Finding Where an Improper Integral Converges

In summary, the integral ∫∞e(ln(x)/x)dx has a conflicting answer between the book and Wolfram's solution. After using L'Hospital's Rule, it is determined that the integral diverges. This is due to the fact that lim as x approaches infinity of lnx/x is an indeterminate, which is a necessary condition for convergence. However, the integral fails to meet the sufficient condition and therefore diverges. It is important to note that while 1/n goes to zero as n goes to infinity, the infinite series of 1/n still diverges. Additional practice is needed to fully understand this concept.
  • #1
rocapp
95
0
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?
 
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  • #3
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
rocapp said:
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
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
rocapp said:
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
But 1/infinity goes to zero, correct?
 
  • #8
rocapp said:
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
Thanks for clarifying that; I need to practice these.
 

Related to Finding Where an Improper Integral Converges

1. What is an improper integral?

An improper integral is a type of definite integral where one or both of the limits of integration are infinite or the integrand is unbounded at some point within the limits of integration. This type of integral cannot be evaluated using the standard methods of integration.

2. How do you determine if an improper integral converges?

To determine if an improper integral converges, you need to evaluate the integral using a proper limit as the upper or lower bound approaches infinity. If the limit exists, then the integral converges. If the limit does not exist, then the integral diverges.

3. What is the difference between a convergent and a divergent improper integral?

A convergent improper integral is one where the limit as the bound approaches infinity exists and is a finite number. This means that the integral has a defined value. On the other hand, a divergent improper integral is one where the limit does not exist or is infinite, indicating that the integral does not have a defined value.

4. How do you handle discontinuities in an improper integral?

If the integrand has a discontinuity within the limits of integration, the integral can be separated into two or more integrals, each of which can be evaluated separately. The final result is the sum of the individual integrals.

5. Can an improper integral converge at both limits?

Yes, an improper integral can converge at both the upper and lower limits. This is known as a doubly convergent improper integral. In this case, the integral is evaluated by taking the limit as the upper and lower bounds approach infinity simultaneously.

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