Finding Where an Improper Integral Converges

[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.

*= lim_a->∞(1/x)/1
= 0

Which would be correct?

 

[pwsnafu]

According to Wolfram the integral diverges.

 

[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!

 

[SammyS]

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 $\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?

 

[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?

 

[SammyS]

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

 

[rocapp]

But 1/infinity goes to zero, correct?

 

[SammyS]

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, $\displaystyle \sum \frac{1}{n}\ ,$ diverges.

 

[rocapp]

Thanks for clarifying that; I need to practice these.