# Evalulate $\int_{-\infty}^{\infty}\frac{1}{(1+x^{2n})^2}dx$ by using residue theorem

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 $n\in\mathbb{Z}$?

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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.

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$\Gamma_3$ 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

let$$I(a,n)=\int_{-\infty}^{\infty}\frac{1}{a^{2n}+x^{2n}}dx$$

$$I(a,n)=\int_{-\infty}^{\infty}\frac{1}{a^{2n}+x^{2n}}dx=\frac{1}{a^{2n}}\int_{-\infty}^{\infty}\frac{1}{1+(\frac{x}{a})^{2n}}dx$$

$$I(a,n)\stackrel{x=au}{=}\frac{1}{a^{2n-1}}\int_{-\infty}^{\infty}\frac{1}{1+(u)^{2n}}du=\frac{1}{a^{2n-1}}\frac{\pi}{n\sin(\frac{\pi}{2n})}$$

$$(I(1,n))'_a=-\int_{-\infty}^{\infty}\frac{2n}{(1+x^{2n})^2}dx=-\frac{2n-1}{n}\frac{\pi}{\sin(\frac{\pi}{2n})}$$

$$I=\int_{-\infty}^{\infty}\frac{1}{(1+x^{2n})^2}dx=\frac{2n-1}{2n^2}\frac{\pi}{\sin(\frac{\pi}{2n})}$$

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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.

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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: $$\int_0^{\infty} x^{\alpha-1}Q(x)dx =\frac{\pi}{sin(\alpha \pi)} \sum_{i=1}^{n} Res_i((-z)^{\alpha-1}Q(z))$$

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