From 84543029d30985c34c7b8427ec0ac0fd4ee57356 Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Wed, 7 Oct 2015 09:04:52 +0200
Subject: Fix 5.4.d

---
 assignment5.tex | 9 +--------
 1 file changed, 1 insertion(+), 8 deletions(-)

(limited to 'assignment5.tex')

diff --git a/assignment5.tex b/assignment5.tex
index 26eb79c..cf7d376 100644
--- a/assignment5.tex
+++ b/assignment5.tex
@@ -80,14 +80,7 @@
 
                 \setcounter{enumii}{3}
 
-            \item \begin{align*}
-                    \int_0^\infty e^{-x^2}\diff x &= \lim\limits_{t\to\infty} \int_0^t e^{-x^2}\diff x \qquad\text{where $x=z$} \\
-                        &= \lim\limits_{t\to\infty} \int_0^t 1\cdot e^{-x^2}\diff x \\
-                        &= \lim\limits_{t\to\infty} \left(x\cdot e^{-x^2} - \int_0^t x\cdot-2\cdot e^{-x^2} \diff x\right) \\
-                        &= \lim\limits_{t\to\infty} \left(\left.x\cdot e^{-x^2}\right|_0^t + 2\int_0^t xe^{-x^2}\diff x\right) \\
-                        &= \lim\limits_{t\to\infty} \left.xe^{-x^2}\right|_0^t + 1 \qquad\text{($-x^2$ is dominant over $x$, and see 4.a)} \\
-                        &= 0 + 1 = 1.
-                \end{align*}
+            \item In 4.c we saw that $\int_{-\infty}^\infty e^{-z^2}\diff z = \sqrt\pi$. We know that $f(x)=e^{-x^2}$ is even because of the power of two, therefore $\int_0^\infty e^{-x^2}\diff x = \frac12 \int_{-\infty}^\infty e^{-z^2}\diff z = \frac12\sqrt\pi$.
         \end{enumerate}
 
     \item \begin{enumerate}
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