Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know the answer of the integral $$\int_{-\infty}^{\infty}\frac{1}{1+x^{2n}}dx=\frac{\pi}{n\sin\left(\frac{\pi}{2n}\right)}$$where $n\in\mathbb{N}$.

But how to evalulate $$\int_{-\infty}^{\infty}\frac{1}{(1+x^{2n})^2}dx$$ where $n\in\mathbb{Z}$?

share|cite|improve this question
up vote 5 down vote accepted

First of all, write the integral (call it $J$) like this: $$ J = \int_{-\infty}^{+\infty} \frac{1}{(1+x^{2n})^2} dx = 2 \int_0^\infty \frac{1}{(1+x^{2n})^2} dx.$$ Denote the integral on the right by $I$. The latter can be evaluated using a slice contour that consists of three parts: the segment $\Gamma_0$, which runs along the x axis from $0$ to some large $R$ then turns and appends the arc $$\Gamma_1: R e^{i t} \quad \text{where} \quad 0\le t \le \frac{\pi}{n}$$ and finally $\Gamma_2$, which runs straight from $R e^{\frac{i\pi}{n}}$ back to the origin to form a slice. Now the integral along $\Gamma_1$ is bounded by $R \frac{\pi}{n} \frac{1}{R^{4n}}$ and hence it disappears as $R$ goes to infinity because $\lim_{R\to\infty} \frac{1}{R^{4n-1}} = 0$. As for the integral along $\Gamma_2$, it can be parameterized setting $x = e^{\frac{i\pi}{n}} t$, giving $$ \int_R^0 \frac{1}{(1+e^{\frac{i\pi}{n}2n} t^{2n})^2} e^{\frac{i\pi}{n}} dt = - e^{\frac{i\pi}{n}}\int_0^R \frac{1}{(1+e^{2i\pi} t^{2n})^2} dt =- e^{\frac{i\pi}{n}}\int_0^R \frac{1}{(1+ t^{2n})^2} dt$$

Now apply the Cauchy Residue theorem to the closed contour $\Gamma_0 - \Gamma_1 - \Gamma_2.$ There is one pole inside the contour (a double pole at $x = e^{i\pi/2/n}$). The residue is $$ \operatorname{Res}_{x=e^{i\pi/2/n}}\frac{1}{(1+x^{2n})^2} = - \frac{2n-1}{(2n)^2}e^{i\pi/2/n} $$ Putting it all together we obtain $$I \left(1 - e^{\frac{i\pi}{n}}\right) = - 2\pi i \frac{2n-1}{(2n)^2}e^{i\pi/2/n}$$ or $$I = 2\pi i \frac{2n-1}{(2n)^2} \frac{e^{i\pi/2/n}}{e^{i\pi/n} -1} = 2\pi i \frac{2n-1}{(2n)^2} \frac{1}{e^{i\pi/2/n} - e^{-i\pi/2/n}} = \frac{2n-1}{(2n)^2} \frac{\pi}{\sin\left(\frac{\pi}{2n}\right)}.$$

It follows that the original integral is $$J = \frac{2n-1}{2n^2} \frac{\pi}{\sin\left(\frac{\pi}{2n}\right)}.$$

Edit. As to the question about how we bound $\int_{\Gamma_1} f(x) dx$ where $f(x) = \frac{1}{(1+x^{2n})^2}$, this is done as follows: $$ \left| \int_{\Gamma_1} f(x) dx \right| = \left| \int_0^{\pi/n} \frac{1}{(1+R^{2n} e^{2nit})^2} R i e^{it} dt\right| \le \int_0^{\pi/n} \frac{R}{(R^{2n} -1)^2} dt = \frac{\pi}{n} \frac{R}{(R^{2n} -1)^2}.$$ This term is $\theta(1/R^{4n-1})$ and disappears as claimed.

share|cite|improve this answer
$\Gamma_3$ is a typo of mine, which I'll fix right now. For the bounds I can write it up later. Thanks for pointing me to those typos. – Marko Riedel Nov 20 '12 at 19:50






share|cite|improve this answer

For $n<0$, denote $m=-n$. Then $$\int_{-\infty}^{\infty}\frac{1}{(1+x^{-2m})^2}dx=\int_{-\infty}^{\infty}\frac{x^{4m}}{(1+x^{2m})^2}dx$$ Which you can compute using the residue theorem, the same way you found the first integral.

share|cite|improve this answer
can you show me a few steps for n>0 and n<0 ??Thanks a lot : ) – cwk709394 Nov 20 '12 at 12:21
No residue theorem is needed here. The integral trivially diverges. – Sasha Nov 20 '12 at 13:00
@Sasha: Oh, your'e right - $\lim_{x\to\infty}\frac{x^{4m}}{(1+x^{2m})2}=1\neq 0$ – Dennis Gulko Nov 20 '12 at 13:03

close format for this type of integrals: $$ \int_0^{\infty} x^{\alpha-1}Q(x)dx =\frac{\pi}{sin(\alpha \pi)} \sum_{i=1}^{n} Res_i((-z)^{\alpha-1}Q(z))$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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