Keyhole Integration Complex Analysis For the following equation 
$$
I = \int_{0}^{\infty }\frac{\log(x)}{1+x^{3}}dx
$$ 
using 
$$
f(z) = \frac{\log(z)}{1+z^{3}}
$$ 


*

*Draw the contour and any poles (Hint: It is a keyhole contour, draw and label it)

*Prove that the residue is $= \frac{-4\sqrt{3}}{9}*\pi ^{2}i $ (Hint: Calculate the residue $\operatorname{Res}(f(z)$)

*Show that it is a function of a real space (Principal Value) from $0 > \infty  \oint \frac{1}{1+x^{3}}$   the answer is $= \frac{2\sqrt{3}}{9} $ (Hint: Just show that the Principal Value of the orginal function is equal to$ \frac{2\sqrt{3}}{9})$
Context: My professor did this in a lecture a while ago and he didn't really take it up. I went to his office hours and he didn't explain it properly. I don't understand how to deal with the $\log(x)/(1 + x^3)$ as usually it isn't that complex. He gave us the answers which I put as the hints. Additionally, this is a keyhole contour problem. He said assume regular bounds as well. 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Lets consider $\ds{\int_{\large z\ =\ R\expo{\ic\pars{-\pi + \epsilon,\pi - \epsilon}}}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z}$
  where $\ds{R > 1}$ and $\ds{0 < \epsilon < \pi}$.
  The $\ds{\ln}$-branch-cut is defined by
  $\ds{\ln\pars{z} = \ln\pars{\verts{z}} + \arg\pars{z}\ic\,,\quad
\arg\pars{z} \in \pars{-\pi,\pi}\,,\quad z \not = 0}$.

Then,
\begin{align}
&\int_{\large z\ =\ R\expo{\ic\pars{-\pi + \epsilon,\pi - \epsilon}}}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
\oint_{\mc{C}}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z
\\[2mm] &\
-\int_{-R}^{-\epsilon}{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over
x^{3} - 1}\,\dd x
- \int_{-\epsilon}^{-R}{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over
x^{3} - 1}\,\dd x
\\[2mm] & -
\int_{\pi}^{-\pi}{\bracks{\ln\pars{\epsilon} + \ic\theta}^{2} \over
\epsilon^{3} - 1}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta
\end{align}



*

*The integral in the LHS $\ds{\to 0}$ as $\ds{R \to \infty}$.

*$\mc{C}$ is a key-hole contour wich "takes cares" of the above mentioned branch-cut.  

*The last integral, in the RHS, $\ds{\to 0}$ as $\ds{\epsilon \to 0^{+}}$.

*Poles of $\ds{\ln^{2}\pars{z} \over z^{3} - 1}$ belong to
$\ds{\braces{\expo{-2\pi\ic/3},\expo{2\pi\ic/3}}}$.

We are left with
\begin{align}
0 & = 2\pi\ic\sum\mrm{Res}
\bracks{\ln^{2}\pars{z} \over z^{3} - 1} +
\int_{0}^{\infty}{\bracks{\ln\pars{x} + \ic\pi}^{\, 2} -
\bracks{\ln\pars{x} - \ic\pi}^{\, 2} \over x^{3} + 1}\,\dd x
\\[5mm] & =
2\pi\ic\sum_{r\ \in\ \braces{\expo{-2\pi\ic/3},\expo{2\pi\ic/3}}}
{1 \over 3}\,r\ln^{2}\pars{r} +
4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1}\,\dd x
\\[5mm] & =
2\pi\ic\
\underbrace{\bracks{{1 \over 3}\, 2\Re\pars{\expo{2\pi\ic/3}}
\pars{2\pi\ic \over 3}^{2}}}_{\ds{4\pi^{2} \over 27}}\  +
4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1}\,\dd x
\\[8mm] & \implies
\bbx{\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1}\,\dd x =
-\,{2\pi^{2} \over 27}} \approx -0.7311
\end{align}
