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\ifsolutions
\title[Math 618 (Spring 2024): Solutions~5]{Math 618 (Spring 2024,
        Phillips): Solutions to Homework~5}
\date{8~May 2024}
\else
\title[Math 618 (Spring 2024): Homework~5]{Math 618 (Spring 2024,
        Phillips): Homework~5}
\date{1~May 2024}
\fi
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\begin{document}

%  \ifsolutions
%  \else
%  \vspace*{-1ex}
%  \fi

\maketitle

% \setcounter{section}{-1}

\indent

\ifsolutions
This assignment is due on Canvas on Wednesday 8~May 2024 at 9:00 pm.

Problems and all other items use two independent numbering sequences.
This is annoying, but necessary to preserve
the problem numbers in the solutions files.

Little proofreading has been done.

Some parts of problems have several different solutions.

Solutions are written to be read independently.
Arguments used in more than one solution are therefore
repeated in each one.
\fi


%DIF > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DIFaddbegin 

\DIFaddend \begin{pbm}[Problem~12
  in Chapter~10 of Rudin's book]\label{Pb_2024_5_10_12}
For $t \in \R$, use the Residue Theorem to compute
\[
\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x.
\]
\end{pbm}

\DIFdelbegin %DIFDELCMD < \ifsolutions
%DIFDELCMD < \bigskip
%DIFDELCMD < 

%DIFDELCMD < %%%
\DIFdel{No full solution yet written,
but here is an outline.
(Also, compare }\DIFdelend \DIFaddbegin \DIFadd{Compare }\DIFaddend with Rudin Chapter~9 Problem~2.
\DIFdelbegin \DIFdel{)
}\DIFdelend 


\DIFdelbegin %DIFDELCMD < \begin{proof}[Solution (sketch)]
%DIFDELCMD < %%%
\DIFdel{For $a \in (1, \I)$, }\DIFdelend \DIFaddbegin \ifsolutions
%DIF >  \bigskip

\begin{proof}[Solution (with a few steps just sketched)]
\DIFadd{for $t \in \R$ }\DIFaddend define
\DIFdelbegin \DIFdel{paths by $\sm_a (t) = a e^{i t}$ for $t \in [0, \pi]$,
$\ta_a (t) = a e^{i t}$ for $t \in [\pi, \, 2 \pi]$,
$\rh (t) = e^{i t}$ for $t \in [\pi, \, 2 \pi]$,
}\DIFdelend \DIFaddbegin \[
\DIFadd{t_t (z)
 = \begin{cases}
   \left( \frac{\sin (z)}{z} \right)^2 e^{i t z} & \hspace*{1em}
        z \in \C \SM \{ 0 \}
        \\
   1 & \hspace*{1em} z = 0.
\end{cases}
}\]
\DIFadd{Then $f_t$ is an entire function.
Also, $f_t$ is integrable on~$\R$
because $| f_t (x) | \leq x^{- 2}$ when $| x | \geq 1$
and $f_t$ is bounded on $[-1, 1]$ (since $[-1, 1]$ is compact).
Further, for $s \in \R$ define $g_s \colon \C \SM \{ 0 \} \to \C$
by $g_s (z) = e^{i s z} / z^2$.
Using the relation
}\[
[ \DIFadd{\sin (z) }]\DIFadd{^2 = \left[ \frac{1}{2 i} (e^{i z} - e^{- i z} )\right]^2,
}\]
\DIFadd{one checks that
}\[
\DIFadd{f_t (z) = - \left( \frac{1}{4} \right) g_{t + 2} (z)
            + \left( \frac{1}{2} \right) g_{t} (z)
            - \left( \frac{1}{4} \right) g_{t - 2} (z)
}\]
\DIFadd{for $z \in \C \SM \{ 0 \}$.
}

\DIFadd{For $a > 1$ define (one of these does not depend on~$a$):
%DIF > 
}\begin{enumerate}
%DIF >  [label=$\mathrm{(\arabic*)}$]
%DIF > 
\item\label{I_2024_5_10_12_sm}
\DIFadd{$\sm_a (\te) = a e^{i \te}$ for $\te \in [0, \pi]$
%DIF > 
}\item\label{I_2024_5_10_12_ta}
\DIFadd{$\ta_a (\te) = a e^{i \te}$ for $\te \in [\pi, \, 2 \pi]$.
%DIF > 
}\item\label{I_2024_5_10_12_rh}
\DIFadd{$\rh (\te) = e^{i \te}$ for $\te \in [0, \, \pi]$.
%DIF > 
}\item\label{I_2024_5_10_12_af}
\DIFaddend $\af_a (t) = t$ for $t \in [1, a]$\DIFdelbegin \DIFdel{,
and }\DIFdelend \DIFaddbegin \DIFadd{.
%DIF > 
}\item\label{I_2024_5_10_12_bt}
\DIFaddend $\bt_a (t) = t$ for $t \in [-a, \, - 1]$.
\DIFdelbegin \DIFdel{Let $\gm_a$ be the chain $\af_a + \bt_a + \rh$.
Set
}\[
\DIFdel{f_t (z) = \left( \frac{\sin (z)}{z} \right)^2 e^{i t z}
}\]%DIFAUXCMD
\DIFdelend %DIF > 
\DIFaddbegin \item\label{I_2024_5_10_12_io}
\DIFadd{$\io_a (t) = t$ }\DIFaddend for \DIFdelbegin \DIFdel{$z \in \C \SM \{ 0 \}$, which we can rewrite as }\DIFdelend \DIFaddbegin \DIFadd{$t \in [-a, \, a]$.
%DIF > 
}\end{enumerate}
%DIF > 
\DIFadd{Define $\Gm_a = [\af_a] - [\rh] + \bt_a]$.
Then
}\[
\DIFadd{\Gm_a + }[\DIFadd{\sm_a}]\DIFadd{, \qquad
\Gm_a - }[\DIFadd{\ta_a}]\DIFadd{, \andeqn
\Gm_a - }[\DIFadd{\io_a}]
\DIFadd{}\]
\DIFadd{are immediately seen to be cycles.
}

\DIFadd{For $a > 1$ and $s \in \R$, define
}\[
\DIFadd{\ph_a (s) = \int_{\Gm_a} g_s (z) \, d z.
}\]
\DIFadd{We claim that $\ps (s) = \limi{a} \ph_s (a)$ exists and is given by
}\[
\DIFadd{\ps (s)
 = \begin{cases}
   0 & \hspace*{1em} s > 0
        \\
   - 2 \pi s & \hspace*{1em} s < 0.
\end{cases}
}\]

\DIFadd{We prove the claim for $s \geq 0$.
By considering the negative imaginary axis,
one sees that $0$ is in the unbounded component of
$\Ran (\Gm_a) \cup \Ran (\sm_a)$.
Therefore $\Ind_{\Gm_a + [\sm_a]} (0) = 0$.
(Something must be said here.)
Using Cauchy's Theorem at the fist step, we then get
%DIF > 
}\begin{equation}\DIFadd{\label{Eq_24611_s_pos}
\begin{split}
\int_{\Gm_a} g_s (z) \, d z
& = - \int_{\sm_a} g_s (z) \, d z
  = - \int_0^{\pi}
   \frac{\exp (i s \sm_a (\te) ) i a e^{i \te}}{\sm_a (\te)^2}
       \sm_a' (\te) \, d \te
\\
& = - \int_0^{\pi}
   \frac{\exp (i s a e^{i \te} ) i a e^{i \te}}{(a e^{i \te})^2} \, d \te
  = - \int_0^{\pi} a^{-1} \exp (i s a e^{i \te} ) i a e^{i \te} \, d \te.
\end{split}
}\end{equation}
%DIF > 
\DIFadd{Since $\sin (\te) \geq 0$ for $\te \in [0, \pi]$, and $a, s \geq 0$,
we get
}\[
\DIFadd{| \exp (i s a e^{i \te} ) |
  = | \exp (i s a }[\DIFadd{\cos (\te) + i \sin (\te) }] \DIFadd{) |
  = \exp (- a s \sin (\te))
  \leq 1.
}\]
\DIFadd{Therefore the integrand in the last expression in~(\ref{Eq_24611_s_pos})
convereges uniformly to~$0$ as $a \to \I$.
So $\ps (s) = 0$.
}

\DIFadd{To prove the claim for $s < 0$,
we first write
}\DIFaddend \[
\DIFdelbegin \DIFdel{f_t }\DIFdelend \DIFaddbegin \DIFadd{\int_{\Gm_a} g_s }\DIFaddend (z) \DIFaddbegin \DIFadd{\, d z\
 }\DIFaddend = \DIFdelbegin \DIFdel{\frac{ - e^{(t + 2) i z} + 2 e^{i t z} - e^{(t - 2) i z}}{4 z^2}.
}%DIFDELCMD < \]
%DIFDELCMD < %%%
\DIFdel{Define
}%DIFDELCMD < \[
%DIFDELCMD < %%%
\DIFdel{\ph_a }\DIFdelend \DIFaddbegin \DIFadd{\int_{\ta_a} g_s }\DIFaddend (\DIFdelbegin \DIFdel{s}\DIFdelend \DIFaddbegin \DIFadd{z}\DIFaddend ) \DIFdelbegin \DIFdel{= }\DIFdelend \DIFaddbegin \DIFadd{\, d z + }\DIFaddend \int\DIFdelbegin \DIFdel{_{\gm_a} \frac{e^{i s z}}{z^2} }\DIFdelend \DIFaddbegin \DIFadd{_{\Gm_a - \ta_a} g_s (z) }\DIFaddend \, d z.
\]
\DIFdelbegin \DIFdel{Then
}\DIFdelend \DIFaddbegin \DIFadd{One checks (details omitted, but something must be said) that
$\Ind_{\Gm_a - \ta_a} (0) = - 1$.
The expansion
}\DIFaddend \[
\DIFdelbegin \DIFdel{\int_{- a}^{a} f_t }\DIFdelend \DIFaddbegin \DIFadd{g_s }\DIFaddend (\DIFdelbegin \DIFdel{x}\DIFdelend \DIFaddbegin \DIFadd{z}\DIFaddend ) \DIFdelbegin \DIFdel{d x
 }\DIFdelend = \DIFdelbegin \DIFdel{- }%DIFDELCMD < \tfrac{1}{4} %%%
\DIFdel{\ph_a (t }\DIFdelend \DIFaddbegin \DIFadd{\frac{1}{z^2} }\DIFaddend + \DIFdelbegin \DIFdel{2)
   }\DIFdelend \DIFaddbegin \DIFadd{\frac{i s}{z} }\DIFaddend + \DIFdelbegin %DIFDELCMD < \tfrac{1}{2} %%%
\DIFdel{\ph_a }\DIFdelend \DIFaddbegin \DIFadd{\frac{(i s)^2}{2!}
   + \frac{(i s)^3 z}{3!} + \cdots
}\]
\DIFadd{shows that $\Res (g_s; 0) = i s$.
Similar methods to the case $s \geq 0$ show that
}\[
\limi{a} \DIFadd{\int_{\ta_a} g_s }\DIFaddend (\DIFdelbegin \DIFdel{t}\DIFdelend \DIFaddbegin \DIFadd{z}\DIFaddend ) \DIFdelbegin \DIFdel{- }%DIFDELCMD < \tfrac{1}{4} %%%
\DIFdel{\ph_a (t - 2).
}\DIFdelend \DIFaddbegin \DIFadd{\, d z = 0.
}\DIFaddend \]
\DIFdelbegin \DIFdel{The residue of $e^{i s z} / z^2$ at $0$ is $i s$.
It should follow (most easily using direct estimation
of the integrals of $e^{i s z} / z^2$ over $\sm_a$ and $\ta_a$;
the Dominated Convergence Theorem should not be needed)that $\limi{a} \ph_a (s) = - 2 \pi s$
when $s \geq 0$
and $\limi{a} \ph_a (s) = 0$
when $s < 0$}\DIFdelend \DIFaddbegin \DIFadd{Therefore
}\[
\DIFadd{\limi{a} \int_{\Gm_a} g_s (z) \, d z
   = \limi{a} \int_{\ta_a} g_s (z) \, d z + 2 \pi i \Res (g_s; 0)
   = 2 \pi s,
}\]
\DIFadd{as desired.
The completes the proof of the claim.
}

\DIFadd{With the last step justified because $f_t$ is entire
and $\Gm_a - [\io_a]$ is a cycle, we have
}\[
\DIFadd{\begin{split}
\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x
& = \int_{- \I}^{\I} f_t (x) \, d x
  = \limi{a} \int_{- a}^{a} f_t (x) \, d x
\\
& = \limi{a} \int_{\io_a} f_t (x) \, d x
  = \limi{a} \int_{\Gm_a} f_t (x) \, d x
\\
& = - \left( \frac{1}{4} \right) \ph (t + 2)
            + \left( \frac{1}{2} \right) \ph (t)
            - \left( \frac{1}{4} \right) \ph (t - 2).
\end{split}
}\]
\DIFadd{For $t \geq 2$, all terms are zero.
For $t \leq - 2$, we get
}\[
\DIFadd{\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x
 = - \left( \frac{1}{4} \right) 2 \pi (t + 2)
            + \left( \frac{1}{2} \right) 2 \pi t
            - \left( \frac{1}{4} \right) 2 \pi (t - 2)
 = 0.
}\]
\DIFadd{For $t \in [0, 2]$, we get
}\[
\DIFadd{\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x
 = - \left( \frac{1}{4} \right) \cdot 0
            + \left( \frac{1}{2} \right) \cdot 0
            - \left( \frac{1}{4} \right) 2 \pi (t - 2)
 = \frac{\pi}{2} (2 - t).
}\]
\DIFadd{For $t \in [-2, 0]$, we get
}\[
\DIFadd{\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x
 = - \left( \frac{1}{4} \right) \cdot 0
            + \left( \frac{1}{2} \right) 2 \pi t
            - \left( \frac{1}{4} \right) 2 \pi (t - 2)
 = \frac{\pi}{2} (t + 2).
}\]

\DIFadd{One can put this together in one formula (not necessary):
}\[
\DIFadd{\int_{- \I}^{\I} \left( \frac{\sin (x)}{x} \right)^2 e^{i t x} \, d x
 = \frac{\pi}{2} ( 2 - | t |).
}\]
\DIFadd{This completes the solution}\DIFaddend .
\end{proof}
\DIFdelbegin %DIFDELCMD < 

%DIFDELCMD < %%%
\DIFdelend \fi

The next problem counts as two ordinary problems.


%DIF > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DIFaddbegin 

\DIFaddend \begin{pbm}[Problem~8
  in Chapter~10 of Rudin's book]\label{Pb_2024_5_10_8}
Let $P$ and $Q$ be polynomials such that $\deg (Q) \geq \deg (P) + 2$
\DIFaddbegin \DIFadd{and $Q (x) \neq 0$ for all $x \in \R$}\DIFaddend .
Let $R$ be the rational function $R (z) = P (z) / Q (z)$
for $z \in \C$ such that $Q (z) \neq 0$.
\begin{enumerate}
\item\label{Pb_10_8_1}
Prove that $\int_{- \I}^{\I} R (x) \, d x$ is equal to
$2 \pi i$ times the sum of the residues of $R$ in the upper half plane.
(Replace the integral over $[- A, \, A]$ by the integral
over a suitable semicircle, and apply the Residue Theorem.)
\item\label{Pb_10_8_2}
What is the analogous statement for the lower half plane?
\item\label{Pb_10_8_3}
Use this method to compute
\[
\int_{- \I}^{\I} \frac{x^2}{1 + x^4} \, d x.
\]
\end{enumerate}
\end{pbm}


\ifsolutions
It is convenient to begin the solution with a lemma.

\begin{lem}\label{10_8_Est}
Let $p$ be a polynomial of degree~$n$.
Then there exist constants $m_p, M_p, r_p > 0$
such that for all $z \in \C$ with $| z | \geq r_p$,
we have $m_p | z |^n \leq | p (z) | \leq M_p | z |^n$.
\end{lem}

We give a direct proof below.
But one can also derive this lemma by showing,
using algebraic properties of limits,
that if $p (z) = \sum_{k = 0}^{n} a_k z^k$ for $z \in \C$,
then
\[
\lim_{|z| \to \I} \frac{p (z)}{z^n}
  = \lim_{|z| \to \I} \sum_{k = 0}^{n} a_k z^{k - n}
  = a_n.
\]

\begin{proof}[Proof of Lemma~\ref{10_8_Est}]
There are $a_0, a_1, \ldots, a_n \in \C$,
with $a_n \neq 0$,
such that $p (z) = \sum_{k = 0}^{n} a_k z^k$ for all $z \in \C$.
Define
\[
m_p = \frac{| a_n |}{2},
\qquad
M_p = \sum_{k = 0}^{n} | a_k |,
\andeqn
r_p = \max \left( 1, \, \frac{2}{| a_n |} \sum_{k = 0}^{n - 1} | a_k |
         \right).
\]
Let $z \in \C$ satisfy $| z | \geq r_p$.
Then,
using $| z | \geq 1$ at the second step
and $r_p \geq \frac{2}{| a_n |} \sum_{k = 0}^{n - 1} | a_k |$
at the fourth step,
\begin{align*}
| p (z) |
& \geq | a_n | \cdot | z |^n
     - \sum_{k = 0}^{n - 1} | a_k | \cdot | z |^k
  \geq | a_n | \cdot | z |^n
     - | z |^{n - 1} \sum_{k = 0}^{n - 1} | a_k |
   \\
& \geq | a_n | \cdot | z |^n
     - r_p^{-1} | z |^{n} \sum_{k = 0}^{n - 1} | a_k |
   \geq | a_n | \cdot | z |^n
     - \left( \frac{| a_n |}{2} \right) | z |^n
   = m_p | z |^n.
\end{align*}
Also,
using $| z | \geq 1$ at the second step,
\[
| p (z) |
  \leq \sum_{k = 0}^{n} | a_k | \cdot | z |^k
  \leq | z |^n \sum_{k = 0}^{n} | a_k |
  = M_p | z |^n.
\]
This completes the proof.
\end{proof}

\begin{proof}[Solution to part~(\ref{Pb_10_8_1})]
For $A > 0$,
we define curves $\gm_A$, $\rh_A$, and $\sm_A$ in~$\C$ by
$\gm_A (t) = t$ for $t \in [- A, \, A]$,
$\rh_A (t) = A e^{i t}$ for $t \in [0, \pi]$,
and $\sm_A (t) = A e^{i t}$ for $t \in [\pi, 2 \pi]$.
Then \DIFdelbegin \DIFdel{$\gm_A + \rh_A$, $\gm_A - \sm_A$,
and $\rh_A + \sm_A$ }\DIFdelend \DIFaddbegin \DIFadd{$[\gm_A + [\rh_A]$, $[\gm_A] - [\sm_A]$,
and $[\rh_A] + [\sm_A]$ }\DIFaddend are cycles.

We further let $Z_{+}$ be the set of $z$ in the upper half plane
such that $Q (z) = 0$,
and we let $Z_{-}$ be the set of $z$ in the lower half plane
such that $Q (z) = 0$.
Thus $Z_{+}$ and $Z_{-}$ are finite sets.
Let $m_P, M_P, r_P, m_Q, M_Q, r_Q$
be the constants of Lemma~\ref{10_8_Est}
for the polynomials $P$ and~$Q$.
Also set $L = \sup_{z \in Z_{+} \cup Z_{-}} | z |$.

We first claim that if $z \in Z_{-}$ and $A > L$,
then $\Ind_{\gm_A + \rh_A} (z) = 0$.
Indeed,
the path $t \mapsto z - i t$,
for $t \in [0, \I)$,
does not intersect $\Ran (\gm_A + \rh_A)$,
so $z$ is in the unbounded component of $\Ran (\gm_A + \rh_A)$.

We next claim that if $z \in Z_{+}$ and $A > L$,
then $\Ind_{\gm_A + \rh_A} (z) = 1$.
Indeed,
by Theorem 10.11 of Rudin,
we know that $\Ind_{\rh_A + \sm_A} (z) = 1$,
since $\rh_A + \sm_A$ is essentially the circle of radius~$A$
and center~$0$,
and $| z | < A$.
Moreover, consideration of the path $t \mapsto z + i t$,
for $t \in [0, \I)$,
which does not intersect $\Ran (\gm_A - \sm_A)$,
shows that $z$ is in the unbounded component of $\Ran (\gm_A - \sm_A)$.
Thus $\Ind_{\gm_A - \sm_A} (z) = 0$.
Since integration of a fixed function
is additive in the chains over which one is integrating,
it follows that
\[
\Ind_{\gm_A + \rh_A} (z)
 = \Ind_{\gm_A - \sm_A} (z) + \Ind_{\gm_A + \rh_A} (z)
 = 1.
\]
The claim is proved.

The Residue Theorem now implies that
if $A > L$ then
%
\begin{equation}\label{Eq_10_8_UseReq}
\int_{- A}^{A} R (x) \, d x
 = 2 \pi i \sum_{z \in Z_{+}} \Res (R; z)
    - \int_{\rh_A} R (z) \, d z.
\end{equation}
%

We now claim that $\limi{A} \int_{\rh_A} R (z) \, d z = 0$.
For $A \geq \max (r_P, r_Q)$,
we have,
using the choices of $m_Q$ and $M_P$
and the estimates from Lemma~\ref{10_8_Est},
\begin{align}
\left| \int_{\rh_A} R (z) \, d z \right|
& = \left| \int_{0}^{\pi}
    \frac{P (A e^{- i t}) i A e^{- i t}}{Q (A e^{- i t})} \, d t \right|
  \leq \int_{0}^{\pi}
     \frac{| P (A e^{- i t}) | A | e^{- i t} |}{| Q (A e^{- i t}) |}
         \, d t
      \label{Eq_10_8_IntEst}
\\
& \leq \int_{0}^{\pi}
     \frac{M_P A^{\deg (P) + 1}}{m_Q A^{\deg (Q)}} \, d t
  \leq \left( \frac{\pi M_P}{m_Q} \right) A^{\deg (P) - \deg (Q) + 1}.
  \notag
\end{align}
Since $\deg (P) - \deg (Q) + 1 < 0$,
the claim follows. 

Substituting the claim into~(\ref{Eq_10_8_UseReq}),
we deduce that $\limi{A} \int_{- A}^{A} R (x) \, d x$ exists
and is equal to $2 \pi i \sum_{z \in Z_{+}} \Res (R; z)$.
\end{proof}

It isn't sufficient to prove that $\limi{z} R (z) = 0$.
Knowing this sets one up to use the Dominated Convergence Theorem,
but one must still produce a dominating function.

It is easy to use Lemma~\ref{10_8_Est}
to prove directly that the function $R$ is Lebesgue integrable
on $(- \I, \, \I)$.

It is not hard to compute the relevant winding numbers
using Theorem 10.37 of Rudin.
But some justification {\emph{does}} need to be given.

\begin{proof}[Solution to part~(\ref{Pb_10_8_2}) (sketch)]
Let the notation be the same as in the
solution to part~(\ref{Pb_10_8_1}).
Methods similar to those used there
show that if $A > L$
then $\Ind_{\gm_A - \sm_A} (z) = 0$ for $z \in Z_{+}$,
while $\Ind_{\gm_A - \sm_A} (z) = - 1$ for $z \in Z_{-}$.
So the Residue Theorem gives
\[
\int_{- A}^{A} R (x) \, d x
 = - 2 \pi i \sum_{z \in Z_{-}} \Res (R; z)
    + \int_{\sm_A} R (z) \, d z.
\]
Using the same methods as used to get~(\ref{Eq_10_8_IntEst}),
one shows that $\limi{A} \int_{\sm_A} R (z) \, d z = 0$.
Therefore
$\int_{- \I}^{\I} R (x) \, d x
 = - 2 \pi i \sum_{z \in Z_{-}} \Res (R; z)$.
\end{proof}

Instead of repeating all the work,
one can reduce part~(\ref{Pb_10_8_2}) to part~(\ref{Pb_10_8_1}).

\begin{proof}[Second solution to part~(\ref{Pb_10_8_2})]
Let the notation be the same as in the
solution to part~(\ref{Pb_10_8_1}).

We claim that $\limi{A} \int_{\rh_A + \sm_A} R (z) \, d z = 0$.
For $A \geq \max (r_P, r_Q)$,
we have,
using the choices of $m_Q$ and $M_P$
and the estimates from Lemma~\ref{10_8_Est},
\begin{align*}
\left| \int_{\rh_A + \sm_A} R (z) \, d z \right|
& = \left| \int_{0}^{2 \pi}
    \frac{P (A e^{- i t}) i A e^{- i t}}{Q (A e^{- i t})} \, d t \right|
  \leq \int_{0}^{2 \pi}
     \frac{| P (A e^{- i t}) | A | e^{- i t} |}{| Q (A e^{- i t}) |}
         \, d t
\\
& \leq \int_{0}^{2 \pi}
     \frac{M_P A^{\deg (P) + 1}}{m_Q A^{\deg (Q)}} \, d t
  \leq \left( \frac{2 \pi M_P}{m_Q} \right) A^{\deg (P) - \deg (Q) + 1}.
\end{align*}
Since $\deg (P) - \deg (Q) + 1 < 0$,
the claim follows. 

For $A > L$,
by Theorem 10.11 of Rudin
we have $\Ind_{\rh_A + \sm_A} (z) = 1$ for all $z \in Z_{+} \cup Z_{-}$.
Therefore
\[
\int_{\rh_A + \sm_A} R (z) \, d z
 = 2 \pi i \sum_{z \in Z_{+} \cup Z_{-}} \Res (R; z).
\]
Combining this fact with the claim,
we get
\[
2 \pi i \sum_{z \in Z_{+} \cup Z_{-}} \Res (R; z) = 0.
\]
Therefore
\[
\limi{A} \int_{- A}^{A} R (x) \, d x
 = 2 \pi i \sum_{z \in Z_{+}} \Res (R; z)
 = - 2 \pi i \sum_{z \in Z_{-}} \Res (R; z).
\]
This completes the proof.
\end{proof}

The following lemma is convenient for the computation of the residues
needed in part~(\ref{Pb_10_8_3}).
\DIFaddbegin \DIFadd{It isn't in Chapter~10 of Rudin's book,
but it was proved in class this year.
}\DIFaddend 

\begin{lem}\label{10_8_CompRes}
Let $\Om \S \C$ be an open set, let $a \in \Om$,
and let $f$ be a \hfn{} on~$\Om \SM \{ a \}$
which has a simple pole at~$a$.
Then $\Res (f; a) = \lim_{z \to a} (z - a) f (z)$.
\end{lem}

\begin{proof}
Since $f$ has a simple pole at~$a$,
by definition there are $c \in \C \SM \{ 0 \}$
and a \hfn{} $g$ on~$\Om$
such that
\[
f (z) = g (z) + \frac{c}{z - a}
\]
for all $z \in \Om \SM \{ a \}$.
Moreover, by definition,
$\Res (f; a) = c$.
Now
\[
\lim_{z \to a} (z - a) f (z)
 = \lim_{z \to a} \big( (z - a) g (z) + c \big)
 = 0 \cdot g (a) + c
 = c.
\]
This completes the proof.
\end{proof}


\begin{proof}[Solution to part~(\ref{Pb_10_8_3})]
Set $\om = \exp (\pi i / 4)$.
Then
\[
1 + z^4 = (z - \om) (z - \om^3) (z - \om^5) (z - \om^7).
\]
So the function $R (z) = \frac{z^2}{1 + z^4}$ has two poles
in the upper half plane,
namely simple poles at $\om$ and at $\om^3$.
By part~(\ref{Pb_10_8_1}) and Lemma~\ref{10_8_CompRes},
we therefore have,
factoring out powers of $\om$ and repeatedly using $\om^2 = i$
at the fourth step,
\begin{align*}
\int_{- \I}^{\I} \frac{x^2}{1 + x^4} \, d x
 & = 2 \pi i ( \Res (R; \om) + \Res (R; \om^3) )
  \\
 & = 2 \pi i \left( \lim_{z \to \om} (z - \om) R (z)
                   + \lim_{z \to \om^3} (z - \om\DIFdelbegin \DIFdel{^) }\DIFdelend \DIFaddbegin \DIFadd{^3) }\DIFaddend R (z) \right)
  \\
 & = 2 \pi i \left( \frac{\om^2}{(\om - \om^3) (\om - \om^5)
                           (\om - \om^7)}
           + \frac{\om^6}{(\om^3 - \om) (\om^3 - \om^5)
                   (\om^3 - \om^7)} \right)
  \\
 & = 2 \pi i \left( \frac{\om^{-1}}{(1 - i) (1 - (- 1))
                           (1 - (- i))}
           + \frac{\om^{-3}}{(1 - (- i)) (1 - i)
                   (1 - (-1))} \right)
  \\
 & = \left( \frac{\pi i}{2} \right) ( \om^{-1} + \om^{-3} )
   = \left( \frac{\pi i}{2} \right) \big(- i \sqrt{2} \big)
   = \frac{\pi}{\sqrt{2}}.
\end{align*}
This completes the solution.
\end{proof}
\DIFaddbegin 

\begin{proof}[Alternate residue computation for part~(\ref{Pb_10_8_3})
  (sketch)]
\DIFadd{The residues can be read off directly from the partial fraction
decomposition
}\[
\DIFadd{\frac{z^2}{1 + z^4}
 = \frac{1}{4} \left( \frac{\om^3}{z - \om} + \frac{\om^5}{z - \om^3}
             + \frac{\om^7}{z - \om^5} + \frac{\om}{z - \om^7} \right).
}\]
\DIFadd{One can also use a partial fraction decomposition for $(1 + z^4)^{-1}$
and multiply it by~$z^2$.
}\end{proof}
\DIFaddend \fi


%DIF > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DIFaddbegin 

\DIFaddend \begin{pbm}[Problem~11
  in Chapter~10 of Rudin's book]\label{Pb_2024_5_10_11}
Let $\af \in \C$ satisfy $| \af | \neq 1$.
Calculate
\[
\int_{0}^{2 \pi} \frac{1}{1 - 2 \af \cos (\te) + \af^2} \, d \te
\]
by integrating
$(z - \af)^{-1} (z - 1 / \af)^{-1}$ around the unit circle.
\end{pbm}


\ifsolutions
We will use the following lemma to compute residues.
\DIFdelbegin \DIFdel{(This has also been used previously.
}\DIFdelend It isn't in Chapter\DIFaddbegin \DIFadd{~}\DIFaddend 10 of Rudin's book,
but it was proved in class \DIFdelbegin \DIFdel{.)
}\DIFdelend \DIFaddbegin \DIFadd{this year.
}\DIFaddend For the residues needed in this problem,
a different calculation is given in Remark~\ref{R_3610_PartialF}.

\begin{lem}\label{L_3610_CompRes}
Let $\Om \S \C$ be an open set, let $a \in \Om$,
and let $f$ be a \hfn{} on~$\Om \SM \{ a \}$
which has a simple pole at~$a$.
Then $\Res (f; a) = \lim_{z \to a} (z - a) f (z)$.
\end{lem}

\begin{proof}
Since $f$ has a simple pole at~$a$,
by definition there are $c \in \C \SM \{ 0 \}$
and a \hfn{} $g$ on~$\Om$
such that
\[
f (z) = g (z) + \frac{c}{z - a}
\]
for all $z \in \Om \SM \{ a \}$.
Moreover, by definition,
$\Res (f; a) = c$.
Now
\[
\lim_{z \to a} (z - a) f (z)
 = \lim_{z \to a} \big( (z - a) g (z) + c \big)
 = 0 \cdot g (a) + c
 = c.
\]
This completes the proof.
\end{proof}

\begin{proof}[Solution]
Define a closed curve $\gm$ in~$\C$ by
$\gm (\te) = e^{i \te}$ for $\te \in [0, \, 2 \pi]$.
Define a meromorphic function $f_{\af}$ on~$\C$ by
\[
f_{\af} (z) = \frac{1}{(z - \af) \left( z - \frac{1}{\af} \right)}.
\]
Then $f_{\af}$ has simple poles at $\af$ and at $\af^{-1}$.

We have
\begin{align*}
\int_{\gm} f (z) \, d z
& = \int_{0}^{2 \pi}
  \frac{1}{(e^{i \te} - \af) \left( e^{i \te} - \frac{1}{\af} \right)}
       i e^{i \te} \, d \te
\\
& = \int_{0}^{2 \pi}
  \frac{- i \af}{(e^{i \te} - \af) ( e^{- i \te} - \af )}
       \, d \te
  = \int_{0}^{2 \pi}
  \frac{- i \af}{1 - 2 \af \cos (\te) + \af^2}
       \, d \te.
\end{align*}
We now compute this integral by the residue theorem.

Suppose $| \af | < 1$.
Then $\Ind_{\gm} (\af) = 1$ and $\Ind_{\gm} (1 / \af) = 0$
by Theorem 10.11 of Rudin.
Lemma~\ref{L_3610_CompRes}
gives
\[
\Res (f_{\af}, \af)
 = \frac{1}{\af - \frac{1}{\af} }
 = \frac{\af}{\af^2 - 1}.
\]
Therefore
\begin{align*}
\int_{0}^{2 \pi}
  \frac{1}{1 - 2 \af \cos (\te) + \af^2}
       \, d \te
& = \left( \frac{1}{- i \af} \right) \int_{\gm} f_{\af} (z) \, d z
\\
& = \left( \frac{1}{- i \af} \right) 2 \pi i \Res (f_{\af}, \af)
  = - \frac{2 \pi}{\af^2 - 1}.
\end{align*}

Suppose now $| \af | > 1$.
Then, using the result for $1 / \af$ at the second step,
we have
\begin{align*}
\int_{0}^{2 \pi}
  \frac{1}{1 - 2 \af \cos (\te) + \af^2}
       \, d \te
& = \int_{0}^{2 \pi}
  \frac{\af^{- 2}}{1 - 2 \af^{- 1} \cos (\te) + \af^{- 2}}
       \, d \te
\\
& = - \frac{2 \pi \af^{- 2}}{\af^{- 2} - 1}
  = \frac{2 \pi}{\af^2 - 1}.
\end{align*}
This completes the solution.
\end{proof}

\begin{rmk}\label{R_3610_PartialF}
The residues
\[
\Res (f_{\af}, \af)
 = \frac{\af}{\af^2 - 1}
\andeqn
\Res (f_{\af}, \af^{-1})
 = - \frac{\af}{\af^2 - 1}
\]
can be read directly off the partial fraction decomposition
\[
f_{\af} (z)
 = \left( \frac{\af}{\af^2 - 1} \right)
  \left( \frac{1}{z - \af} - \frac{1}{z - \af^{-1}} \right),
\]
without the need for Lemma~\ref{L_3610_CompRes}.
\end{rmk}
\fi


%DIF > %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\DIFaddbegin 

\DIFaddend \begin{pbm}[Problem~13
  in Chapter~10 of Rudin's book]\label{Pb_2024_5_10_13}
Prove that
\[
\int_0^{\I} \frac{1}{1 + x^n} \, d x
 = \frac{ \pi / n }{\sin ( \pi / n )}
\]
for $n \in \N$ with $n \geq 2$.
\end{pbm}


\ifsolutions
The following lemma is convenient for the computation of the residues
needed here.
It isn't in Chapter~10 of Rudin's book,
but it was proved in class \DIFaddbegin \DIFadd{this year}\DIFaddend .

\begin{lem}\label{10_8_CompRes}
Let $\Om \S \C$ be an open set, let $a \in \Om$,
and let $f$ be a \hfn{} on~$\Om \SM \{ a \}$
which has a simple pole at~$a$.
Then $\Res (f; a) = \lim_{z \to a} (z - a) f (z)$.
\end{lem}

\begin{proof}
Since $f$ has a simple pole at~$a$,
by definition there are $c \in \C \SM \{ 0 \}$
and a \hfn{} $g$ on~$\Om$
such that
\[
f (z) = g (z) + \frac{c}{z - a}
\]
for all $z \in \Om \SM \{ a \}$.
Moreover, by definition,
$\Res (f; a) = c$.
Now
\[
\lim_{z \to a} (z - a) f (z)
 = \lim_{z \to a} \big( (z - a) g (z) + c \big)
 = 0 \cdot g (a) + c
 = c.
\]
This completes the proof.
\end{proof}


\begin{proof}[Solution]
Set $\om = \exp (\pi i / n)$.
For $r \in (1, \I)$, define paths $\rh_r, \sm_r \colon [0, r] \to \C$
by $\rh_r (t) = t$ and $\sm_r (t) = t \om^2$ for $t \in [0, r]$.
Also define $\gm_r \colon [0, \, 2 \pi / n] \to \C$
and $\bt_r \colon [2 \pi / n, \, 2 \pi] \to \C$
by $\gm_r (t) = r e^{i t}$ for $t \in [0, \, 2 \pi / n]$
and $\bt_r (t) = r e^{i t}$ for $t \in [2 \pi / n, \, 2 \pi]$.
Then \DIFdelbegin \DIFdel{$\gm_r + \bt_r$, $\rh_r + \gm_r - \sm_r$,
and $\sm_r + \bt_r - \rh_r$
}\DIFdelend \DIFaddbegin \DIFadd{$[\gm_r] + [\bt_r]$, $[\rh_r] + [\gm_r] - [\sm_r]$,
and $[\sm_r] + [\bt_r] - [\rh_r]$ }\DIFaddend are cycles.

The formula
\[
f (z) = \frac{1}{1 + z^n}
\]
defines a meromorphic function on~$\C$,
with poles at $\om, \om^3, \ldots, \om^{2 n - 1}$.

Using $r > 1$, we get $\Ind_{\gm_r + \bt_r} (\om) = 1$
by Theorem 10.11 of Rudin.
Also, the path $t \mapsto t \om$,
for $t \in [1, \I)$,
is continuous, goes to~$\I$ as $t \to \I$,
and has range disjoint from $\Ran (\sm_r + \bt_r - \rh_r)$,
so $\Ind_{\sm_r + \bt_r - \rh_r} (\om) = 0$.
Therefore
\[
\Ind_{\rh_r + \gm_r - \sm_r} (\om)
  = \Ind_{\gm_r + \bt_r} (\om) - \Ind_{\sm_r + \bt_r - \rh_r} (\om)
  = 1.
\]
On the other hand,
for $k = 2, 3, \ldots, n$,
the path $t \mapsto t \om^k$,
for $t \in [1, \I)$,
is continuous, goes to~$\I$ as $t \to \I$,
and has range disjoint from $\Ran (\rh_r + \gm_r - \sm_r)$.
So $\Ind_{\rh_r + \gm_r - \sm_r} (\om^k) = 0$.
We can now apply the Residue Theorem
using the cycle $\rh_r + \gm_r - \sm_r$.
The condition
$\Ind_{\rh_r + \gm_r - \sm_r} (z) = 0$ for $z \not\in \C$
is vacuous,
so we get
\[
\int_{\rh_r + \gm_r - \sm_r} f (z) \, d z
 = 2 \pi i \Res (f; \om).
\]

Since $n \geq 2$,
\[
\limi{r} \int_{\rh_r} f (z) \, d z
= \limi{r} \int_0^{r} \frac{1}{1 + t^n} \, d t
= \int_0^{\I} \frac{1}{1 + t^n} \, d t
\]
exists and is finite.
Similarly
\[
\limi{r} \int_{\sm_r} f (z) \, d z
= \limi{r} \int_0^{r} \frac{1}{1 + (\om^2 t)^n} \om^2 \, d t
= \om^2 \int_0^{\I} \frac{1}{1 + t^n} \, d t.
\]
We claim that
\[
\limi{r} \int_{\gm_r} f (z) \, d z = 0.
\]
For $| z | > 2$, we have
\[
\left| \frac{1}{1 + z^n} \right|
  \leq \frac{1}{| z |^n - 1}
  \leq \frac{2}{| z |^n},
\]
so for $r > 2$ we have,
since the length of $\gm_r$ is $2 \pi r / n$,
\[
\left| \int_{\gm_r} f (z) \, d z \right|
  \leq \left( \frac{2 \pi r}{n} \right) \left( \frac{2}{r^n} \right)
  = \frac{4 \pi}{n r^{n - 1}}.
\]
Since $n \geq 2$,
the claim follows.
So
\[
(1 - \om^2) \int_0^{\I} \frac{1}{1 + t^n} \, d t
 = 2 \pi i \Res (f; \om).
\]

We next calculate $\Res (f; \om)$.
We use Lemma~\ref{10_8_CompRes}.
% We first factor:
% \[
% z^n + 1 = \prod_{k = 0}^{n - 1} (z - \om^{2 n + 1}).
% \]
% Therefore
% \[
% \lim_{z \to \om} \frac{z^n + 1}{z - \om}
%  = \lim_{z \to \om} \prod_{k = 1}^{n - 1} (z - \om^{2 n + 1})
%  = \prod_{k = 1}^{n - 1} (\om - \om^{2 n + 1})
%  = \om^{n - 1} \prod_{k = 1}^{n - 1} (1 - \om^{2 n}).
% \]
We have, using $\om^n = - 1$ at the second step,
\begin{align*}
\Res (f; \om)
& = \lim_{z \to \om} \frac{z - \om}{z^n + 1}
  = \lim_{z \to 1} \frac{\om z - \om}{(\om z)^n + 1}
  = - \om \lim_{z \to 1} \frac{z - 1}{z^n - 1}
     \\
& = - \om \lim_{z \to 1} \frac{1}{z^{n - 1} + z^{n - 2} + \cdots + 1}
  = - \frac{\om}{n}.
\end{align*}
We conclude
\begin{align*}
\int_0^{\I} \frac{1}{1 + t^n} \, d t
& = \frac{2 \pi i \Res (f; \om)}{1 - \om^2}
  = - \frac{2 \pi i \om}{n (1 - \om^2)}
    \\
& = \frac{2 \pi i}{n (\om - \om^{-1})}
  = \frac{\pi / n}{ (\om - \om^{-1}) / (2 i)}
  = \frac{\pi / n}{ \sin ( \pi / n)}.
\end{align*}
This completes the proof.
\DIFaddbegin \end{proof}

\begin{proof}[Alternate residue computation]
\DIFadd{The residues can be read off directly from the partial fraction
decomposition
}\[
\DIFadd{\frac{1}{1 + z^n}
 = - \frac{1}{n} \sum_{k = 1}^n \frac{\om^{2 k - 1}}{z - \om^{2 k - 1}}.
}\]
\DIFadd{(This partial fraction decomposition has not been checked.)
}\DIFaddend \end{proof}
\fi

\end{document}
