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 June 14th, 2017, 07:45 AM #1 Newbie   Joined: Jun 2017 From: Yaounde-cameroon Posts: 2 Thanks: 0 Functions. We are given a function which is continuous, positive decreasing and integrable on [1,$\infty$[. We are required to show that as x tends to infinity, xf(x) tends to 0. How do we proceed? Last edited by Micromike; June 14th, 2017 at 07:55 AM. Reason: An omission
 June 14th, 2017, 07:52 AM #2 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1386 Consider what happens if it tends to a non-zero value. Thanks from Micromike
 June 15th, 2017, 05:07 PM #3 Global Moderator   Joined: May 2007 Posts: 6,307 Thanks: 526 Use fact integral of 1/x diverges as x becomes infinite. Thanks from Micromike
 June 16th, 2017, 03:26 AM #4 Newbie   Joined: Jun 2017 From: Yaounde-cameroon Posts: 2 Thanks: 0 Where does the 1/x come from? Also as x tends to inf, 1/x tends to 0 Last edited by skipjack; June 17th, 2017 at 05:21 AM.
June 17th, 2017, 03:45 AM   #5
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Quote:
 Originally Posted by Micromike Where does the 1/x come from? Also as x tends to inf, 1/x tends to 0
Yes, it does. But what is the integral of 1/x? What is $\displaystyle \int_1^\infty \frac{1}{x}dx$?

"1/x" is sort of the "border" of "integrable functions". $\displaystyle \frac{1}{x^\alpha}$ is integrable, from 1 to infinity, as long as $\displaystyle \alpha> 1$.

Last edited by skipjack; June 17th, 2017 at 05:21 AM.

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