Homework Help: Another test for convergence

Bipolarity · Sep 17, 2012

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

2. Relevant equations

3. The attempt at a solution

So far I have brought it down to

[tex] \int^{2}_{0}\frac{1}{1-x}+\frac{1}{1+x} dx[/tex]

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

LCKurtz · Sep 17, 2012

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.

Dick · Sep 17, 2012

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 · Sep 17, 2012

LCKurtz said: ↑

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

[tex] \lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|) [/tex]

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

Dick · Sep 17, 2012

Bipolarity said: ↑

Thanks!

The problem has been reduced to

[tex] \lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|) [/tex]

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.

Zondrina · Sep 17, 2012

Bipolarity said: ↑

Thanks!

The problem has been reduced to

[tex] \lim_{x→1-}ln(|1-x|)-\lim_{x→1+}ln(|1-x|) [/tex]

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.

Dick · Sep 17, 2012

Zondrina said: ↑

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.

Zondrina · Sep 17, 2012

Dick said: ↑

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 · Sep 18, 2012

Zondrina said: ↑

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

Dick · Sep 18, 2012

Bipolarity said: ↑

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 · Sep 18, 2012

Dick said: ↑

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

Dick · Sep 18, 2012

Bipolarity said: ↑

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

BiP

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?