Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.3 - The Integral Test and Estimates of Sums - 11.3 Exercises - Page 726: 32

Answer

Convergent for $p\gt 1$.

Work Step by Step

$\int_{1}^{\infty}\frac{lnx}{x^{p}}dx=\int_{0}^{\infty}ue^{-u(p-1)}du=[\frac{ue^{-u(p-1)}}{p-1}-\frac{e^{-u(p-1)}}{(p-1)^{2}}]_{0}^{\infty}$ $=\frac{\infty e^{-\infty(p-1)}}{p-1}-\frac{e^{-\infty (p-1)}}{(p-1)^{2}}$ which is convergent for $p\gt 1$.
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