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This question is part of a assignment of complex analysis and I was unable to solve the following question.

Using contour integration evaluate the following : $\int_{0}^{\infty} \frac{1} {1+ x^6} dx$.

By De Moivres's theorem one can find roots of $1+x^6$ but I am unable to decided how to choose the contour. Actually, I am new to such questions which involve contour integration.

I searched MSE for quite some time and found some questions on contour integration like this: Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$ but I am not able to understand the solution of it.

Before I found it in the assignment I had a question of contour integration like this : $\int_{-\infty}^{\infty} \frac{1}{1+x^2}$ and I evaluated that sucessfully but I am unable to apply that here due to number of roots being 6 and I am having trouble in choosing contour .

So, can you please help with choosing the contour in this question and what details should be kept in mind while choosing contour?

phi-rate
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  • @KaviRamaMurthy This is the same question which I mentioned in my question in paragraph 3 whose solution i am not able to understand? –  Dec 06 '20 at 09:19
  • The contour for that integral (parametrized by $R \to \infty$) as per attached answer, is as follows : you start from $0$ and go to $R$, then from $R$ travel along the circle of radius $R$ around the origin for $\frac{2\pi}{n}$ degrees, then travel back to $0$. This breaks into three pieces (the real line, the curved part and the returning real line). The integral of the curved part goes to $0$ and the rest combines to give something times $\int_0^R \frac 1{1+z^n}$. But the argument theorem tells you the answer for the LHS because there's only one pole inside this contour. – Sarvesh Ravichandran Iyer Dec 06 '20 at 09:55
  • @TeresaLisbon I am not able to understand how to visualize " you start from 0 and go to R, then from R travel along the circle of radius R around the origin for 2πn degrees, then travel back to 0. " Can you please explain or tell it by a diagram ? –  Sep 17 '21 at 06:01
  • See here. The suitability of this contour is clear when you take the linked answer and replace $n$ by $6$ everywhere. – Sarvesh Ravichandran Iyer Sep 17 '21 at 06:26
  • Choose a sector of a circle with the angle $\pi/3$. On the question of how to choose a contour, do more examples and maybe check out this. – Tab1e Sep 17 '21 at 07:47
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    @Tab1e And how does "circle with the angle π/3" look like. Can you add a diagram as answer if you have time for that –  Sep 18 '21 at 07:10
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    It is a sector of a circle. Check the graph of the comment by @TeresaLisbon. – Tab1e Sep 18 '21 at 07:13
  • @No-One The diagram I drew in the comment I wrote above, is the exact contour you need , after you substitute $n=6$. The coordinates I've given, are basically the usual coordinates for the Argand plane i.e. $(x,y)$ representing the complex number $x+iy$. The explanation for why this is a good contour, is exactly the same as the explanations that you have in the answer you linked. Basically, the contour contains only pole inside it,so the argument principle applies. It also breaks into three separate contours, which behave predictably as $R \to \infty$, so equating these gives the result. – Sarvesh Ravichandran Iyer Sep 18 '21 at 07:20

1 Answers1

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$$I=\int\limits_0^\infty\dfrac{\text dx}{1+x^6} =\dfrac12\int\limits_{-\infty}^\infty f(x)\,\text dx,$$ where $\,f(x)=\dfrac{1}{1+x^6}\,$ is the even function.

Let us consider the closed contour $\,\Gamma=L+C,\,$ where $\,L\,$ is the interval $\,[-R,R]\,$ (purple line) and $\,C\,$ is the half-circle with the radius $\,R>1,\,$ considered by increasing of polar angle $\,\varphi\,$ from $\,0\,$ to $\,\pi\,$ (blue line).

Integration contour

The function $\,f(z),\;z\in\mathbb C,\,$ has six poles, which can be founded from the equation $\,z^6=-1=e^{i\pi}.\,$ Assuming a polar notation $\,z=\rho e^{i\varphi},\;\rho\in(0,\infty),\;\varphi\in[0,2\pi),\,$ easily to get $$\rho^6=1,\quad 6\varphi =\pi+2\pi k,\quad k\in\mathbb Z,$$ with the set of solutions $$\rho=1,\;\varphi\in\left\{\dfrac{\pi}6, \dfrac{\pi}2, \dfrac{5\pi}6, \dfrac{7\pi}6, \dfrac{3\pi}2, \dfrac{11\pi}6\right\},$$ $$z\in\left\{\dfrac{\sqrt3}2+\dfrac12i, i, -\dfrac{\sqrt3}2+\dfrac12i, -\dfrac{\sqrt3}2-\dfrac12i, -i, \dfrac{\sqrt3}2-\dfrac12i\right\},$$ wherein the poles $$z_0=e^{i\,\large^\pi/_6}=\dfrac{\sqrt3}2+\dfrac12i,\quad z_1=e^{i\,\large^\pi/_2}=i,\quad z_2=e^{i\,\large^{5\pi}/_6}=-\dfrac{\sqrt3}2+\dfrac12i$$ belong to the inner part of the contour $\,\Gamma.\,$ Since the poles $\,z_0,z_1,z_2\,$ are simple and the contour $\,\Gamma\,$ does not contain additional singularities, then for arbitrary $\,R\in(1,\infty)\,$ $$J(R)=\oint\limits_\Gamma f(z)\text dz =2\pi i\sum\limits_{j=0,1,2}\underset{z_j}{\text{Res}}\,\dfrac1{1+z^6} =2\pi i\sum\limits_{j=0,1,2}\lim\limits_{z\to z_j}\,\dfrac1{6z^5} =\dfrac\pi3 i\sum\limits_{j=0,1,2}\dfrac {z_j}{z_j^6},$$ $$J(R)=-\dfrac\pi3 i\sum\limits_{j=0,1,2}z_j =-\dfrac\pi3 i\left(\dfrac{\sqrt3}2+\dfrac12i+i-\dfrac{\sqrt3}2+\dfrac12i\right)=\dfrac23\pi.$$ On the other hand, $$J(R)=\int\limits_{-R}^R f(x)\,\text dx + iR\int\limits_0^\pi f(Re^{i\varphi})e^{i\varphi}\,\text d\varphi,$$ wherein $$\lim\limits_{R\to\infty}\left|iR\int\limits_0^\pi f(Re^{i\varphi})e^{i\varphi}\,\text d\varphi\right| \leq \lim\limits_{R\to\infty}\pi\dfrac R{R^6-1}=0,$$ $$J(\infty)=2I,$$ $$\color{green}{\mathbf{I=\dfrac\pi3.}}$$

Yuri Negometyanov
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