Evaluate $\int_1^\infty \frac {dx}{x^3+1}$

Question

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?

Answer 1

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.

Answer 2

Hint:

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

Answer 3

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, $$ \int_1^\infty\frac{\mathrm{d}x}{1+x^3}=\frac\pi{3\sqrt3}-\frac{\log(2)}3 $$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{1}^{\infty}{\dd x \over x^{3} + 1} = {\root{3} \over 9}\,\pi - {\ln\pars{2} \over 3} \approx {\tt 0.3736}:\ {\large ?}}$

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