% Homework for Physical Mathematics I
% Ewan Stewart
% KAIST
% 2011/12/21

\documentclass[a4paper,12pt]{article}

\usepackage{pm1}

\setcounter{homework}{5}
\setcounter{question}{3}

\begin{document}

\begin{homework}
[Optional extra question.]

\question{factorial}

Use contour integration to determine
\begin{equation}
\int_{-i\infty}^{i\infty} z! \d{z}
\end{equation}

\answer{factorial}

Using Section~2.3.1, $z! = \fn{\Gamma}{z+1}$ has simple poles at the negative integers
\begin{eqnarray}
z! & = & \frac{(z+n)!}{(z+1)\cdots(z+n-1)(z+n)}
\\
& = & \frac{(-1)^{n-1}(z+n)!}{[n-1-(z+n)]\cdots[1-(z+n)](z+n)}
\\
& \sim & \frac{(-1)^{n-1}}{(n-1)!\,(z+n)}
\qquad \textrm{for } z \simeq -n
\end{eqnarray}
and, using Section~2.3.2, the asymptotic behaviour
\begin{eqnarray}
z! & = & z \fn{\Gamma}{z}
\eqnarrayrelation\sim \sqrt{2\pi}\, z^\frac{1}{2} \exp \left\{ z \left( \ln z - 1 \right) \right\}
\\
& \sim & \sqrt{2\pi}\, z^\frac{1}{2}
\exp \left\{ i \left[ \Im z \left( \ln |z| - 1 \right) + \Re z \arg z \right] \right\}
\nonumber \\ && \times
\exp \left\{ \left[ \Re z \left( \ln |z| - 1 \right) - \Im z \arg z \right] \right\}
\end{eqnarray}
with branch cut along the negative real axis.
\begin{figure}
\newcommand{\rmax}{4.5}
\centering
\begin{tikzpicture}[x=0.05\textwidth,y=0.05\textwidth]
% axes
\draw[very thick] (-\rmax,0) -- (-4,0);
\draw[dotted] (-3.5,0) -- (-4,0);
\draw[->] (-3.5,0) -- (1,0) node[right]{$\Re z$};
\draw[->] (0,-\rmax) -- (0,\rmax) node[above]{$\Im z$};
% $C$
\draw[blue,labelled arrow={2/3}{node[right]{$C$}}] (0,-\rmax) -- (0,\rmax);
% poles
\foreach \n in {1,2,3} {
\fill (-\n,0) circle(0.1) ++(0,-0.2) node[below]{$-\n$};
\draw[red,labelled arrow={0.2501}{node[above]{$C_{\n}$}}] (-\n,0) circle(0.2);
}
% $C_-$
\draw[red,labelled arrow={1/2}{node[above right]{$C_-$}}] (0,-\rmax) arc(-90:-177.5:\rmax);
\draw[red] (-177.5:\rmax) -- (-4,-0.2);
% $C_+$
\draw[red] (-4,0.2) -- (177.5:\rmax);
\draw[red,labelled arrow={1/2}{node[below right]{$C_+$}}] (177.5:\rmax) arc(177.5:90:\rmax);
\end{tikzpicture}
\caption{ \label{poles}
Contour integration of $\int_{-i\infty}^{i\infty} z! \d{z}$.}
\end{figure}
Therefore, deforming the contour as shown in Figure~\ref{poles},
\begin{equation}
\int_{-i\infty}^{i\infty} \hspace{-1em} z! \d{z}
= \int_{C_-} \hspace{-0.5em} z! \d{z}
+ \sum_{n=1}^\infty \int_{C_n} \hspace{-0.5em} z! \d{z}
+ \int_{C_+} \hspace{-0.5em} z! \d{z}
= 0 + 2\pi i \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n-1)!} + 0
= \frac{2\pi i}{e}
\end{equation}

\end{homework}

\end{document}

