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Another test for convergence

  • Thread starter Bipolarity
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  • #1
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Homework Statement


Test the following integral for convergence
[tex] \int^{2}_{0}\frac{dx}{1-x^{2}} [/tex]


Homework Equations





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
 

Answers and Replies

  • #2
LCKurtz
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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
Dick
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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
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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
 
  • #5
Dick
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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.
 
  • #6
Zondrina
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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.
 
  • #7
Dick
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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.
 
  • #8
Zondrina
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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.
 
  • #9
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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
 
  • #10
Dick
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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?
 
  • #11
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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
 
  • #12
Dick
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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?
 

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