Improper integral convergence from 0 to 1

In summary: To prove continuity, you need to show that the limit of the function as x approaches a certain value is equal to the value of the function at that point. This can often be done by using algebraic manipulations or by using the definition of continuity. To show that a function is continuous on an interval, you need to show that it is continuous at every point within that interval.
  • #1
Cathr
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Homework Statement


I have to prove that the improper integral ∫ ln(x)/(1-x) dx on the interval [0,1] is convergent.

Homework Equations


I split the integral in two intervals: from 0 to 1/2 and from 1/2 to 1.

The Attempt at a Solution


The function can be approximated to ln(x) when it approaches zero. It is continuous therefore the integral is convergent on [0, 1/2].
We can also write that ln(x)=ln(1+x-1) ≅x-1, this may be applied when the finction is close to zero too.
However I have no idea how to prove that it is convergent on [1/2, 1].
I will greatly appreciate any help.
 
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  • #2
Cathr said:

Homework Statement


I have to prove that the improper integral ∫ ln(x)/(1-x) dx on the interval [0,1] is convergent.

Homework Equations


I split the integral in two intervals: from 0 to 1/2 and from 1/2 to 1.

The Attempt at a Solution


The function can be approximated to ln(x) when it approaches zero. It is continuous therefore the integral is convergent on [0, 1/2].
We can also write that ln(x)=ln(1+x-1) ≅x-1, this may be applied when the finction is close to zero too.
However I have no idea how to prove that it is convergent on [1/2, 1].
I will greatly appreciate any help.

That function is not continuous on [0, 1/2]. It's not even bounded as ##x \rightarrow 0##. But you do have a good idea to compare it with a function like ##\ln(x)##. On [1/2, 1] the function is bounded. Show it has a finite limit as ##x \rightarrow 1##.
 
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  • #3
Define ##F(x)=\int_\frac12^x \frac{\log u}{1-u}du##. Then by definition, ##\int_\frac12^1 \frac{\log x}{1-x}dx = \lim_{x\to 1} F(x)##, if the limit exists.

Try to show that the integrand ##f(x)=\frac{\log x}{1-x}## is continuous on ##[\frac12, 1]## and always has the same sign. If it is continuous, it is bounded, since the interval is compact, so the integral will be bounded, so ##F(x)## must converge (how does the 'integrand never changes sign' come into this?).
 
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  • #4
Cathr said:

Homework Statement


I have to prove that the improper integral ∫ ln(x)/(1-x) dx on the interval [0,1] is convergent.

Homework Equations


I split the integral in two intervals: from 0 to 1/2 and from 1/2 to 1.

The Attempt at a Solution


The function can be approximated to ln(x) when it approaches zero. It is continuous therefore the integral is convergent on [0, 1/2].
We can also write that ln(x)=ln(1+x-1) ≅x-1, this may be applied when the finction is close to zero too.
However I have no idea how to prove that it is convergent on [1/2, 1].
I will greatly appreciate any help.
If your function is ##f(x) = \ln(x)/(1-x)##, it is helpful to write
$$f(x) = f_1(x) + f_2(x), \: \text{where} \: f_1(x) = \ln (x) , \: f_2(x) = x \ln (x)/(1-x).$$ This is true for all ##x## in the open interval ##(0,1)##. The ##f_1## part integrates easily, while the ##f_2## part behaves nicely as ##x \to 0+## and behaves the same as ##f(x)## itself when ##x \to 1-.##
 
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  • #5
Dick said:
That function is not continuous on [0, 1/2]. It's not even bounded as ##x \rightarrow 0##. But you do have a good idea to compare it with a function like ##\ln(x)##. On [1/2, 1] the function is bounded. Show it has a finite limit as ##x \rightarrow 1##.
Using l'Hôpital's rule, I obtained that the limit is equal to -1, so it means that, if the function is continuous, it is integrable on the interval.
Is it right to use l'Hôpital's rule in this case?
 
  • #6
andrewkirk said:
Define ##F(x)=\int_\frac12^x \frac{\log u}{1-u}du##. Then by definition, ##\int_\frac12^1 \frac{\log x}{1-x}dx = \lim_{x\to 1} F(x)##, if the limit exists.

Try to show that the integrand ##f(x)=\frac{\log x}{1-x}## is continuous on ##[\frac12, 1]## and always has the same sign. If it is continuous, it is bounded, since the interval is compact, so the integral will be bounded, so ##F(x)## must converge (how does the 'integrand never changes sign' come into this?).
I suppose that is the function changes sign, we may have a problem in its zero. Is this right?
Also, how can I prove the function is continuous? We didn't study any methods and I didn't find any helpful sources on the internet, may you please give me an insigt on that?
Thanks a lot.
 
  • #7
Cathr said:
Using l'Hôpital's rule, I obtained that the limit is equal to -1, so it means that, if the function is continuous, it is integrable on the interval.
Is it right to use l'Hôpital's rule in this case?

Sure, you can use l'Hopital. For the first interval you know the integral of ##\ln(x)## is convergent, yes?
 
  • #8
Ray Vickson said:
If your function is ##f(x) = \ln(x)/(1-x)##, it is helpful to write
$$f(x) = f_1(x) + f_2(x), \: \text{where} \: f_1(x) = \ln (x) , \: f_2(x) = x \ln (x)/(1-x).$$ This is true for all ##x## in the open interval ##(0,1)##. The ##f_1## part integrates easily, while the ##f_2## part behaves nicely as ##x \to 0+## and behaves the same as ##f(x)## itself when ##x \to 1-.##
Thank you, this is a good idea. To show that the function behaves nicely close to zero is it enough to say that any function multiplied by 0 equals to 0, therefore we don't have to worry about the value of ln(x) in zero?
And when the function approaches 1, using l'Hôpital I found that the limit is equal to -1, therefore there is no more problem in that point.
 
  • #9
Dick said:
Sure, you can use l'Hopital. For the first interval you know the integral of ##\ln(x)## is convergent, yes?
Yes, I may first calculate the integral from ε to 1/2, then take the limit of what I obtain when ε approaches zero.
 
  • #10
Cathr said:
Yes, I may first calculate the integral from ε to 1/2, then take the limit of what I obtain when ε approaches zero.

Right. So you can use that to show the first integral is convergent by a comparison with a multiple of that.
 
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  • #11
Dick said:
Right. So you can use that to show the first integral is convergent by a comparison with a multiple of that.
Thank you! Do you know any tips to prove the continuity of functions? For the non-trivial ones.
 
  • #12
Cathr said:
Thank you! Do you know any tips to prove the continuity of functions? For the non-trivial ones.

I think what you are really asking is what to do with a function that's badly behaved at a point in an interval. You can show it has a finite limit using l'Hopital (as you did at ##x=1##) or you can show it's bounded by a function you know how to integrate. If ##f(x)## converges over an interval and ##|g(x)| \le |f(x)|## then ##g(x)## converges, assuming the functions are continuous in the interior.
 
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  • #13
Dick said:
I think what you are really asking is what to do with a function that's badly behaved at a point in an interval. You can show it has a finite limit using l'Hopital (as you did at ##x=1##) or you can show it's bounded by a function you know how to integrate. If ##f(x)## converges over an interval and ##|g(x)| \le |f(x)|## then ##g(x)## converges, assuming the functions are continuous in the interior.
Thank you!
 
  • #14
Cathr said:
Thank you, this is a good idea. To show that the function behaves nicely close to zero is it enough to say that any function multiplied by 0 equals to 0, therefore we don't have to worry about the value of ln(x) in zero?
And when the function approaches 1, using l'Hôpital I found that the limit is equal to -1, therefore there is no more problem in that point.

No, you need to know that ##x \ln x## does not "blow up" too badly as ##x \to 0+## (that is, that it either has a finite limit or does not go to ##\infty## too quickly).
 

Related to Improper integral convergence from 0 to 1

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite, or where the integrand function has a vertical asymptote in the interval of integration. It is called "improper" because it does not meet the criteria for a standard Riemann integral.

How do you determine if an improper integral from 0 to 1 converges or diverges?

To determine convergence or divergence of an improper integral from 0 to 1, one can use the limit comparison test, comparison test, or the integral test. These tests involve evaluating the limit of the integrand function as it approaches the limits of integration, or comparing the integrand to a known convergent or divergent function.

What are some common examples of improper integrals from 0 to 1?

Some common examples of improper integrals from 0 to 1 include integrals with unbounded integrands, such as 1/x or 1/sqrt(x), as well as integrals with infinite limits of integration, such as 1/(x^2+1) from 0 to ∞. Other examples can involve trigonometric functions or logarithmic functions.

Can an improper integral from 0 to 1 converge even if the integrand function is unbounded?

Yes, it is possible for an improper integral from 0 to 1 to converge even if the integrand function is unbounded. This can happen if the unbounded region of the integrand is "small enough," meaning that the integral still has a finite value when the unbounded part is omitted.

How can the convergence or divergence of an improper integral from 0 to 1 affect its applications in real-world problems?

The convergence or divergence of an improper integral from 0 to 1 can greatly affect its applications in real-world problems. If the integral converges, it can be used to calculate the area under a curve, the volume of a solid, or other physical quantities. However, if the integral diverges, it means that the quantity being calculated is infinite, which can have significant implications in real-world scenarios.

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