Re: [analysis] I'm halfway this integral

From: The World Wide Wade (waderameyxiii_at_comcast.remove13.net)
Date: 03/17/05 

Date: Thu, 17 Mar 2005 11:37:37 -0800

In article <lxk_d.85095$Pj6.66695@amsnews02.chello.com>,
 "Steve Liem \(Ritzi Lee\)" <info@underground-liberation.nl>
 wrote:

> By the way thanks for helping me with the last problem....
>
>
>
> Given the integral from 1 to 2 on the function log(x) / ((x - 1) sqrt(-x^2 +
> 3x - 2)).
> So we have to see how this one converges.
>
> Anyway, we have 1 and 2 as critical points, so we have to write in 2
> integrals from 1 to c, and from c to 2, where c is in the interval (1,2).
>
>
> The first part: Integral from 1 to c over log(x) / ((x - 1) sqrt(-x^2 + 3x -
> 2)).
> Here x = 1 is the critical point.
> So around x = 1:
>
> log(x) ~ x - 1
> and sqrt(-x^2 + 3x - 2) ~ sqrt(x - 1)
>
> ==> |log(x) / ((x - 1) sqrt(-x^2 + 3x - 2))| <= 1 / sqrt(x-1)
> and the integral from 1 to c on 1 / sqrt(x-1) is convergent.
> So the integral from 1 to c over log(x) / ((x - 1) sqrt(-x^2 + 3x - 2)) also
> converges.
>
>
>
> Now i'm bussy with the second part:
> The integral from c to 2 over log(x) / ((x - 1) sqrt(-x^2 + 3x - 2)).
>
>
> I've tried everything from estimating the function around x = 2, to trying
> to find testfunctions where I can aply the limit criterium. Nothing works
> because this sqrt(-x^2 + 3x - 2)) keeps bugging me. (sounds funny!)
>
> Is there a nice estimation for this?
> Around x = 2:
> log(x) ~ 1/2 x - 2.

log(x) is bounded near 2; forget about it. same with 1/(x-1).