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\begin{document}

{\bf Question}

Verify the following integral.

$\ds\int_0^{\infty} \ds\frac{dx}{x^6+1}=\ds\frac{\pi}{3}$
\medskip

{\bf Answer}

Consider $\ds\oint_C \ds\frac{dz}{z^6+1}$

PICTURE \vspace{2in}

Now $z^6+1=0$ when

$z=e^{\frac{i\pi}{6}},\ e^{\frac{3i\pi}{6}},\
e^{\frac{5i\pi}{6}},\ e^{\frac{7i\pi}{6}},\ e^{\frac{9i\pi}{6}},\
e^{\frac{11i\pi}{6}}$

being simple poles of the integrand.

Only $e^{\frac{i\pi}{6}},\ e^{\frac{i\pi}{2}},\
e^{\frac{5i\pi}{6}}$ are included in $C$.

Then we have

$$J=2\pi i \sum\ \rm{residues\ at}\ e^{\frac{i\pi}{6}},\
e^{\frac{i\pi}{2}},\ e^{\frac{5i\pi}{6}}$$

\begin{itemize}
\item

${}$

$\begin{array} {rcl} \rm{Residue}(e^{\frac{i\pi}{6}}) & = &
\lim_{z \to e^{\frac{i\pi}{6}}}
\left[(z-e^{\frac{i\pi}{6}}\ds\frac{1}{z^6+1}\right]\\ & &
\rm{Use\ l'Hopital!}\\ & = & \lim_{z \to
e^{\frac{i\pi}{6}}}\left(\ds\frac{1}{6z^5}\right)\\ & = &
\ds\frac{1}{6}e^{\frac{-5i\pi}{6}} \end{array}$

\item

${}$

$\begin{array} {rcl} \rm{Residue}(e^{\frac{i\pi}{2}}) & = &
\lim_{z \to e^{\frac{i\pi}{2}}}
\left[(z-e^{\frac{i\pi}{2}}\ds\frac{1}{z^6+1}\right]\\ & &
\rm{Use\ l'Hopital!}\\ & = & \lim_{z \to
e^{\frac{i\pi}{2}}}\left(\ds\frac{1}{6z^5}\right)\\ & = &
\ds\frac{1}{6}e^{\frac{-5i\pi}{2}} \end{array}$

\item

${}$

$\begin{array} {rcl} & & \rm{Residue}(e^{\frac{5i\pi}{6}})\\ & = &
\lim_{z \to e^{\frac{5i\pi}{6}}}
\left[(z-e^{\frac{5i\pi}{6}}\ds\frac{1}{z^6+1}\right]\\ & &
\rm{Use\ l'Hopital!}\\ & = & \lim_{z \to
e^{\frac{5i\pi}{6}}}\left(\ds\frac{1}{6z^5}\right)\\ & = &
\ds\frac{1}{6}e^{\frac{-25i\pi}{6}} \end{array}$

\end{itemize}
Thus

$$\ds\oint\ds\frac{dz}{z^6+1}=2\pi i \left[\ds\frac{1}{6}
e^{\frac{-5i\pi}{6}}+\ds\frac{1}{6}e^{\frac{-5i\pi}{2}}
+\ds\frac{1}{6}e^{\frac{-25i\pi}{6}}\right]=\ds\frac{2\pi}{3}$$

Now let $R \to \infty$ and contribution from semicircle, $k$ is
given by

\begin{eqnarray*} |k| & = & iR\ds\int_0^{\pi}
\ds\frac{d\theta}{R^6e^{i6\theta}+1}
\\ & \leq & R\ds\int_0^{\pi}\ds\frac{d\theta}{R^6e^{6i\theta+1}}\\ &
\leq & \ds\frac{R}{|R^6-1|}\ds\int_0^{\pi}d\theta\\ & = &
\ds\frac{\pi R}{R^6-1}\end{eqnarray*}

Thus $\lim_{R \to \infty} |k| \leq \lim_{R \to \infty}
\ds\frac{\pi R}{R^6-1}=0$ so no contribution from semicircle.

Thus

\begin{eqnarray*} \ds\int_{-\infty}^{+\infty}
\ds\frac{dz}{z^6+1}=\ds\frac{2\pi}{3} & \Rightarrow &
2\ds\int_0^{\infty} \ds\frac{dx}{x^6+1}=\ds\frac{2\pi}{3}\\ &
\Rightarrow & \ds\int_0^{\infty}
\ds\frac{dx}{x^6+1}=\ds\frac{\pi}{3} \end{eqnarray*}

\end{document}