hôpital, $$\lim_{z\to z_0}\frac{z-z_0}{1+z^n}=\frac1{nz_0^{n-1}}=-\frac{z_0}n$$ – robjohn Aug 29 '20 at 23:03

Supposed that you have no idea of how to do integration over the complex, this problem can also be solved neatly.

Denote $t:=x^n$, and we have \begin{align} \int_0^\infty\frac{\mathrm{d} x}{1+x^n} &= \frac1n\int_0^\infty\frac{t^{\frac1n-1}}{1+t}\mathrm{d} t\\ &=\frac1n\:\mathrm{B}(\frac1n, 1-\frac1n) \\ &=\frac1n\:\Gamma(\frac1n)\cdot\Gamma(1-\frac1n)\\ &=\frac{\pi/n}{\sin \pi/n} \end{align}

Here we used a famous result, $\displaystyle\Gamma(z)\cdot\Gamma(1-z)=\frac{\pi}{\sin{\pi z}} \quad (0<\mathrm{Re}(z)<1)$