Improper Integral

• Feb 16th 2010, 09:58 AM
lysserloo
Improper Integral
The initial problem is to find whether

from 0 to $\displaystyle \pi$ $\displaystyle \int\frac{7 - sin(\alpha)}{\alpha^2}$ converges or diverges. I found that it diverges.

Then the second part of the problem asks:

Which of the following inequalities can be used to help prove your conclusion?

http://i193.photobucket.com/albums/z...l/problems.jpg

I think it's the first option because the integral I'm comparing to has to be bigger than the original one for it to have any meaning, but should it be

$\displaystyle \frac{6}{\alpha^2}$?

Or is that entirely wrong? I have no idea how to determined which inequality it should be.

Thank you for any help, and if you could explain a little of how you determined which one it was, I'd really appreciate it.
• Feb 16th 2010, 11:42 AM
Krizalid
just a limit comparison test with $\displaystyle \int_0^\pi\frac{d\alpha}\alpha.$
• Feb 16th 2010, 11:44 AM
lysserloo
I don't understand what that means, sorry. Could you explain?
• Feb 16th 2010, 11:47 AM
Krizalid
have you ever taught the limit comparison test? it's pretty useful when you do not find the appropriate inequality for your integral.
• Feb 16th 2010, 02:47 PM
lysserloo
Is that where you replace infinity with "b" and take the limit as b goes to infinity of the integral? If so, yes I know that. That's partially how I determined that the integral diverges.

I just don't understand the second part...