Finding Where an Improper Integral Converges

  • Thread starter rocapp
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  • #1
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Hi all,

This is a case of a book answer going against Wolfram's and my answer.

The problem is ∫e(ln(x)/x)dx

The book claims the answer is ∞.

I would think it is a case of ∞/∞ and use L'Hospital's Rule. Wolfram has the same solution.

*= lima->∞(1/x)/1
= 0

Which would be correct?
 

Answers and Replies

  • #3
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Good call. I must have typed it in incorrectly.

Is it correct that the integral diverges because lim as x approaches infinity of lnx/x is an indeterminate?

Thanks!
 
  • #4
SammyS
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Good call. I must have typed it in incorrectly.

Is it correct that the integral diverges because lim as x approaches infinity of lnx/x is an indeterminate?

Thanks!
No. Being indeterminate has nothing to do with it.

It's true that [itex]\displaystyle \lim_{x\,\to\,\infty} \frac{\ln(x)}{x}=0\,,\ [/itex] as well as [itex]\displaystyle \lim_{x\,\to\,\infty} \frac{1}{x}=0\ .[/itex]

Those results are necessary conditions (but not sufficient) for [itex]\displaystyle \int_{e}^{\infty} \frac{\ln(x)}{x}\,dx\ [/itex] to converge.

Does [itex]\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx\ [/itex] converge?
 
  • #5
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No, but I'm not sure why.


EDIT:

It's because ∫1/x dx = ln(x)

and ln(∞) = ∞

So since

∫ln(x)/x = ln(x)2/2, and ln(∞)2/2= ∞,

the integral diverges.

Correct?
 
Last edited:
  • #6
SammyS
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No, but I'm not sure why.


EDIT:

It's because ∫1/x dx = ln(x)

and ln(∞) = ∞

So since

∫ln(x)/x = ln(x)2/2, and ln(∞)2/2= ∞,

the integral diverges.

Correct?
We know that [itex]\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx[/itex] diverges, because we know that [itex]\displaystyle \sum_{n=3}^{\infty} \frac{1}{n}[/itex] diverges.
 
  • #7
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But 1/infinity goes to zero, correct?
 
  • #8
SammyS
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But 1/infinity goes to zero, correct?
Yes 1/n goes to zero as n goes to ∞ . That's basically looking at the sequence [1/n] .

However, the infinite series, [itex]\displaystyle \sum \frac{1}{n}\ ,[/itex] diverges.
 
  • #9
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Thanks for clarifying that; I need to practice these.
 

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