<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}
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\newcommand\ds{\displaystyle}
QUESTION Use the keyhole contour to evaluate
$$\int_0^\infty\frac{x^\frac{1}{2}dx}{1+x^3}$$

\bigskip

ANSWER
 $$I=\int_0^\infty\frac{x^\frac{1}{2}}{1+x^3}\,dx,\ \
 J=\int_C\frac{z^\frac{1}{2}}{1+z^3}\,dz$$

 \begin{center}
 \epsfig{file=362-2-4.eps, width=110mm}
 \end{center}

 Simple poles at $\ds z_0=1,e^{\pm i\frac{\pi}{3}}\\
 \textrm{Res}\left(z_0\right)=\frac{z_0^\frac{1}{2}}{3z_0^2}=\frac{z_0^{-\frac{1}{2}}}{3z_0^3}
 =-\frac{1}{3}z_0^{-\frac{1}{2}}$

$\ds \int_{\gamma_1}=I,\ \left|\int_{\gamma_2}\right|\leq
 \frac{R^\frac{1}{2}}{R^3-1}\pi R\sim R^{-\frac{3}{2}}\to 0 \textrm{
 as}  R\to \infty$


$\ds\int_{\gamma_3}=\int_\infty^0\frac{-x^\frac{1}{2}}{1+x^3}\,dx=I$

$ \ds\left|\int_{\gamma_4}\right|\leq
\frac{r^\frac{1}{2}}{1-r^3}\pi r \sim
 r^\frac{3}{2} \to 0 \textrm{ as }r \to 0$


$\ds\Rightarrow J=2I=2 \pi
 i\left(-\frac{1}{3}\right)\left(\left(e^\frac{i\pi}{3}\right)^\frac{1}{2}
 +\left(e^{i\pi}\right)^\frac{1}{2}
 +\left(e^\frac{5i\pi}{3}\right)^\frac{1}{2}\right)$

 $\ds =-\frac{2\pi
 i}{3}\left(e^\frac{i\pi}{6}+i-e^{-\frac{i\pi}{6}}\right)=\frac{2\pi}{3}+\frac{4\pi}{3}\sin
 \frac{\pi}{6}=\frac{2\pi}{3}+\frac{4\pi}{3}\frac{1}{2}=\frac{\pi}{3}$

 (Note that we measure all arguments from the positive real axis.)

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