# Condition of an exponent for convergence of an integral

• Jul 16th 2008, 11:27 AM
arbolis
Condition of an exponent for convergence of an integral
I must find for which values of $\displaystyle p \in \mathbb{R}$ the following integral converges : $\displaystyle \int_e^{+\infty} \frac{dx}{x\ln^p|x|}$. Mathstud28 already done that (well, something pretty similar) in one of my earlier threads, but I didn't understand a step he did, so I would like a little bit more explanations :
Hmm, sorry the quote doesn't work. I copy and past it, but it only shows the first line and all the other is cut, so I quote it by parts :[quote].
Still doesn't work!
Anyway, he was working with $\displaystyle \int_2^{+\infty}\frac{dx}{x\ln^p|x|}$. He said that if $\displaystyle 0<p<1$, then the indefinite integral is equal to $\displaystyle \frac{\ln^{-p+1}(x)}{-p+1}$. This is what I missed to understand. Did he do integration by parts? How could he reach this result? Thanks!!
• Jul 16th 2008, 11:29 AM
Mathstud28
Quote:

Still doesn't work!
Anyway, he was working with $\displaystyle \int_2^{+\infty}\frac{dx}{x\ln^p|x|}$. He said that if $\displaystyle 0<p<1$, then the indefinite integral is equal to $\displaystyle \frac{\ln^{-p+1}(x)}{-p+1}$. This is what I missed to understand. Did he do integration by parts? How could he reach this result? Thanks!!
If $\displaystyle p\ne{-1}$

Then $\displaystyle \int\frac{dx}{x\ln^p(x)}=\int\frac{\frac{dx}{x}}{\ ln^p(x)}$

Now let $\displaystyle u=\ln(x)$
• Jul 16th 2008, 11:35 AM
galactus
This is mathstud's baby, so I will leave him answer. I am sure he is typing as I write this. Anyway, here is a site with a list of integrals you may find helpful instead of deriving them each time. Unless you want to.

Definite Integrals, General Formulas Involving Definite Integrals
• Jul 16th 2008, 11:36 AM
arbolis
Thanks mathstud28, I reached it! I thought it would have been much more complicated.
• Jul 16th 2008, 11:38 AM
Mathstud28
Quote:

Originally Posted by galactus
I am sure he is typing as I write this.

Or finishing before you (Sun)
Quote:

Originally Posted by arbolis
Thanks mathstud28, I reached it! I thought it would have been much more complicated.

Yeah, sometimes the obvious alludes even the best of us.
• Jul 16th 2008, 11:53 AM
galactus
Hey mathstud, you know what might be fun?. Go to the link I posted above and actually derive the solutions they give for the integrals. You have probably done that already. (Happy)

Some of them are probably very challenging.