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I would like some help with the following integral.

I would like to find a contour line to evaluate $$\int_1^\infty \frac {dx}{x^3+1}$$

So one can see that on any circumference it goes to $0$, but how do I connect it to $1$?

I have tried a number of things but without results. Can somebody offer any help?

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    $\begingroup$ Note that this integral can be handled without too much trouble via the Method of Partial Fractions. $\endgroup$ – Travis Dec 16 '14 at 13:46
  • $\begingroup$ Does a third of circle will do the job? $\endgroup$ – Krokop Dec 16 '14 at 14:07
  • $\begingroup$ @Krokop I think that would work if one was looking for the integral all the way form $0$. $\endgroup$ – Jonas Dahlbæk Dec 16 '14 at 14:28
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    $\begingroup$ Generally, contour integration is helpful to evaluate a definite integral if the integral is evaluated between two singularities of the integrand (or at least points special in some way for the integrand). Here, $1$ is not a special point for the integrand. Which means, perhaps: evaluating using a contour is the same work as evaluating without using a contour. $\endgroup$ – GEdgar Dec 16 '14 at 15:02
  • $\begingroup$ @GEdgar I begin to see it, I still don't have a lot of confidence with contour analysis. Even though the answers below are pretty nice! :D $\endgroup$ – Ant Dec 17 '14 at 12:02
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Since you want to evaluate this integral using contour integration method, although, I am not the expert of this method, I'll try to help you. Let $y=x-1$, then $$I=\int_1^\infty\frac{\mathrm dx}{x^3+1}=\int_0^\infty\frac{\mathrm dy}{(y+2)(y^2+y+1)}$$ Now, we can consider the contour integral \begin{align} \int_\Gamma\underbrace{\frac{\ln z}{(z+2)(z^2+z+1)}}_{f(z)}\frac{\mathrm dz}{2\pi i} =&\int^\infty_0\frac{\ln{x}-(\ln{x}+2\pi i)}{(x+2)(x^2+x+1)}\frac{\mathrm dx}{2\pi i}\\ =&-I\\ =&\color{#E2062C}{{\rm \ Res}\left(f(z),-2\right)}+\color{#00A000}{{\rm Res}\left(f(z),e^{2\pi i/3}\right)}+\color{#21ABCD}{{\rm Res}\left(f(z),e^{\color{red}{4\pi} i\color{red}{/3}}\right)}\\ =&\color{#E2062C}{\frac{1}{3}\ln{2}+\frac{\pi i}{3}}+\color{#00A000}{\frac{\pi}{3\sqrt{3}}-\frac{\pi i}{9}}\color{#21ABCD}{-\frac{2\pi}{3\sqrt{3}}-\frac{2\pi i}{9}}\\ =&-\frac{\pi}{3\sqrt{3}}+\frac{1}{3}\ln{2} \end{align} where $\Gamma$ is a keyhole contour. You may also refer to the following sources $[1]$, $[2]$, $[3]$, and $[4]$ for the rest calculation. I hope this helps.

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  • $\begingroup$ @M.N.C.E. Thanks. I fix my answer. Anyway, can we avoid the real pole at $z=-2$ using this contour? $\endgroup$ – Venus Dec 16 '14 at 16:10
  • $\begingroup$ The pole at $z=-2$ isn't a problem as the keyhole contour encloses it and doesn't go through it. $\endgroup$ – M.N.C.E. Dec 16 '14 at 16:31
  • $\begingroup$ So we don't take the residue at that pole, right? And the contribution residue only comes from the imaginary poles, right? $\endgroup$ – Venus Dec 16 '14 at 16:34
  • $\begingroup$ All three poles contribute as all are enclosed within the contour. One thing to take extra care of though is making sure that the arguments of the imaginary poles lie between $0$ and $2\pi$, not $-\pi$ and $\pi$ (which is what Wolfram&Mathematica use as their convention.) $\endgroup$ – M.N.C.E. Dec 16 '14 at 17:07
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    $\begingroup$ @Venus I really hope you don't mind me adding a little more detail to your answer. $\endgroup$ – M.N.C.E. Dec 16 '14 at 17:27
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Let's use the branch cut $\lim\limits_{R\to\infty}[1,R]$.

Now consider the integral $$ \int_\gamma\frac{\log(z-1)\,\mathrm{d}z}{1+z^3} $$ along the contour $$ \gamma=1+[\epsilon,R]e^{i\epsilon}\cup1+Re^{i[\epsilon,2\pi-\epsilon]}\cup1+[R,\epsilon]e^{-i\epsilon}\cup1+\epsilon e^{i[2\pi-\epsilon,\epsilon]} $$ As $\epsilon\to0$ and $R\to\infty$, the difference between the integral along the upper and lower lines is that $\log(z-1)$ is $2\pi i$ greater on the lower contour. Furthermore, the integral along the circular curves goes to $0$. This means that $$ \int_\gamma\frac{\log(z-1)\,\mathrm{d}z}{1+z^3}=-2\pi i\int_1^\infty\frac{\mathrm{d}x}{1+x^3} $$ $\gamma$ circles the three singularities of $\frac1{1+z^3}$ counterclockwise. Thus, the integral on the left is $2\pi i$ times the sum of the residues of $\frac{\log(z-1)}{1+z^3}$ $$ 2\pi i\left[\vphantom{\frac{\frac\pi3}3}\right.\overbrace{\frac{\frac{\pi i}3(-1-\sqrt3i)}3}^{z=e^{i\pi/3}}+\overbrace{\frac{\vphantom{\frac{\pi i}3}\log(2)+\pi i}3}^{z=-1}+\overbrace{\frac{\frac{2\pi i}3(-1+\sqrt3i)}3}^{z=e^{-i\pi/3}}\left.\vphantom{\frac{\frac\pi3}3}\right]=2\pi i\left[\frac{\log(2)}3-\frac\pi{3\sqrt3}\right] $$ Therefore, $$ \int_1^\infty\frac{\mathrm{d}x}{1+x^3}=\frac\pi{3\sqrt3}-\frac{\log(2)}3 $$

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  • $\begingroup$ Nice and clear! Thank you :D $\endgroup$ – Ant Dec 17 '14 at 11:59
  • $\begingroup$ On second thoughts, there is something which is not clear; Your curve passes through the point $1$, but it is a branch point for the logarithm; shouldn't you "avoid" it with a little circle around it? And in this case, one should show that the integral over the little circle goes to $0$, but setting $z = 1 + re^{i[\epsilon, 2\pi - \epsilon]}, r \to 0$ results in the $\log (z - 1) $ to blow up. What am I missing? $\endgroup$ – Ant Dec 17 '14 at 12:18
  • $\begingroup$ @Ant: I've added a small circle about $1$. Since $\log(x)$ is locally integrable, integrating up to the singularity can usually be ignored, but there's nothing wrong with being a bit more careful. $\endgroup$ – robjohn Dec 17 '14 at 14:02
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    $\begingroup$ @Ant $\log(z−1)$ blows up (in the little circle) as $\log r$ but the length of the integration path decreases as $r$, hence the integral tends to zero. $\endgroup$ – leonbloy Dec 17 '14 at 14:06
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{1}^{\infty}{\dd x \over x^{3} + 1} = {\root{3} \over 9}\,\pi - {\ln\pars{2} \over 3} \approx {\tt 0.3736}:\ {\large ?}}$

\begin{align}&\overbrace{\color{#66f}{\large% \int_{1}^{\infty}{\dd x \over x^{3} + 1}}} ^{\ds{\dsc{x}=\dsc{1 \over t}\ \imp\ \dsc{t}=\dsc{1 \over x}}}\ =\ \int_{0}^{1}{t\,\dd x \over 1 + t^{3}} =\sum_{n\ =\ 0}^{\infty}\pars{-1}^{n}\int_{0}^{1}t^{3n + 1}\,\dd t =\sum_{n\ =\ 0}^{\infty}{\pars{-1}^{n} \over 3n + 2} \\[5mm]&=\sum_{n\ =\ 0}^{\infty}\pars{{1 \over 6n + 2} - {1 \over 6n + 5}} =3\sum_{n\ =\ 0}^{\infty}{1 \over \pars{6n + 2}\pars{6n + 5}} \\[5mm]&={1 \over 12}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{n + 1/3}\pars{n + 5/6}} ={1 \over 12}\,{\Psi\pars{1/3} - \Psi\pars{5/6} \over 1/3 - 5/6} \\[5mm]&=\color{#66f}{\large% {1 \over 6}\bracks{\Psi\pars{5 \over 6} - \Psi\pars{1 \over 3}}} \end{align}

With Gauss Digamma Theorem: \begin{align} \Psi\pars{5 \over 6}& =-\gamma + {\root{3} \over 2}\,\pi - 2\ln\pars{2} - {3 \over 2}\,\ln\pars{3} \\[5mm] \Psi\pars{1 \over 3}& =-\gamma - {\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3} \end{align}

Finally, $$\color{#66f}{\large% \int_{1}^{\infty}{\dd x \over x^{3} + 1}} =\color{#66f}{\large% {\root{3} \over 9}\,\pi - {\ln\pars{2} \over 3}} $$

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  • $\begingroup$ Now, while this does not help with a contour line I still find it very cool. Only thing, I did not understand this passage ${1 \over 12}\sum_{n\ =\ 0}^{\infty}{1 \over \pars{n + 1/3}\pars{n + 5/6}} ={1 \over 12}\,{\Psi\pars{1/3} - \Psi\pars{5/6} \over 1/3 - 5/6}$ $\endgroup$ – Ant Dec 30 '14 at 21:55
  • $\begingroup$ @Ant For example, compute $\Psi\left(\, a\,\right) - \Psi\left(\, b\,\right)$ with identity ${\bf 6.3.16}$. Indeed, that difference is a well known identity. $\endgroup$ – Felix Marin Dec 30 '14 at 22:00
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Hint:

$$\frac{1}{x^3+1} = \frac{1}{(x+1)(x^2-x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-x+1}$$

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    $\begingroup$ Does this help to find a contour line? $\endgroup$ – Krokop Dec 16 '14 at 13:54
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    $\begingroup$ @Krokop It helps to calculate the integral. Contour lines are not needed. $\endgroup$ – 5xum Dec 16 '14 at 14:00
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    $\begingroup$ Exactly but the op needs a contour line.. $\endgroup$ – Krokop Dec 16 '14 at 14:02

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