# Evaluating this integral

• Jul 28th 2013, 05:48 PM
wondering
Evaluating this integral
$\displaystyle {\int^\infty_1} \frac{lnx}{x^2}dx$

I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.
• Jul 28th 2013, 06:38 PM
Plato
Re: Evaluating this integral
Quote:

Originally Posted by wondering
$\displaystyle {\int^\infty_1} \frac{lnx}{x^2}dx$
I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.

Have a look at this.
• Jul 28th 2013, 06:57 PM
wondering
Re: Evaluating this integral
Is integration by parts the only way. The solution manual doesn't show any steps, but they will usually mention if they have used by parts, and they don't. Thank you for the link.
• Jul 28th 2013, 07:03 PM
Plato
Re: Evaluating this integral
Quote:

Originally Posted by wondering
Is integration by parts the only way. The solution manual doesn't show any steps, but they will usually mention if they have used by parts, and they don't. Thank you for the link.

On the page I posted there is a "show steps" option. Click it and see the complete solution.
• Jul 29th 2013, 04:52 PM
HallsofIvy
Re: Evaluating this integral
Quote:

Originally Posted by wondering
$\displaystyle {\int^\infty_1} \frac{lnx}{x^2}dx$

I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.

You say you "tried u substitution" but don't say what substitution. Was it "u= ln(x)"? That would seem the obvious substitution. Of course then $\displaystyle du= (1/x)dx$ so the integral becomes $\displaystyle \int_1^\infty \frac{ln(x)}{x}(1/x)dx= \int_0^\infty \frac{u}{x}du$.

And are you saying "it didn't seem to get me anywhere" because of that "x" still in the integrand? Well, if u= ln(x), then $\displaystyle x= e^u$ so the integral becomes
$\displaystyle \int_0^\infty \frac{u}{e^u}du= \int_0^\infty ue^{-u}du$ which can be integrated "by parts".
• Jul 29th 2013, 05:12 PM
wondering
Re: Evaluating this integral
Quote:

Originally Posted by HallsofIvy
You say you "tried u substitution" but don't say what substitution. Was it "u= ln(x)"? That would seem the obvious substitution. Of course then $\displaystyle du= (1/x)dx$ so the integral becomes $\displaystyle \int_1^\infty \frac{ln(x)}{x}(1/x)dx= \int_0^\infty \frac{u}{x}du$.

And are you saying "it didn't seem to get me anywhere" because of that "x" still in the integrand? Well, if u= ln(x), then $\displaystyle x= e^u$ so the integral becomes
$\displaystyle \int_0^\infty \frac{u}{e^u}du= \int_0^\infty ue^{-u}du$ which can be integrated "by parts".

Yes, it was the x in the denominator that I didn't know what to do with. Thanks for the reply, it makes sense now. I should have written out all of my steps, but it takes me forever to get the latex right.