Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 23


$-\displaystyle \frac{\pi}{8}.\quad $ (converges)

Work Step by Step

$\displaystyle \int\frac{z}{z^{4}+4}dz=\left[\begin{array}{l} u=z^{2}\\ du=2zdz \end{array}\right]=\frac{1}{2}\int\frac{du}{u^{2}+2^{2}}du$ Table of integrals: 17. $\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$ The integral is type I improper integral, so $\displaystyle \int_{-\infty}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\int_{t}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\frac{1}{2}\left[\frac{1}{2}\tan^{-1}\left(\frac{z^{2}}{2}\right)\right]_{t}^{0}\quad $ $=\displaystyle \lim_{t\rightarrow-\infty}\left[0-\frac{1}{4}\tan^{-1}\left(\frac{t^{2}}{2}\right)\right]$ $=-\displaystyle \frac{1}{4}\left(\frac{\pi}{2}\right)$ $=-\displaystyle \frac{\pi}{8}.\quad $ (converges)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.