# Another test for convergence

1. Sep 17, 2012

### Bipolarity

1. The problem statement, all variables and given/known data
Test the following integral for convergence
$$\int^{2}_{0}\frac{dx}{1-x^{2}}$$

2. Relevant equations

3. 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

2. Sep 17, 2012

### LCKurtz

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.

3. Sep 17, 2012

### Dick

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.

4. Sep 17, 2012

### Bipolarity

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

5. Sep 17, 2012

### Dick

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

6. Sep 17, 2012

### Zondrina

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.

7. Sep 17, 2012

### Dick

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

8. Sep 17, 2012

### Zondrina

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

9. Sep 18, 2012

### Bipolarity

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

BiP

10. Sep 18, 2012

### Dick

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

11. Sep 18, 2012

### Bipolarity

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

BiP

12. Sep 18, 2012

### Dick

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?