Another test for convergence

Have you noticed there is a problem at ##x=1##? Also, for what it's worth,$$
\int \frac 1 u\, du = \ln|u| + C$$which works when the argument is negative.

 

For one thing, the integrand is singular at x=1. You don't want to integrate directly from 0 to 2. That's dangerous. Try the integral from 0 to 1 first. And for another, you can take the integral of 1/(1-x) to be -log(|1-x|). Note the absolute value.

 

Don't do that. Just work on the interval [0,1) first. If it doesn't converge, then the whole integral can't converge.

 

Consider when x→1-. So you have the possibility of all values less than one being considered here, including the negative ones.

As for x→1+, note that ln|-x| = ln(x). You can apply the abs before taking the ln I believe.

 

Both limits are negative infinity. That's what's confusing Bipolarity. The difference is quite undefined.

 

I was just stating that they could be computed :P, but yeah what you said would actually help him get the answer he wants.

 

The difference of the two terms can't be computed. You are clear on that, right?

 

Almost. The form (+infinity)-(+infinity) is not defined. On the other hand you can define (+infinity)+(+infinity) to be +infinity. So you can define (+infinity)-(-infinity) to be +infinity. Do you see how that applies to your example when you have the other sign on the infinite limits?