<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
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\begin{document}

{\bf Question}

Verify the following integral.

$\ds\int_0^1
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}=\ds\frac{2\pi}{\sqrt{3}}$

(Hint: There is a branch cut between $x=0,\ x=1$.)
\medskip

{\bf Answer}

Consider $J=\ds\int_C \ds\frac{dz}{(z^2-z^3)^\frac{1}{3}}$

PICTURE \vspace{2in}

This is equal to $\ds\int \ds\frac{dz}{(z^2-z^3)^{\frac{1}{3}}}$
by Cauchy's theorem.

PICTURE \vspace{1in}

\begin{eqnarray*} \ds\int_{C_1} &amp; = &amp; \ds\int_1^0
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}+\undb{\ds\int_0^{e^{-2\pi
i}} \ds\frac{dz}{(z^2-z^3)^{\frac{1}{3}}}}\\ &amp; &amp; \ \ \ \ \ \ \ \ \
z=xe^{-2\pi i}\ (\rm{defining}\ z\ \rm{on}\ -2\pi&lt;\arg(z) \leq 0\
(\star))\\ &amp; = &amp; \ds\int_1^0
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}+\ds\int_0^1
\ds\frac{dx}{x^{\frac{2}{3}}(1-x)^{\frac{1}{3}}e^{\frac{-2\pi
i}{3}}}\\ &amp; = &amp; (e^{\frac{2\pi i}{3}}-1)\ds\int_0^1
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}\\ &amp; = &amp;
e^{i\frac{\pi}{3}}(e^{i\frac{\pi}{3}}-
e^{-i\frac{\pi}{3}})\ds\int_0^1\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}
\\ &amp; = &amp; 2ie^{i\frac{\pi}{3}}\sin\ds\frac{\pi}{3} \times \ds
\int_0^1 \ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}} \\ &amp; = &amp; i \sqrt{3}
e^{i\frac{\pi}{3}} \ds\int_0^1
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}} \end{eqnarray*}

Now $\ds\int_C dz=\ds\int_0^{2\pi} i\ds\frac{Re^{i\theta}\
d\theta}{(R^2e^{2i\theta}-R^3e^{3i\theta})^{\frac{1}{3}}}$

Now let $R \to \infty$ we still have $\lim_{R \to\infty}
J=\ds\int_{C_1} \ds\frac{dz}{(z^2-z^3)^{\frac{1}{3}}}$

\begin{eqnarray*} \lim_{R \to \infty}J &amp; = &amp; \lim_{R \to \infty}
\ds\frac{iR}{R} \ds\int_0^{2\pi} \ds\frac{d\theta\
e^{i\theta}}{e^{i\theta}}\left(\frac{e^{-i\theta}}{R}-1\right)^{\frac{1}{3}}\\
&amp; = &amp; i\ds\int_0^{2\pi}d\theta\ \undb{e^{+i\frac{\pi}{3}}} = 2\pi
i e^{+i \frac{\pi}{3}} \end{eqnarray*}

\hspace{1in} $\ds\frac{1}{(-1)^{\frac{1}{3}}}$ with
$(-1)=e^{i\pi}$. Why? This is consistent with branch $(\star)$
chosen

Thus

$i\sqrt{3}e^{\frac{\pi i}{3}}\ds\int_0^21
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}=2\pi ie^{+\frac{i\pi}{3}}$

$\Rightarrow \ds\int_0^1
\ds\frac{dx}{(x^2-x^3)^{\frac{1}{3}}}=\ds\frac{2\pi}{\sqrt{3}}$

\end{document}
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