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: 57

Answer

The integral $\int_{1}^{\infty} e^{-x^2} dx$ converges to $e$. Hence, the integral $\int_0^\infty e^{-x^2} dx$ converges as well by the comparison test.

Work Step by Step

The given integral can be re-written as sum of two integral as: $\int_{0}^{\infty} e^{-x^2} dx=\int_{0}^{1} e^{-x^2} dx+\int_{1}^{\infty} e^{-x^2} dx$ We can see that the second integral above defines an improper integral and so, we compare it with $e^{x}$ For $x\geq 1$, we have: $e^{-x} \geq e^{-x^2}$ and $\int_{1}^{\infty} e^{-x^2} dx \leq \int_1^\infty e^{-x} dx \\=\lim\limits_{R \to \infty}[-e^{-x}]_1^R\\=e-\lim\limits_{R \to \infty} e^{-R} \\=e$. So, the integral $\int_{1}^{\infty} e^{-x^2} dx$ converges to $e$. Hence, the integral $\int_0^\infty e^{-x^2} dx$ converges as well by the comparison test.
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