Keyhole contour for $\frac{(\log{x})^2}{1+x^3}$

Question

I am trying to evaluate the integral $\displaystyle \int_0^{\infty}\frac{\log x}{x^3+1}dx$ and I was trying the keyhole contour with $\displaystyle f(z)=\frac{(\log z)^2}{z^3+1}$

The large circular contour $\gamma_R:t\mapsto Re^{it}$ and the small circular contour $\gamma_{\epsilon}:t\to \epsilon e^{it}$ around $0$ both have integrals which tend to $0$ as we let $R\to \infty$ and $\epsilon \to 0$.

The three residues at $z=-1, e^{i\pi/3}, e^{-i\pi/3}$ are

$\displaystyle \frac{(\log i)^2}{3i^2}=-\frac{\pi^2}{3}, \frac{(\log e^{i\pi/3})^2}{3(e^{i\pi/3})^2}=-\frac{\pi^2}{27}e^{-2i\pi/3}$ and $\displaystyle -\frac{\pi^2}{27}e^{2i\pi/3}$

The path above the branch (i.e. the positive real axes) converges to $\displaystyle \int_0^{\infty}\frac{(\log x)^2}{x^3+1}dx$ and the path below the branch $\displaystyle -\int_0^{\infty}\frac{(\log x+2i\pi)^2}{x^3+1}dx$

So when we add everything up we get $\displaystyle -\int_0^{\infty}\frac{4i\pi\log x-4\pi^2}{x^3+1}dx=2i\pi[-\pi^2/3-\pi^2/27(e^{-2\pi i/3}+e^{2\pi i/3})]=2i\pi\left(-\frac{\pi^2}{3}+\frac{\pi^2}{27}\right)$

By taking the imaginary part, this would (incorrectly) seem to imply that $\displaystyle 4\pi\int_0^{\infty}\frac{\log x}{x^3+1}dx=-2\pi\frac{8\pi^2}{27}$

$\displaystyle \int_0^{\infty}\frac{\log x}{x^3+1}dx=\frac{4\pi^2}{27}$

(The real answer is $\frac{2\pi^2}{27}$)

This also (again incorrectly) implies that

$\displaystyle \int_0^{\infty}\frac{1}{x^3+1}dx=0$

(The real answer is $\frac{2\pi}{3\sqrt{3}}$)

as the real part is $0$ on the right hand side.

I have verified that these answers are in fact wrong.

I appreciate that there are alternative contours to evaluate this integral but I would like to know what I missed in this particular computation.

Answer 1

Your problem is that the pole at $z=e^{-i \pi/3}$ should be $z=e^{i 5 \pi/3}$, because you chose your branch cut such that $\arg{z} \in [0,2 \pi)$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

• Hereafter, $\ds{\ln}$ is the logarithm principal branch and $\ds{\cal C}$ is a key-hole contour which "takes care" of the above mentioned branch-cut.
• Contributions from the big arc and the smal arc around $\ds{z = 0}$ vanish out in the,respectively, radius $\ds{\to \infty}$ and radius $\ds{\to 0^{+}}$.
• Poles of the integrand are, $\underline{according}$ to the above mentioned branch-cut, are $\ds{r = \exp\pars{\pm{2\pi \over 3}\,\ic}}$.

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Namely, \begin{align} &\bbox[5mm,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z} \\ = &\ 2\pi\ic\sum_{r = \exp\pars{\pm 2\pi\ic/3}}\,\,\,\, \lim_{z \to r}\bracks{\pars{z - r}\,{\ln^{2}\pars{z} \over z^{3} - 1}} \\[5mm] = &\ 2\pi\ic\sum_{r = \exp\pars{\pm 2\pi\ic/3}}\,\,\,\, \bracks{-4\pi^{2}/9 \over 3r^{2}} = -\,{8\pi^{3} \over 27}\,\ic \sum_{r = \expo{\pm 2\pi\ic/3}}\,\,\,\,r \\[5mm] = &\ -\,{16\pi^{3} \over 27}\,\ic\cos\pars{2\pi \over 3} = {8\pi^{3} \over 27}\,\ic\label{1}\tag{1} \end{align}

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\begin{align} &\bbox[5mm,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over z^{3} - 1}\,\dd z} \\ = &\ \int_{-\infty}^{0}{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over x^{3} - 1}\,\dd x + \int_{0}^{-\infty}{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over x^{3} - 1}\,\dd x \\[3mm] = &\ -\int_{0}^{\infty}{\bracks{\ln\pars{x} + \ic\pi}^{2} \over x^{3} + 1}\,\dd x + \int_{0}^{\infty}{\bracks{\ln\pars{x} - \ic\pi}^{2} \over x^{3} + 1}\,\dd x \\[5mm] = & -4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1} \,\dd x\label{2}\tag{2} \end{align}

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(\ref{1}) and (\ref{2}) yield: \begin{align} &\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1} \,\dd x = {8\pi^{3}\,\ic\,/27 \over -4\pi\ic} = \bbx{-\,{2\pi^{2} \over 27}} \approx -0.7311 \\ & \end{align}