TITLE: Evaluate $\int_1^\infty \frac {dx}{x^3+1}$
QUESTION [16 upvotes]: 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?

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

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