Let $n$ be a natural number. I want to calculate $$\int_0^{\pi/2} \sqrt[n]{(\tan x)}dx=\int_0^{\infty} \frac{x^n}{1+x^{2n}}dx$$ using contour integration.

I declare $f(z) = \frac{z^n}{1+z^{2n}}$. This function has $2n$ simple poles of the form $exp(i\frac{-\pi}{2n} + i\frac{\pi k}{n})$ for $k = 0,1,2, ... , 2n-1$.

My problem is to find which contour to use. I thought about using the following:

$\gamma_1(t) = t$, $t \in [0,R]$

$\gamma_2(t) = Re^{it}$, $t \in [0,\frac{\pi}{n}]$

$\gamma_3(t) = te^{i\frac{\pi}{n}}$, $t \in [0,R]$

This way I have only one pole in the contour and I can use the residue theorem. Is this a good approach? I somehow have problem with the calculations here.

Help would be appreciated

  • $\begingroup$ $\gamma_3= t e^{\frac{i \pi}{n}}$. Yes, I believe this is a straightforward approach (surround one pole with a sector of a circle). $\endgroup$
    – mjw
    Jun 11, 2019 at 22:39
  • $\begingroup$ This has already been solved on this site plenty of times. Please search "Approach0" online and see if you can find it yourself $\endgroup$ Jun 11, 2019 at 22:51

2 Answers 2


As you have defined $\gamma_3$, that line contour is beginning at $t= R$ and comming back to 0. We are going to need to flip the sign.

$Res_{z=e^{\frac{\pi i }{2n}}} f(z) = \int_{\gamma_1} f(z) \ dz + \int_{\gamma_2} f(z) \ dz - \int_{\gamma_3} f(z) \ dz$

We are hoping to solve for $\int_{\gamma_1} f(z) \ dz$

You will need to show that $\lim_\limits{R\to\infty} \int_{\gamma_2} f(z) \ dz= 0.$ I will leave that to you.

$Res_{z=e^{\frac{\pi i }{2n}}} f(z) = \int_0^\infty f(t) \ dt - \int_0^\infty f(e^\frac{\pi i}{n} t)e^\frac{\pi i}{n} \ dt$

$Res_{z=e^{\frac{\pi i }{2n}}} f(z) = (1+e^\frac{\pi i}{n})\int_0^\infty f(t) \ dt\\ 2e^{\frac{\pi i}{2n}} \cos \frac{\pi}{2n}\int_0^\infty f(t) \ dt = Res_{z=e^{\frac{\pi i }{2n}}} f(z)\\ \int_0^\infty f(t) \ dt = \frac{1}{2e^{\frac{\pi i}{2n}}\cos \frac{\pi}{2n}} Res_{z=e^{\frac{\pi i }{2n}}} f(z)$

$Res_{z=e^{\frac{\pi i }{2n}}} f(z) = 2\pi i \frac{i}{2n e^{\frac{(2n-1)\pi i }{2n}}} = \frac{\pi e^\frac{\pi i}{2n}}{n}$

$\int_0^\infty f(t) \ dt = \frac{\pi}{2n\cos \frac{\pi}{2n}}$

  • $\begingroup$ You've lost a factor $2$ in the third line from the bottom. $\endgroup$ Jun 11, 2019 at 23:16
  • $\begingroup$ I saw that, thanks. $\endgroup$
    – Doug M
    Jun 11, 2019 at 23:18
  • $\begingroup$ Thanks, it helped me to finalise my answer $\endgroup$
    – Gabi G
    Jun 12, 2019 at 7:58

Substitute $x^{2n} = z$ to get $$ \int_0^\infty \frac{x^{n}}{1+x^{2n}}dx = \frac{1}{2n}\int_0^\infty \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz$$ Using the keyhole contour, you can find that for $n>1$: $$\oint_C \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz = (1-e^{2\pi i(-\frac12+\frac{1}{2n})}) \int_0^\infty \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz = (1+e^{\frac{i\pi }{n}}) \int_0^\infty \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz$$ On the other hand $$\oint_C \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz = 2\pi i \,{\rm Res}_{-1}\frac{z^{-\frac12+\frac{1}{2n}}}{1+z} =2\pi i e^{i\pi(-\frac12+\frac{1}{2n})} = 2\pi e^{\frac{i\pi }{2n}}$$ That means that $$ \int_0^\infty \frac{z^{-\frac12+\frac{1}{2n}}}{1+z} dz = \frac{2\pi e^{\frac{i\pi }{2n}}}{1+e^{\frac{i\pi }{n}}} = \frac{\pi}{\cos\frac{\pi}{2n}}$$ $$ \int_0^\infty \frac{x^{n}}{1+x^{2n}}dx = \frac{\pi}{2n \cos\frac{\pi}{2n}}$$

Your approach is equivalent, but without the initial change of variables.


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.