Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 441: 58


The integral $\int_{-\infty}^\infty e^{-x^2} dx$ converges by the comparison test.

Work Step by Step

The given integral can be re-written as sum of two integrals as: $\int_{-\infty}^{\infty} e^{-x^2} dx=\int_{-\infty}^{1} e^{-x^2} dx+\int_{-1}^{1} e^{-x^2} dx+\int_{1}^{\infty} e^{-x^2} dx$ We can see that the first and third integral above defines an improper integral and so, we compare it with $e^{-|x|}$ For $x\geq 1$, or, $x \leq -1$, we have: $e^{-x^2} \leq e^{-|x|}$ and $\int_{1}^{\infty} e^{-x^2} dx \leq \int_1^\infty e^{-|x|} dx $ Now, $\int_{-\infty }^{-1} e^{-|x|} dx =\int_{\infty}^1 e^{-|x|} (-dx) \\=\int_1^{\infty} e^{-|x|} dx\\= \lim\limits_{R \to \infty}[-e^{-x}]_1^R\\=e-\lim\limits_{R \to \infty} e^{-R} \\=e$. So, the integral $\int_{1}^{\infty} e^{-|x|} dx$ converges to $e$. Hence, the integral $\int_{-\infty}^\infty e^{-x^2} dx$ converges as well by the comparison test.
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