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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Give the Laurent series expansions in powers of $z$ for

$$ f(z)=\frac{1}{z^2(z-2)}$$

in each of the two regions $0<|z|<2$ and $|z|>2$.

Hence, or otherwise, evaluate $\ds\int_{C_r}f(z)dz$, where $C_r$
is the circle $|z|=r$ in the clockwise sense in each of the cases
$0<r<2$ and $2<r$.


\item[b)]
Use the calculus of residues to show that

$$\int_0^\infty\frac{dx}{(x^2+1)^2}=\frac{\pi}{4}.$$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
For $0<|z|<2$

$\ds\frac{1}{z^2(z-2)}=\frac{1}{z^2}
\left(\frac{-1}{2\left(1-\frac{z}{2}\right)}\right)=
\frac{-1}{2z^2}\left(1+\frac{z}{2}+\left(\frac{z}{2}\right)^2
+\cdots\right)$

$\ds\hspace{0.5in}=\frac{-1}{2z^2}-\frac{1}{4z}-\frac{1}{8}-
\frac{z}{16}-\cdots-\frac{z^n}{2^{n+3}}-\cdots$

${}$

For $|z|>2$

$\ds\frac{1}{z^2(z-2)}=
\frac{1}{z^3\left(1-\frac{2}{z}\right)}=\frac{1}{z^3}
\left(1+\frac{2}{z}+\left(\frac{2}{z}\right)^2+\cdots\right)$

$\ds\hspace{0.5in}=\frac{1}{z^3}+\frac{2}{z^4}+{4}{z^5}+
\cdots+\frac{2^{n-3}}{z^n}+\cdots$

$\ds\int_{C_r}f(z)dz=2\pi i\sum$residues within $C_r$

The residue at $z=2$ is $\ds\lim_{z\to2}z-2f(z)=\frac{1}{4}$

The residue at $z=0$ is $-\frac{1}{4}$ from the first expansion.

So if $\ds r<2, \oint_{C_r}f(z)dz=-2\pi
i\frac{1}{4}=-\frac{\pi}{2}$ (going clockwise)

Thus $\oint_{C_r}f(z)dz=\frac{\pi i}{2}$ (going anticlockwise)

In the case $r>2$ the sum of the residues is zero,

So $\int_{C_r}f(z)dz=0$ in both directions.


\item[b)]
Let $\ds f(z)\frac{1}{(z^2+1)^2}$, now
$\ds\frac{1}{(x^2+1)^2}\leq\frac{1}{x^4}$ and
$\ds\int^\infty\frac{1}{x^4}$ converges.

Integrate $f(z)$ round $\Gamma$, with $R>1$

$f(z)$ has a pole of order 2 at $z=i$ inside $\Gamma$, with
residue $-\frac{1}{4}$ using the diffn formula.

Thus $\ds\int_{\Gamma}f(z)dz=\frac{\pi}{2}$

$\ds\left|\int_{{\rm semicircle}}f(z)dz\right|\leq\frac{\pi
R}{(R^2-1)^2}\to0$ as $R\to\infty$

Thus $\ds\int_{-\infty}^\infty f(x)dx=\frac{\pi}{2}$ and so
$\ds\int_0^\infty\frac{1}{(x^2+1)^2}dx=\frac{\pi}{4}$

\end{itemize}

\end{document}
