From: W. Trevor King <wking@drexel.edu>
Date: Tue, 29 Mar 2011 20:51:37 +0000 (-0400)
Subject: Fix (a^2-z^2)^2 typo in integrals.tex.
X-Git-Tag: v1.0~361
X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=bbf7247da06b39dacea2a8f6e33d618146fc44b8;p=thesis.git

Fix (a^2-z^2)^2 typo in integrals.tex.
---

diff --git a/tex/src/cantilever-calib/integrals.tex b/tex/src/cantilever-calib/integrals.tex
index 3d10bb8..afac3bc 100644
--- a/tex/src/cantilever-calib/integrals.tex
+++ b/tex/src/cantilever-calib/integrals.tex
@@ -22,7 +22,7 @@ There are simple poles at $u = \pm i$.
 We will show that, for any $(a,b > 0) \in \Reals$,%
 \nomenclature[aR]{\Reals}{Real numbers}
 \begin{equation}
-  I = \iInfInf{z}{\frac{1}{(a^2-z^2) + b^2 z^2}} = \frac{\pi}{b a^2} \;.
+  I = \iInfInf{z}{\frac{1}{(a^2-z^2)^2 + b^2 z^2}} = \frac{\pi}{b a^2} \;.
 \end{equation}
 
 First we note that $|f(z)| \rightarrow 0$ like $|z^{-4}|$ for $|z| \gg 1$,
@@ -38,7 +38,7 @@ into $(A+iB)(A-iB)$. % thanks Prof. Yuan
   (a^2-z^2)^2 + b^2 z^2
       = (a^2-z^2 \colA{+} ibz)(a^2-z^2 \colA{-} ibz)
 \end{equation}
-And the roots of $z^2 \colA{\pm} ibz - a^2$
+The roots of $z^2 \colA{\pm} ibz - a^2$ are given by
 \begin{equation}
   z_{r\colB{\pm}}
        = \colA{\pm}\frac{ib}{2} \left(