Improper Integral: Convergence and Divergence Analysis for 1/(1-x) from 0 to 2

In summary, the integral from 0 to 2 of 1/(1-x)dx is divergent. The second example of the integral from -1 to 1 of 1/x converges to 0 because it is a special case known as the Cauchy Principle Value integral. The improper integral of 1/x does not exist because the limits do not exist. The Cauchy Principle Value integral requires taking the limits on both sides at the same rate.
  • #1
member 508213
Use of the homework template is mandatory when posting in the homework forums.

State whether the integral converges or diverges and if it converges state the value it converges to.

Integral from 0 to 2 of 1/(1-x)dx



I broke it up into 2 integrals (0 to 1) and (1 to 2) set up the limit for both using variables instead of 1 and I evaluated the integral to equal 0 so I figured it converges to 0 but in my book it says it diverges...it also says it diverges on wolfram alpha so I'm sure that the book is correct, therefore I must be missing something.

Additionally, I took a look at the integral from -1 to 1 of 1/x (which seems very similar to the first problem I stated), yet wolfram alpha says this one is 0.
Why does the second converge to 0 and the first one diverge?

I just learned about improper integrals today so I figure it is likely I am unaware of some detail of problems like these so I would like some help. Thanks.
 
Physics news on Phys.org
  • #2
I think wolfram should tell you the integral of -1 to 1 of 1/x diverges as well. In fact that integral is formally divergent, but there is a special case in which you can make it converge, called the Cauchy principle value integral.

Describe in more detail how you set up "the limit for both using variables instead of 1".
 
  • #3
I'll try to explain how I attempted to solve it:

I saw that there would be an infinite discontinuity at 1 so I first broke the integral up into lim b-> 1- of the integral from o to b of 1/(1-x)dx + lim a->1+ of the integral from a to 2 of 1/(1-x)dx I determined the anti derivative to be -ln(1-t) and then plugged in the bounds of as I would for a normal integral problem. After I had done that I took the limit and what I had at that point was lim b->1- of -ln(1-b) + lim a->1+ of ln(1-a). I figured that since it looked like these expressions would reach -infinity and infinity respectively at the same rate then they would cancel out to be 0. This is what I did to solve the problem.

Additionally, the graph looks like it should be 0 I think...It looks like they cancel out perfectly, so I am very confused.
 
  • #4
Austin said:
I'll try to explain how I attempted to solve it:

I saw that there would be an infinite discontinuity at 1 so I first broke the integral up into lim b-> 1- of the integral from o to b of 1/(1-x)dx + lim a->1+ of the integral from a to 2 of 1/(1-x)dx I determined the anti derivative to be -ln(1-t) and then plugged in the bounds of as I would for a normal integral problem. After I had done that I took the limit and what I had at that point was lim b->1- of -ln(1-b) + lim a->1+ of ln(1-a). I figured that since it looked like these expressions would reach -infinity and infinity respectively at the same rate then they would cancel out to be 0. This is what I did to solve the problem.

Additionally, the graph looks like it should be 0 I think...It looks like they cancel out perfectly, so I am very confused.

What you've done is essentially calculate the Cauchy Principle Value (CPV). This is, in fact, different from an improper integral. For an improper integral, all limits must exist (and be finite) otherwise the integral is not defined. For the CPV, you can combine limits in the way you did.

One good example is:

##lim_{a \rightarrow \infty} \int_{-a}^{+a}sin(x)dx = 0##

This is the CPV.

But, the improper integral would be defined as:

##\int_{-\infty}^{+\infty}sin(x)dx = lim_{a \rightarrow -\infty} \int_{a}^{0}sin(x)dx + lim_{a \rightarrow \infty} \int_{0}^{a}sin(x)dx##

As neither of these limits exists (and is finite), the improper integral is not defined.
 
  • #5
Austin said:
I'll try to explain how I attempted to solve it:

I saw that there would be an infinite discontinuity at 1 so I first broke the integral up into lim b-> 1- of the integral from o to b of 1/(1-x)dx + lim a->1+ of the integral from a to 2 of 1/(1-x)dx I determined the anti derivative to be -ln(1-t) and then plugged in the bounds of as I would for a normal integral problem. After I had done that I took the limit and what I had at that point was lim b->1- of -ln(1-b) + lim a->1+ of ln(1-a). I figured that since it looked like these expressions would reach -infinity and infinity respectively at the same rate then they would cancel out to be 0.

You can't cancel infinities like that.

You have to take the limits separately; if either limit diverges then the integral diverges.
 
  • #6
What you are describing (that you are attempting to do) is the CPV of the integral and, unfortunately, Cauchy beat you to it.
If you look closely at the definition of Riemann integrals in terms of limits of Riemann sums over ever finer partitions, you should see that the Riemann integral does not exist as defined.

This speaks directly to your example:
http://en.wikipedia.org/wiki/Cauchy_principal_value
 
  • #7
Thank you for the replies this is much more clear to me now
 
  • #8
I'd like to add that the CPV integral requires you to take the limit on both sides at the same rate. So instead of 2 limits with a and b separately approaching 1, you need to have limit as a approaches 1 and as (2-a) approaches 1. Otherwise, the two "infinities" will not cancel at every step and you end up with a divergence. The Cauchy Principle value says you take the limits in such a way that the two pieces cancel at every step and arrive at an answer of 0. But like others have said, a CPV integral is not the same as a regular integral.
 

FAQ: Improper Integral: Convergence and Divergence Analysis for 1/(1-x) from 0 to 2

1. What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or the integrand is unbounded at one or more points within the interval of integration.

2. How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you must evaluate the integral using the appropriate techniques for each type of improper integral. For the integral 1/(1-x) from 0 to 2, you would need to use the limit comparison test or the direct comparison test.

3. What is the limit comparison test?

The limit comparison test is a method used to determine if an improper integral converges or diverges by comparing it to a known integral with known convergence or divergence. For the integral 1/(1-x) from 0 to 2, you would compare it to the integral 1/x from 0 to 2, which is known to diverge. If the limit of the ratio of the two integrals is a finite nonzero number, then the original integral also diverges.

4. What is the direct comparison test?

The direct comparison test is a method used to determine if an improper integral converges or diverges by comparing it to a known integral with known convergence or divergence. For the integral 1/(1-x) from 0 to 2, you would compare it to the integral 1/x from 0 to 2, which is known to diverge. If the original integral is always less than or equal to the known integral, and the known integral diverges, then the original integral also diverges.

5. What is the result of the improper integral 1/(1-x) from 0 to 2?

Using the limit comparison test or the direct comparison test, it can be determined that the integral 1/(1-x) from 0 to 2 diverges. This means that the integral does not have a finite value and cannot be evaluated as a definite integral.

Similar threads

Replies
3
Views
6K
Replies
5
Views
1K
Replies
13
Views
2K
Replies
7
Views
1K
Replies
16
Views
3K
Replies
2
Views
1K
Replies
6
Views
10K
Back
Top