# Another test for convergence

Bipolarity

## Homework Statement

Test the following integral for convergence
$$\int^{2}_{0}\frac{dx}{1-x^{2}}$$

## The Attempt at a Solution

So far I have brought it down to

$$\int^{2}_{0}\frac{1}{1-x}+\frac{1}{1+x} dx$$

However, it seems that this integral produces a natural logarithm, which then must be evaluated with a negative logerand, making it a complex number.
My textbook is introductory calculus, so it deals strictly with real numbers, so I am puzzled.

BiP

Homework Helper
Gold Member
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.

Homework Helper
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.

Bipolarity
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.

Thanks!

The problem has been reduced to

$$\lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|)$$

I can't fathom how I would compute this though. I am not very good with one-sided limits being subtracted from each other.

BiP

Homework Helper
Thanks!

The problem has been reduced to

$$\lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|)$$

I can't fathom how I would compute this though. I am not very good with one-sided limits being subtracted from each other.

BiP

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

Homework Helper
Thanks!

The problem has been reduced to

$$\lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|)$$

I can't fathom how I would compute this though. I am not very good with one-sided limits being subtracted from each other.

BiP

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.

Homework Helper
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.

Homework Helper
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.

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

I get how to do the problem but can that limit actually be computed? If so, how?

BiP

Homework Helper
I get how to do the problem but can that limit actually be computed? If so, how?

BiP

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

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

I see. Is it always the case that the difference between two limits, each of which does not by itself exist, is incalculable?

BiP