%====================================================================
% This is a LaTeX-2e document
%====================================================================

%\documentclass[12pt]{amsart}
\documentclass[10pt]{article}
\parindent=0pt \parskip=8pt

\usepackage{verbatim}
\usepackage{fullpage}

\usepackage{amssymb,amsmath,amsthm}
\usepackage[mathscr]{eucal}

\usepackage{makeidx}
\input{../isomath}
\input{../mathenv}
\input{../syms}
\usepackage{graphicx}

%--------------------------------------------------------------------
\begin{document}

Daniel:

I am unable to decipher a step in the Qmath paper.  You write that
``a calculation'' shows that
$$
	V(\vecx,-\vecx,0,0) = \frac{2a}{\|\vecx\|} + O(a^2).
$$

I have
$$
	V(\vecx,-\vecx,0,0) = \int \left( 1 - e^{
		-\frac{1}{4} \int_0^{4\beta} U(\omega(s)) ds
	} \right) dW^{4\beta}_{\vecx,-\vecx}(\omega).
$$
I dare not Taylor-expand $1-e^{-I}$ (where $I$ is the integral) into $I +
O(I^2)$ since the integral is infinite in the case of the hard-core potential.
I really need the integral to remain exponentiated.

Using the definition of the stochastic integral, I have
$$
	V(\vecx,\vecy,0,0) =
	\int 1 \; dW^{4\beta}_{\vecx,\vecy}(\omega)
	-
	\int e^{
		-\frac{1}{4} \int_0^{4\beta} U(\omega(s)) ds
	} dW^{4\beta}_{\vecx,\vecy}(\omega).
$$
The first term is known to be $1/g_{4\beta}(\vecx-\vecy)$; the second is
\begin{equation*}
\begin{aligned}
	\lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		&
		\int_{\R^{d}}
		g_{4\beta/n}(\vecx-\vecz_1)
		g_{4\beta/n}(\vecz_1-\vecz_2)
		\cdots
		g_{4\beta/n}(\vecz_{n-2}-\vecz_{n-1})
		g_{4\beta/n}(\vecz_{n-1}-\vecy)
		\\
%
	&
		\exp\left\{\frac{-\beta}{n} \left(
			\sum_{j=1}^{n-1} U(\vecz_i) \right)
		\right\}
		\; d\vecz_1 \cdots d\vecz_{n-1} \\
%%
	= \lim_{n\to\infty}
		\int_{\R^{d}}
		\cdots
		&
		\int_{\R^{d}}
%
		g_{4\beta/n}(\vecx-\vecz_1)
		g_{4\beta/n}(\vecz_1-\vecz_2)
		\cdots
		g_{4\beta/n}(\vecz_{n-2}-\vecz_{n-1})
		g_{4\beta/n}(\vecz_{n-1}-\vecy)
		\\
%
	&
		\prod_{j=1}^{n-1}
		\exp\left\{\frac{-\beta}{n} U(\vecz_i) \right\}
		\; d\vecz_1 \cdots d\vecz_{n-1}.
\end{aligned}
\end{equation*}
Now, $e^{-(\beta/n) U(\vecz_i)}$ is 1 when $\|\vecz_i\| \ge a$ and
0 when $\|\vecz_i\| < a$.  So this becomes
\begin{equation*}
\begin{aligned}
	\lim_{n\to\infty}
		\int_{\|\vecz_1\| \ge a}
		\cdots
		&
		\int_{\|\vecz_{n-1}\| \ge a}
%
		g_{4\beta/n}(\vecx-\vecz_1)
		g_{4\beta/n}(\vecz_1-\vecz_2)
		\cdots
		g_{4\beta/n}(\vecz_{n-2}-\vecz_{n-1})
		g_{4\beta/n}(\vecz_{n-1}-\vecy)
		\\
%
	&
		\; d\vecz_1 \cdots d\vecz_{n-1}.
\end{aligned}
\end{equation*}
I don't see how $\vecy=-\vecx$ permits any simplification \ldots.

Am I ignorant of some basic fact about Brownian motion?

\end{document}
