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QUESTION

Evaluate the following real integrals by using the
residue theorem:

$$\textrm{(i)}\int_{-\infty}^\infty \frac{dx}{1+x^8}\ \
\textrm{(ii)}\ \ \int_{-\infty}^\infty \frac{dx}{(1+x^2)^2}\ \
\textrm{(iii)}\ \ \int_{-\infty}^\infty
\frac{dx}{(x^2+4)(x^2+9)}$$


You can close the contour either in the lower or upper half here,
so take the upper half.

\bigskip

ANSWER

\begin{description}

\item[(i)]
 $\ds I=\int_{-\infty}^\infty\frac{dx}{1+x^8}=\int_C\frac{dz}{1+z^4}$

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 \epsfig{file=272-2-2.eps, width=60mm}
 \end{center}

 The integration contour encloses simple poles at
 $$z_0=e^{\frac{i\pi}{8}},\ e^\frac{3i \pi}{8},\ e^\frac{5i \pi}{8},\ e^\frac{7 i
 \pi}{8}$$

 $$ \textrm{Res}\left(\frac{1}{1+z^8},z_0\right)  =  \frac{1}{8z_0^7}  =  \frac{z_0}{8z_0^8}
 =  -\frac{1}{8}z_0$$

 \begin{eqnarray*}
 I&=&2 \pi i \left(-\frac{1}{8}\right)\left(e^\frac{i \pi}{8}+e^\frac{3 i
 \pi}{8}++e^\frac{5 i \pi}{8}+e^\frac{7 i \pi}{8}\right)\\
 &=&-\frac{\pi i}{4}\left(e^{i \pi}{8}+e^\frac{3 i \pi}{8}-e^{-\frac{3i
 \pi}{8}}-e^{-\frac{i \pi}{8}}\right)\\
 &=&\frac{\pi}{2}\left(\sin \frac{\pi}{8}+\sin \frac{3 \pi}{8}\right)
 \end{eqnarray*}

\item[(ii)]
 $\ds I=\int_{-\infty}^\infty
 \frac{dx}{(1+x^2)^2}=\int_{\gamma}\frac{dz}{(z+i)^2(z-i)^2}$
 has double poles at $z=\pm i$, but only $z=i$ is inside the
 contour.

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 \begin{eqnarray*}
 \textrm{Res}\left(\frac{1}{(1+z^2)},i\right)&=&\lim_{z \to
 i}\frac{d}{dz}\frac{1}{(z+i)^2}\\
 &=&\lim_{z \to i}-\frac{2}{(z+i)^3}=-\frac{i}{4}\\
 I&=&2 \pi i \left(-\frac{i}{4}\right)=\frac{\pi}{2}
 \end{eqnarray*}

\item[(iii)]
 $\ds I=\int \frac{dx}{(x^2+4)(x^2+9)}$ has simple poles inside the
 contour at $z=2i$ and $3i$.
 \begin{eqnarray*}
 \textrm{Res}(2i)&=&\frac{1}{(2i+2i)(2i+3i)(2i-3i)}=-\frac{i}{20}\\
 \textrm{Res}(3i)&=&\frac{1}{(3i+3i)(3i+2i)(3i-2i)}=-\frac{i}{30}\\
 I&=&2 \pi i\left(-\frac{i}{20}-\frac{i}{30}\right)=\frac{\pi}{30}
 \end{eqnarray*}

\end{description}

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