# Finding Where an Improper Integral Converges

1. Oct 17, 2012

### rocapp

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?

2. Oct 17, 2012

### pwsnafu

3. Oct 17, 2012

### rocapp

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. Oct 17, 2012

### SammyS

Staff Emeritus
No. Being indeterminate has nothing to do with it.

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

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

Does $\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx\$ converge?

5. Oct 19, 2012

### rocapp

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: Oct 19, 2012
6. Oct 20, 2012

### SammyS

Staff Emeritus
We know that $\displaystyle \int_{e}^{\infty} \frac{1}{x}\,dx$ diverges, because we know that $\displaystyle \sum_{n=3}^{\infty} \frac{1}{n}$ diverges.

7. Oct 21, 2012

### rocapp

But 1/infinity goes to zero, correct?

8. Oct 21, 2012

### SammyS

Staff Emeritus
Yes 1/n goes to zero as n goes to ∞ . That's basically looking at the sequence [1/n] .

However, the infinite series, $\displaystyle \sum \frac{1}{n}\ ,$ diverges.

9. Oct 21, 2012

### rocapp

Thanks for clarifying that; I need to practice these.