Improper integral convergence from 0 to 1

[Cathr]

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.

 

[Dick]

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 $x \rightarrow 1$.

 

[andrewkirk]

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?).

 

[Ray Vickson]

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 $x \to 1-.$

 

[Cathr]

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?

 

[Cathr]

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.

 

[Dick]

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?

 

[Cathr]

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.

 

[Cathr]

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.

 

[Dick]

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.

 

[Cathr]

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.

 

[Dick]

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.

 

[Cathr]

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!

 

[Ray Vickson]

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).