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.

  • $\begingroup$ The sum will only have one term (there is only one pole inside the contour). $\endgroup$ – mrf Nov 30 '12 at 7:19
  • $\begingroup$ 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. $\endgroup$ – 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 $\frac{\pi}{\sin(\pi)}=\frac\pi0$.

  • $\begingroup$ 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. $\endgroup$ – Bayerischer May 19 '16 at 6:29
  • $\begingroup$ It is unclear where you are getting the negative. $(6)$ says that $$\int_0^\infty\frac1{1+x^7}\,\mathrm{d}x=\frac{\pi/7}{\sin(\pi/7)}$$ $\endgroup$ – robjohn May 19 '16 at 6:37
  • $\begingroup$ @robjohn How did you find the residue of $\frac{1}{1+z^n}$ at $z_0$? $\endgroup$ – Twink Aug 29 '20 at 20:33
  • 2
    $\begingroup$ @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, $$\lim_{z\to z_0}\frac{z-z_0}{1+z^n}=\frac1{nz_0^{n-1}}=-\frac{z_0}n$$ $\endgroup$ – 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, $\displaystyle\Gamma(z)\cdot\Gamma(1-z)=\frac{\pi}{\sin{\pi z}} \quad (0<\mathrm{Re}(z)<1)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.