<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

Use the calculus of residues to show that

\begin{itemize}
\item[a)]
$\ds\int_0^\infty\frac{dx}{1+x^4}=\frac{\pi}{2\sqrt2}$


\item[b)]
$\ds\int_0^\infty\frac{x^\frac{1}{2}}{(1+x)^2}dx=\frac{\pi}{2}$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $\ds f(z)=\frac{1}{1+z^4}$, this has simple poles at the
fourth roots of $-1$,

i.e. $\ds\frac{\pm1\pm i}{\sqrt2}$

Using $\Gamma$, the poles inside $\Gamma$ are at
$\ds\frac{1+i}{\sqrt2}$ and $\ds\frac{-1+i}{\sqrt2}$ $(R&gt;1)$

Res$\ds\left(\frac{1+i}{\sqrt2}\right)=\frac{-1-i}{4\sqrt2}
\hspace{0.5in}$
Res$\ds\left(\frac{-1+i}{\sqrt2}\right)=\frac{1-i}{4\sqrt2}$

$\ds\int_{\Gamma}f(z)dz=2\pi i\left(\frac{-1-i}{4\sqrt2}+
\frac{1-i}{4\sqrt2}\right)=\frac{\pi}{\sqrt2}$

$\ds\left|\int_{{\rm
semicircle}}\frac{1}{1+z^4}dz\right|\leq\frac{\pi R}{R^4-1}\to0$
as $R\to\infty$.

Thus $\ds\int_{-\infty}^\infty\frac{dx}{1+x^4}=\frac{\pi}{\sqrt2}$
and so $\ds\int_0^\infty\frac{dx}{1+x^4}=\frac{\pi}{2\sqrt2}$


\item[b)]
Let $y=x^\frac{1}{2} \hspace{0.3in} y&gt;0$ so $x=y^2$ and $dx=2ydy$

$\ds\int_0^\infty\frac{x^\frac{1}{2}}{(1+x)^2}dx=
\int_0^\infty\frac{2y^2}{(1+y^2)^2}dy$

Let $\ds f(z)=\frac{2z^2}{(1+z^2)^2}dz$, this has a pole of order
2 at $z=i$,

with residue $\ds\frac{-i}{2}$, using diffn formula.

So using $\Gamma$ with $R&gt;1$,

$\ds\int_{\Gamma}f(z)dz=2\pi i\left(\frac{-i}{2}\right)=\pi$

Now on the semicircle $\ds|f(z)|\leq\frac{2R^2}{(R^2-1)^2}$

So $\ds\left|\int_{{\rm semicircle}}f(Z)dz\right|\leq\frac{\pi
R2R^2}{(R^2-1)^2}\to0$ as $R\to\infty$.

Thus we have
$\ds\int_{-\infty}^\infty\frac{2y^2}{(1+y^2)^2}dy=\pi$

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

\end{itemize}

\end{document}
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