% Lecture notes for Physical Mathematics I
% Ewan Stewart
% KAIST
% 2011/10/13

\documentclass[a4paper,12pt]{report}

\usepackage{pm1}

\setcounter{chapter}{2}
\setcounter{section}{1}
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\begin{document}

\section{Holomorphic integration}

\subsection{Holomorphic integrals}

A holomorphic integral
\begin{equation}
\int_C \fn{f}{z} \d{z}
\end{equation}
cannot depend on the curve $C$ since a curve cannot be defined holomorphically.
Rather, it depends on the end points, if the curve is open, and possibly on the winding number of the curve around any singularities.

\begin{figure}[h]
\centering
\begin{tikzpicture}[x=0.05\textwidth,y=0.05\textwidth,]
% axes
\draw[->] (-2,0) -- (2,0) node[right]{$\Re z$};
\draw[->] (0,-2) -- (0,2) node[above]{$\Im z$};
% loop
\draw[thick,blue,labelled arrow={0.125}{node[above right]{$C$}}] (0,0) circle(1);
\end{tikzpicture}
\caption{
A loop $C$ with winding number $+1$ around the origin.
}
\end{figure}

The winding number dependence can be seen by explicit integration of a Laurent series term around a closed curve $C$ with winding number $N$ around the origin
\begin{equation} \label{pole}
\int_C z^n \d{z} = \left\{
\begin{array}{cclr}
\displaystyle
\left[ \frac{z^{n+1}}{n+1}  \right]_{\partial C} & = & 0
& \textrm{for $n \in \Z$, $n \neq -1$}
\\[4ex]
\displaystyle
\Bigg[ \ln z \Bigg]_{\partial C} & = & 2\pi i N
& \textrm{for $n = -1$}
\end{array}
\right.
\end{equation}
Thus only the winding number around simple poles, i.e.\ those with $n=-1$, is relevant.
Note that if $n \notin \Z$ then the branch cut prevents the curve $C$ from being closed and the integral will depend on the end points of the curve.

\subsection{Contour integration}

Holomorphic integration provides a powerful method to evaluate definite integrals, exploiting the fact that the curve can be deformed to a form convenient for evaluation.

\subsubsection{Example with a pole}

Consider the integral
\begin{equation}
I = \int_0^\infty \frac{\d{x}}{1+x^2}
\end{equation}
\begin{figure}
\centering
\tikzset{pole/.style={black}}
\tikzset{original/.style={thick,blue}}
\tikzset{deformed/.style={thick,red}}
\begin{tikzpicture}[x=0.1\textwidth,y=0.1\textwidth]
% poles
\fill[pole] (0,1) circle (0.05) node[right]{pole};
\fill[pole] (0,-1) circle (0.05) node[right]{pole};
% C
\draw[original,labelled arrow={0.5}{node[below]{$C$}}] (0,0) node[below,black]{$0$} -- (2,0);
% C1
\draw[deformed,labelled arrow={0.5}{node[below]{$C_1$}}] (0,0) -- (-2,0);
% C2
\draw[deformed,labelled arrow={0.33}{node[above left]{$C_2$}}] (180:2) arc(180:0:2);
% C3
\draw[deformed,labelled arrow={0.5}{node[left]{$C_3$}}] (0,1) circle(0.1);
\end{tikzpicture}
\caption{ \label{polecontour}
Contour integration of $\int_0^\infty \frac{\d{z}}{1+z^2}$.
}
\end{figure}
Deforming the contour as shown in Fig.~\ref{polecontour}, we get
\begin{eqnarray}
I & = & \int_C \frac{\d{z}}{1+z^2}
\nonumber \\ \label{pC123}
& = & \int_{C_1} \frac{\d{z}}{1+z^2} + \int_{C_2} \frac{\d{z}}{1+z^2} + \int_{C_3} \frac{\d{z}}{1+z^2}
\end{eqnarray}
The last term comes from a loop around the pole, left behind as the contour is deformed across the pole.

By symmetry
\begin{equation} \label{pC1}
\int_{C_1} \frac{\d{z}}{1+z^2} = - I
\end{equation}
On $C_2$, the integrand $\sim |z|^{-2}$ and so the integral $\sim |z|^{-1}$.
Taking the arc to infinity gives
\begin{equation} \label{pC2}
\int_{C_2} \frac{\d{z}}{1+z^2} = 0
\end{equation}
On $C_3$, $z \sim i$ therefore
\begin{equation}
\frac{1}{1+z^2} \sim \frac{1}{2i} \left( \frac{1}{z-i} \right)
\end{equation}
and so from \eq{pole}
\begin{equation} \label{pC3}
\int_{C_3} \frac{\d{z}}{1+z^2} = \frac{1}{2i} \left( 2\pi i \right) \left( +1 \right) = \pi
\end{equation}

Substituting Eqs.~(\ref{pC1}), (\ref{pC2}) and (\ref{pC3}) into \eq{pC123} gives
\begin{equation}
I = -I + 0 + \pi
\end{equation}
and so
\begin{equation}
\int_0^\infty \frac{\d{x}}{1+x^2} = \frac{\pi}{2}
\end{equation}

\subsubsection{Example with a branch point}

Consider the integral
\begin{equation}
I = \int_0^\infty \frac{x^\frac{1}{2}\d{x}}{1+x^2}
\end{equation}
\begin{figure}
\centering
\tikzset{branchpoint/.style={black}}
\tikzset{branchcut/.style={thick,black,dashed}}
\tikzset{pole/.style={black}}
\tikzset{original/.style={thick,blue}}
\tikzset{deformed/.style={thick,red}}
\begin{tikzpicture}[x=0.1\textwidth,y=0.1\textwidth]
% axes
%\draw[->] (-2,0) -- (2,0) node[right]{$\Re z$};
%\draw[->] (0,-2) -- (0,2) node[above]{$\Im z$};
% branch point
\fill[branchpoint] (0,0) circle (0.05) node[left]{\begin{tabular}{r} branch \\ point \end{tabular}};
% branch cut
\draw[branchcut] (0,0) -- node{\begin{tabular}{c} branch \\ cut \end{tabular}} (1.5,0) -- (2,0);
% poles
\fill[pole] (0,1) circle (0.05) node[left]{\begin{tabular}{r} pole \end{tabular}};
\fill[pole] (0,-1) circle (0.05) node[left]{\begin{tabular}{r} pole \end{tabular}};
% C
\draw[original,labelled arrow={0.75}{node[above]{$C$}}] (30:0.1) -- (2,0.05);
% C1
\draw[deformed,labelled arrow={0.8}{node[below]{$C_1$}}] (30:0.1) arc(30:330:0.1);
% C2
\draw[deformed,labelled arrow={0.75}{node[below]{$C_2$}}] (330:0.1) -- (2,-0.05);
% C3
\draw[deformed,labelled arrow={0.67}{node[above left]{$C_3$}}] (358.6:2) arc(358.6:1.4:2);
% C4
\draw[deformed,labelled arrow={1}{node[right]{$C_4$}}] (0,1) circle(0.1);
% C5
\draw[deformed,labelled arrow={1}{node[right]{$C_5$}}] (0,-1) circle(0.1);
\end{tikzpicture}
\caption{ \label{branchcontour}
Contour integration of $\int_0^\infty \frac{z^\frac{1}{2}\d{z}}{1+z^2}$.
}
\end{figure}
Deforming the contour as shown in Fig.~\ref{branchcontour}, we get
\begin{eqnarray}
I & = & \int_C \frac{z^\frac{1}{2}\d{z}}{1+z^2}
\nonumber \\ \label{bC12345}
& = & \int_{C_1} \frac{z^\frac{1}{2}\d{z}}{1+z^2} + \int_{C_2} \frac{z^\frac{1}{2}\d{z}}{1+z^2} + \int_{C_3} \frac{z^\frac{1}{2}\d{z}}{1+z^2} + \int_{C_4} \frac{z^\frac{1}{2}\d{z}}{1+z^2} + \int_{C_5} \frac{z^\frac{1}{2}\d{z}}{1+z^2}
\end{eqnarray}
The last two terms come from loops around the poles, left behind as the contour is deformed across the poles.

On $C_1$, the integrand $\sim |z|^\frac{1}{2}$ and so the integral $\sim |z|^\frac{3}{2}$.
Taking the arc to zero gives
\begin{equation} \label{bC1}
\int_{C_1} \frac{z^\frac{1}{2}\d{z}}{1+z^2} = 0
\end{equation}
$C_2$ is on the opposite side of the branch cut from $C$ and so
\begin{equation}
\left. z^\frac{1}{2} \right|_{C_2} = - \left. z^\frac{1}{2} \right|_C
\end{equation}
Therefore
\begin{equation} \label{bC2}
\int_{C_2} \frac{z^\frac{1}{2}\d{z}}{1+z^2} = - I
\end{equation}
On $C_3$, the integrand $\sim |z|^{-\frac{3}{2}}$ and so the integral $\sim |z|^{-\frac{1}{2}}$.
Taking the arc to infinity gives
\begin{equation} \label{bC3}
\int_{C_3} \frac{z^\frac{1}{2}\d{z}}{1+z^2} = 0
\end{equation}
On $C_4$, $z \sim e^\frac{\pi i}{2}$ therefore
\begin{equation}
\frac{z^\frac{1}{2}}{1+z^2} \sim \frac{e^\frac{\pi i}{4}}{2i} \left( \frac{1}{z-i} \right)
\end{equation}
and so from \eq{pole}
\begin{equation} \label{bC4}
\int_{C_4} \frac{z^\frac{1}{2}\d{z}}{1+z^2} = \frac{e^\frac{\pi i}{4}}{2i} \left( 2\pi i \right) \left( +1 \right) = \pi e^\frac{\pi i}{4}
\end{equation}
On $C_5$, $z \sim e^\frac{3\pi i}{2}$ therefore
\begin{equation}
\frac{z^\frac{1}{2}}{1+z^2} \sim \frac{e^\frac{3\pi i}{4}}{-2i} \left( \frac{1}{z+i} \right)
= \frac{e^{-\frac{\pi i}{4}}}{2i} \left( \frac{1}{z+i} \right)
\end{equation}
and so from \eq{pole}
\begin{equation} \label{bC5}
\int_{C_5} \frac{z^\frac{1}{2}\d{z}}{1+z^2} = \frac{e^{-\frac{\pi i}{4}}}{2i} \left( 2\pi i \right) \left( +1 \right) = \pi e^{-\frac{\pi i}{4}}
\end{equation}

Substituting Eqs.~(\ref{bC1}), (\ref{bC2}), (\ref{bC3}), (\ref{bC4}) and (\ref{bC5}) into \eq{bC12345} gives
\begin{equation}
I = 0 - I + 0 + \pi e^\frac{\pi i}{4} + \pi e^{-\frac{\pi i}{4}}
\end{equation}
and so
\begin{equation}
\int_0^\infty \frac{x^\frac{1}{2}\d{x}}{1+x^2} = \frac{\pi}{\sqrt{2}\,}
\end{equation}

\end{document}
