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| -rw-r--r-- | assignment5.tex | 9 |
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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} |