Condition of an exponent for convergence of an integral

**arbolis** · July 16th 2008, 11:27 AM

I must find for which values of $p \in \mathbb{R}$ the following integral converges : $\int_e^{+\infty} \frac{dx}{x\ln^p|x|}$ . 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 :
[quote]How about when
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 $\int_2^{+\infty}\frac{dx}{x\ln^p|x|}$ . He said that if $0<p<1$ , then the indefinite integral is equal to $\frac{\ln^{-p+1}(x)}{-p+1}$ . This is what I missed to understand. Did he do integration by parts? How could he reach this result? Thanks!!

**Mathstud28** · July 16th 2008, 11:29 AM

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

If $p\ne{-1}$

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

Now let $u=\ln(x)$

**galactus** · July 16th 2008, 11:35 AM

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

**arbolis** · July 16th 2008, 11:36 AM

Thanks mathstud28, I reached it! I thought it would have been much more complicated.

**Mathstud28** · July 16th 2008, 11:38 AM

Originally Posted by galactus

I am sure he is typing as I write this.

Or finishing before you

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.

**galactus** · July 16th 2008, 11:53 AM

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.

Some of them are probably very challenging.

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